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AWS::EC2::VPCCidrBlock

The AWS::EC2::VPCCidrBlock resource associates a single Amazon-provided IPv6
                        CIDR block or a single user-specified IPv4 CIDR block with a Virtual Private Cloud
                        (VPC).

Topics

Syntax

Properties

Examples

Syntax

To declare this entity in your AWS CloudFormation template, use the following syntax:

JSON

{
   "Type" : "AWS::EC2::VPCCidrBlock",
   "Properties" : {
      "AmazonProvidedIpv6CidrBlock" : Boolean,
      "CidrBlock" : String,
      "VpcId" : String
   }
}

YAML

Type: AWS::EC2::VPCCidrBlock
Properties:
  AmazonProvidedIpv6CidrBlock: Boolean
  CidrBlock: String
  VpcId: String

Properties

AmazonProvidedIpv6CidrBlock

Whether to request an Amazon-provided IPv6 CIDR block with a /56 prefix length
                                 for the VPC. You can't specify the range of IPv6 addresses or the size of the CIDR
                                 block.

Required: No

Type: Boolean

Update requires: Replacement

CidrBlock

An IPv4 CIDR block to associate with the VPC.

Required: No

Type: String

Update requires: Replacement

VpcId

The ID of the VPC to associate the Amazon-provided IPv6 CIDR block with.

Required: Yes

Type: String

Update requires: Replacement

Examples

Associate an Amazon-provided IPv6 CIDR block

The following snippet associates an Amazon-provided IPv6 CIDR block (with a prefix
                        length of /56) with the TestVPCIpv6 VPC.

JSON

{
  "Ipv6VPCCidrBlock": {
    "Type": "AWS::EC2::VPCCidrBlock",
    "Properties": {
      "AmazonProvidedIpv6CidrBlock": true,
      "VpcId": { "Ref" : "TestVPCIpv6" }
    }
  }
}

YAML

Ipv6VPCCidrBlock:
  Type: AWS::EC2::VPCCidrBlock
  Properties:
    AmazonProvidedIpv6CidrBlock: true
    VpcId: !Ref TestVPCIpv6

Associate an IPv4 CIDR block and Amazon-provided IPv6 CIDR block

The following example associates an IPv4 CIDR block and an Amazon-provided IPv6 CIDR
                        block with a VPC. It also outputs the
                        list of IPv4 CIDR block association IDs and IPv6 CIDR blocks that are associated with
                        the VPC.

JSON

{
    "Resources": {
        "VPC": {
            "Type": "AWS::EC2::VPC",
            "Properties": {
                "CidrBlock": "10.0.0.0/24"
            }
        },
        "VpcCidrBlock": {
            "Type": "AWS::EC2::VPCCidrBlock",
            "Properties": {
                "VpcId": {
                    "Ref": "VPC"
                },
                "CidrBlock": "192.0.0.0/24"
            }
        },
        "VpcCidrBlockIpv6": {
            "Type": "AWS::EC2::VPCCidrBlock",
            "Properties": {
                "VpcId": {
                    "Ref": "VPC"
                },
                "AmazonProvidedIpv6CidrBlock": true
            }
        }
    },
    "Outputs": {
        "VpcId": {
            "Value": {
                "Ref": "VPC"
            }
        },
        "PrimaryCidrBlock": {
            "Value": {
                "Fn::GetAtt": [
                    "VPC",
                    "CidrBlock"
                ]
            }
        },
        "Ipv6CidrBlock": {
            "Value": {
                "Fn::Select": [
                    0,
                    {
                        "Fn::GetAtt": [
                            "VPC",
                            "Ipv6CidrBlocks"
                        ]
                    }
                ]
            }
        },
        "CidrBlockAssociation": {
            "Value": {
                "Fn::Select": [
                    0,
                    {
                        "Fn::GetAtt": [
                            "VPC",
                            "CidrBlockAssociations"
                        ]
                    }
                ]
            }
        }
    }
}

YAML

Resources:
  VPC:
    Type: AWS::EC2::VPC
    Properties:
      CidrBlock: 10.0.0.0/24
  VpcCidrBlock:
    Type: AWS::EC2::VPCCidrBlock
    Properties:
      VpcId: !Ref VPC
      CidrBlock: 192.0.0.0/24
  VpcCidrBlockIpv6:
    Type: AWS::EC2::VPCCidrBlock
    Properties:
      VpcId: !Ref VPC
      AmazonProvidedIpv6CidrBlock: true

Outputs:
  VpcId:
    Value: !Ref VPC
  PrimaryCidrBlock:
    Value: !GetAtt VPC.CidrBlock
  Ipv6CidrBlock:
    Value: !Select [ 0, !GetAtt VPC.Ipv6CidrBlocks ]
  CidrBlockAssociation:
    Value: !Select [ 0, !GetAtt VPC.CidrBlockAssociations ]


text:
Q: Google Drive - How to Autopopulate Google Spreadsheet with Concatenate functions when users submit data via webform? I have a Google Drive Spreadsheet in which I'm trying to have a formula autopopulate all the way down the rows, as users submit data via a webform attached to the spreadsheet. The tricky part is that I use a CONCATENATE function already, to perform a concatenation of  the data submitted on several columns into one single cell.
However, for the CONCATENATE function to work, I have to "apply" it to newly submitted rows.
Is there a way to automate the filling of this formula down the rows in the spreadsheet?
I've tried to place an ArrayFormula function to it, even setting the range (A1:A), but I couldn't find the proper syntax for it work, if it may like this.
The function goes:
=CONCATENATE(CHAR(10)&X14&V14&Y14&J14&" "&K14&" "&L14&M14&N14&O14&" "&P14&" "&Q14&R14&S14&CHAR(10)&T14&"."&CHAR(10)&U14&"."&CHAR(10)&W14&CHAR(10)&CHAR(10)&CHAR(10)&I14&CHAR(10)&Z14&" "&AA14&CHAR(10)&AB14&AC14&AD14&AE14&AF14&AG14&AH14&"."&CHAR(10)&AI14&AJ14)

Any suggestion will be greatly appreciated.

A: (Answered by the OP in a question edit. Transformed into a community wiki answer. See Question with no answers, but issue solved in the comments (or extended in chat) )
The OP wrote:

I've found a little script from the Script Gallery of Google Spreadsheet (AutoFormulas, by tuxincarnate[@]gmail[dot]com), which solves the trick! Just tested it with a dozen of submissions from the webform and it does what it promises, autopopulate with formulas down the columns where applied.

// Updates all rows except the last to have the same formulas as row 3 (allowing for a header and different row 2 formula)
// To activate this functionality, the last row in the column should have the value '(auto)'
function UpdateFormulas() {
  var ss = SpreadsheetApp.getActiveSpreadsheet();
  var sheetCount = ss.getNumSheets();

  for (var sheetIndex = 0; sheetIndex < sheetCount; ++sheetIndex) {
    var sheet = ss.getSheets()[sheetIndex];

    var rowCount = sheet.getMaxRows();
    var columnCount = sheet.getMaxColumns();

    if (rowCount < 5) continue;

    for (var columnIndex = 1; columnIndex <= columnCount; ++columnIndex) {
      if (sheet.getRange(rowCount, columnIndex).getValue() == '(auto)') {
        var row3Range = sheet.getRange(3, columnIndex);
        for (var rowIndex = 4; rowIndex < rowCount; ++rowIndex) {
          if (sheet.getRange(rowIndex, columnIndex).isBlank()) {
            row3Range.copyTo(sheet.getRange(rowIndex, columnIndex));
          }
        }
      }
    }
  }
}

@David Tew wrote:

Whilst you have a solution already, you might like to consider for the future the following formula, which is what you were originally looking for (You don't need to use CONCATENATE since you are choosing the columns to join together with & symbol)

=arrayformula(CHAR(10)&X2:X&V2:V&Y2:Y&J2:J&" "&K2:K&" "&L2:L&M2:M&N2:N&O2:O&" "&P2:P&" "&Q2:Q&R2:R&S2:S&CHAR(10)&T2:T&"."&CHAR(10)&U2:U&"."&CHAR(10)&W2:W&CHAR(10)&CHAR(10)&CHAR(10)&I2:I&CHAR(10)&Z2:Z&" "&AA2:AA&CHAR(10)&AB2:AB&AC2:AC&AD2:AD&AE2:AE&AF2:AF&AG2:AG&AH2:AH&"."&CHAR(10)&AI2:AI&AJ2:AJ)

this should go in the second row

The OP wrote:

@DavidTew showed me a clear and easy way to have it solved. The ArrayFormula should go alone, telling it to put together the ranges for every column into every single row. Works like charm. Proper syntax is: =arrayformula(A2:A&B2:B&C2:C), in order to have contents from cells A2, B2 and C2 "concatenated" into the cell where the ArrayFormula is applied. The most important issue is that by using this function, it autpopulates all the way down the rows as users submit data via a webform attached to the spreadsheet.

redpajama_set_name: RedPajamaStackExchange


text:
\section{Introduction}

This paper is mainly divided into two parts. First, we are interested in studying the regularity properties of the following Stochastic Differential Equation (SDE)
\begin{align}\label{SDE0}
dX_t = b(t,X_t)dt+dB_t, \ 0\leq t \leq T, \ X_0 =x\in \mathbb{R}^d,
\end{align}
where $B_t$, $t\in [0,T]$ is a $d$-dimensional Brownian motion and $b$ is a measurable function such that a unique strong solution exists. Our goal is to analyse the regularity of strong solutions to (\ref{SDE0}) both in space and in the Malliavin sense. We give a condition based on the regularity properties of $b$ to obtain regularity properties of $X_t$, $t\in [0,T]$. Then we study the consequences of the aforementioned properties and we take two different directions. On one hand, the Malliavin regularity allows us to improve the regularity of densities of strong solutions. On the other hand, the regularity in space entitles us to study the associated Stochastic Transport Equation and gain more regularity on the solution. Namely, for $b$ Lipschitz we are able to show that one obtains a classical solution to the Stochastic Transport Equation.

Considerable research in the direction of regularity of densities of solutions to SDEs has been done in the past years. There are well-known results on conditions for a density to be smooth when the coefficients are smooth, for example, in \cite{Nua10} or in the case of SDEs with boundary conditions in \cite{KHSS97} under the so-called H\"{o}rmander's condition.

We highlight the work by S. Kusuoka and D. Stroock in \cite{Kus82} where the authors show that if \mbox{$b\in C_b^{n+2}(\mathbb{R}^d)$} then the density lies in $C_b^n(\mathbb{R}^d)$ using Sobolev inequalities associated to the \mbox{$H$-derivative} of the solution. Here, we improve the regularity of the density and skip the boundedness of $b$, instead we consider additive noise and provide an extension to a class of non-degenerate diffusion coefficients. In \cite{Kus10} S. Kusuoka also gives conditions for the law to be absolutely continuous with respect to the Lebesgue measure when drift coefficients are non-Lipschitz, his work is closely related to the findings of N. Bouleau and F. Hirsch in \cite{BH91} where the authors show that for global Lipschitz coefficient a density exists, here we show that such density is H\"{o}lder continuous with exponent $\alpha<1$. An improvement was given in dimension one in \cite{FP10} where they show that for a H\"{o}lder continuous drift of, at most, linear growth and H\"{o}lder continuous diffusion coefficient the solution admits densities at any given time.

Our technique is mainly based on Malliavin calculus and on a sharp estimate on the moments of the derivative of the flow associated to the solution motived by a previous work in \cite{MNP14} where a very similar estimate is used to prove that solutions of SDE's with irregular drift coefficients are once Sobolev differentiable in the initial condition. SDE's with irregular drift coefficients has been a very active topic of research in the last years. For example, it is known, see \cite{Men11}, where the authors prove that solutions of SDEs with merely bounded drift are Malliavin differentiable. Nevertheless, once Malliavin differentiability is not enough to guarantee regularity of the densities. A related result on this matter is \cite{AP14} where the drift is assumed to be irregular and a representation for the derivative is also obtained, even in the higher dimensional case. Also, \cite{Zhang11} deals with the same problem as in \cite{MNP14} but here, the diffusion is non-trivial. Here instead, we look at the regularity of the drift coefficient and connect it to regularity of the solution in both the Malliavin and Sobolev sense.

For the study of densities, we use a powerful result by V. Bally and L. Caramellino in \cite{Bally2011} which allows us directly to improve the regularity of densities of solutions with with sufficient Malliavin regularity. In addition, we also look at the regularity in space. As a consequence of the relationship between the Malliavin and Sobolev derivatives we are also able to give a condition to determine the regularity  of solutions to (\ref{SDE0}) in the Sobolev sense (locally) and show that such derivatives admit moments of any order. At the end of the section we also give an extension to more general diffusions.

The last application of the paper is devoted to the study of the Stochastic Transport Equation (STE) since it is closely related to the SDE (\ref{SDE0}) by the inverse of the flow of the solution. We use the results obtained in the first part of the paper to show that, for $b$ Lipschitz, the solution is classical. Work in the direction of SPDE's and in particular the Stochastic Transport Equation has brought a lot of interests in the last years. In \cite{FGP10} F. Flandoli, M. Gubinelli and E. Priola study the well-posedness for H\"{o}lder-continuous drifts and show pathwise uniqueness of the weak solution. In \cite{Nil13}, in dimension one, it is shown that when the drift is a step function then the solution to the transport equation is even once continuously differentiable.

\section{Framework}

In this section we recall some facts from Malliavin calculus and Sobolev spaces, which we aim at employing in Section \ref{main results} to analyse the regularity of densities of strong solutions of SDEs. See \cite{Nua10, Mall78, Mall97, DOP08} for a deeper insight on Malliavin Calculus. As for theory on Sobolev spaces the reader is referred to \cite{Leo09, Evan10}.

\subsection{Basic elements of Malliavin Calculus}\label{somedefmalcal}

In this Section we briefly elaborate a framework for Malliavin calculus.

Let $H$ be a real separable Hilbert space and $W=\{W(h), h\in H\}$ an isonormal Gaussian process, see \cite[Definition 1.1.1]{Nua10}. Assume $W$ is defined on a complete probability space $\left( \Omega ,\mathcal{F},P\right)$ and $\mathcal{F}$ is generated by $W$.

Denote by $D$ the derivative
operator acting on elementary smooth random variables in the sense that%
\begin{equation*}
D(f(h_{1},\ldots,h_{n}))=\sum_{i=1}^{n}\partial
_{i}f(h_{1},\ldots,h_{n})h_{i},\quad h_{i}\in H, \quad f\in C_{b}^{\infty }(\mathbb{R%
}^{n}).
\end{equation*}
Further let $\mathbb{D}^{k,p} (\Omega)$, $k,p\geq 1$ be the closure of the family of elementary smooth random variables with respect to the norm%
\begin{equation*}
\left\Vert F\right\Vert _{\mathbb{D}^{k,p}(\Omega)}:=\left\Vert F\right\Vert _{L^{p}(\Omega
)}+\sum_{i=1}^k \left\Vert D\overset{i)}{\cdots} DF\right\Vert _{L^{p}(\Omega ;H\otimes \overset{i)}{\cdots} \otimes H)}.
\end{equation*}

Our framework will rely on the special case where $H$ is isometric to $L^2([0,T];\mathbb{R}^d)$ endowed with the Lebesgue measure, then we have that the Malliavin derivative is a process $\{D_t F\}_{t\in [0,T]}$ in $L^2(\Omega\times [0,T];\mathbb{R}^d)$ defined as
$$D_tF = \sum_{i=1}^n \frac{\partial}{\partial x_i} f\left(\int_0^T h_1(u)dW_u, \dots, \int_0^T h_n(u)dW_u \right) h_i(t)$$
and again we take the closure w.r.t. the norm
$$\|F\|_{\mathbb{D}^{k,p}(\Omega)} = E[|F|^p]^{1/p}+ \sum_{i=1}^k E\left[ \int_0^T\cdots \int_0^T \|D_{t_1}\cdots D_{t_i} F\|^p dt_1\cdots t_i\right]^{1/p}$$
where $\|\cdot\|$ denotes any norm in $\mathbb{R}^{d\times \overset{i)}{\cdots} \times d}$.

