Community anchor institutions are essential partners to increasing broadband adoption.
The Internet is driving innovation in community and economic development, education, health care, and government services. But residential broadband adoption1 has stalled. Community anchor institutions (CAIs) are improving residential broadband adoption in several ways: providing digital literacy training, educating consumers about government programs to promote broadband adoption, leading community planning efforts, lending wireless "hot spots," and, in some cases, providing wireless broadband services directly to consumers. For these efforts to have the greatest impact, however, policymakers must provide CAIs and their community partners with the right resources and incentives. Solutions should be locally customized to meet the needs of specific populations.
Community anchor institutions are addressing broadband access and adoption gaps in many ways. Communities, led by their anchor institutions, are often engaged in broadband planning to achieve digital inclusion. For instance, the Kansas City Public Library (KCPL) coordinates and hosts the efforts of Digital Inclusion KC, which is facilitating collaboration among local organizations and initiatives in order to maximize resources for the greatest community impact.
NYPL is not the only CAI loaning out "hot spots." Public libraries in states from Maine to Kansas to Washington are "checking out" hundreds of hotspots to community residents of all ages. For example, in Missouri, the KCPL is piloting a hotspot program with Kansas City Public Schools. The program has a holistic purpose that brings connectivity and digital skills to entire families. A family can borrow a tablet and a hotspot device and receive digital literacy training. Students are required to perform 40 hours of community service, which may include training parents to use the Internet.
Some CAIs are extending their broadband networks to reach low-income homes and public spaces outside of CAI buildings, including through experimental use of TV White Space (TVWS). Communities across the country have begun to install TVWS units to support new remote public library Wi-Fi access points in parks, community centers, shelters, kiosks, underserved library branches, and other publicly accessible places. One of the libraries piloting the use of TVWS to extend library-supported public wireless broadband is Manhattan Public Library in Kansas, which is now providing community Wi-Fi hotspots in four remote locations.
The disparity between information "haves" and "have-nots" can be dangerous not just for individuals, but to the future of democracy. Closing the "digital divide" is of critical importance to our nation, and anchor institutions have a central role in helping solve this problem.
Angela Siefer is the Director of the National Digital Inclusion Alliance (NDIA). Angela envisions a world in which all members of society have the skills and resources to use the Internet for the betterment of themselves and their communities. Since 1997, Angela has worked on digital inclusion issues with local community organizations, the National Telecommunications and Information Administration, state governments, and the Schools, Health & Libraries Broadband (SHLB) Coalition. This work led Angela to co-found the National Digital Inclusion Alliance, a unified national voice for local technology training, home broadband access, and public broadband access programs. A profile of her written work is at angelasiefer.com.


Recently SAP lauched their SAP Mobile Platform 3.0 (more about that here ) and I decide to write about how one can develop an Android Client App that can consume SAP SOAP Web Services with the ksoap2 library. I am maybe a bit late but it still deserves some attention.
Even though recently I had the chance to try out for myself the SAP Mobile Platform 2.2 and develop some Demo Apps testing both MBO and OData concepts I still personally feel that there are many people out there that are still developing mobile apps for SAP the old-fashion-way: ask for their SAP ABAP developers to prepare a couple of web services and develop mobile clients for these.
I have one very old post about consuming SAP web services (or it should be consuming .NET web services that call RFC-enabled Function Modules from an SAP-Backend) where the landscape I had for development was SAP Backend, Visual Studio and SAP .NET connector library. Well, this time I had SAP Backend with enabled Webservices Runtime and as any normal developer would do: I made the use of the web services that you can build with SAP from any Remote-enabled Function module and develop clients that would consume this services. The best library out there that one can make use of at the moment is ksoap2 and since it is free you can use it and adapt it to your needs. I actually came to the idea of this blog post after reading many forums where people complained that they didn't manage to get it work or they had to change the source and rebuild it after changing the names of the tags or something in the XML bulding logic.
How am I using the ksoap2 library to consume SAP Web services? Very elegant and smooth actually.
1. I developed an wrapper class for the SAPSerializationEnvelope that inherits from the SoapSerializationEnvelope where I just make it create XML-Requests in the SAP manner.
2. Made a helper class for making calls.
3. And used it later in calls. A simple call is as follows.
This is a simple code that calls a method to return a list of Routes that one has to visit and read the meters in homes and upload them later to the backend.
With this simple "trick" I managed to build a Demo app for an SAP Backend with ISU Component that was displaying the Routes and devices from which the Meter data were read and uploaded to the backend for later billing purposes.
In the following posts I will try to keep up with the trends and write something about what I learned from the course that SAP was offered at openSAP about Mobile development with the SAP Mobile Platform and maybe other ways how to achieve this as well.


Can I make green tea lotion with just green tea and regular lotion?
Answers to Everyday Questions : Can I make green tea lotion with just green tea and regular lotion?
I have severe facial redness and have heard that green tea is very beneficial. I want to make a green tea moisturizer but all the recipes I've looked at are complicated and contain things like beeswax, which I don't think I want to put on my face. Can I make a more simple version with just some brewed tea and regular unscented lotion? Thanks for the input!
Is drinking green tea just as good as taking the green tea diet pills?
Is green tea really good for you or is it just hype?
Is regular tea as good for you as green tea?
Is there caffeine in green tea and the new white tea?
Is it okay to replace herbal laxative tea with green tea during the lemon diet?
no, a green tea lotion/moisturizer is made from extracts and like you said complicated procedures, I imagine if you put green tea in regular lotion, it'll just make it smell funny, or if you put the actual tea leaves in it, you'll have a gritty mess.


Below is a series of links that will take you to our flicker page. We have uploaded a selection of images collected over the last few years and present them on that site. They are a selection from particularly notable events we have been part of, from our tours overseas, and from tours within Aotearoa New Zealand with friends from off shore.
AUGUST 29 SATURDAY : Dunedin : Venue – Musos Club/Crown Hotel – Onslaught 11 : SO MANY GREAT BANDS!!!!
NOVEMBER 28 SATURDAY : Auckland :Venue – Audio Foundation : Artists Talk- 2pm.
FEBRUARY 6 – FEBRUARY 9 : Waitangi Waste Fest : Wellington : HEAPS of bands!!!
MARCH 27 THURSDAY : Auckland, NZ : Venue – Wharekai of the University of Auckland's Waipapa Marae. 16 Wynard Street. Beyond the Ohlala Mountains: Alan Brunton / Poems 1968-2002 Book Launch . Performers include Ruby Brunton, Anne Kennedy, Arthur Baysting, Barry Saunders, Bob Orr, Brian Potiki, International Superstars of Westlynn, Jean McAllister, Jeff Henderson, John Davies, John Newton, Kilda Northcott, Ksenija Chobanovich, Leila Adu, Madeline McNamara, Mr Sterile Assembly, Murray Edmond, Nisha Madhan, Peter Simpson, Russell Haley, Stephen Bain and Tony McMaster.
APRIL 4 FRIDAY : Wellington, NZ : Venue – The Pyramid Club w/Flit Nintron and Swamp Doctor.
MARCH 24 SATURDAY : Invercargill : venue Players, in the backroom, 25 Tay St : w/Bazooka, Bolshy McBard, & Mince On Toast.
APRIL 1 FRIDAY Venue: The Factory, Palmerston North w/Neckstretchers, Project: Blood, & Crackpot Theory.
APRIL 22 FRIDAY: Venue: KOF Gallery 68 Pink Oak Cr, Flemington, Melbourne. w/Onion Engines + Sohei + Go Genre Everything!
– w/ Atomgevitter (Scotland), Family Man (Germany), GeenNaam, Crimescene, SMG, Prisma Fosil!


text:
\section{Introduction}
Let $w:\mathbb{R}_+\rightarrow\mathbb{R}_+$ be a continuous decreasing function,
\begin{equation}\label{e:0}
\lim_{y\rightarrow0+}w(y) = \infty,\; w(y) = 1,\, y>1,
\end{equation}
that satisfies the doubling condition
\begin{equation}\label{e:wdc}
w(y) \leq Dw(2y),\quad y\in\mathbb{R}_+,
\end{equation}
for some constant $D>0$.