The operator $D\overset{k)}{\cdots} D$ is then a closed operator from $\mathbb{D}^{k,p} (\Omega)$ to $L^p(\Omega \times [0,T]^k ; \mathbb{R}^{d\times \overset{k)}{\cdots} \times d})$ for all $p\geq 1$. Moreover, for $p\leq q$ and $k\leq l$ we have
$$\|F\|_{\mathbb{D}^{k,p}(\Omega)} \leq \|F\|_{\mathbb{D}^{l,q}(\Omega)}$$
and as a consequence
$$\mathbb{D}^{k+1,p}(\Omega) \hookrightarrow \mathbb{D}^{k,q} (\Omega)$$
if $k\geq 0$ and $p>q$.

We shall say that a random variable is $k$-times \emph{Malliavin differentiable} with derivatives in $L^p(\Omega)$, $p\geq 1$ if it lies on $\mathbb{D}^{k,p}(\Omega)$.

Finally, we have the chain-rule for the Malliavin derivative. Let $\varphi :\mathbb{R}^m\rightarrow \mathbb{R}^m$ be a function such that
$$|\varphi(x) - \varphi(y)| \leq K |x-y|$$
for any $x,y \in \mathbb{R}^m$. Suppose $F=(F^1,\dots,F^m)$ is a random vector whose components belong to the space $\mathbb{D}^{1,2}(\Omega)$. Then $\varphi(F) \in \mathbb{D}^{1,2}(\Omega)$ and there exists a random vector $G=(G_1,\dots,G_m)$ bounded by $K$ such that
$$D \varphi(F) = \sum_{i=1}^mG_i DF^i.$$

In particular if $\varphi'$ exists, then $G= \varphi'(F)$.

For a stochastic process $v \in \mbox{Dom}( \delta)$ (not necessarily adapted to $\{\mathcal{F}_t\}_{t\in [0,T]}$) we denote by
\begin{align}\label{skorokhod}
\delta(v) := \int_0^T v_t \delta B_t
\end{align}
the action of $\delta$ on $v$. The above expression (\ref{skorokhod}) is known as the Skorokhod integral of $v$ and it is an anticipative stochastic integral. It turns out that all $\{\mathcal{F}_t\}_{t\in [0,T]}$-adapted processes in $L^2(\Omega\times[0,T])$ are in the domain of $\delta$ and for such processes $v_t$ we have
$$\delta( v ) =\int_0^T v_t dB_t,$$
i.e.the Skorokhod and It\^{o} integrals coincide. In this sense, the Skorokhod integral can be considered to be an extension of the It\^{o} integral to non-adapted integrands.

The dual relation between the Malliavin derivative and the Skorokhod integral implies the following important formula:

\begin{thm}[Duality formula]\label{duality}
Let $F\in \mathbb{D}^{1,2}$ and $v\in \mbox{Dom}(\delta)$. Then
\begin{align}
E\left[ F\int_0^T v_t \delta B_t\right] = E\left[ \int_0^T v_t D_t F dt\right].
\end{align}
\end{thm}

\subsection{Basic facts of theory on Sobolev spaces}

In this section we concisely review some basic facts about theory on Sobolev spaces.

Let $U$ be an open bounded subset of $\mathbb{R}^d$. Fix $1\leq p \leq \infty$ and let $k\geq 0$ an integer. The Sobolev space $W^{k,p}(U)$ is composed by all locally integrable functions $u: U \rightarrow \mathbb{R}$ such that for any multiindex $\alpha$ with $|\alpha|\leq k$, then $D^\alpha u$ exists in the weak sense and belongs to $L^p(U)$.

We endow the space $W^{k,p}(U)$ with the topology generated by the norm
$$\|u\|_{W^{k,p}(U)} := \left(\sum_{|\alpha|\leq k} \int_U |D^\alpha u|^p dx \right)^{1/p}, \ \ 1\leq p <\infty$$
or
$$\|u\|_{W^{k,\infty}(U)}:= \sum_{|\alpha|\leq k} {\mathrm{ess}\sup}_{U} |D^\alpha u|, \ \ p=\infty. $$

The following relations will be of high relevance for our purposes. For $1\leq p < q \leq \infty$, $k>l$ such that $(k-l)p<d$ and
$$\frac{1}{q} = \frac{1}{p}- \frac{k-l}{d}$$
then we have the following continuous embedding
\begin{align}
W^{k,p}(\mathbb{R}^d) \hookrightarrow W^{l,q}(\mathbb{R}^d).
\end{align}

Also, we have the following embedding as a consequence of Morrey's inequality; if $\frac{k-r-\alpha}{d}=\frac{1}{p}$ with $\alpha\in (0,1)$ then
\begin{align}\label{embed}
W^{k,p}(\mathbb{R}^d) \hookrightarrow C^{r,\alpha}(\mathbb{R}^d).
\end{align}
Essentially, this means that if we have enough Sobolev-regularity then we may expect some continuous classical derivatives up to some order.

We will though use $\frac{\partial}{\partial x}$ to denote differentiation in both the weak and classical sense when the context is clear.

\subsection{Shuffles}\label{shuffles}
Let $m,n\in \mathbb{N}_0$ and denote by $S_{m} =\{\sigma: \{1,\dots, m\}\rightarrow \{1,\dots,m\} \}$ the set of permutations of $m$ elements. Define the set of \emph{shuffle permutations} of length $n+m$ as
$$S(m,n) := \{\sigma\in S_{m+n}: \, \sigma(1)<\cdots <\sigma(m), \, \sigma(m+1)<\cdots <\sigma(m+n)\}.$$

Fix $s,t\in [0,T]$ with $s <t$ and define the $m$-dimensional subset of $[0,1]^m$
$$\Lambda_{s,t}^m:=\{(u_1,\dots,u_m)\in [0,T]^m : \, s<u_1<\cdots < u_m<t\}.$$

Denote $u=(u_1,\dots,u_m)\in [0,T]^m$ and for given a permutation $\sigma \in S_m$ let $u_{\sigma(1:m)} := (u_{\sigma(1)},\dots, u_{\sigma(m)})$.

Let $f:[0,T]^m \rightarrow \mathbb{R}$ and $g:[0,T]^n\rightarrow \mathbb{R}$ be two measurable functions. Then
\begin{align}\label{shuffle}
\left(\int_{\Lambda_{s,t}^m} f(u)du \right)\left( \int_{\Lambda_{s,t}^n} g(u)du \right)= \sum_{\sigma\in S(m,n)} \int_{\Lambda_{s,t}^{m+n}} f(u_{\sigma(1:m)})g(u_{\sigma(m+1:m+n)})du.
\end{align}

We will also need the following formula. Let $f_i:[0,T] \rightarrow \mathbb{R}$, $i=1,\dots,m+n$ be measurable functions and $k=0,\dots, m$ be fixed. Then
\begin{align}\label{shuffle2}
\begin{split}
\int_{\Lambda_{s,t}^m} \int_{\Lambda_{s,u_k}^n} \prod_{i=1}^n f_i(v_i) \prod_{i=n+1}^{n+m} f_i(u_i) dvdu = \sum_{\sigma\in S_k(m,n)} \int_{\Lambda_{s,t}^{m+n}} \prod_{i=1}^{m+n} f_i(w_{\sigma(i)}) dw
\end{split}
\end{align}
where
$$S_k (m,n) = \{\sigma \in S(m,n): \, \sigma(j)=j, \, j=k,\dots,m\}.$$
If we consider the case $u_k=t$ in the last formula we indeed obtain \eqref{shuffle}. Observe also that the number of terms in the sum given in \eqref{shuffle2} is at most of order $C^{m+n}$ for some finite constant $C>0$.

\section{Malliavin and flow regularity of strong solutions of SDEs}\label{main results}

Consider the \emph{stochastic differential equation} (SDE) given by
\begin{align}\label{SDE}
\begin{cases}
dX_t = b(t,X_t) dt + dB_t,\\
X_0=x \in \mathbb{R}^d,
\end{cases}
\end{align}
where the drift coefficient $b:[0,T]\times \mathbb{R}^d \rightarrow \mathbb{R}^d$ is a Borel measurable function and $B_t$ is a \mbox{$d$-dimensional} Brownian motion defined on the filtered probability space $(\Omega, \mathcal{F}, \{\mathcal{F}_t\}_{t\in[0,T]}, P)$ where the filtration $\{\mathcal{F}_t\}_{t\in [0,T]}$ is the one generated by $B_t$, $t\in [0,T]$ augmented by the $P$-null sets.

If $b$ is of linear growth and Lipschitz continuous it is well-known that there exists a unique global strong solution to the SDE (\ref{SDE}) which belongs to $\mathbb{D}^{1,2}(\Omega)$. In fact, under more relaxed conditions on $b$ one has the same result, see for instance \cite{Men11}, \cite{Nil13}.

In this section we are concerned with the regularity of the solution in the Malliavin sense in terms of the regularity of $b$. We will assume the following hypotheses for $b$, for every $(t,x) \in [0,T]\times \mathbb{R}^d$
\begin{align}\label{bcondition}
\begin{split}
\sup_{t\in [0,T]}|b(t,x)|\leq C(1+|x|), C>0,\\
D b(t,\cdot), D^2 b(t,\cdot),\dots, D^k b(t,\cdot)\in L^{\infty}(\mathbb{R}^d)
\end{split}\tag{H}
\end{align}
for some $k\geq 1$ where here, the derivatives are understood in the weak sense. In particular, $b$ is $k-1$ times continuously differentiable in virtue of the Sobolev embedding (\ref{embed}) and equation (\ref{SDE}) admits a unique strong solution.

Before we proceed to the main statements of this section we need two preliminary results which are essential for our targets.

\begin{lemm}\label{epsilonbound}
Let $\{b_n\}_{n\geq 0}$ be a sequence of compactly supported smooth functions approximating $b$ a.e. in $t \in [0,T]$ and $x\in \mathbb{R}^d$ such that $\sup_{n\geq 0}|b_n(t,x)| \leq C(1+|x|)$, all $x\in \mathbb{R}^d$ and $t\in [0,T]$. Then for any compact subset $K\subset \mathbb{R}^d$ there exists an $\varepsilon>0$ such that
\begin{align}\label{epsilonboundn}
\sup_{x\in K} \sup_{n\geq 0}E\left[ \mathcal{E}\left( \int_0^T b_n(u,B_u^x)dB_u\right)^{1+\varepsilon} \right] <\infty.
\end{align}
where $B_t^x := x+B_t$ and
$$\mathcal{E}(M_t):= \exp \left\{M_t- M_0 - \frac{1}{2}[M]_t\right\}$$
denotes the Dol\'{e}ans-Dade exponential of a martingale $M$ and $[M]$ the quadratic variation of $M$. In particular we also have
\begin{align}\label{epsilonbound}
\sup_{x\in K} E\left[ \mathcal{E}\left( \int_0^T b(u,B_u^x)dB_u\right)^{1+\varepsilon} \right] <\infty.
\end{align}
\end{lemm}
\begin{proof}
Indeed, write
\begin{align*}
E&\left[\mathcal{E}\left( \int_0^T b_n(u,B_u^x)dB_u \right)^{1+\varepsilon} \right] =\\
&= E\left[\exp \left\{\int_0^T (1+\varepsilon)b_n(u,B_u^x)dB_u - \frac{1}{2}\int_0^T (1+\varepsilon)|b_n(u,B_u^x)|^2du \right\}\right]\\
&= E\Bigg[\exp \Bigg\{\int_0^T (1+\varepsilon)b_n(u,B_u^x)dB_u - \frac{1}{2}\int_0^T (1+\varepsilon)^2|b_n(u,B_u^x)|^2du\\
&+\frac{1}{2}\int_0^T \varepsilon(1+\varepsilon) |b_n(u,B_u^x)|^2du \Bigg\}\Bigg]\\
&= E\left[\exp \left\{\frac{1}{2}\int_0^T  \varepsilon(1+\varepsilon) |b_n(u,X_u^{\varepsilon,x})|^2du \right\}\right]\\
\end{align*}
where the last step follows from Girsanov's theorem and here $X_t^{\varepsilon,x}$ is a solution of the following SDE
$$
\begin{cases}
dX_t^{\varepsilon,x} = (1+\varepsilon)b_n(t,X_t^{\varepsilon,x}) dt + dB_t, \ \ t\in [0,T] \\
X_0^{\varepsilon,x} = x.
\end{cases}
$$
Observe that, since $b$ has at most linear growth, we have
$$
|X_t^{\varepsilon,x}| \leq |x| + C(1+\varepsilon)\int_0^t(1+|X_u^{\varepsilon,x}|)du + |B_t|
$$
for every $t\in [0,T]$. Then Gronwall's inequality gives
\begin{align}\label{gronw}
|X_t^{\varepsilon, x}| \leq \left( |x| + C(1+\varepsilon)T +|B_t|\right)e^{C(1+\varepsilon)T},
\end{align}
and the sublinearity of $b$ and the estimate (\ref{gronw}) give
$$|b_n(u,X_u^{\varepsilon,x})| \leq C_{\varepsilon,T}\left(1+|x| + |B_t|\right)$$
where $C_{\varepsilon,T}$ denotes the collection of all constants depending on $\varepsilon,T$.