Given a Lipschitz function $\phi:\mathbb{R}^n\rightarrow \mathbb{R}$, we denote by $\Omega_{\phi}$ the domain above the graph of $\phi$,
\begin{equation}\nota
\Omega_{\phi} = \{(x,y): x\in\mathbb{R}^n, y\geq \phi(x)\}.
\end{equation}
We consider harmonic functions in $\Omega_{\phi}$ with the growth restriction
\begin{equation}\nota
|u(x,y)| \leq Kw(\dist((x,y),\partial\Omega_{\phi})),\quad (x,y) \in \Omega_{\phi}.
\end{equation}
The space of these functions is denoted by $h^{\infty}_{w}(\Omega_{\phi})$ and the smallest $K$ for which this inequality is satisfied is called the norm of $u$ in $h^{\infty}_{w}(\Omega_{\phi})$. We denote it by $\|u\|_{w,\infty}$.\par
In \cite{EML:13} the following result was obtained
\begin{theoremm}
Let $u$ be a function in $h^{\infty}_w(\mathbb{R}^{n+1}_+)$. For $x\in\mathbb{R}^n$ and $0<\delta\leq1$ put
\begin{equation}\notag
I_0(x,\delta) = \int_{\delta}^1u(x,y)\,d\left(\frac{1}{w(y)}\right).
\end{equation}
Then the following inequality holds
\begin{equation}\label{e:inot}
\limsup_{\delta\rightarrow0}\frac{I_0(x,\delta)}{\sqrt{\log w(\delta)\log\log\log w(\delta)}} \leq C\|u\|_{w,\infty},\quad a.e.\, x\in\mathbb{R}^n,
\end{equation}
where the constant $C = C(n)$ depends only on the dimension $n$.
\end{theoremm}
Our main goal is to extend this result to the domains in $\mathbb{R}^{n+1}_+$ above the graphs of Lipschitz functions.
\begin{theorem}\label{t:th1}
Let $\phi$ be a Lipschitz function on $\mathbb{R}^n$ and let $u$ be a function in $h^{\infty}_w(\Omega_{\phi})$. For $x\in\mathbb{R}^n$ and $0<\delta\leq1$ put
\begin{equation}\label{e:idd}
I(x,\delta) = \int_{\delta}^1u(x,\phi(x)+y)\,d\left(\frac{1}{w(y)}\right).
\end{equation}
Then the following LIL holds
\begin{equation}\label{e:alil}
\limsup_{\delta\rightarrow0}\frac{I(x,\delta)}{\sqrt{\log w(\delta)\log\log\log w(\delta)}} \leq C\|u\|_{w,\infty},\quad a.e.\, x\in\mathbb{R}^n,
\end{equation}
where $C$ depends only on the function $\phi$, weight $w$ and dimension $n$.
\end{theorem}
We see that the condition $u \in h^{\infty}_w(\mathbb{R}^{n+1}_+)$ implies immediately that $I_0(x,\delta) \leq C\log w(\delta)$. In addition, the weighted average $I_0$ enjoys some nice cancellation properties, so, with the help of the Law of Iterated Logarithm (LIL) techniques, a better estimate \eqref{e:inot} was obtained. In doing this we relied heavily on the wavelet decomposition of $u(\cdot,\delta)$. Unfortunately these methods work only on nice domains (such as the disc or the upper halfspace), and since we need to extend this result to domains  $\Omega_{\phi}$, we have to use a different approach. This approach is based on the ideas by J. Llorente and A. Nicolau, and in the simplest case of the upper halfplane it goes as follows. First we approximate $I_0$ by a Bloch function $H$ that also happens to belong to $h^{\infty}_{\log w}(\mathbb{R}^2_+)$. Recall that $H$ is a Bloch function in $\mathbb{R}^2_+$ if it is harmonic there and $|\nabla H|(x,y) \leq \frac{C}{y}$ for $(x,y)\in \mathbb{R}^2_+$. Here we would like to mention two of the LILs for harmonic functions, namely the Makarov-Llorente LIL for the Bloch functions, \cite[Corollary \textbf{3.2}]{Mak} \cite[Theorem \textbf{1}]{Ll:98}, and the LIL of Ba\~{n}uelos-Moore, \cite[Theorem \textbf{3.04}]{BM:98}. Unfortunately, we could not use either of those directly, since the former does not provide the desired estimate for slow growing weights $w$, and the latter involves the Lusin area integral, which, we believe, can not be properly estimated by the weight. Therefore we modify the ideas used in the proof of those LILs, so we proceed by approximating $H$ by a (super)dyadic martingale and estimating its quadratic function by $w$. Then it remains to apply the LIL for the martingales.\\ We would also like to note that Theorem \ref{t:th1} remains true if we replace $\Omega_{\phi}$ with some star-like Lipschitz domain.\par
As a corollary of Theorem \ref{t:th1} we have the local version of Theorem \textbf{5} from \cite{EML:13}
\begin{theorem}\label{t:th2}
Let $u$ be a harmonic function in $\mathbb{R}^{n+1}_+$. Assume that there exists a set $\Sigma\subset\mathbb{R}^n$ of positive $n$-dimensional Lebesgue measure such that for every $x_0\in \Sigma$
\begin{equation}\label{e:cgr}
|u(x,y)| \leq Kw(y),\quad |x-x_0|\leq My,
\end{equation}
where $M$ is some positive constant. Then
\begin{equation}\label{e:th2st}
\limsup_{\delta\rightarrow0}\frac{I_0(x,\delta)}{\sqrt{\log w(\delta)\log\log\log w(\delta)}} \leq CK,\quad a.e.\, x\in\Sigma,
\end{equation}
where the constant $C$ depends only on $M$, $w$ and $n$.
\end{theorem}
Note that the condition \eqref{e:cgr} restricts the \textit{non-tangential} growth of $u$ near the boundary. We do not know if this result remains true if we replace \eqref{e:cgr} by a \textit{radial} growth condition.\par
The plan of the paper is as follows: in Section \ref{s:not} we collect some notation and known results about dyadic martingales, and in Sections \ref{ss:ml}-\ref{ss:dma} we prove Theorem \ref{t:th1}. The proof itself consists of three parts. In Section \ref{ss:hir} we approximate $I$ by a Bloch function $H$ that belongs to $h^{\infty}_{\log w}(\Omega_{\phi})$. Then, in Section \ref{ss:dma}, we approximate $H$ by a superdyadic martingale. After that we can apply the LIL for martingales to finish the proof (Section \ref{ss:mtprf}). Theorem \ref{t:th2} is proven in Section \ref{ss:crlr}. In Section \ref{s:ex} we consider the question  whether \textit{every} Bloch function in $h^{\infty}_{\log w}(\Omega_{\phi})$ satisfies the LIL \eqref{e:alil}. Using the technique of dyadic martingales we construct an example of a Bloch function that provides the negative answer to this question.
\section{Notation}\label{s:not}
By $\lambda_n$ we denote the Lebesgue measure in $\mathbb{R}^n,\; n\in\mathbb{N}$.