As a result,
\begin{align*}
E\Big[ \exp\bigg\{\varepsilon(1+&\varepsilon) \int_0^T |b_n(u,X_u^{\varepsilon,x})|^2 du\bigg\}\Big] \leq E\left[\exp\left\{\tilde{C}_{\varepsilon,T} \int_0^T \left( 1+|x|+|B_u|\right)^2 du\right\}\right]\\
&\leq e^{\tilde{C}_{\varepsilon,T} (1+|x|)^2}E\left[\exp\left\{\tilde{C}_{\varepsilon,T}(1+|x|) \int_0^T (|B_u|+|B_u|^2) du\right\}\right]
\end{align*}
where $\tilde{C}_{\varepsilon,T}>0$ is a constant such that $\lim_{\varepsilon \searrow 0} \tilde{C}_{\varepsilon,T}=0$. Clearly, for every compact set $K\subset \mathbb{R}^d$ we can choose $\varepsilon>0$ small enough such that
$$\sup_{x\in K} \sup_{n\geq 0} E\Big[ \exp\bigg\{\varepsilon(1+\varepsilon) \int_0^T |b_n(u,X_u^{\varepsilon,x})|^2 du\bigg\}\Big] <\infty.$$
\end{proof}

\begin{rem}\label{remdensity}
We point out that the finite dimensional laws of the strong solution of (\ref{SDE}) are absolutely continuous with respect to the Lebesgue measure. To see this, let $A$ denote a set with null Lebesgue measure. Then since $b$ is of, at most, linear growth, by Girsanov's theorem, see e.g. \cite[Proposition 5.3.6]{Kar98}, and Lemma \ref{epsilonbound} one has
\begin{align*}
P(X_t \in A) &\leq E\left[\textbf{1}_{\{B_t^x \in A\}} \mathcal{E} \left( \int_0^T b (u,B_u^x)dB_u\right)\right]\\
&\leq C_{\varepsilon} P(B_t^x \in A)^{\frac{\varepsilon}{1+\varepsilon}}\\
&=0.
\end{align*}
\end{rem}

Next, we give a crucial estimate for the proof of our main results.

\begin{prop} \label{mainEstimate}
Let $B$ be a $d$-dimensional Brownian motion starting from $z_0\in \mathbb{R}^d$ and $b_1, \dots , b_m$ be compactly supported continuously differentiable functions $b_i : [0,T] \times \mathbb{R}^d \rightarrow \mathbb{R}$ for $i=1,2, \dots m$. Let $\alpha_i\in \{0,1\}^d$ be a multiindex such that $| \alpha_i| \leq 1$ for $i = 1,2, \dots, m$. Then there exists a universal constant $C$ (independent of $\{ b_i \}_i$, $m$, and $\{ \alpha_i \}_i$) such that
\begin{equation}
 \label{estimate}
\left| E \left[ \int_{t_0 < t_1 < \dots < t_m < t} \left( \prod_{i=1}^m D^{\alpha_i}b_i(t_i,B_{t_i})  \right) dt_1 \dots dt_m \right] \right| \leq \frac{C^m \prod_{i=1}^m \|b_i \|_{\infty} (t-t_0)^{m/2}}{\Gamma(\frac{m}{2} + 1)}
\end{equation}
for every $t_0,t\in [0,T]$ where $\Gamma$ is the Gamma-function. Here, $\alpha_i=(0,\dots, 1,\dots, 0)$ is a multiindex where $1$ is placed in position $1\leq  j\leq d$ and thus $D^{\alpha_i}$ denotes the partial derivative with respect to the $j'$th space variable.
\end{prop}
\begin{proof}
See \cite[Proposition 3.7]{Men11} for a detailed proof when $|\alpha_i|=1$ for all $i=1,\dots,m$. For the case, $|\alpha_i|\leq 1$ the proof is fairly similar.
\end{proof}

We turn now to one of the main results of this section.

\begin{thm}\label{mainprop}
Let $X_t$, $t\in[0,T]$ denote the solution to equation (\ref{SDE}) with \mbox{$b:[0,T]\times \mathbb{R}^d \rightarrow \mathbb{R}^d$} a function satisfying hypotheses $(H)_1$, i.e. of linear growth with bounded weak derivative, then we have $X_t \in \mathbb{D}^{2,p}(\Omega)$ for all $p\geq 1$. In particular the result holds if $b$ is (globally) Lipschitz.
\end{thm}
\begin{proof}
In order to carry out the proof of Theorem \ref{mainprop}, we use the following result in \mbox{\cite[Proposition 1.5.5.]{Nua10}}.
\begin{prop}
Let $\{X_n\}_{n\geq 0}$ a sequence of random variables such that $X_n \rightarrow X$ in $L^p(\Omega)$, $p\geq 1$ and such that for $k\geq 1$
$$\sup_{n\geq 0}\|X_n\|_{\mathbb{D}^{k,p}(\Omega)}<\infty,$$
then $X\in \mathbb{D}^{k,p}(\Omega)$.
\end{prop}

We start with the proof of Theorem \ref{mainprop} by showing that the solution $X_t$ of (\ref{SDE}) can be approximated by random variables in $L^p(\Omega)$ for every $t\in[0,T]$.

We have that the weak derivative of $b$ lies in $L^{\infty}(\mathbb{R}^d)$ and $b$ has linear growth, i.e. there is $C>0$ such that
$$|b(t,x)|\leq C(1+|x|)$$
for every $t\in [0,T]$ and $x\in \mathbb{R}^d$. Then we can approximate $b$ a.e. in $t\in [0,T]$ and $x\in \mathbb{R}^d$ by a sequence of functions $\{b_n\}_{n\geq 1} \subset \mathcal{C}^{2}(\mathbb{R}^d)$ such that \mbox{$\sup_{n\geq 0} |b_n(t,x)| \leq C(1+|x|)$} and \mbox{$\sup_n \|b_n'\|_{\infty}<\infty$}. For each $t\in[0,T]$, denote by $X_t^n$ the sequence of random variables in $L^p(\Omega)$ solution to equation (\ref{SDE}) with drift coefficient $b_n$. Then

$$X_t^n = x + \int_0^t b_n(u,X_u^n)du + B_t.$$

Denote by $p_{X_t}$ the density of $X_t$ for a fixed $t\in [0,T]$ from Remark \ref{remdensity}. Denote by $|\cdot|$ the Euclidean norm in $\mathbb{R}^d$, then
\begin{align*}
E\big[|X_t^n &- X_t|^p\big] = E\left[\bigg| \int_0^t \left( b_n(u,X_u^n) - b(u,X_u)\right) du \bigg|^p\right]\\
&\leq (2t)^{p-1} E\left[\int_0^t |b_n(u,X_u)-b(u,X_u)|^p du\right] + (2t)^{p-1} E\left[\int_0^t |b_n(u,X_u^n)- b_n(u,X_u)|^p du\right]\\
&\leq (2t)^{p-1} E\left[\int_0^t |b_n(u,X_u)-b(u,X_u)|^p du\right] + (2t)^{p-1} \|b_n'\|_{\infty}^p E\left[\int_0^t |X_u^n - X_u|^p du\right].
\end{align*}
Using Gronwall's inequality we obtain
\begin{align*}
E\left[|X_t^n - X_t|^p\right] &\leq (2t)^{p-1} \exp\left\{ (2t)^{p-1}t \sup_{k} \|b_k'\|_{\infty}^p\right\} E\left[\int_0^t |b_n(u,X_u)-b(u,X_u)|^p du\right] \\
&\leq C \int_0^t \int_{\mathbb{R}^d} |b_n(u,x)-b(u,x)|^p p_{X_u}(x) dx du
\end{align*}
for a constant $C>0$ independent of $n$. Then Lebesgue's dominated convergence theorem gives the $L^p(\Omega)$-convergence.

Let us now proceed with the proof that the random variables $X_t^n$ are bounded in $\mathbb{D}^{2,p}(\Omega)$ for every $p\geq 1$: Fix $s_1,t \in [0,T]$, $s_1,\leq t$. Then
\begin{align}\label{malliavin1}
D_{s_1} X_t^n = \mathcal{I}_d + \int_{s_1}^t b_n'(u,X_u^n)D_{s_1}X_u^ndu.
\end{align}
The above equations for $D_{s_1}X_u^n$, $n\geq 1$, are linear equations with matrix-valued unknowns. Since each $b_n$ is smooth we have a unique solution of (\ref{malliavin1}). Again, for notational convenience we denote by \mbox{$\Lambda_m(s,t):= \{(u_1,\dots,u_m)\in [0,T]^m: \ s<u_1<\cdots < u_m <t\}$} the $m$. Then using a Picard iteration argument we may write the solution of (\ref{malliavin1}) as a series expansion as follows
\begin{align}\label{picard}
D_{s_1} X_t^n = \mathcal{I}_d + \sum_{m\geq 1} \int_{\Lambda_m(s_1,t)} b_n'(u_1,X_{u_1}^n)\cdots b_n'(u_m ,X_{u_m}^n) du_1\cdots du_m.
\end{align}
To see that the above expression is indeed the solution of (\ref{malliavin1}) just make the following observation
$$\frac{d}{dt} D_{s_1} X_t^n = b_n'(t,X_t^n)\left(\mathcal{I}_d+ \sum_{m\geq 1} \int_{\Lambda_m(s_1,t)}b_n'(u_1,X_{u_1}^n)\cdots b_n '(u_{m}, X_{u_m}^n)du_1\cdots du_{m}\right).$$

Take now $s_2\in [0,t]$. Then
\begin{align}\label{malliavin21}
D_{s_2} D_{s_1} X_t^n =  \sum_{m\geq 1} \int_{\Lambda_m(s_1 \vee s_2,t)} D_{s_2}\left[b_n'(u_1,X_{u_1}^n)\cdots b_n'(u_m, X_{u_m}^n)\right] du_1\cdots du_m.
\end{align}
We expand the integrand of (\ref{malliavin21}) using Leibniz's rule as follows
$$ D_{s_2}\left[b_n'(u_1,X_{u_1}^n)\cdots b_n'(u_m,X_{u_m}^n)\right] = \sum_{r=1}^m b_n'(u_1,X_{u_1}^n)\cdots b_n''(u_r,X_{u_r}^n)D_{s_2}X_{u_r}^n\cdots b_n'(u_m,X_{u_m}^n).$$

\begin{rem}
We recall here that $b:[0,T]\times \mathbb{R}^d \rightarrow \mathbb{R}^d$ so $Db(t,\cdot): \mathbb{R}^d\rightarrow L(\mathbb{R}^d,\mathbb{R}^d)$ and so $Db(t,x)\in L(\mathbb{R}^d,\mathbb{R}^d)$. The second derivative is then $D^2 b(t,\cdot):\mathbb{R}^d \rightarrow L(\mathbb{R}^d,L(\mathbb{R}^d, \mathbb{R}^d))$ so $D^2b(t,x): L(\mathbb{R}^d, L(\mathbb{R}^d,\mathbb{R}^d)) \cong L^2(\mathbb{R}^d \times \mathbb{R}^d, \mathbb{R}^d)$ denoting by $L^2(\mathbb{R}^d \times \mathbb{R}^d, \mathbb{R}^d )$ the bilinear forms from $\mathbb{R}^d\times \mathbb{R}^d$ into $\mathbb{R}^d$.
\end{rem}

Inserting the representation (\ref{picard}) for $D_{s_2}X_{u_r}^n$ in this case we have that the above quantity can be written as
\begin{align*}
\sum_{r=1}^m & b_n'(u_1,X_{u_1}^n)\cdots b_n''(u_r,X_{u_r}^n)\\
&\times\left(\mathcal{I}_d+ \sum_{m\geq 1} \int_{\Lambda_m(s_2,u_r)} b_n'(v_1, X_{v_1}^n)\cdots b_n'(v_m, X_{v_m}^n) dv_1\cdots dv_m\right)b_n'(u_{r+1},X_{u_{r+1}}^n)\cdots b_n'(u_m, X_{u_m}^n).
\end{align*}
Altogether
\begin{align*}
D_{s_2} D_{s_1} X_t^n &=  \sum_{m_1\geq 1} \int_{\Lambda_{m_1}(s_1\vee s_2,t)} \sum_{r=1}^{m_1} b_n'(u_1, X_{u_1}^n)\cdots b_n''(u_r, X_{u_r}^n)\\
&\times \left(\mathcal{I}_d+ \sum_{m_2\geq 1} \int_{\Lambda_{m_2}(s_2,u_r)} b_n'(v_1, X_{v_1}^n)\cdots b_n'(v_{m_2} ,X_{v_{m_2}}^n) dv_1\cdots dv_{m_2}\right)\\
&\times b_n'(u_{r+1},X_{u_{r+1}}^n) \cdots b_n'(u_{m_1}, X_{u_{m_1}}^n) du_1\cdots du_{m_1}. \\
&= \sum_{m_1\geq 1} \int_{\Lambda_{m_1}(s_1\vee s_2,t)} \sum_{r=1}^{m_1} b_n'(u_1, X_{u_1}^n)\cdots b_n''(u_r, X_{u_r}^n) \cdots b_n'(u_{m_1}, X_{u_{m_1}}^n)du_1\cdots du_{m_1}  \\
&+ \sum_{m_1\geq 1} \int_{\Lambda_{m_1}(s_1\vee s_2,t)} \sum_{r=1}^{m_1} b_n'(u_1, X_{u_1}^n)\cdots  b_n''(u_r, X_{u_r}^n) \\
&\times  \left(\sum_{m_2\geq 1} \int_{\Lambda_{m_2}(s_2,u_r)} b_n'(v_1, X_{v_1}^n)\cdots b_n'(v_{m_2}, X_{v_{m_2}}^n) dv_1\cdots dv_{m_2}\right)\\
&\times b_n'(u_{r+1},X_{u_{r+1}}^n)\cdots b_n'(u_{m_1}, X_{u_{m_1}}^n) du_1\cdots du_{m_1}.
\end{align*}

We reallocate terms by dominated convergence and respecting the order of matrices
\begin{align}\label{malliavin22}
D_{s_2} D_{s_1} & X_t^n = \sum_{m_1\geq 1}\sum_{r=1}^{m_1} \int_{\Lambda_{m_1}(s_1\vee s_2,t)} b_n'(u_1, X_{u_1}^n)\cdots b_n''(u_r, X_{u_r}^n) \cdots b_n'(u_{m_1}, X_{u_{m_1}}^n)du_1\cdots du_{m_1}  \\
&+ \sum_{m_1\geq 1} \sum_{r=1}^{m_1}\sum_{m_2\geq 1} \int_{\Lambda_{m_1}(s_1\vee s_2,t)} \int_{\Lambda_{m_2}(s_2,u_r)} b_n'(u_1, X_{u_1}^n)\cdots  b_n''(u_r, X_{u_r}^n) \notag\\
&\times b_n'(v_1, X_{v_1}^n)\cdots b_n'(v_{m_2}, X_{v_{m_2}}^n) b_n'(u_{r+1},X_{u_{r+1}}^n)\cdots b_n'(u_{m_1}, X_{u_{m_1}}^n) dv_1\cdots dv_{m_2} du_1\cdots du_{m_1}.\notag\\ &=: I_1^n + I_2^n.\notag
\end{align}
Denote by $\|\cdot\|$ the maximum norm on $\mathbb{R}^{d\times d \times d}$. Then Minkowski's inequality gives
\begin{align*}
E\|D_{s_2}D_{s_1}X_t^n\|^p = E\| I_1^n + I_2^n \|^p\leq 2^{p-1}\left(E\|I_1^n\|^p + E\|I_2^n\|^p\right)
\end{align*}
Let $p\geq 1$ and choose $p_1,p_2\in [1,\infty)$ such that $pp_1=2^q$ for some integer $q$ and $\frac{1}{p_1}+\frac{1}{p_2}=1$. We focus now on the term $I_2^n$. Then by Girsanov's theorem we have

\begin{align}
E\|I_2^n\|^p &= E\Bigg[\bigg\| \sum_{m_1\geq 1} \sum_{r=1}^{m_1}\sum_{m_2\geq 1} \int_{\Lambda_{m_1}(s_1\vee s_2,t)} \int_{\Lambda_{m_2}(s_2,u_r)} b_n'(u_1, B_{u_1}^x)\cdots  b_n''(u_r, B_{u_r}^x) \notag\\
&\times b_n'(v_1, B_{v_1}^x)\cdots b_n'(v_{m_2}, B_{v_{m_2}}^x) b_n(u_{r+1}, B_{u_{r+1}}^x)\cdots b_n'(u_{m_1}, B_{u_{m_1}}^x) dv_1\cdots dv_{m_2} du_1\cdots du_{m_1}\bigg\|^p \notag \\
&\times \mathcal{E}\left( \sum_{i=1}^d \int_0^T b_n^{(i)}(u,B_u^x)dB_u^{(i)}\right)\Bigg] \notag.
\end{align}