Given a point $x = (x_1,x_2,\dots,x_n)\in \mathbb{R}^n$ and $r>0$ we denote by $Q(x,r)$ the cube of radius $r$ centered at $x$
\begin{equation}\notag
Q(x,r) = \prod_{i=1}^n(x_i-r,x_i+r],
\end{equation}
we also put $Q(x,\frac12) := Q(x)$. Further, given a cube $Q$ we denote its center by $x_Q$, so that $Q = Q(x_Q,r)$ for some positive $r$.\par Fix $x\in\mathbb{R}^n$. If $2^kx_i-\frac12\in\mathbb{Z}$ for every $i=1\dots n$, and $r = 2^{-k-1}$ for some $k\in \mathbb{Z}_+$, we call the cube $Q = Q(x,r)$ dyadic of rank $k$. For $x\in \mathbb{R}^n$ and $k\in\mathbb{Z}_+$ we denote by $\Delta_k(x)$ the collection of all (shifted) dyadic cubes of rank $k$ in $Q(x)$,
\begin{multline*}
\Delta_k(x)=\\ = \{\prod_{i=1}^n(x-\frac12+m_i2^{-k},x-\frac12+(m_i+1)2^{-k}),\; m_i\in\mathbb{Z}_+,\; 0\leq m_i \leq 2^{k}-1\},
\end{multline*}
we also put $\Delta(x) = \bigcup_{k=0}^{\infty}\Delta_k(x)$. If $Q(x) = (0,1]^n$, we write $\Delta_k$ and $\Delta$ respectively. By $\mathcal{F}_k(x)$ we denote the sigma-algebra generated by dyadic cubes of rank $k$ in $Q(x)$,
\begin{equation}\notag
\mathcal{F}_k(x) = \sigma(\Delta_k).
\end{equation}
Given a probability Borel measure $\mu$ on $Q(x)$ and an increasing sequence $\{\alpha_k\}_{k=0}^{\infty}\subset \mathbb{Z}_+$ we can consider the (super)dyadic martingales on $Q(x)$ with respect to the filtration $\{\mathcal{F}_{\alpha_k}(x)\}_{k=0}^{\infty}$, they are usually denoted by $\Lambda = \{\Lambda_{k}, \mathcal{F}_{\alpha_k}(x),\mu\}$. This means that $\Lambda_{k}$ is a piecewise constant function on the (shifted) dyadic cubes of rank $\alpha_k$, and if $q$ is a dyadic cube in $\Delta_{\alpha_{k-1}}(x)$, then
\begin{equation}\notag
\frac{1}{\mu(q)}\int_{q}\Lambda_{k}(t)\,d\mu(t) = \Lambda_{k-1}(x_q).
\end{equation}
In particular, if $n=1$ and $\mu = \lambda_1$ is the Lebesgue measure on $Q\left(\frac{1}{2}\right) = (0,1]$ then $\Lambda$ has a following truncated wavelet representation
\begin{equation}\label{e:hpwr}
\Lambda_{k}(t) = L + \sum_{j=0}^{\alpha_k}\sum_{i=0}^{2^j-1}b_{ij}\psi_{ij}(t),\quad t\in (0,1],
\end{equation}
where $L = \mathbb{E}\Lambda_{k} = \int_{Q_0}\Lambda_k(t)\,d\lambda_1(t)$, $\psi_{ij}(t) = \psi(2^{j}t-i),\; t\in\mathbb{R}$ , and $\psi$ is the Haar wavelet, $\psi = \chi_{[0,1]} -2\chi_{[0,\frac12]}$ (instead of the usual $L^2$ scaling we use $L^{\infty}$ one here, it is more convenient for our purposes). For any interval $I\subset (0,1]$ of length $2r$, $I = [x_I-r,x_I+r]$, we put
\begin{equation}\notag
\psi_I(t) = \psi\left(\frac{t-x_I+r}{2r}\right),\quad t\in\mathbb{R}.
\end{equation}
Then \eqref{e:hpwr} can be written as follows
\begin{equation}\notag
\Lambda_{k}(t) = L + \sum_{j=0}^{\alpha_k}\sum_{I\in\Delta_{j}}b_I\psi_{I}(t),\quad t\in (0,1],
\end{equation}
where $b_I = \frac{1}{\lambda_1(I)}\int_{\mathbb{R}}\Lambda_k(t)\psi_I(t)$.\par
By $\langle\Lambda\rangle_k$ we denote the quadratic function of $\Lambda$,
\begin{equation}\nota
\langle\Lambda\rangle_k^2 = \sum_{j=1}^{k}\mathbb{E}[|\Lambda_{j}-\Lambda_{j-1}|^2\vert\mathcal{F}_{\alpha_{j-1}}].
\end{equation}
If $\alpha_k = k$ and we use the Haar representation of $\Lambda$, we can write the quadratic function in the following way
\begin{equation}\label{e:hmqf}
\langle\Lambda\rangle_k^2(t) = L^2 + \sum_{I\in\Delta:t\in I, \lambda_1(I)\geq 2^{-k}}b_I^2.
\end{equation}
Let $u$ be a harmonic function in $\Omega_{\phi}$. We say that $u$ belongs to the Bloch class in $\Omega_{\phi}$, if there exists a constant $D>0$  such that
\begin{equation}\notag
|\nabla u(x,y)| \leq \frac{D}{\dist((x,y),\partial \Omega_{\phi})},\quad (x,y) \in \Omega_{\phi}.
\end{equation}
We denote the space of such functions by $\mathcal{B}(\Omega_{\phi})$  and the smallest $D$ for which this inequality is satisfied by $\|u\|_{\mathcal{B}}$.\\
The connection between Bloch functions and dyadic martingales is well established, see, for example, \cite{Mak} for the unit disc case and \cite{Ll:98} for Lipschitz domains. Here, however, we use a superdyadic martingale, which is, essentially, a thinned dyadic martingale. It means that instead of the usual dyadic filtration $\mathcal{F}_k$ we use some subsequence of dyadic sigma-algebras $\mathcal{F}_{\alpha_k}$ where $\alpha_k$ depends on the weight $w$ (and is lacunary for slow growing $w$). The main reason for the transition from the dyadic to the superdyadic martingale approximation here is that the quadratic function of the superdyadic martingale is much easier to estimate (similar ideas were used in \cite{LM:12}).
\section{Proof of Theorems \ref{t:th1} and \ref{t:th2}}\label{s:prf}
\subsection{Main approximation lemma}\label{ss:ml}
Fix a Lipschitz function $\phi:\mathbb{R}^n\mapsto\mathbb{R}$, a doubling weight $w$ and a function $u$ in $h^{\infty}_w(\Omega_\phi)$.
Given two functions $f$ and $g$, we say that $f\lesssim g$ if there is a positive constant $C = C(w,n,\|\phi'\|_{\infty}, \|u\|_{w,\infty})$ such that $f\leq Cg$. We write $f\sim g$ if $f\lesssim g$ and $g\lesssim f$ simultaneously. Consider a positive decreasing sequence $\{s_k\}_{k=0}^{\infty}$ such that
\begin{equation}\notag
w(s_k) = 2^k,\quad k\in \mathbb{Z}_+,
\end{equation}
and put
\begin{equation}\notag
\alpha_0 = 0,\;\alpha_k = -\left[\frac{\log s_k}{\log2}\right],\quad k\in\mathbb{N}.
\end{equation}
It follows from the doubling property \eqref{e:wdc} that $w(2^{-\alpha_k})\sim 2^{k}$. Consider $I(x,\delta)$ defined in \eqref{e:idd}.
The approximation of $I(x,\delta)$ by martingales is provided by the following lemma
\begin{lemma}\label{l:ml}
Assume that $u \in h^{\infty}_w(\Omega_{\phi})$. Then for every $x_0\in\mathbb{R}^n$ there exists a probability measure $\mu$ on $Q(x_0)$ and a (super)dyadic martingale $\Lambda = \{\Lambda_k, \mathcal{F}_{\alpha_k},\mu\}_{k=0}^{\infty}$ on $Q(x_0)$ such that $\mu$ is absolutely continuous with respect to the Lebesgue measure on $Q(x_0)$ and for every $k\in\mathbb{Z}_+$
\begin{subequations}
\begin{eqnarray}
\label{e:li.1}&|\Lambda_k(x) - I(x,s_k)| \lesssim 1,\\
\label{e:li.2}&|\Lambda_k(x) - \Lambda_{k+1}(x)| \lesssim 1,\quad x\in Q(x_0).
\end{eqnarray}
\end{subequations}
\end{lemma}
\subsection{How to deduce Theorem \ref{t:th1}}\label{ss:mtprf}
Assuming that Lemma \ref{l:ml} holds, we proceed by the standard argument. Fix any $x_0\in \mathbb{R}^n$ and put
\[E = \{x\in Q(x_0): \lim_{m\rightarrow\infty}|\langle\Lambda\rangle_m|(x) < \infty\}.
\]
The inequality \eqref{e:li.2} implies that $\langle\Lambda\rangle_m^2 \lesssim m,\; m\geq 1$. Applying the law of the iterated logarithm for martingales to $\Lambda$ (see, for example, Theorem \textbf{3.0.2} in \cite{BM:98}), we get
\begin{equation}\notag
\limsup_{m\rightarrow\infty}\frac{|\Lambda_m|(x)}{\sqrt{m\log\log m}} \lesssim 1\quad \;\mu\; a.e.\; x\in Q(x_0)\setminus E.
\end{equation}
It is well known that for $\mu$ almost every $x\in E$ the sequence $\{\Lambda_m(x)\}$ is bounded, so \eqref{e:li.1} implies that the sequence $\{I(x,s_m)\}$ is bounded $\mu\; a.e.$ on $E$ as well. It follows that
\begin{equation}\label{e:ilil}
\limsup_{m\rightarrow\infty}\frac{|I(x,s_m)|}{\sqrt{m\log\log m}} \lesssim 1\quad \mu\; a.e.\; x\in Q(x_0).
\end{equation}
Now for $s_{m}\leq \delta\leq s_{m-1}$ and $x\in Q(x_0)$ we have
\begin{multline*}
\left|I(x,s_m) - I(x,\delta)\right| \leq \int_{s_m}^{\delta}|u(x,\phi(x)+y)|\,d\left(\frac{1}{w(y)}\right)\\ \lesssim \log w(s_m) - \log w(s_{m-1}) = 1,
\end{multline*}
also, clearly, $w(\delta) \geq w(s_{m-1}) = \frac{1}{2}w(s_m)$. Combined with \eqref{e:ilil} and the fact that $\mu$ is absolutely continuous with respect to the Lebesgue measure, it gives us
\begin{equation}\notag
\limsup_{\delta\rightarrow\infty}\frac{|I(x,\delta)|}{\sqrt{\log w(\delta)\log\log\log w(\delta)}} \lesssim 1, \quad a.e.\; x\in Q(x_0).
\end{equation}
The inequality \eqref{e:alil} follows immediately.
\subsection{Proof of the lemma \ref{l:ml}: auxiliary function $H$}\label{ss:hir}
The approximation of $I(x,\theta)$ by a Bloch function is covered by the following lemma
\begin{lemma}\label{l:bl}
Assume that $u \in h^{\infty}_w(\Omega_{\phi})$. Put
\begin{equation}\label{e:had}
H(x,t) = \int_{0}^1u(x,t+y)\,d\left(\frac{1}{w(y)}\right),\quad (x,t)\in \Omega_{\phi}.
\end{equation}
Then $H$ belongs to $\mathcal{B}(\Omega_{\phi})$ and $\|H\|_{\mathcal{B}}\lesssim 1$. Moreover
\begin{equation}\label{e:hia}
|H(x,\phi(x)+\theta) - I(x,\theta)| \lesssim 1,\quad x\in\mathbb{R}^n, 0<\theta\leq 1.
\end{equation}
\end{lemma}
\begin{proof}
First we note that $H$ is harmonic in $\Omega_{\phi}$ (it is the average of harmonic functions). We proceed by proving the following inequality
\begin{equation}\label{e:gru}
|\nabla u|(x,\phi(x)+\theta) \lesssim \frac{w(\theta)}{\theta},\quad x\in\mathbb{R}^n,\; \theta >0.
\end{equation}
Fix any positive $\theta$. Since $\phi$ is a Lipschitz function, we see that $\dist((x,\theta+\phi(x)),\partial\Omega_{\phi})\sim \theta$ for any $x\in\mathbb{R}^n$ and positive $\theta$. It follows from \eqref{e:wdc} that for $y\geq \frac{\theta}{2}$ we have
\begin{equation}
|u(x,\phi(x)+y)| \leq w(\dist(x,y+\phi(x)),\partial\Omega_{\phi})) \lesssim w\left(\frac{\theta}{2}\right),
\end{equation}
so there exists a constant $C = C(n,\phi,w)$ such that
\begin{equation*}
0\leq u(x,y) + Cw\left(\frac{\theta}{2}\right) \leq (C+1)w\left(\frac{\theta}{2}\right),\quad (x,y)\in \Omega_{\phi+\frac{\theta}{2}}.
\end{equation*}
Then the estimate \eqref{e:gru} follows from \eqref{e:0}, \eqref{e:wdc}, and the Harnack inequality.
For $(x,\theta)\in \Omega_{\phi}$ \eqref{e:gru} implies
\begin{multline*}
\left|\nabla H\right|(x,\phi(x) + \theta) \leq \int_{0}^1|\nabla u|(x,\phi(x) + \theta+y)\,d\left(\frac{1}{w(y)}\right)\\ \lesssim \int_{0}^1\frac{w(\theta+y)}{\theta+y}\,d\left(\frac{1}{w(y)}\right)   =
\int_{0}^{\theta}\frac{w(\theta+y)}{\theta+y}\,d\left(\frac{1}{w(y)}\right) + \int_{\theta}^1\frac{w(\theta+y)}{\theta+y}\,d\left(\frac{1}{w(y)}\right).
\end{multline*}
Since the function $\frac{w(y)}{y}$ is decreasing, we have
\begin{equation}\notag
\int_{0}^{\theta}\frac{w(\theta+y)}{\theta+y}\,d\left(\frac{1}{w(y)}\right) \leq \int_{0}^{\theta}\frac{w(\theta)}{\theta}\,d\left(\frac{1}{w(y)}\right) = \frac{1}{\theta}.
\end{equation}
On the other hand,
\begin{multline}\label{e:intm}
\int_{\theta}^1\frac{w(\theta+y)}{\theta+y}\,d\left(\frac{1}{w(y)}\right) \leq \int_{\theta}^1\frac{w(y)}{y}\,d\left(\frac{1}{w(y)}\right)\\ \leq
\sum_{k=0}^{[\log\frac{1}{\theta}]}\int_{2^k\theta}^{2^{k+1}\theta}\frac{1}{y}d\log\frac{1}{w(y)} \leq \sum_{k=0}^{[\log\frac{1}{\theta}]}\frac{1}{2^{k}\theta}\int_{2^k\theta}^{2^{k+1}\theta}d\log\frac{1}{w(y)} \\ \leq
\frac{1}{\theta}\sum_{k=0}^{[\log\frac{1}{\theta}]}2^{-k}(\log w(2^k\theta) - \log w(2^{k+1}\theta))\\ \leq \frac{1}{\theta}\sum_{k=0}^{[\log\frac{1}{\theta}]}2^{-k}(\log \left(Dw(2^{k+1}\theta)\right) - \log w(2^{k+1}\theta)) \\ \leq \frac{\log D}{\theta}\sum_{k=0}^{[\log\frac{1}{\theta}]}2^{-k} \lesssim \frac{1}{\theta}.
\end{multline}
Gathering the estimates, we arrive at
\begin{equation}\notag
\left|\nabla H\right|(x,\phi(x)+\theta) \lesssim \frac{1}{\theta} \sim \frac{1}{\dist((x,\phi(x)+\theta), \partial\Omega_{\varphi})},
\end{equation}
and we get the first part of the lemma.\par
To prove \eqref{e:hia} we write
\begin{multline*}
H(x,\phi(x)+\theta) - I(x,\theta)\\ = \int_{0}^1u(x,\phi(x) + \theta+y)\,d\left(\frac{1}{w(y)}\right) - \int_{\theta}^1u(x,y)\,d\left(\frac{1}{w(y)}\right)\\ =
\int_{0}^{\theta}u(x,\phi(x) + \theta+y)\,d\left(\frac{1}{w(y)}\right)\\ + \int_{\theta}^1\left(u(x,\phi(x) + \theta+y) - u(x,\phi(x+y))\right)\,d\left(\frac{1}{w(y)}\right).