Then choose $p_2=1+\varepsilon$ and $p_1=\frac{1+\varepsilon}{\varepsilon}$ with $\varepsilon>0$ sufficiently small and apply Lemma \ref{epsilonbound} to obtain

\begin{align}\label{I2}
\begin{split}
E&\|I_2^n\|^p \leq C_{\varepsilon} \bigg\| \sum_{m_1\geq 1} \sum_{r=1}^{m_1}\sum_{m_2\geq 1} \int_{\Lambda_{m_1}(s_1\vee s_2,t)} \int_{\Lambda_{m_2}(s_2,u_r)} b_n'(u_1,B_{u_1}^x)\cdots  b_n''(u_r,B_{u_r}^x)b_n'(v_1, B_{v_1}^x) \\
&\times \cdots b_n'(v_{m_2}, B_{v_{m_2}}^x) b_n(u_{r+1},B_{u_{r+1}}^x)\cdots b_n'(u_{m_1}, B_{u_{m_1}}^x) dv_1\cdots dv_{m_2} du_1\cdots du_{m_1} \bigg\|_{L^{2^q}(\Omega; \mathbb{R}^{d\times d\times d})}^p
\end{split}
\end{align}

Now we carry out the product of linear and bilinear forms in the integrand of (\ref{I2}). Recall that $b''(u,B_u^x)=\left(\frac{\partial^2}{\partial x_j \partial x_k} b^{(i)}(u,B_u^x)\right)_{i,j,k=1,\dots,d}$ and $b'(u,B_u^x)=\left(\frac{\partial}{\partial x_j} b^{(i)}(u,B_u^x) \right)_{i,j=1,\dots,d}$ where the superscript $b^{(i)}(u,B_u^x)$ here denotes the $i$-th component of the vector $b(u,B_u^x)$ and $\frac{\partial}{\partial x_j}$, resp. $\frac{\partial^2}{\partial x_j \partial x_k}$, denote the weak derivative of $b^{(i)}(u,B_u^x)$ with respect to the $j$-th space component, resp. with respect to the $j$-th and $k$-th space components. So we represent the second order derivatives as a matrix of matrices in this case, i.e. \mbox{$b''(t,x) = \nabla\otimes \nabla b(t,x)$} where $\otimes$ denotes the Kronecker tensor product.

Hence we can represent the second order derivatives in the integrand in (\ref{I2}) in this manner
\begin{align}\label{D2b}
b''(u,B_u^x)=\begin{pmatrix}
\frac{\partial}{\partial x_1}\begin{pmatrix} \frac{\partial}{\partial x_1} b^{(1)}(u,B_u^x) & \cdots & \frac{\partial}{\partial x_d} b^{(1)}(u,B_u^x) \\ \vdots & \ddots & \vdots \\ \frac{\partial}{\partial x_1} b^{(d)}(u,B_u^x) & \cdots & \frac{\partial}{\partial x_1} b^{(d)}(u,B_u^x)\end{pmatrix}\\
~\\
 \vdots\\
~\\
\frac{\partial}{\partial x_d}\begin{pmatrix} \frac{\partial}{\partial x_1} b^{(1)}(u,B_u^x) & \cdots & \frac{\partial}{\partial x_d} b^{(1)}(u,B_u^x) \\ \vdots & \ddots & \vdots \\ \frac{\partial}{\partial x_1} b^{(d)}(u,B_u^x) & \cdots & \frac{\partial}{\partial x_1} b^{(d)}(u,B_u^x)\end{pmatrix}\\
\end{pmatrix}
\end{align}

The product of $b''(B_{u})$ with $b'(B_v)$ is then
\begin{align}\label{D2bDb}
b''(u,B_u^x)b'(v,B_v^x)=\begin{pmatrix}
\frac{\partial}{\partial x_1}\begin{pmatrix} \frac{\partial}{\partial x_1} b^{(1)}(u,B_u^x) & \cdots & \frac{\partial}{\partial x_d} b^{(1)}(u,B_u^x) \\ \vdots & \ddots & \vdots \\ \frac{\partial}{\partial x_1} b^{(d)}(u,B_u^x) & \cdots & \frac{\partial}{\partial x_1} b^{(d)}(u,B_u^x)\end{pmatrix} b'(v,B_v^x)\\
~\\
 \vdots\\
~\\
\frac{\partial}{\partial x_d}\begin{pmatrix} \frac{\partial}{\partial x_1} b^{(1)}(u,B_u^x) & \cdots & \frac{\partial}{\partial x_d} b^{(1)}(u,B_u^x) \\ \vdots & \ddots & \vdots \\ \frac{\partial}{\partial x_1} b^{(d)}(u,B_u^x) & \cdots & \frac{\partial}{\partial x_1} b^{(d)}(u,B_u^x)\end{pmatrix} b'(v,B_v^x)\\
\end{pmatrix}
\end{align}
As a result
\begin{align*}
& &b''(u,B_u^x)b'(v,B_v^x)= \left( \sum_{l=1}^d \frac{\partial^2}{\partial x_k \partial x_l}b^{(i)}(u,B_u^x)\frac{\partial}{\partial x_j} b^{(l)}(v,B_v^x) \right)_{i,j,k=1}^d
\end{align*}
Hence, taking maximum norm over all products
\begin{align}\label{I2eq2}
\begin{split}
E\|I_2^n\|^p &\leq C_p \ \Bigg( \sum_{m_1\geq 1} \sum_{r=1}^{m_1}\sum_{m_2\geq 1}\sum_{i,j,k=1}^d \sum_{l_1,\dots,l_{m_1+m_2-1}=1}^d \bigg\| \int_{\Lambda_{m_1}(s_1\vee s_2,t)}\int_{\Lambda_{m_2}(s_2,u_r)} \frac{\partial}{\partial x_{l_1}}b_n^{(i)}(u_1,B_{u_1}^x) \\
&\times \frac{\partial}{\partial x_{l_2}}b_n^{(l_1)}(u_2,B_{u_2}^x)\cdots  \frac{\partial}{\partial x_{l_{r-1}}}b_n^{(l_{r-2})}(u_{r-1},B_{u_{r-1}}^x)\frac{\partial}{\partial x_k}\frac{\partial}{\partial x_{l_r}}b_n^{(l_{r-1})}(u_r,B_{u_r}^x) \\
&\times \frac{\partial}{\partial x_{l_{r+1}}}b_n^{(l_r)}(v_1,B_{v_1}^x)\cdots  \frac{\partial}{\partial x_{ l_{r+m_2} } }b_n^{ (l_{r+m_2-1} ) }(v_{m_2},B_{v_{m_2}}^x)\frac{\partial}{\partial x_{l_{r+m_2+1}}}b_n^{(l_{r+m_2})}(u_{r+1},B_{u_{r+1}}^x)\\
&\times  \cdots  \frac{\partial}{\partial x_j}b_n^{(l_{m_1+m_2-1})}(u_{m_1},B_{u_{m_1}}^x)  dv_1\cdots dv_{m_2} du_1\cdots du_{m_1}  \bigg\|_{L^{2^q}(\Omega; \mathbb{R})}\Bigg)^p.
\end{split}
\end{align}
Observe the second order partial derivatives in the integrand.

The following step is to apply expectation and get rid of the second order derivatives. To do so, we will use the estimate from Proposition \ref{mainEstimate}.

Before applying Proposition \ref{mainEstimate} we need to make the following observation on the integrating regions in connection to (\ref{I2eq2}): the iterated integrals of (\ref{I2eq2}) can be split up as a sum of integrals where the regions which we integrate over are ordered. Indeed, using formula \eqref{shuffle2} we express the term in (\ref{I2eq2}) as follows
\begin{align}\label{I2eq3}
\begin{split}
E&\|I_2^n\|^p\\
\leq&  C_p \ \Bigg( \sum_{m_1\geq 1} \sum_{r=1}^{m_1}\sum_{m_2\geq 1}\sum_{i,j,k=1}^d \sum_{l_1,\dots,l_{m_1+m_2-1}=1}^d \sum_{\sigma \in S_r(m,n)}\bigg\| \int_{\Lambda_{m_1+m_2}(s_1\vee s_2,t)} \frac{\partial}{\partial x_{l_1}}b_n^{(i)}(w_{\sigma(1)},B_{w_{\sigma(1)}}^x) \\
&\times \frac{\partial}{\partial x_{l_2}}b_n^{(l_1)}(w_{\sigma(2)},B_{w_{\sigma(2)}}^x)\cdots  \frac{\partial}{\partial x_{l_{r-1}}}b_n^{(l_{r-2})}(w_{\sigma(r-1)},B_{w_{\sigma(r-1)}}^x)\frac{\partial}{\partial x_k}\frac{\partial}{\partial x_{l_r}}b_n^{(l_{r-1})}(w_{\sigma(r)},B_{w_{\sigma(r)}}^x) \\
&\times \frac{\partial}{\partial x_{l_{r+1}}}b_n^{(l_r)}(w_{\sigma(r+1)},B_{w_{\sigma(r+1)}}^x)\cdots  \frac{\partial}{\partial x_{ l_{r+m_2} } }b_n^{ (l_{r+m_2-1} ) }(w_{\sigma(r+m_2)},B_{w_{\sigma(r+m_2)}}^x)\\
&\times\frac{\partial}{\partial x_{l_{r+m_2+1}}}b_n^{(l_{r+m_2})}(w_{\sigma(r+m_2+1)},B_{w_{\sigma(r+m_2+1)}}^x) \cdots \\
&  \cdots  \frac{\partial}{\partial x_j}b_n^{(l_{m_1+m_2-1})}(w_{\sigma(m_1+m_2)},B_{w_{\sigma(m_1+m_2)}}^x)  dw_1\cdots dw_{m_1+m_2} \bigg\|_{L^{2^q}(\Omega; \mathbb{R})}\Bigg)^p.
\end{split}
\end{align}

Now that the sets over which we integrate are symmetric we can use deterministic integration by parts to write the integrals in (\ref{I2eq3}) to the power two as a sum of at most $2^{2m}$ summands of the form
$$\int_{\Lambda_{2m}^s (s_1\vee s_2,t)} g_1(w_1)\cdots g_{2m}(w_{2m})dw_1\cdots dw_{2m}$$
where $m:=m_1+m_2$ and $g_l \in \left\{ \frac{\partial}{\partial x_j} b^{(i)}(\cdot, B_{\cdot}^x), \frac{\partial^2}{\partial x_{l_1} \partial x_{l_2}} b^{(k)}(\cdot,B_{\cdot}^x), i,j,k,l_1,l_2=1,\dots d\right\}$ and \mbox{$l=1,\dots,2m$}. Once more, we can write the integrals to the power four as a sum of at most $2^{8m}$ summands of the form
$$\int_{\Lambda_{4m}^s (s_1\vee s_2,t)} g_1(w_1)\cdots g_{4m}(w_{4m})dw_1\cdots dw_{4m}.$$
Repeating this principle, one can write the integrals to the power $2^q$ as a sum of at most $2^{q2^q m}$ summands of the form
$$ \int_{\Lambda_{2^q m}^s (s_1\vee s_2,t)} g_1(w_1)\cdots g_{2^q m}(w_{2^q m})dw_1\cdots dw_{2^q m}.$$

Combining this with Proposition \ref{mainEstimate} we obtain
\begin{align}\label{I2eq4}
\begin{split}
&\bigg\| \int_{\Lambda_{m_1+m_2}(s_1\vee s_2,t)} \frac{\partial}{\partial x_{l_1}}b_n^{(i)}(w_{\sigma(1)},B_{w_{\sigma(1)}}^x) \\
&\times \frac{\partial}{\partial x_{l_2}}b_n^{(l_1)}(w_{\sigma(2)},B_{w_{\sigma(2)}}^x)\cdots  \frac{\partial}{\partial x_{l_{r-1}}}b_n^{(l_{r-2})}(w_{\sigma(r-1)},B_{w_{\sigma(r-1)}}^x)\frac{\partial}{\partial x_k}\frac{\partial}{\partial x_{l_r}}b_n^{(l_{r-1})}(w_{\sigma(r)},B_{w_{\sigma(r)}}^x) \\
&\times \frac{\partial}{\partial x_{l_{r+1}}}b_n^{(l_r)}(w_{\sigma(r+1)},B_{w_{\sigma(r+1)}}^x)\cdots  \frac{\partial}{\partial x_{ l_{r+m_2} } }b_n^{ (l_{r+m_2-1} ) }(w_{\sigma(r+m_2)},B_{w_{\sigma(r+m_2)}}^x)\\
&\times\frac{\partial}{\partial x_{l_{r+m_2+1}}}b_n^{(l_{r+m_2})}(w_{\sigma(r+m_2+1)},B_{w_{\sigma(r+m_2+1)}}^x) \cdots \\
&  \cdots  \frac{\partial}{\partial x_j}b_n^{(l_{m_1+m_2-1})}(w_{\sigma(m_1+m_2)},B_{w_{\sigma(m_1+m_2)}}^x)  dw_1\cdots dw_{m_1+m_2} \bigg\|_{L^{2^q}(\Omega; \mathbb{R})}  \\
&\leq \left(\frac{2^{q2^q (m_1+m_2)}C_{d,p,T}^{2^q (m_1+m_2)} \|b_n'\|_{\infty}^{2^q (m_1+m_2)}|t-(s_1 \vee s_2)|^{2^{q-1}(m_1+m_2)}}{\Gamma\left( 2^{q-1}(m_1+m_2)+1\right)}\right)^{2^{-q}} \\
&=\frac{2^{q(m_1+m_2)}C_{d,p,T}^{m_1+m_2}\|b_n'\|_{\infty}^{m_1+m_2}|t-(s_1 \vee s_2)|^{(m_1+m_2)/2}}{\left[\left(2^{q-1}(m_1+m_2)\right)!\right]^{2^{-q}}}
\end{split}
\end{align}

Using the bound in (\ref{I2eq4}) we get
\begin{align*}
E&\|I_2^n\|^p \leq\\
&\leq \Bigg( \sum_{m_1\geq 1} \sum_{r=1}^{m_1}\sum_{m_2\geq 1} C^{m_1+m_2}\frac{d^{m_1+m_2+2}2^{q(m_1+m_2)}C_{d,p,T}^{m_1+m_2} \|b_n'\|^{m_1+m_2}|t-(s_1 \vee s_2)|^{(m_1+m_2)/2}}{\left[\left(2^{q-1}(m_1+m_2)\right)!\right]^{2^{-q}}} \Bigg)^p\\
&\leq \Bigg( \sum_{m_1\geq 1}\sum_{m_2\geq 1} m_1C^{m_1+m_2}\frac{d^{m_1+m_2+2}2^{q(m_1+m_2)}C_{d,p,T}^{m_1+m_2} \|b_n'\|^{m_1+m_2}|t-(s_1 \vee s_2)|^{(m_1+m_2)/2}}{\left[\left(2^{q-1}(m_1+m_2)\right)!\right]^{2^{-q}}} \Bigg)^p\\
&\leq \Bigg( \sum_{m\geq 1} mC^m\frac{d^{m+2}2^{qm}C_{d,p,T}^{m} \|b_n'\|_{\infty}^{m}|t-(s_1 \vee s_2)|^{m/2}}{\left[\left(2^{q-1}m\right)!\right]^{2^{-q}}} \Bigg)^p\\
&\leq C_{d,p,T} f(\|b_n'\|_{\infty})
\end{align*}
for some continuous function $f$ only depending on $d$, $p$ and $T$. As a result,
$$\sup_{n\geq 0}\sup_{s_1,s_2\in [0,T]} E\|I_2^n\|^p \leq C_{d,p,T} \sup_{n\geq 0}\sup_{s_1,s_2\in [0,T]} f(\|b_n'\|_{\infty}) < \infty.$$