\end{multline*}
Following the same reasoning as above, we see that
\begin{equation}\notag
\left|\int_{0}^{\theta}u(x,\phi(x) + \theta+y)\,d\left(\frac{1}{w(y)}\right)\right| \leq w(\theta)\int_{0}^{\theta}\,d\left(\frac{1}{w(y)}\right) = 1.
\end{equation}
Again, \eqref{e:gru} implies that
\begin{multline*}
\left|\int_{\theta}^1\left(u(x,\phi(x) + \theta+y) - u(x,\phi(x+y))\right)\,d\left(\frac{1}{w(y)}\right)\right|\\ \leq \int_{\theta}^1\int_{y}^{y+\theta}|\nabla u|(x,\phi(x)+s)\,ds\,d\left(\frac{1}{w(y)}\right)\\\leq
\int_{\theta}^1\int_{y}^{y+\theta}\frac{w(s)}{s}\,ds\,d\left(\frac{1}{w(y)}\right)\leq \int_{\theta}^1w(y)\frac{\theta}{y}\,d\left(\frac{1}{w(y)}\right) \lesssim 1,
\end{multline*}
just like in \eqref{e:intm}. Combining these two inequalities we get \eqref{e:hia}.
\end{proof}
\subsection{Proof of Lemma \ref{l:ml}: dyadic martingale}\label{ss:dma}
Once we obtained the intermediate approximation of $I$ by a Bloch function, we can proceed to martingales. It is well known (see, for example, \cite{Mak}) that the Bloch functions in the unit disc can (up to a constant error) be viewed as dyadic martingales. The case of Lipschitz domains was considered by Llorente, Corollary \textbf{2} in \cite{Ll:98} is the main instrument in the following argument.

Fix any point $x_0\in\mathbb{R}^n$ and let
\begin{equation}\notag
\begin{split}
&A = \|\phi'\|_{\infty}\sqrt{n},\;\lambda = 8+ \frac{1}{A},\\
&\Omega_1 = \{(x,y): x\in \lambda Q(x_0):\phi(x)\leq y \leq \phi(x) + \lambda A\}.
\end{split}
\end{equation}
The following proposition holds
\begin{proposition}[Corollary \textbf{2}, \cite{Ll:98}]\label{p:c2ll}
If $v\in \mathcal{B}(\Omega_1)$ then there exists a dyadic martingale $\mathcal{M} = \{M_k,\mathcal{F}_k(x_0),\omega\}$ in $Q(x_0)$ and a positive constant $C = C(\phi,n)$ such that $\omega$ is absolutely continuous with respect to the Lebesgue measure on $Q(x_0)$, and if $A2^{-(k+1)}\leq t\leq A2^{-k}$ then for every $k\in\mathbb{N}$ and $x\in Q(x_0)$
\begin{subequations}
\begin{eqnarray}
\label{e:mv.1}& |M_k(x) - v(x,\phi(x)+t)| \leq C\|v\|_{\mathcal{B}},\\
\label{e:mv.2}&|M_{k+1}(x) - M_k(x)| \leq C\|v\|_{\mathcal{B}}.
\end{eqnarray}
\end{subequations}
\end{proposition}
We apply this proposition to $H$ and put
\begin{equation}\notag
\Lambda = \{\Lambda_k,\mathcal{F}_{\alpha_k}(x_0),\omega\} := \{M_{\alpha_k},\mathcal{F}_{\alpha_k}(x_0),\omega\}.
\end{equation}
Now we prove \eqref{e:li.1} and \eqref{e:li.2}. It follows from \eqref{e:mv.1} and \eqref{e:hia} that
\begin{multline*}
|\Lambda_k(x) - I(x,s_k)| = |M_{\alpha_k}(x) - I(x,s_k)| \\ \leq
|M_{\alpha_k}(x) - H(x,\phi(x) + A2^{-\alpha_k})| + |H(x,\phi(x) + A2^{-\alpha_k}) - H(x,\phi(x) + s_k)| \\+ |H(x,\phi(x) + s_k) - I(x,s_k)| \lesssim
1 + \int_{s_k} ^{A2^{-\alpha_k}}\left|\nabla H(x,\phi(x)+y)\right|\,dy \\ \lesssim 1,\quad x\in Q(x_0),
\end{multline*}
since $\|H\|_{\mathcal{B}} \lesssim 1$, and we get \eqref{e:li.1}. To obtain \eqref{e:li.2} we note that
\begin{multline*}
|\Lambda_k(x) - \Lambda_{k+1}(x)| = |M_{\alpha_k}(x) - M_{\alpha_{k+1}}(x)| \leq
|M_{\alpha_k}(x) - I(x,s_k)|\\ + |I(x,s_k) - I(x,s_{k+1})|
 + |I(x,s_{k+1}) - M_{\alpha_{k+1}}(x)|\\ \lesssim 1 + |I(x,s_k) - I(x,s_{k+1})|.
\end{multline*}
Clearly,
\begin{equation}\notag
|I(x,s_k) - I(x,s_{k+1})| \leq w(s_k)\int_{s_{k+1}}^{s_k}\,d\left(\frac{1}{w(y)}\right) = 2^k(2^{-k} - 2^{-k-1}) = 2,
\end{equation}
and the inequality \eqref{e:li.2} follows.
\subsection{Proof of Theorem \ref{t:th2}}\label{ss:crlr} The proof is standard. We apply the usual ice-cream cone construction to $\Sigma$, i.e. consider the domain
\begin{equation}\notag
\Omega = \bigcup_{x\in \Sigma} \Gamma(x,M),
\end{equation}
where $\Gamma(x,M)$ is the cone with vertex $x$ and aperture $M$, $\Gamma(x,M) = \{(\tilde{x},y)\in\mathbb{R}^{n+1}_+: |\tilde{x}-x|\leq My \}$. Clearly $\Omega$ is the area above the graph of some Lipschitz function $\phi$ with $\|\phi'\|_{\infty} = \frac{1}{M}$, so that $\Omega = \Omega_{\phi}$. The condition \eqref{e:cgr} then implies that
\begin{equation}\notag
|u(x,y)| \leq Kw(y) \lesssim w(\dist((x,y),\partial \Omega)),\quad (x,y)\in \Omega,
\end{equation}
and we can apply Theorem \ref{t:th1} to obtain
\begin{equation}\notag
\limsup_{\delta\rightarrow0}\frac{I(x,\delta)}{\sqrt{\log w(\delta)\log\log\log w(\delta)}} \leq C,\quad a.e.\; x\in\mathbb{R}^n.
\end{equation}
Theorem \ref{t:th2} follows immediately.
\section{An example}\label{s:ex}
\subsection{}
In the proof of Theorem \ref{t:th1} we introduced the harmonic function $H$ which is shown to be a Bloch function. In addition, the estimate \eqref{e:hia} implies that $H \in h^{\infty}_{\log w}(\Omega_{\phi})$, and that the LIL  in \eqref{e:alil} holds for $H$ as well,
\begin{equation}\label{e:blil}
\limsup_{\delta\rightarrow0}\frac{H(x,\delta)}{\sqrt{\log w(\delta)\log\log\log w(\delta)}} \lesssim 1, \quad a.e.\; x\in\mathbb{R}^n.
\end{equation}
To obtain this estimate we used the special nature of $H$, namely that it was constructed on $u\in h^{\infty}_w(\Omega_{\phi})$. It is then natural to ask if an \textit{arbitrary} function $v \in h^{\infty}_{w_0}(\Omega_{\phi})\bigcap \mathcal{B}(\Omega_{\phi})$ satisfies the following LIL
\begin{equation}\label{e:clil}
\limsup_{\delta\rightarrow0}\frac{H(x,\delta)}{\sqrt{w_0(\delta)\log\log w_0(\delta)}} \lesssim 1,\quad a.e. \;x\in\mathbb{R}^n.