Finally, one can bound $E\|I_1^n\|^p$ using exactly the same steps as for $I_2^n$.
\end{proof}

We are now in a position to state one of the main results of this section on the Malliavin regularity of the solution to SDE (\ref{SDE}).

\begin{thm}\label{mainthm}
Assume that $b$ satisfies condition (\ref{bcondition}) for some $k\geq 1$. Let $X_t$, $t\in[0,T]$ denote the solution to equation (\ref{SDE}). Then
$$X_t \in \bigcap_{p\geq 1} \mathbb{D}^{k+1,p}(\Omega).$$
\end{thm}
\begin{proof}
The proof of this more general result relies on Theorem \ref{mainprop} by iterating all arguments up to $k+1$. Similarly as before, let $\{b_n\}_{n\geq 1} \subset \mathcal{C}^{k+1}(\mathbb{R}^d)$ be an approximating sequence of functions such that $b_n \to b$ a.e. in $t\in [0,T]$ and $x\in \mathbb{R}^d$ as $n\to \infty$ and $\sup_{n\geq 0} |b_n(t,x)|\leq C(1+|x|)$ and $\sup_n \|b_n^{(j)}\|<\infty$, $j\leq k$ where $\|\cdot\|$ denotes any norm in $\mathbb{R}^d \times \overset{j)}{\dots} \times \mathbb{R}^d$, $j\leq k$. For each $t\in[0,T]$, denote by $X_t^n$ the sequence of random variables in $L^p(\Omega)$ solution to equation (\ref{SDE}) with drift coefficient $b_n$. Then we wish to compute the Malliavin derivative of $X_t^n$ up to order $k+1$. This becomes a large expression where the terms increase at a binomial speed. We saw in the proof of Proposition (\ref{mainprop}) that the second order Malliavin derivative of $X_t^n$ can be written as $D_{s_2}D_{s_1}X_t^n = I_1^n+I_2^n$ where the integrals in $I_2^n$ are doubled. If we fix $s_3\in[0,t]$ then $D_{s_3}D_{s_2}D_{s_1} X_t^n = D_{s_3} I_1^n + D_{s_3} I_2^n = I_1^n + I_2^n + I_3^n + I_4^n$ and so on. Each term $I_i^n$, $i=1,2,3,4$ is a sum of integrals of the form (\ref{malliavin22}) with at most one factor $b_n^{(3)}$. Iterating this argument, we have that for fixed $s_1,\dots,s_{k+1}\in [0,t]$
$$D_{s_{k+1}}\cdots D_{s_1}X_t^n = I_1^n + \dots + I_{2^{k}}^n$$
where each $I_i^n$, $i=1,2,\dots, 2^k$ is an integral over at most $\Lambda_{m_1+\dots+m_{k+1}}$ with at most one factor $b_n^{(k+1)}$ and the rest $b_n^{(j)}$, $j \leq k$. This can be readily checked by looking at expression (\ref{malliavin22}). Then, estimating $I_{2^{k}}^n$ implies that all former terms are also bounded.

To illustrate $I_{2^{k}}^n$ we use expression (\ref{malliavin22}) and apply $D_{s_3}\cdots D_{s_{k+1}}$ and focus on the last term. In order to simplify notation and make the reading clearer we consider indices $m_1,\dots,m_{k+1},r_1,\dots, r_{k}\in \mathbb{N}\setminus \{0\}$ and denote
$$\sum_{\substack{m_1,\dots,m_{k+1}\\r_1,\dots,r_{k}}} :=\sum_{m_1\geq 1}\sum_{r_1=1}^{m_1} \sum_{m_2\geq 1}\sum_{r_2=1}^{m_1+m_2}\cdots \sum_{r_{k}=1}^{m_1+\cdots+m_{k}}\sum_{m_{k+1}\geq 0},$$
as well as,
$$\int_{\Delta} := \int_{\Lambda_{m_1}(s_1\vee\dots\vee s_{k+1},t)}\int_{\Lambda_{m_2}(s_2\vee\dots\vee s_{k+1},t)}\cdots \int_{\Lambda_{m_{k+1}}(s_{k+1},t)}.$$
Then $I_{2^{k}}^n$ will take the following form
\begin{align*}
I_{2^{k}}^n &= \sum_{\substack{m_1,\dots,m_{k+1}\\r_1,\dots,r_{k}}} \int_{\Delta} \mathcal{A}(u_1^1,\dots,u_{m_1}^1,\dots, u_{1}^{k+1},\dots, u_{m_{k+1}}^{k+1}) du_1^{k+1}\cdots du_{m_{k+1}}^{k+1} \cdots du_1^1 \cdots du_{m_1}^1
\end{align*}
with integrand
\begin{align*}
\mathcal{A}:= g_n(u_1^1)\cdots g_n(u_{r_1}^1)\bigg[g_n(u_1^2)\cdot g_n(u_{r_2}^2)&\bigg[\cdots g_n(u_1^{k+1})\cdots\\
&\cdots g_n(u_{m_{k+1}}^{k+1})\bigg]g_n(u_{r_{k}+1}^{k})\cdots g_n(u_{m_2}^2)\bigg]g_n(u_{r_1+1}^1)\cdots g_n(u_{m_1}^1)
\end{align*}
where the functions $g_n$ denote an element in the set
$$g_n\in \{D b_n, D^2 b_n, \cdots, D^{k+1} b_n\}.$$
Then, using exactly the same procedure as for $I_1^n$ and $I_2^n$, mutatis mutandis, we obtain an integral of products of partial derivatives of at most order $k+1$, this together with Proposition \ref{mainEstimate} one is able to get rid of the $k+1$-th derivative as we did for $I_2^n$ in Theorem \ref{mainthm}.
\end{proof}

To emphasize that the solution depends on the initial point $x$ we write $X_t^x$. Next result gives a condition for the regularity of $x\mapsto X_t^x$ in the space variable in the Sobolev sense.

\begin{thm}\label{sobreg}
Assume that $b$ satisfies condition (\ref{bcondition}) for some $k\geq 1$. Let $U\subset \mathbb{R}^d$ be an open bounded set and $X_t$, $t\in[0,T]$ denote the solution to equation (\ref{SDE}). Then
$$X_t^{\cdot} \in \bigcap_{p\geq 1} L^2\big(\Omega, W^{k+1,p}(U)\big).$$
\end{thm}
\begin{proof}
This result actually follows by observing that the process $\frac{\partial}{\partial x} X_t^x$ satisfies the following linear ODE
$$\frac{\partial}{\partial x} X_t^x = \mathcal{I}_d + \int_0^t b'(u, X_u^x)\frac{\partial}{\partial x}X_u^x du.$$

This equation is the same as (\ref{malliavin1}) when $s=0$. Using this observation, in connection with the same method employed in the proof of Theorem \ref{mainthm} by replacing the Malliavin derivative of $X_t$ with $\frac{\partial}{\partial x}X_t^x$ we get that for the approximating sequence of solutions $X_t^{n,x}$, $n\geq 0$ described in Theorem \ref{mainthm} we have
$$\sup_{n\geq 0}\sup_{x\in \overline{U}} E\left[\|\frac{\partial^{j}}{\partial x^{j}} X_t^{n,x}\|^p \right] <\infty$$
for all $j=0,\dots,k+1$ and any $p\geq 1$ so $X_t^{n,\cdot}$ is bounded in the Sobolev norm $ L^2(\Omega, W^{k+1,p}(U))$ for each $n\geq 0$. Indeed
$$\sup_{n\geq 0} \|X_t^{n,\cdot}\|_{L^2(\Omega, W^{k+1,p}(U))}^2 = \sup_{n\geq 0}\sum_{i=0}^{k+1} E\left[\|\frac{\partial^i}{\partial x^i}X_t^{n,\cdot}\|_{L^p(U)}^2\right]\leq \sum_{i=0}^{k+1} \int_{U} \sup_{n\geq 0}E\left[ \|\frac{\partial^i}{\partial x^i}X_t^{n,x}\|^p \right]dx <\infty.$$
Since $L^2(\Omega, W^{k+1,p}(U))$ is reflexive, by Banach-Alaoglu's theorem we get that the set $\{X_t^{n,x}\}_{n\geq 0}$ is weakly compact in the $L^2(\Omega, W^{k+1,p}(U))$ topology. Thus, there exists a subsequence $n(j)$, $j\geq 0$ such that
$$X_t^{n(j),\cdot}  \xrightarrow[j\to \infty]{w} Y \in L^2(\Omega, W^{k+1,p}(U)).$$

On the other hand, we have that $X_t^{n,x} \to X_t^x$ strongly in $L^p(\Omega)$, so by uniqueness of the limit we can conclude that
$$X_t^{\cdot} =Y, \ \ P-a.s.$$
\end{proof}

\begin{rem}
The previous result actually gives classical derivatives of the solution up to order $k+\alpha$ with $\alpha\in (0,1)$ as a consequence of the Sobolev embedding (\ref{embed}).
\end{rem}

\section{Application to the regularity of densities}

As mentioned in the introduction one implication of improving the Malliavin regularity of SDEs with drift coefficient satisfying hypotheses \textbf{(H)} is that the finite dimensional laws have $k$-times differentiable densities due to a result by V.Bally and L.Caramellino, see \cite{Bally2011}. We see this as an improvement of the regularity condition given by \cite{Kus82} for the additive noise case. In addition, we see that the boundedness of $b$ is not needed.

The following is a consequence of Theorem \ref{mainthm} for the special case $d=1$ and illustrates how we may gain regularity of the densities of solutions to (\ref{SDE}) and provide with an explicit expression for the density and its derivatives. Later on, we will show it for higher dimensions.

\begin{cor}
For $d=1$, let $p_{X_t}$ denote the density of the solution $X_t$ to equation (\ref{SDE}) for a given \mbox{$t\in [0,T]$}. If $b$ satisfies (\ref{bcondition}) for some $k\geq 1$ then $p_{X_t} \in C^{k-1}(\mathbb{R})$ for every $t\in [0,T]$.
\end{cor}
\begin{proof}
Let $G_0, G_1,G_2,\dots,G_k$ be the random variables defined as $G_0=1$ and for each $i=1,\dots,k$
$$G_i=\delta \left(G_{i-1}\cdot\left(\int_0^T D_s X_t ds\right)^{-1}\right),$$
where $\delta$ denotes the Skorokhod integral as introduced in \eqref{skorokhod}.

It is known that if $X_t\in \mathbb{D}^{1,2}(\Omega)$, $\int_0^T D_s X_t ds \neq 0$, $P-a.s.$ and \mbox{$G_i\left( \int_0^T D_s X_t ds\right)^{-1}\in \mbox{Dom}(\delta)$} for each $i=0,\dots,k$ then $X_t$ has a density of class $C^k(\mathbb{R})$ and
\begin{align}\label{density}
\frac{d^i}{dy^i}p_{X_t}(y) = (-1)^i E\left[ \textbf{1}_{\{X_t>y\}} G_{i+1}\right]
\end{align}
for each $i=0,\dots,k$. See \cite[p115]{Nua10}.

We will then prove that $G_i \left(\int_0^T D_s X_t ds\right)^{-1}\in \mbox{Dom}(\delta)$ for $i=0,\dots,k-1$. First, observe that for dimension $d=1$ we can easily solve the linear SDE for $D_s X_t$ and write
\begin{align}\label{mallexp}
D_s X_t = \exp\left\{\int_s^t b'(u,X_u)du\right\}
\end{align}
where $b'$ denotes the weak derivative of $b$ (one may also use local time to express (\ref{mallexp}) independently of $b'$ if $b$ is non-regular, see \cite{Ein2000}). Hence, for any $t\in [0,T]$ there is an $\varepsilon >0$ such that \mbox{$\int_0^T D_s X_tds \geq \varepsilon >0$}. Since $x\mapsto \frac{1}{x}$ is smooth on the domain $(\varepsilon, \infty)$ we see that \mbox{$\left(\int_0^T D_s X_t ds\right)^{-1} \in \mathbb{D}^{k,2}(\Omega)$} since $X_t \in \mathbb{D}^{k+1,2}(\Omega)$ by Theorem \ref{mainthm}. Denote $F:=\left(\int_0^T D_s X_t ds\right)^{-1}$.

Now, since $F\in \mathbb{D}^{1,2}(\Omega)$ we have $F\in \mbox{Dom}(\delta)$ and $G_1=\delta(F) = FW(T) + \int_0^T D_s F ds$. Then we see that $G_1\in \mathbb{D}^{1,2}(\Omega)$ and hence $G_1F\in \mathbb{D}^{1,2}(\Omega)$ therefore $G_1 F\in \mbox{Dom}(\delta)$ with \mbox{$G_2= \delta(G_1F)= G_1FW(T) - \int_0^T [D_s G_1 F + G_1 D_sF]ds$}. Again, it is readily checked that \mbox{$G_2\in \mathbb{D}^{1,2}(\Omega)$} since \mbox{$G_1,F\in \mathbb{D}^{1,2}(\Omega)$} so $G_2F\in \mbox{Dom}(\delta)$. For a fixed $i=0,\dots,k-1$ we have $G_i,F\in \mathbb{D}^{1,2}(\Omega)$ therefore $G_iF\in \mbox{Dom}(\delta)$ with $G_{i+1}=\delta(G_iF)= G_iFW(T) - \int_0^T [D_sG_i F + G_i D_sF]ds$. So $G_i$ is well-defined for $i=0,\dots,k$ but we can not say anything about $G_{k+1}$ so $p_{X_t}$ is at least $k-1$-times differentiable with derivatives given by (\ref{density}).
\end{proof}

As a consequence of the Malliavin regularity we have shown for SDEs of the form (\ref{SDE}) we may apply the results by V.Bally and L.Caramellino, see \cite{Bally2011}, to be able to obtain regularity of the densities, also in higher dimension. In order to do so, we need to study integrability properties of the Malliavin covariance matrix. Let us denote
$$\gamma_{X_t}^{ij} := \langle D_{\cdot} X_t^{(i)}, D_{\cdot} X_t^{(j)}\rangle_{H}, \ \ i,j=1,\dots,d,$$
the Malliavin covariance matrix of the process $X_t$, given $t\in [0,T]$. We will say that \mbox{$\gamma_{X_t} = (\gamma_{X_t}^{ij})_{i,j=1,\dots,d}$} satisfies the \emph{non-degeneracy condition} whenever
\begin{align}\label{nondeg}
(\det \gamma_{X_t})^{-1} \in \bigcap_{p\geq 1} L^p(\Omega).
\end{align}

Next, we invoke a result by \cite[Proposition 23]{Bally2011} which gives us the desired properties on the density of $X_t$, $t\in [0,T]$.

\begin{prop}
Let $F=(F^1,\dots,F^d)$ with $F^1,\dots,F^d \in \bigcap_{p\geq 1} \mathbb{D}^{k+1,p}(\Omega)$. Assume that condition (\ref{nondeg}) holds for $\gamma_F$. Denote by $p_F$ the density of $F$. Then $p_F \in C^{k-1,\alpha}(\mathbb{R}^d)$ with $\alpha<1$, i.e. $p_F$ is $k-1$-times differentiable with H\"{o}lder continuous derivatives of exponent $\alpha<1$.
\end{prop}

In view of the above result we only need to check that the non-degeneracy condition (\ref{nondeg}) is fulfilled. To do so, we use the following intermediate result.

\begin{lemm}\label{lemmaepsilon}
Let $Z:\Omega \rightarrow E$ be a random variable taking values on a separable Banach space with norm $\|\cdot\|_E$. Fix $p>0$. Then the following are equivalent
\begin{itemize}
\item[(i)]
\begin{align}
E\left[\|Z\|_E^{-p}\right] <\infty.
\end{align}
\item[(ii)] There exists $\varepsilon_0 >0$, depending on $p$, such that
$$\int_0^{\varepsilon_0} \varepsilon^{-(p+1)} P(\|Z\|_E^2< \varepsilon) d\varepsilon <\infty.$$
\end{itemize}
\end{lemm}
\begin{proof}
We have that, for any positive integrable random variable $Y$,
$$E[Y] = \int_0^\infty P(Y>\eta) d\eta.$$

Condition $(i)$ implies that $\|Z\|_E >0$ $P$-a.s. so
\begin{align*}
E\left[\|Z\|_E^{-2p}\right] &= \int_0^{\eta_0} P(\|Z\|_E^{-2p}>\eta) d\eta  + \int_{\eta_0}^\infty P(\|Z\|_E^{-2p}>\eta) d\eta\\
&\leq \eta_0 +  \int_{\eta_0}^\infty P(\|Z\|_E^{-2p}>\eta) d\eta\\
&= \eta_0 + p\int_0^{\eta_0^{-1/p}} \varepsilon^{-(p+1)} P(\|Z\|_E^2 <\varepsilon)d\varepsilon
\end{align*}
where in the last step we have used the change of variables $\eta = \varepsilon^{-p}$.
\end{proof}

Now we are in a position to prove the non-degeneracy condition for the Malliavin matrix associated to the solution of the SDE (\ref{SDE}). The proof of this result is much inspired in Proposition 8.1 from \cite{MSS05}.
\begin{prop}
Let $X_t$, $t\in [0,T]$ be the solution to SDE (\ref{SDE}) with drift coefficient $b$ satisfying condition (\ref{bcondition}) for $k=1$. Then the Malliavin covariance matrix $\gamma_{X_t}$ satisfies
$$(\det \gamma_{X_t})^{-1} \in \bigcap_{p\geq 1} L^p(\Omega)$$
\end{prop}
\begin{proof}
Consider $X_t^n$ with drift coefficient $b_n$ approximating $b$ a.e. such that $\sup_{n\geq 0} \|b_n'\|_{\infty}<\infty$.

It suffices to show that
$$\sup_{n\geq 0} E \left[\bigg| \int_0^T \|D_s X_t^n\|_{\infty}^2 ds \bigg|^{-p}\right]<\infty$$
for any $p\geq 1$.

Recall that for $0\leq s\leq t$, $t\in [0,T]$ we have
$$D_s X_t^n = \mathcal{I}_d + \sum_{m\geq 1} \int_{s<u_1<\cdots <u_m<t} b_n'(u_1,X_{u_1}^n) \cdots b_n'(u_m, X_{u_m}^n)du_1 \cdots du_m.$$

Then for any $\delta >0$, $t-\delta >0$ one has
\begin{align*}
\int_0^T \|D_s X_t^n \|_\infty^2 ds \geq \int_{t-\delta}^t \|D_s X_t^n\|_\infty^2 ds \geq \frac{\delta}{2} - I_n(t,\delta)
\end{align*}
where
$$I_n(t,\delta):= \int_{t-\delta}^t \bigg\|\sum_{m\geq 1} \int_{s<u_1<\cdots <u_m<t} b_n'(u,X_{u_1}^n)\cdots b_n'(u,X_{u_m}^n)du_1\cdots du_m\bigg\|_{\infty}^2ds.$$

Clearly, we have
\begin{align}\label{deltap}
\sup_{n\geq 0} E\left[|I_n(t,\delta)|^p\right] \leq C \delta^p
\end{align}
since $b_n'$, $n\geq 0$ are uniformly bounded.

Then by the previous estimates
\begin{align*}
P\left(\|D_{\cdot} X_t^n\|_{L^2(\Omega, \mathbb{R}^{d\times d})}^2 <\varepsilon\right) &\leq P\left( \int_{t-\delta}^t \|D_s X_t^n\|_\infty^2 ds <\varepsilon\right) \\
&\leq P\left( I_n(t,\delta) \geq \frac{\delta}{2} - \varepsilon\right)\\
&\leq \left(\frac{\delta}{2}-\varepsilon\right)^{-p} E[|I_n(t,\delta)|^p]
\end{align*}
for any $p\geq 1$ due to Chebyshev's inequality. Now, by estimate (\ref{deltap}) we obtain that
\begin{align*}
\sup_{n\geq 0} P\left(\|D_{\cdot} X_t^n\|_{L^2(\Omega, \mathbb{R}^{d\times d})}^2 <\varepsilon\right)&\leq C \left(\frac{\delta}{2}-\varepsilon\right)^{-p}\delta^p.
\end{align*}

By virtue of Lemma \ref{lemmaepsilon} we can conclude if we find $\delta:(0,\infty) \rightarrow \mathbb{R}$, $\varepsilon \mapsto \delta(\varepsilon)$ such that $\lim_{\varepsilon \searrow 0} \delta(\varepsilon) = 0$ and
$$\int_0 \varepsilon^{-(p+1)}\left(\frac{\delta (\varepsilon)}{2}-\varepsilon\right)^{-p}\delta (\varepsilon)^p d\varepsilon < \infty$$
for an arbitrary large $p\geq 1$.

We claim that
$$\delta (\varepsilon) := \left|\frac{2\varepsilon^{\frac{1}{2p}+2}}{\varepsilon^{\frac{1}{2p}+1}-2}\right|$$
does the job.
\end{proof}

Finally, we are able to state our conditions to determine the regularity of densities of solutions to SDEs.

\begin{cor}\label{maincor}
Let $X_t^x$, $t\in [0,T]$ be the strong solution to SDE (\ref{SDE}). Assume $b$ satisfies condition (\ref{bcondition}) for some integer $k\geq 1$. Then the density $p_{X_t}$ belongs to $C^{k-1,\alpha}(\mathbb{R}^d)$, $\alpha<1$, i.e. $k-1$-times continuously differentiable with H\"{o}lder continuous derivatives with exponent $\alpha<1$.
\end{cor}

We end this section by giving an example that shows that the Malliavin regularity we obtained in Theorem \ref{mainthm} is optimal when $k=1$, for the general condition we conjecture it is also optimal.

\begin{exam}
\normalfont
In this example we show that Theorem \ref{mainprop} is an optimal result in the sense that, if $b$ is of linear growth and one time weakly differentiable with bounded derivative then $X_t\in \mathbb{D}^{2,p}(\Omega)$ for all $p\geq 1$ and $X_t\notin \mathbb{D}^{3,p}(\Omega)$ for any $p\geq 1$. Just choose $b$, in dimension $d=1$, to be such that
$$b'(x) = \textbf{1}_{(0,\infty)}(x), \ x\in \mathbb{R}.$$
Then fix $t\in [0,T]$ and for $s_1\leq t$
$$D_{s_1} X_t = \exp\left\{\int_{s_1}^t b'(X_u)du\right\}.$$
Denote by $\tilde{b}(x) := b(a) + \int_a^x b(y)dy$, $a\in \mathbb{R}$ a primitive of $b$. It\^{o}'s formula implies
$$D_{s_1} X_t = \exp\left\{2\tilde{b}(X_t) - 2\tilde{b}(X_{s_1})-2\int_{s_1}^t b^2(X_u)du-2\int_{s_1}^t b(X_u)dB_u\right\}.$$
Then by Theorem \ref{mainthm}, $D_{s_1}X_t \in \mathbb{D}^{1,2}(\Omega)$ for all $t\in [0,T]$. So for $s_2\leq t$
\begin{align}\label{twoMall}
\begin{split}
D_{s_2}D_{s_1} X_t &=  D_{s_1}X_t \bigg(2b(X_t)D_{s_2}X_t - 2b(X_{s_1}) D_{s_2}X_{s_1})-4\int_{s_1\vee s_2}^t b(X_u)b'(X_u)D_{s_2}X_u du -2 b(X_{s_2})\\
&-2\int_{s_1\vee s_2}^t b'(X_u)D_{s_2}X_u dB_u ) \bigg).
\end{split}
\end{align}
Now observe that $b(X_t)\in \cap_{p\geq 1} \mathbb{D}^{1,p}(\Omega)$ for all $t\in [0,T]$ so all terms are \mbox{immediately} Malliavin differentiable with all moments except from maybe $\int_{s_1\vee s_2}^t b(X_u)b'(X_u)D_{s_2}X_u du$ and $\int_{s_1\vee s_2}^t b'(X_u)D_{s_2}X_u dB_u$. The stochastic integral is in fact not Malliavin differentiable. Indeed, by \cite[Lemma 1.3.4]{Nua10}
\begin{align}\label{stochint}
\int_{s_1\vee s_2}^t b'(X_u)D_{s_2}X_u dB_u \in \mathbb{D}^{1,2}(\Omega)
\end{align}
if, and only if
$$b'(X_u)D_{s_2}X_u \in \mathbb{D}^{1,2}(\Omega).$$
On the other hand we have $b'(X_u)=\textbf{1}_{(0,\infty)}(X_u) \notin \mathbb{D}^{1,2}(\Omega)$ since $0<P(0<X_u<\infty)<1$, see \mbox{\cite[Proposition 1.2.6]{Nua10}}, and so $D_{s_1}X_t\int_{s_1\vee s_2}^t b'(X_u)D_{s_2}X_u dB_u \notin \mathbb{D}^{1,2}(\Omega)$.

Let us finally prove that
$$Y_t := \int_{s_1\vee s_2}^t b(X_u)b'(X_u)D_{s_2}X_u du \in \mathbb{D}^{1,2}(\Omega).$$
Let $\{b_n\}_{\{n\geq 0\}}$ be a sequence of smooth functions such that $b_n(x) \to b(x)$ a.e. in $x\in \mathbb{R}$ as $n\to \infty$ and $\sup_{n\geq 0}\|b_n'\|_{\infty}<\infty$ and $b_n(x),b_n'(x),b_n''(x)\geq 0$ for all $x\in \mathbb{R}$, we claim that this is trivially possible by the very concrete shape of the function $b$ in this example. Define

$$Y_t^n := \int_{s_1\vee s_2}^t b(X_u)b_n'(X_u)D_{s_2}X_u du.$$
Clearly, $Y_t^n \rightarrow Y_t$ in $L^2(\Omega)$ for all $t\in [0,T]$. We only need to bound $\|D_{\cdot} Y_t^n\|_{L^2([0,T]\times \Omega)}$ uniformly in $n\geq 0$. Then
\begin{align*}
D_{s_3} Y_t^n &= \int_{s^\ast}^t b'(X_u)D_{s_3} X_u b_n'(X_u)D_{s_2}X_u du\\
&+ \int_{s^\ast}^t b(X_u)b_n''(X_u)D_{s_3}X_u D_{s_2}X_u du+\int_{s^\ast}^t b(X_u)b_n'(X_u)D_{s_3}D_{s_2}X_u du
\end{align*}
where $s^\ast := \max\{s_1,s_2,s_3\}$.

Then the critical term is
\begin{align*}
I_n := E\left[\int_0^T\left(\int_{s^\ast}^t b(X_u)b_n''(X_u)D_{s_3}X_u D_{s_2}X_u du\right)^2 ds_3 \right]
\end{align*}
Denote $\tilde{B}_{s,t}:= \exp\left\{\int_s^t b'(B_u^x)du \right\}$. Then, by Girsanov's theorem and Lemma \ref{epsilonbound} we have for a suitable $\varepsilon>0$
\begin{align*}
I_n&=\int_0^T E\left[\left(\int_{s^\ast}^t b(B_u^x)b_n''(B_u^x)\tilde{B}_{s_3,u} \tilde{B}_{s_2,u} du\right)^2 \ \mathcal{E}\left(\int_0^T b(B_u^x)dB_u\right)\right]ds_3\\
&\leq C_{\varepsilon} \int_0^T E\left[\left(\int_{s^\ast}^t b(B_u^x)b_n''(B_u^x)\tilde{B}_{s_3,u} \tilde{B}_{s_2,u} du\right)^{2\frac{1+\varepsilon}{\varepsilon}}\right]^{\frac{\varepsilon}{1+\varepsilon}} ds_3.
\end{align*}
Now we focus on the expectation. Choose $\varepsilon>0$ so that $p:= 2\frac{1+\varepsilon}{\varepsilon}$ is a natural number. Then since $|\tilde{B}_{s_3,{u_i}} \tilde{B}_{s_2,{u_i}}|\leq e^{(|t-s_3|+|t-s_2|)\|b'\|_{\infty}}\leq C<\infty$ and since $b$ and $b_n''$ are positive we have
\begin{align*}
\Bigg|E\bigg[&\int_{s^\ast}^t\cdots \int_{s^\ast}^t \prod_{i=1}^p b(B_{u_i}^x)b_n''(B_{u_i}^x)\tilde{B}_{s_3,{u_i}} \tilde{B}_{s_2,{u_i}} du_1 \cdots du_p\bigg]\Bigg| \leq \\
&\leq E\bigg[\int_{s^\ast}^t\cdots \int_{s^\ast}^t \prod_{i=1}^p \big| b(B_{u_i}^x)b_n''(B_{u_i}^x)\tilde{B}_{s_3,{u_i}} \tilde{B}_{s_2,{u_i}}\big| du_1 \cdots du_p\bigg]\\
&\leq C E\bigg[\int_{s^\ast}^t\cdots \int_{s^\ast}^t \prod_{i=1}^p \big| b(B_{u_i}^x)b_n''(B_{u_i}^x)\big| du_1 \cdots du_p\bigg]\\
&= C E\bigg[\int_{s^\ast}^t\cdots \int_{s^\ast}^t \prod_{i=1}^p b(B_{u_i}^x)b_n''(B_{u_i}^x) du_1 \cdots du_p\bigg].
\end{align*}
Then since $(u_1,\dots,u_p)\mapsto b(B_{u_1}^x)b_n''(B_{u_1}^x)\cdots b(B_{u_p}^x)b_n''(B_{u_p}^x)$ is symmetric we may write
\begin{align*}
E\bigg[\int_{s^\ast}^t\cdots \int_{s^\ast}^t \prod_{i=1}^p b(B_{u_i}^x)b_n''(B_{u_i}^x)& du_1 \cdots du_p\bigg]\\
&\leq p! E\bigg[\int_{s^\ast<u_1<\cdots <u_p<t} \prod_{i=1}^p b(B_{u_i}^x)b_n''(B_{u_i}^x) du_1 \cdots du_p\bigg]
\end{align*}
and the last may be bounded independently of $b_n''$ by using Proposition \ref{mainEstimate}. In fact,
$$\sup_{s^\ast \in [0,T]} \sup_{n\geq 0} E\bigg[\int_{s^\ast<u_1<\cdots <u_p<t} \prod_{i=1}^p b(B_{u_i}^x)b_n''(B_{u_i}^x) du_1 \cdots du_p\bigg] \leq C$$
for a finite constant $C$. So
$$\sup_{n\geq 0} I_n <\infty$$
being thus $Y_t \in \mathbb{D}^{1,2}(\Omega)$ for every $t\in [0,T]$.