\end{equation}
The answer to this question is negative as provided by the following proposition.
\begin{proposition}\label{p:ex}
Let $w_0(y) = \log\log\frac{e}{y}+1,\; y\in (0,1]$. There exists a constant $A>0$, a function $v\in h^{\infty}_{w_0}(\mathbb{R}^2_+)\bigcap \mathcal{B}(\mathbb{R}^2_+) $, a number $k_0\in\mathbb{N}$ and a sequence $\{y_k\}_{k=k_0}^{\infty}\rightarrow 0$ such that
\begin{equation}\label{e:ex}
\lambda_1\left(\{t\in[0,1]:\, |v(t,y_k)| \geq \frac{w_0(y_k)}{A}\}\right) \geq \frac{1}{10}.
\end{equation}
\end{proposition}
It is known that a function in $h^{\infty}_w(\mathbb{R}^2_+)$ can grow as fast as $w$ only on small part of vertical rays $\{x+iy\},\; y\in \mathbb{R}_+$, however it can attain the maximal growth on the subsets of those rays for a.e. $x\in \mathbb{R}$ (see \cite{LM:12}, \cite{BLMT:10}). Unfortunately, we can not use the example provided there, since it is constructed as a lacunary trigonometric series, for which, as it can be shown, \eqref{e:blil} holds if it belongs to the Bloch class. \\
\begin{proof}
Given two real-valued functions $f,g\in L^2(\mathbb{R}^n)$ we denote the scalar product $\int_{\mathbb{R}^n}f(t)g(t)dt$ by $\langle f,g\rangle$. Consider a function $\varphi:\mathbb{R}\mapsto\mathbb{R}$ such that $\supp \varphi\subset [0,1],\;\varphi\in C^{10},\; \|\varphi\|_{\infty}\leq 1$. We also require that $\int_{\mathbb{R}}\varphi(t)\,dt =0$ and $\langle\varphi,\psi\rangle \neq 0$ (where $\psi$ is a Haar wavelet mentioned earlier). For example we can take the suitable renormalization of the Daubechies wavelet (or any other smooth wavelet with compact support that satisfies our conditions). By $P_y$ we denote the Poisson kernel for the halfplane, $P_y(t) = \frac{y}{\pi(y^2+t^2)},\; y>0,\,t\in\mathbb{R}$. \par
The idea is to obtain a functional series of the form
\begin{equation}\label{e:wavjs}
\sum_{j=0}^{k}\sum_{I\in \Delta_j}c_{I}\varphi_{I}(t) := \Phi_k(t),\quad t\in\mathbb{R},
\end{equation}
that satisfies properties similar to those in the statement, and then prove that the corresponding Bloch function provides the required example. To elaborate we first construct $\Phi_k$ and an increasing sequence $\{\beta_j\}_{j=1}^{\infty}\subset \mathbb{Z}_+$ in such a way that  we have
\begin{subequations}\label{e:mblgr}
\begin{eqnarray}
\label{e:mblgr.1}&\|\Phi_k - \Phi_{k-1}\|_{\infty} \leq 1,\\
\label{e:mblgr.2}&\|\Phi_{k}\|_{\infty} \leq w_0(2^{-k})+2,\\
\label{e:mblgr.3}&\lambda_1\left(\{t\in(0,1]: |\Phi_{\beta_{k}}|(t) \geq \frac{w_0(2^{-\beta_{k}})}{4}\}\right) \geq \frac{1}{10},
\end{eqnarray}
\end{subequations}
for any integer $k\geq k_0$.\par
The property \eqref{e:mblgr.1} is an analogue of the Bloch condition, \eqref{e:mblgr.2} is the growth restriction, and \eqref{e:mblgr.3} corresponds to \eqref{e:ex} (so that there is no LIL for $\Phi_k$ with $w_0$).
\subsection{Construction of $\{\beta_j\}$ and $\Phi_k$}\label{ss:sbeta}
First we chose $a\in\mathbb{N}$ such that
 $2^{-a+1}\|\varphi'\|_{\infty} \leq \frac14\left|\langle\varphi,\psi\rangle\right|$.
Now chose a natural $j_0 \geq 4\|\varphi'\|_{\infty}+4$ and an increasing sequence $\beta_j\in\mathbb{N}$ in such a way that
\begin{equation}\label{e:jn}
\begin{split}
&\beta_1 = 0,\;\frac{\beta_j}{a}\in\mathbb{N},\\
&j-1\leq w_0(2^{-\beta_j})\leq j,\\
&\left(\frac{\beta_j - \beta_{j-1}}{a}-1\right)\langle\varphi,\psi\rangle^2 \geq 4j^2, \quad j\geq j_0.
\end{split}
\end{equation}
It is not hard to verify that such choice is possible (we remind that $w_0(y) = \log\log\frac{e}{y}+1$).\par
The functions $\Phi_k$ are constructed via double induction, first on $j$, and then on $m$ between $\beta_j$ and $\beta_{j+1}$. Put $\Phi_0(t) = \varphi(t)$. Assume now that we obtained $\Phi_{\beta_j}$ for some $j\in\mathbb{N}$. Consider all the intervals $I\in \Delta_{\beta_j}$ such that $\sup_{t\in I}|\Phi_{\beta_j}|(t) > j$, we denote the set of these intervals by $\mathcal{E}^{\beta_j}_j$. Now suppose that we have constructed $\Phi_{m-1}$ and $\mathcal{E}_{j}^{m-1}$ for some $m,\, \beta_{j}+1\leq m \leq \beta_{j+1}$. If $\frac{m}{a}\in\mathbb{Z}$, then for $I\in \Delta_m$ and $t\in I$ let
\begin{subequations}\label{e:lmdr}
\begin{eqnarray}
\label{e:lmdr.3}&\Phi_{m}(t) = \Phi_{m-1}(t),\quad t\in \bigcup_{J\in \mathcal{E}^{m-1}_j}J,\\
\label{e:lmdr.2}&\Phi_{m}(t) = \Phi_{m-1}(t) + \varphi_{I}(t),\quad t\notin \bigcup_{J\in \mathcal{E}^{m-1}_j}J,\\
\label{e:lmdr.1}&\mathcal{E}^m_j = \mathcal{E}^{m-1}_j\bigcup\{J\in\Delta_m: \sup_{t\in J}|\Phi_m(t)| >j\}.
\end{eqnarray}
\end{subequations}
Otherwise we put
\begin{subequations}\label{e:lmdr1}
\begin{eqnarray}
\label{e:lmdr1.3}&\Phi_{m}(t) = \Phi_{m-1}(t),\quad t\in (0,1],\\
\label{e:lmdr1.1}&\mathcal{E}^m_j = \mathcal{E}^{m-1}_j.
\end{eqnarray}
\end{subequations}
Finally, put
\begin{equation}\notag
\mathcal{E}_j = \bigcup_{m=\beta_j}^{\beta_{j+1}-1}\mathcal{E}^{m}_j = \mathcal{E}^{\beta_{j+1}-1}_j.
\end{equation}
What we do here is, essentially, a stopping time procedure applied (instead of martingales as usual) to the functional series of the form like in \eqref{e:wavjs}. We see that if $I \in \mathcal{E}_j$, then the construction is stopped at this interval, and $\Phi_{\beta_{j+1}}(t) = \Phi_{m}(t),\; t\in I,\; m = \rank I$. If, on the other hand, $t \in (0,1]\setminus \bigcup_{J\in\mathcal{E}_j}J$, then the construction happens on every step (divisible by $a$) up until $\beta_{j+1}$.  The set $(0,1]\setminus\bigcup_{J\in\mathcal{E}_j}J$ can be decomposed into a disjoint union of intervals from $\Delta_{\beta_{j+1}}$, we denote the set of these intervals by $\mathcal{G}_j$.\\
Clearly $\Phi_m$ is of the form like in \eqref{e:wavjs}, moreover,
\begin{equation}\notag
\Phi_{\beta_{j+1}}(t) = \Phi_{\beta_j}(t) + \sum_{m=\beta_{j}+1}^{\beta_{j+1}}\sum_{J\in\Delta_m}c_J\varphi_J(t),\quad t\in (0,1],
\end{equation}
where $c_J =1$ only if $\frac{\rank J}{a} \in\mathbb{Z}$ \textit{and} there is no interval $I\in\mathcal{E}_j$ such that $I\supset J$, $c_J = 0$ otherwise. We also see that
\begin{equation}\notag
\begin{split}
&\sup_{t\in I}|\Phi_{\beta_{j+1}}|(t) > j,\quad I\in \mathcal{E}_j;\\
&|\Phi_{\beta_{j+1}}| \leq j+1, \quad j\in\mathbb{N}.