In a summary, we have in (\ref{twoMall}) a sum of Malliavin differentiable terms except for the last one $-4D_{s_1} X_t \int_{s_1\vee s_2}^t b'(X_u)D_{s_2}X_u dB_u$ . In conclusion $D_{s_2}D_{s_1} X_t \notin \mathbb{D}^{1,2}(\Omega)$.
\end{exam}

Finally, we give an extension of Theorem \ref{mainthm} and Corollary \ref{maincor} to a class of non-degenerate $d-$dimensional It\^{o}-diffusions$.$
\begin{thm}
\label{generalsde}Consider the time-homogeneous $\mathbb{R}^{d}-$valued SDE%
\begin{equation}
dX_{t}=b(X_{t})dt+\sigma (X_{t})dB_{t},\,\,\text{ }X_{0}=x\in \mathbb{R}^{d},%
\text{ }\,\,0\leq t\leq T,  \label{SDE1}
\end{equation}%
where the coefficients $b:\mathbb{R}^{d}\longrightarrow \mathbb{R}^{d}$ and $%
\sigma :\mathbb{R}^{d}\longrightarrow \mathbb{R}^{d}\times $ $\mathbb{R}^{d}$%
are Borel measurable. Require that there exists a bijection $\Lambda :%
\mathbb{R}^{d}\longrightarrow \mathbb{R}^{d}$, which is twice continuously
differentiable. Let \mbox{$\Lambda _{x}:\mathbb{R}^{d}\longrightarrow L\left(
\mathbb{R}^{d},\mathbb{R}^{d}\right) $} and $\Lambda _{xx}:\mathbb{R}%
^{d}\longrightarrow L\left( \mathbb{R}^{d}\times \mathbb{R}^{d},\mathbb{R}%
^{d}\right) $ be the corresponding derivatives of $\Lambda $ and assume that%
\begin{equation*}
\Lambda _{x}(y)\sigma (y)=id_{\mathbb{R}^{d}}\text{ for }y\text{ a.e.}
\end{equation*}%
as well as%
\begin{equation*}
\Lambda ^{-1}\text{ is Lipschitz continuous.}
\end{equation*}%
Suppose that the function $b_{\ast }:\mathbb{R}^{d}\longrightarrow \mathbb{R}%
^{d}$ given by
\begin{align*}
b_{\ast }(x)&:=\Lambda _{x}\left( \Lambda ^{-1}\left( x\right) \right)
\left[ b(\Lambda ^{-1}\left( x\right) )\right] \\
&\,\,\,+\frac{1}{2}\Lambda _{xx}\left( \Lambda ^{-1}\left( x\right) \right) \left[
\sum_{i=1}^{d}\sigma (\Lambda ^{-1}\left( x\right) )\left[ e_{i}\right]
,\sum_{i=1}^{d}\sigma (\Lambda ^{-1}\left( x\right) )\left[ e_{i}\right] %
\right]
\end{align*}%
satisfies condition (\ref{bcondition}), where $e_{i},$ $%
i=1,\ldots,d$, is a basis of $\mathbb{R}^{d}.$ Then the conclusions of Theorem \ref{mainthm} and Corollary \ref{maincor} also apply to $X_t$, $t\in [0,T]$ and its density.
\end{thm}

\begin{proof}
The proof can be directly obtained from It\^o's Lemma. See \cite{MBP10}.
\end{proof}

\section{A classical solution to the stochastic transport equation}

The Sobolev regularity of the solution shown in Theorem \ref{mainprop} with respect to the initial condition entitles us to construct a classical solution to the \emph{stochastic transport equation} when the drift is Lipschitz which to our knowledge is not proved.

The Stochastic Transport Equation is written in differential form
\begin{align}\label{STE}
\begin{cases}
d_t u(t,x) + \nabla u(t,x) \cdot b(t,x) dt + \sum_{i=1}^d  e_i \cdot \nabla u(t,x) \circ dB_t^{(i)} =0\\
u(0,x) = u_0(x),
\end{cases}
\end{align}
where $b:[0,T] \times \mathbb{R}^d \rightarrow \mathbb{R}^d$ is a given vector field and $u_0: \mathbb{R}^d \rightarrow \mathbb{R}$ a given initial data. The stochastic integration is understood in the Stratonovich sense.

\begin{defi}[Classical solution]\label{solutionDefinition}
Let $u_0$ and $b$ be given functions. We say that a stochastic process \mbox{$u \in L^{\infty}(\Omega \times [0,T] \times \mathbb{R}^d)$} is a classical solution to (\ref{STE}) if

\begin{enumerate}

\item
There exists a measurable set $\tilde{\Omega} \subset \Omega$ with full measure such that for fixed $t \in [0,T]$ and $p \geq 1$, the mapping $x \mapsto u(\omega,t,x)$ is in $W^{2,p}_{loc}(\mathbb{R}^d)$ on $\tilde{\Omega}$;

\item
For fixed $x \in \mathbb{R}^d$ there are $(\mathcal{F}_t)$-adapted versions of $t \mapsto u(t,x)$ and $t \mapsto \nabla u(t,x)$;

\item
The following integral equation is satisfied
\begin{equation} \label{integralSTE}
u(t,x) + \int_0^t b(s,x) \cdot \nabla u(s,x) ds + \sum_{i=1}^d \int_0^t e_i \cdot \nabla u(s,x) \circ dB_s^{(i)} = u_0(x)
\end{equation}
for a.e. $(\omega, x) \in \Omega \times \mathbb{R}^d$.
\end{enumerate}
\end{defi}

Notice that we are using the Stratonovich integral in our definition, but following the same idea as in  \cite{FGP10}, Lemma 13, we can recast (\ref{integralSTE}) in It\^{o}-form as
\begin{equation} \label{integralSTEito}
u(t,x) + \int_0^t b(s,x) \cdot \nabla u(s,x) ds + \sum_{i=1}^d \int_0^t e_i \cdot \nabla u(s,x) dB_s^{(i)}  + \frac{1}{2} \int_0^t \Delta u(s,x) ds= u_0(x).
\end{equation}
We will use these formulations interchangeably.

Before we proceed further we will introduce the concept of stochastic flow associated to SDE (\ref{SDE}):

\begin{defi}[Stochastic flow of diffeomorphisms]\label{flowDefinition}
A function $\phi: [0,T]\times [0,T] \times \mathbb{R}\times \Omega \rightarrow \mathbb{R}$, $\phi_{s,t}(x,\omega)$ is said to be a stochastic flow of diffeomorphisms of the SDE (\ref{SDE}) if there exists a full-measure set $\tilde{\Omega}\in \mathcal{F}$ such that for any $\omega \in \tilde{\Omega}$ the following holds true:
\begin{itemize}
\item[(i)]  $\phi_{s,t}(x,\omega)$, $s,t\in [0,T]$, $x\in \mathbb{R}$ is a (global) strong solution to the SDE (\ref{SDE}).
\item[(ii)] $\phi_{s,t}(x,\omega)$ is continuous in $(s,t,x)\in [0,T] \times [0,T] \times \mathbb{R}$.
\item[(iii)] $\phi_{s,t}(\cdot, \omega) = \phi_{u,t}(\cdot, \omega) \circ \phi_{s,u}(\cdot,\omega)$ for any $s,u,t\in [0,T]$ .
\item[(iv)] $\phi_{s,s}(x,\omega)= x$ for all $x\in \mathbb{R}$ and $s\in [0,T]$.
\item[(v)] $\phi_{s,t}(\cdot,\omega):\mathbb{R}^d \rightarrow \mathbb{R}^d$ are diffeomorphisms (of class $C^k$) for all $s,t\in [0,T]$.
\end{itemize}
\end{defi}

For the rest of this section we will assume that $b$ satisfies condition (\ref{bcondition}) for $k=1$ which in particular means that $b$ is globally Lipschitz, uniformly in time.

To get a globally defined (i.e. on the entire $\mathbb{R}^d$) stochastic flow of diffeomorphisms of the SDE (\ref{SDE}) we notice that since $b$ is uniformly Lipschitz  there exists a unique solution to
$$
\phi_{s,t}(x,\omega) = x + \int_s^t b(r, \phi_{s,r}(x, \omega)) dr + B_t(\omega) - B_s(\omega)
$$
for \emph{all} $\omega \in \Omega$.

It is easy to check conditions (i) to (iv) in Definition \ref{flowDefinition} holds for all $\omega \in \Omega$.

Fix $p \geq 1$ and $N \in \mathbb{N}$ and invoke Theorem \ref{sobreg} to guarantee that there exists a measurable subset $\Omega_N \subset \Omega$ with full measure such that the local solution
$$
\phi^N_{s,t}(x,\omega) = x + \int_s^t b(r, \phi^N_{s,r}(x, \omega)) dr + B_t(\omega) - B_s(\omega).
$$

satisfies $\phi^N_{s,t}(\cdot , \omega) \in W^{2,p} ( B(0,N))$ for all $\omega \in \Omega_N$ and $x \in B(0,N)$. By uniqueness we have that $\phi_{s,t} |_{B(0,N) \times \Omega_N} = \phi^N_{s,t}$.

If we let $\tilde{\Omega} := \cap_{N=1}^{\infty} \Omega_N$, we get that $\mathbb{R}^d \ni x \mapsto \phi_{s,t}(x,\omega) \in \mathbb{R}^d$ is twice weakly differentiable for every $\omega \in \tilde{\Omega}$, and thus condition (v) in \ref{flowDefinition} is satisfied for $k=1$.

\bigskip

In \cite{FGP10} the authors study (\ref{STE}) under the considerably weaker condition (at least for $d > 1$), $b\in L_{loc}^1([0,T] \times \mathbb{R}^d; \mathbb{R}^d)$,  div$b \in L_{loc}^1([0,T] \times \mathbb{R}^d)$ and $u_0 \in L^{\infty}(\mathbb{R}^d)$. However, in this case, one is restricted to study analytically weak solutions in the sense that for every test function $\theta \in C^{\infty}_0(\mathbb{R}^d)$ one has
\begin{align} \label{weakSTE}
\int_{\mathbb{R}^d} u(t,x)\theta(x)dx &= \int_{\mathbb{R}^d} u_0(x) \theta(x)dx\\
\notag&+ \int_0^t \int_{\mathbb{R}^d} u(s,x)\left[b(t,x)\cdot \nabla \theta(x) + \textrm{div} b(t,x) \theta (x)\right]dxds\\
\notag &+\sum_{i=1}^d \int_0^t \left( \int_{\mathbb{R}^d} u(s,x)\cdot\frac{\partial}{\partial x_i}\theta(x) dx\right) \circ dB_s^{(i)} .
\end{align}

Moreover, the equation is uniquely solved by $u(t,x) = u_0 (\phi_t^{-1}(x))$.

Although we consider more restrictive coefficients, we arrive at an analytically stronger solution:

\begin{thm}
Let $b$ satisfy condition (\ref{bcondition}) for $k=1$ and $u_0 \in C_b^2(\mathbb{R}^d)$. Then there exists a unique classical solution to the stochastic transport equation.

Moreover, the equation is explicitly solved by $u(t,x) = u_0(\phi_t^{-1}(x))$.

\end{thm}

\begin{proof}
By the above discussion we know that for every test function $\theta \in C_0^{\infty}(\mathbb{R}^d)$, the equation (\ref{weakSTE}) is satisfied $P$-a.s. by $u(t,x) = u_0(\phi_t^{-1}(x))$. We now choose $\tilde{\Omega}$ such that $x \mapsto \phi_t^{-1}(x)$ is in $W^{2,p}_{loc}(\mathbb{R}^d)$ on $\tilde{\Omega}$. Then we get that $u$ satisfies condition (i) and (ii) of Definition \ref{solutionDefinition}, and by integration by parts we have
\begin{align*}
\int_{\mathbb{R}^d} u(t,x)\theta(x)dx &= \int_{\mathbb{R}^d} u_0(x) \theta(x)dx\\
\notag&- \int_0^t \int_{\mathbb{R}^d} \nabla u(s,x) \cdot b(t,x) \theta (x)dxds\\
\notag &-\sum_{i=1}^d \int_0^t \left( \int_{\mathbb{R}^d} \nabla u(s,x)\theta(x) dx\right) \circ dB_s^{(i)} .
\end{align*}

or equivalently
\begin{align*}
\int_{\mathbb{R}^d} u(t,x)\theta(x)dx &= \int_{\mathbb{R}^d} u_0(x) \theta(x)dx\\
\notag&- \int_{\mathbb{R}^d}\int_0^t \nabla u(s,x) \cdot b(t,x) \theta (x)dsdx\\
\notag &-\int_{\mathbb{R}^d} \sum_{i=1}^d \int_0^t   \nabla u(s,x)t dB_s^{(i)} \theta(x) dx - \frac{1}{2} \int_{\mathbb{R}^d} \int_0^t \Delta u(s,x) ds \theta(x)dx.
\end{align*}

Since $\theta$ was arbitrary, this proves the claim.
\end{proof}

\textbf{Acknowledgement:} We would like to thank the referees for their valuable comments and suggestions.

redpajama_set_name: RedPajamaArXiv


Breakout 5a: The Earth is surrounded by a thin atmosphere that sustains life.
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The success and vitality of Tamworth town centre has and always will be a top priority for this council, but we have to accept that the role of town centres is changing and we need to make sure ours is fit for purpose. We must look to the future and figure out how we can evolve the town centre to meet the needs of residents and visitors in the 21st century.
Town centre regeneration is high on our agenda and Cabinet has recently approved the development of a new Town Centre Strategy which sets out how this regeneration will be approached in the coming years.
Hopefully by exploring these webpages, you will also gain a greater understanding of what Tamworth Borough Council can and can't influence. Business rates, for example, are set by the Government at a national level and Tamworth Borough Council can have no bearing on them. Most commercial rents in Tamworth are set by private landlords and again we have no bearing on them.
However, what we are doing is investing millions of pounds in the town centre and this presents us with a unique opportunity to begin to change its fortunes - but we have to get it right. We are also planning to submit a bid to the Government's new Future High Streets Fund and your feedback would help support our application.
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Please take the time to fill out the form below and send us your ideas. If you're happy for us to share them, we will put them out for discussion via our social media channels on Facebook and Twitter.