\end{split}
\end{equation}\par
We are left to check \eqref{e:mblgr.1}-\eqref{e:mblgr.3}. The condition \eqref{e:mblgr.1} follows straight from \eqref{e:lmdr}, since $\|\varphi_{I}\|_{\infty} = 1$ for any interval $I$. For any $k\in\mathbb{N}$ there exists $j_k\in \mathbb{N}$ such that $\beta_{j_k}\leq k \leq \beta_{j_k+1}-1$. We therefore have
\begin{equation}\notag
\|\Phi_{k}\|_{\infty} \leq j_k+1 \leq w_0(2^{-\beta_{j_k}}) +2 \leq w_0(2^{-k}) + 2,
\end{equation}
and we obtain \eqref{e:mblgr.2}.

\subsection{Proof of \eqref{e:mblgr.3}: martingale decomposition}\label{ss:mblgr.3}
Pick any $j\geq j_0$ (we remind that $j_0$ was defined in \eqref{e:jn}). Since $j_0 \geq 4\|\varphi'\|_{\infty}+4$, we see that $\frac{j}{2} - \|\varphi'\|_{\infty} \geq \frac{j}{2} - \frac{j_0}{4} \geq \frac{j}{4}$, and, due to \eqref{e:jn} we have $\frac{j}{2} - \|\varphi'\|_{\infty} \geq \frac{w_0(2^{-\beta_j})}{4}$. It follows that to obtain \eqref{e:mblgr.3} it is enough to prove
\begin{equation}\label{e:lgi}
\lambda_1\left(\{t\in (0,1]:|\Phi_{\beta_{j+1}}(t)| \geq \frac{j}{2}-\|\varphi'\|_{\infty}\}\right) \geq \frac{1}{10}.
\end{equation}

The first step is to prove that $|\Phi_{\beta_{j+1}}|$ is "sufficiently large" on the intervals from $\mathcal{E}_j$, namely that for any $I\in \mathcal{E}_j$ we have
\begin{equation}\label{e:phijj}
|\Phi_{\beta_{j+1}}|(t) \geq j - 2\|\varphi'\|_{\infty},\quad t\in I.
\end{equation}
Indeed, for $m = \rank I$ we have
\begin{multline*}
|\Phi_m'|(t) \leq \sum_{J\in\Delta:\, t\in JI,\, \rank J \leq m}c_{J}\|\varphi_{J}'\|_{\infty}\\ = \|\varphi'\|_{\infty}\sum_{J\in\Delta:\, t\in J,\, \rank J \leq m}\frac{c_J}{\lambda_1(J)}\leq 2^{m+1}\|\varphi'\|_{\infty},\quad t\in (0,1],
\end{multline*}
since $|c_J|\leq 1$ for any $J\in \Delta$. Again we see that $\Phi_{\beta_{j+1}}(t) = \Phi_{m}(t)$ on $I$, therefore $\left|\sup_{t\in I}\Phi_m(t) - \inf_{t\in I}\Phi_m(t)\right| \leq \int_{I}|\Phi'_m(t)|\,dt\leq 2\|\varphi'\|_{\infty}$, and we get \eqref{e:phijj}.\par
Now we show that
\begin{equation}\label{e:gj}
\lambda_1\left(\bigcup_{J\in\mathcal{G}_j}J\right) \leq \frac{3}{4},
\end{equation}
combined with \eqref{e:phijj} it implies \eqref{e:lgi}. In order to do this consider the Haar decomposition of $\Phi_{\beta_{j+1}}$,
\begin{equation}\label{e:phihd}
\Phi_{\beta_{j+1}} = \sum_{m=0}^{\infty}\sum_{I\in \Delta_m}b_{I}\psi_{I},
\end{equation}
where $b_{I} = 2^{\rank I}\langle\Phi_{\beta_{j+1}},\psi_{I}\rangle =$ $2^{\rank I}\sum_{k=0}^{\beta_{j+1}}\sum_{J\in \Delta_k}c_J\langle\varphi_J,\psi_I\rangle$, and $c_{J}$ is either $0$ or $1$. Here we sum from $m=0$, since $\supp \Phi_{k}\subset [0,1]$ and $\int_{0}^1\Phi_k(t)\,dt=0$ for any $k\in\mathbb{Z}_+$. If we put
\begin{equation}\notag
\tilde{\Lambda}_k = \sum_{m=0}^{k}\sum_{I\in \Delta_m}b_{I}\psi_{I},
\end{equation}
we see that $\{\tilde{\Lambda}_k,\mathcal{F}_k,\lambda_1\}$ is a dyadic martingale on $(0,1]$. Since $\Phi_{\beta_{j+1}} \in C^{10}(\mathbb{R})$, the sum on the right-hand side in \eqref{e:phihd} converges to $\Phi_{\beta_{j+1}}$ uniformly on $\mathbb{R}$ as $k\rightarrow\infty$. It follows immediately that $\langle\tilde{\Lambda}\rangle_k$ converges uniformly to a bounded limit which we denote by $\langle\tilde{\Lambda}\rangle_{\infty}$.\\
Our goal here is to prove that the quadratic function of $\tilde{\Lambda}$ is "big" on the intervals from $\mathcal{G}_j$, so that we can use the standard dyadic martingale methods to estimate the size of $\bigcup_{J\in\mathcal{G}_j}J$. For any $k\in\mathbb{Z}_+$, the following equality holds
\begin{equation}\label{e:gli}
\int_{0}^1\langle\tilde{\Lambda}\rangle_k^2(t)\,dt = \int_{0}^1\tilde{\Lambda}_k^2(t)\,dt.
\end{equation}
Assume for a moment that we know that
\begin{equation}\label{e:qfl}
\langle\tilde{\Lambda}\rangle_{\infty}^2(t) \geq 4j^2,\quad t\in \bigcup_{J\in\mathcal{G}_j}J.
\end{equation}
Then \eqref{e:gli} implies that
\begin{equation}\notag
\begin{split}
&4j^2\cdot\lambda_1\left(\bigcup_{J\in\mathcal{G}_j}J\right) \leq \int_{\bigcup_{J\in\mathcal{G}_j}J}\langle\tilde{\Lambda}\rangle_{\infty}^2(t)\,dt \leq \int_0^1\langle\tilde{\Lambda}\rangle_{\infty}^2(t)\,dt \\=
&\int_0^1\tilde{\Lambda}_{\infty}^2(t)\,dt = \int_0^1\Phi_{\beta_{j+1}}^2(t)\,dt \leq (j+1)^2,
\end{split}
\end{equation}
and \eqref{e:gj} follows immediately. It remains to prove the estimate \eqref{e:qfl}.
\subsection{Proof of \eqref{e:mblgr.3}: inequality \eqref{e:qfl}}\label{ss:mblgr.4}
First we show that if $c_{I} = 1$ for some $I\in\Delta_m,\, \beta_{j} \leq m\leq \beta_{j+1}-1$, then
\begin{equation}\label{e:hcl}
|b_{I}| \geq \frac12\left|\langle\varphi,\psi\rangle\right|.