For years the comments have been made that show cows look very different in the ring than they do most other days of the year. I have heard breeders make comments about some great show cows that look All-World in the show ring and not even All-County in the barn. One of the biggest changes I have noticed in the past 10 years is how much this situation has changed.
More and more we are seeing cattle with consistent show pedigrees or type winning the shows. Does that mean that there are not as many "freaks" out there? Or, does it mean that show ethics and better management practices have led to the top genetic animals rising to the top? I think the latter is true.
It's obvious Rae Lynn has the complete package. Her dam is an EX 2E Champion sister to the 2011 Royal Winter Fair and Madison Supreme Champion EASTSIDE LEWISDALE GOLD MISSY, from the great STADACONA OUTSIDE ABEL VG-88-4YR-CAN 29*. Rae Lynn is also in the top 10 in Canada for her EBV for Conformation. However, as we all know it takes more than just great genetics to win at the big shows. More is exactly what these two cows have. The care they receive from the team at Quality Holsteins (Read more – Quality Holsteins – Well-Deserved Congratulations) and key team member Don Schwartz (Read more – Don Schwartz: "Love what you do and do the best you can!") is second to none. It's for that reason these cows that have the great genetics are all able to fulfill their potential. It is also the reason why these cows look so amazing day in and day out.
Gone are the days when you could take a complete meat bag cow to a show and expect her to do well. Between the progress in show ethics and in how we evaluate cows, the cream really is rising to the top. For me there is nothing better than going to the herds where these top cows reside and seeing that they look just as good in the barn as they did in the show ring…..well with a little more sh*t on them.


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Hygiene and Housing: How Oregon Harbor of Hope is Bridging the Gap Through the Pandemic
BY: MEGAN LOEB
• ASSOCIATE PROGRAM OFFICER, ECONOMIC VITALITY & HEALTH
Shortly after Oregon's COVID-19 "Stay Home, Save Lives" order took effect, I received a phone call from an unfamiliar number. "Megan, it's Homer Williams. I need to talk to you."
I'd met Homer, a successful real estate developer and founder of nonprofit Oregon Harbor of Hope, a handful of times in the preceding months to discuss solutions to Oregon's homelessness crisis, but I was surprised to hear from him that rainy March day. In the midst of fears of the pandemic and the Governor's orders to stay home, I was even more surprised to learn that Homer was calling me from a truck as he and fellow Harbor of Hope employees were distributing tents and sleeping bags.
He said that he'd been out in Rockwood earlier in the morning, then to Lents, and finally back downtown distributing items to people experiencing homelessness. Everywhere, people desperately needed access to hygiene like laundry and shower facilities.
When day shelters, libraries, coffee shops, and public restrooms were shuttered by COVID-19, people experiencing homelessness lost their access to running water--- which is key to staving off preventable skin infections and other maladies. They also needed sleeping bags and tents to sleep outdoors because shelters were operating at reduced capacity to meet required socially-distanced spacing. Homer expressed a need for more mobile shower units like the one operated by his nonprofit, Oregon Harbor of Hope. "Any chance that OCF can help?" Homer asked.
In the months since that phone call, Homer has reached out a number of times to share updates from the front lines, to recount stories of people grateful for basic services such as showers, and to share Oregon Harbor of Hope's expanding role in helping keep Oregonians safe and housed.
A former real estate developer turned homeless advocate, Homer is passionate about finding creative solutions to our state's growing homelessness crisis. For most of his career, Homer was known in Oregon and beyond for his real estate development work, including Portland projects in the Pearl District, the South Waterfront, and Forest Heights. Homer's journey into his new role as homeless advocate began by happenstance, when he had a few extra hours on a business trip to San Antonio. He noticed an article in the local paper about Haven for Hope, a navigation center providing emergency shelter, transitional housing and wrap-around services under one roof. He phoned the San Antonio mayor's office and asked if they could help him set up a quick visit before his flight departed. Once at Haven for Hope, Homer was impressed by the collaborative model of 40 nonprofits onsite offering people what they needed to obtain and maintain housing.
Returning to Portland, he commissioned a study to better understand the homelessness crisis. The research illuminated the faces of Oregon's homeless. "The vast majority of Oregon's homeless are students, people on social security, and people between jobs," Homer said. Data from OCF's commissioned research on homelessness backs that up. Less than 25% of Oregon's homeless population is chronically homeless; most are homeless for a short period due to an event such as unexpected medical bills or a gap in employment.
Empowered by the research, Homer launched Oregon Harbor of Hope, a nonprofit dedicated to addressing homelessness. Homer recognized that the crisis requires that the private and public sectors work together. Homer also came to a conclusion that was seemingly at odds with his developer background. "I figured out early on that we can't build our way out of this," Homer said. Building takes too long, and people need shelter immediately. With that in mind, Oregon Harbor of Hope's framework provides a range of solutions, from emergency services such as mobile food, laundry and shower trucks; short-term shelter, including the River District Navigation Center in Portland; and longer-term solutions including a home share program and affordable apartments.
In the past few months, Oregon Harbor of Hope, with support from donors including OCF's Oregon Community Recovery Fund, has distributed thousands of tents, sleeping bags, and face masks around the Portland Metro region. Its mobile shower and laundry trucks offer a rare opportunity for people experiencing homelessness to access hygiene services through the pandemic. And they're working on longer-term solutions, too.
In the same collaborative vein that Homer witnessed in San Antonio, Harbor of Hope partners with dozens of other service providers in the region and readily shares procured resources. Tents and sleeping bags are dropped off at Portland Rescue Mission, Rose Haven, Street Roots, and even as far away as the Troutdale Police Department. Surgical grade masks are given to area hospitals, while the non-surgical masks are distributed at campsites and to people working for nonprofit organizations. Homer believes in supporting other organizations engaged in work around homelessness. "There are so many good people on the front lines. The quality of people who work in this space—you just can't believe it," Homer recounted, quickly adding praise for his own team including co-founder Don Mazziotti, Lisa Marandas, Marissa Cade, Matt Bordonaro and Susan Gadotti. "This is a team effort."
COVID has delivered a fierce economic blow, and it's reflected in the increasing need. "We are seeing different faces at the trucks and in the food lines," Homer said. "Some of these are people who have never been in a food line in their lives."
People of color, already disproportionately impacted by housing insecurity, have also been harder hit both economically and by the health impacts of COVID. "We have a structural issue, and we need to address it. Everybody deserves a safe place to sleep, they deserve hygiene, healthcare, and food—those are the basics of life," Homer said.
Homer also recognizes the value of respect and employment opportunities for people experiencing homelessness. Oregon Harbor of Hope's mobile shower and laundry services are operated exclusively by people experiencing homelessness. Over 40% of guests at the River District Navigation Center are employed, but most fall short of making enough money to find affordable housing—another symptom of the expensive housing market that contributes to Oregon's homelessness crisis.
Homer worries that if government support slows as the pandemic continues to unfold, the need for basic services and housing will increase exponentially. With a keen awareness of the real estate market, Homer also recognizes that the hospitality industry faces a long, slow recovery. The confluence presents a silver lining for housing, according to Homer. Oregon Harbor of Hope began to explore the feasibility of repurposing hotels and motels into affordable apartments. At a cost of $85,000 per unit, it's significantly more cost-effective than building affordable housing, which averages $300-$350,000 per unit. And, as Homer said, "it's here today—not in four years." Impressed by the cost effectiveness of this model, the Association of Oregon Counties asked Oregon Harbor of Hope to assist with feasibility studies of hotel and motel properties around the state that could be purchased and converted to affordable apartments.
Home sharing is another solution Oregon Harbor of Hope has developed, preventing foreclosure for homeowners while also providing a room at a reasonable rate for renters. "It's about compassion, compatibility, and aging in place," Homer said. Oregon Harbor of Hope's new HomeShare program matches homeowners and renters through proprietary software tools. The mutually beneficial match has a serious upside. Their study revealed more than one million spare bedrooms in Oregon. At a two percent penetration rate, HomeShare could house 20,000 people while allowing another 20,000 to retain their homes. "We're creating inventory out of thin air and it's already paid for," Homer said.
Oregon Harbor of Hope's framework is a continuum, offering interim solutions to alleviate immediate suffering and need while more permanent housing is being built. "All we're doing is building a bridge," Homer acknowledged. But it's a bridge that's desperately needed while Oregon's underbuilt housing supply catches up with the demand.
"I get more satisfaction out of this than any real estate development I've done. I'm just fortunate to have found this," Homer said. And Oregon is fortunate, too. Homer's vision and collaborative leadership have led to thousands of people accessing critical services.
With so many projects in the works, Homer's team is busier than ever exploring partnerships and seeking funds to continue the work. Funds are needed for immediate basic services like tents and another mobile shower unit. Capital is also needed on a larger scale for building emergency shelter structures and converting hotels to apartments. But Homer has faith that the funds will come. "Generally, I have found that good ideas don't die from lack of funding," Homer said. "And I've always said, 'All fishhooks in the water.'"
Championing Equal Opportunity
Within weeks Blanchet House was serving an average of 2,000 meals a day—double our usual numbers.
Child Care Center Boosted by Grant
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He's one of the most sought-after star in showbiz but actor, Andie Chen, is known for more than his acting chops these days. The father of two is widely perceived as being a caring husband to wife, Kate Pang, and doting dad to their bundles of joy: Aden and Avery.
Who do you think has it harder when it comes to parenting: the dad or the mum?
How has your relationship with your dad affected your own ability as a parent?
Model-actor-television host, Jason Godfrey, wears a lot of hats and he wants to add one more to that list: Father. The recently married hunk doesn't have kids yet but it's definitely on the agenda. Having grown up in a mixed heritage household – his dad is Canadian, his mum is Filipino – the father-in-training can't wait to pass on his family values to the next generation.
Why do you want to be a dad? And what kind of dad do you think you'll be?
Why does society expect more from a woman when she's parent as opposed to a man?
What is the greatest challenge for fathers in the modern day?
Jump onto Daniel Ong's Instagram account and you'll immediately notice that his profile is peppered with photos of one girl in particular: Renee, his eight-year-old daughter. The media personality says there's nothing he hates about being a dad, especially because he grew up without one.
How has your relationship with your father affected your parenting style?
What advice do you have for budding fathers?
What are your thoughts on breastfeeding and that rollercoaster journey for mums?
Do you think mums have it harder when it comes to parenting then?
Fatherhood came naturally to Shane who is such a ball of energy, it's a wonder he hasn't stayed a kid forever. Daddy dearest to two bouncing bundles of joy – Ellie Bee who is four and one-year-old Oscar Bear – Shane has the luxury of spending most of the day with his kiddos due to the nature of his job, which means he takes on more than regular dads do on a day-to-day basis.
Do you have any good advice to share with the fathers-to-be out there?
What are your thoughts on breastfeeding and the journey that some mums struggle through?
Aun is one of those rare dads that effortlessly combines his passions with the art of parenting, and his children, aged seven, three and seven months, are all the better for it. When he's not posting drool-worthy photos of the latest concoction he's whipped up in the kitchen, he's shouting praises for his wife, The Ate Group's Su-Lyn Tan.
Do mums or dads have it tougher when it comes to parenting?
What advice do you have for the budding fathers out there?
What do you hate about being a father?


During the second week of March 1942 the Hove Borough Council ran a National Savings 'Warship Week' campaign to raise funds for the war effort, their target was £425,000, on achieving this goal the town was able to "adopt" a warship. The final tally of funds raised was £521,000 and the submarine H.M.S. Unbeaten was "adopted" by the town.
H.M. S. Unbeaten was a 'U' class submarine built by Vickers Armstrong ship builders in Barrow-in-Furness and she entered service on November 10th 1940 At the time of her "adoption" H.M.S. Unbeaten was operating in the Mediterranean as part of the 10th Submarine Flotilla under the command of Lieutenant-Commander Edward 'Teddy' Woodward, DSO, RN and was engaged in anti-shipping operations against German and Italian vessels.
Shortly after their return to the UK the commanding officer and some of the officers and crew were entertained privately to lunch at Hove. Speaking to the Brighton & Hove Herald about the occasion the Mayor said "as long as our destiny and our fate rest upon the efforts of such men as I met we need have no fear of the ultimate outcome of the present grim struggle." During this meeting the ships 'Jolly Roger' flag was presented to the Mayor. This was a special flag unique to submarines; permission to fly the Jolly Roger was normally granted to a submarine after her first successful patrol by the Flotilla Commanding Officer. The crew would then sew on the appropriate emblems following each successful action; the flag would then be flown when entering harbour at the end of each patrol to signify the success of the patrol. Unfortunately this special gift was lost when the Hove Town Hall was bombed during a German air raid.
Later, on Wednesday December 9th, an impressive ceremony was held in the Hove Town Hall at which a commemorative plaque from the residents of Hove was exchanged for a replica of the ships badge by the Mayer of Hove Councillor A. H. Clarke, and Rear Admiral D. W. Boyd, -C.B.10., D.5 C., representing the Admiralty and the Unbeaten which had left the UK early in November to return to Malta. Hove's plaque, to be placed in the submarine, bore the inscription that the sum raised in Warship Week was equivalent to £10 per head of population, men, women and children.
Sadly at the time of this presentation dinner HMS Unbeaten has been reported lost, a fact that was not made public until December 19th when the Brighton & Hove Herald carried the story that the Admiralty had announced the submarine was "overdue and must be considered lost". Unbeaten had sailed for the Mediterranean via Gibraltar after a refit in Britain under the command of her now C.O. Lieutenant. Donald E. O. Watson, DSC, RN. On 11 November 1942 she was attacked and sunk in error by an RAF Wellington of No. 172 Squadron, Coastal Command while in the Bay of Biscay. She was lost with all hands.
This article was originally published as part of My Brighton and Hove living history community web site in March 2008. It was composed from materials originally supplied to the office of the Mayor of Brighton & Hove by Mrs, Barbara Woodward, widow of the late Commander 'Teddy' Woodward D.S.O.** Unbeaten's first commanding officer. Special thanks to Mrs Pat Dines, Mayoral Secretary for allowing access to this material.
His Majesty's Submarine Unbeaten was fully operational for just over two years. During her short tenure she returned to Malta many times symbolically flying her Jolly Roger. HMS/M Unbeaten was 1 of a number of U-Class, group 1, single hulled coastal submarines, ordered by the Admiralty in 1939. These smaller size boats measured around 190 feet in length or approximately 58 meters in new money. By the end of the war a total of 19 U-Class boats would be lost on active service. Action in the Mediterranean would claim 13 and the Atlantic and North Sea would add another 6 boats to that deadly count.
The U-Class submarine Unbeaten was launched on the 9th of July 1940. Her launch took place a day after the famed submarine of the same class, HMS/M Upholder. The Upholder was commanded by LT David Wanklyn, the soon to be, most highly decorated Royal Navy officer in WWII. Unbeaten and Upholder would be inextricably linked throughout the war. The crew of Unbeaten were the last to see the Upholder before Wanklyn and her crew disappeared forever. Several WWII submarines have had books written about them. Unbeaten's activities were no less daring than Upholders or any other submarine in WWII. Her un-timely loss on Armistice Day 1942 was over shadowed by Operation Torch. This was the code name for the combined British and American invasion of North Africa which started on the 8th of November of that year.
A new and detailed book, by the author and Naval researcher David J.B Smith, follows Unbeaten's meticulously researched true story from Holy Loch to Malta and back. This book endeavours to highlight the varied triumphs and tragedies, events and sacrifices, of submarine life during WWII. This exciting testimonial brings together archived accounts from those who were there and emotional contributions from family members of those who were lost. Included are many never-seen-before photographs, alongside detailed U-Class submarine plans drawn by one of Britain's best-known and respected draftsmen, John Lambert.
The reader can Marvel at the crew's valiant efforts towards ensuring the peace and freedoms we all take for granted today. Exactly 70 years on this compelling and revealing book encompasses an extreme twist of fate which is entwined with an un-substantiated third party report.
These revelations could leave the final chapter open forever, and the last crew of Unbeaten still on patrol.
Would like to find out more..just found out my uncle died in this submarine, aged 24...James bruce.