\end{equation}
Fix such an interval $I$. For any $J\in \Delta_k,\; k\leq m$, the standard calculation gives
\begin{multline}\notag
|\langle\varphi_J,\psi_I\rangle|\\ = \left|\int_{\mathbb{R}}\varphi(2^kt-x_J)\psi(2^mt-x_I)\,dt\right| = \left|\int_{\mathbb{R}}\varphi(2^k(t-2^{-m}x_I)-x_J)\psi(2^mt)\,dt\right|\\ =
2^{-k}\left|\int_{\mathbb{R}}\varphi(t-2^{k-m}x_I - x_J)\psi(2^{m-k}t)\,dt\right|\\ = 2^{-k}\left|\int_{\mathbb{R}}\left(\varphi(t-2^{k-m}x_I - x_J) - \varphi(-2^{k-m}x_I - x_J)\right)\psi(2^{m-k}t)\,dt\right| \\=
2^{-k}\left|\int_{\mathbb{R}}\int_{-2^{k-m}x_I - x_J}^{t-2^{k-m}x_I - x_J}\varphi'(s)\,ds\psi(2^{m-k}t)\,dt\right| \leq 2^{-k}\|\varphi'\|_{\infty}\int_{\mathbb{R}}\left|t\psi(2^{m-k}t)\right|\,dt \\=
2^{k-2m}\|\varphi'\|_{\infty}\int_{\mathbb{R}}|t\psi(t)|\,dt \leq 2^{k-2m}\|\varphi'\|_{\infty}.
\end{multline}
Now we see that if $k> m$, then $\langle\varphi_{J},\psi_{I}\rangle =0$ for any $J\in \Delta_k$, and if  $k\leq m$, then there exists at most one $J\in \Delta_k$ such that $\langle\varphi_{J},\psi_{I}\rangle \neq 0$. We therefore have
\begin{multline}\notag
|b_{I}| = 2^{m}\left|\sum_{k=0}^{\beta_{j+1}}\sum_{J\in \Delta_k}c_J\langle\varphi_J,\psi_I\rangle\right| = 2^{m}\left|\sum_{k\leq m,\,J\in \Delta_k,\, J\supset I}c_J\langle\varphi_J,\psi_I\rangle\right|  \\ \geq
 2^{m}\left|\langle\varphi_{I},\psi_{I}\rangle\right| - 2^{m}\sum_{k\leq m-1,\,J\in \Delta_k,\, J\supset I}|c_J||\langle\varphi_J,\psi_I\rangle|\\ \geq
\langle\varphi,\psi\rangle - 2^m\sum_{k\leq m-1,\,J\in \Delta_k,\, J\supset I}|c_J|\|\varphi'\|_{\infty}2^{k-2m}\\ \geq \langle\varphi,\psi\rangle - \|\varphi'\|_{\infty}\sum_{k\leq m-1,\,J\in \Delta_k,\, J\supset I}2^{k-m}|c_{J}|.
\end{multline}
It follows from \eqref{e:lmdr.2} that if $c_{J}=1$ then $c_{J} = 0$ for $J\in\Delta_k$, $m-a< k \leq m-1$ (the decomposition of $\Phi_{\beta_{j+1}}$ has very sparse coefficients). Combined with the choice of $a$, it gives
\begin{equation}\notag
|\langle\varphi,\psi\rangle| - \|\varphi'\|_{\infty}\sum_{k\leq m-1,\,J\in \Delta_k,\, J\supset I}2^{k-m}|c_{J}| \geq \langle\varphi,\psi\rangle - \|\varphi'\|_{\infty}2^{-a} \geq \frac{1}{2}|\langle\psi,\varphi\rangle|,
\end{equation}
and we have \eqref{e:hcl}.\par
Fix any $I\in \mathcal{G}_j$. Again we note that $c_J=1$ for any $J\in \Delta_m$ such that $\frac{m}{a}\in \mathbb{Z}$, $J\supset I$ and $\beta_j\leq m\leq\beta_{j+1}-1$. Therefore \eqref{e:hcl} implies that $|b_{J}| \geq \frac12\left|\langle\varphi,\psi\rangle\right|$ for such intervals $J$, and  due to \eqref{e:jn} we have
\begin{multline}\notag
\langle\tilde{\Lambda}\rangle_{\infty}^2(t) \geq \sum_{\beta_j\leq m\leq\beta_{j+1}-1,\, \frac{m}{a}\in\mathbb{Z},\,J\in \Delta_m,\, t\in J}|b_{J}|^2\\ \geq \frac14\left(\frac{\beta_{j+1}-\beta_j}{a}-1\right)\left|\langle\varphi,\psi\rangle\right|^2 \geq 100j^2,
\end{multline}
for $t\in I$, and we get \eqref{e:qfl}.
\subsection{How to create a Bloch function from $\Phi_j$}\label{ss:bloch}
Let
\begin{equation}\notag
v_k(x,y) = \left(\Phi_{k}*P_y\right)(x),\quad x\in\mathbb{R},\; k\geq k_0.
\end{equation}
First we show that $v_k\rightarrow v$ as $k\rightarrow\infty$, where $v$ is a harmonic function.\par
 Fix any $y>0$.  Since $c_I $ is either $0$ or $1$, we have for natural $m\leq n$
\begin{multline*}
\left|v_m(x,y) - v_n(x,y)\right| \\= \left|\sum_{j=m+1}^{n}\sum_{I\in \Delta_j}c_{I}\varphi_{I}*P_y\right|(x) \leq \sum_{j=m+1}^{n}\sum_{I\in \Delta_j}|c_{I}|\left|\int_{\mathbb{R}}\varphi_{I}(t)P_y(x-t)\,dt\right|\\ \leq
\sum_{j=m+1}^{n}\sum_{i=0}^{2^j-1}\left|\int_{\mathbb{R}}\varphi(2^jt)P_y\left(x-t-\left(i+\frac12\right)2^{-j}\right)\,dt\right|.
\end{multline*}
A standard calculation gives
\begin{equation}\label{e:est}
\begin{split}
&\sum_{i=0}^{2^j-1}\left|\int_{\mathbb{R}}\varphi(2^jt)P_y\left(x-t-\left(i+\frac12\right)2^{-j}\right)\,dt\right| \\=
&\sum_{i=0}^{2^j-1}\left|\int_{\mathbb{R}}\varphi(2^jt)\left(P_y\left(x-t-\left(i+\frac12\right)2^{-j}\right)-P_y\left(x-\left(i+\frac32\right)2^{-j}\right)\right)\,dt\right|\\ \leq
&\sum_{i=0}^{2^j-1}\left|\int_{0}^{\frac{2^{-j}}{y}}\varphi(2^jyt)\left(P\left(\frac{x}{y}-t-\frac{i+\frac12}{y}2^{-j}\right)-P\left(\frac{x}{y}-\frac{i+\frac32}{y}2^{-j}\right)\right)\,dt\right|\\ \leq&
\sum_{i=0}^{2^j-1}\left|\int_0^{\frac{2^{-j}}{y}}\varphi(2^jyt)\int_{\frac{x}{y}-\frac{i+\frac32}{y}2^{-j}}^{\frac{x}{y}-t-\frac{i+\frac12}{y}2^{-j}}|P'|(s)\,ds\,dt\right|\\ &\leq \int_0^{\frac{2^{-j}}{y}}|\varphi(2^jyt)|\sum_{i=0}^{2^j-1}\int_{\frac{x}{y}-\frac{i+\frac32}{y}2^{-j}}^{\frac{x}{y}-t-\frac{i+\frac12}{y}2^{-j}}|P'|(s)\,ds\,dt\\ \leq
&\int_0^{\frac{2^{-j}}{y}}|\varphi(2^jyt)|\int_{\mathbb{R}}|P'|(s)\,ds\,dt \leq C\frac{2^{-j}}{y},\quad x\in\mathbb{R},\; y>0,\; j\in\mathbb{N}.
\end{split}
\end{equation}
We therefore have
\begin{equation}\label{e:est1}
\left|v_m(x,y) - v_n(x,y)\right|\leq C\sum_{j=m+1}^{n}\frac{2^{-j}}{y},\quad x\in\mathbb{R},\; y>0,
\end{equation}
and the uniform convergence follows immediately.\par
Next we show that $v$ satisfies the $h^{\infty}_{w_0}$ growth condition.
For $y\geq 2^{-k}$ \eqref{e:est1} implies that
\begin{equation}\notag
|v(x,y) - v_k(x,y)| \leq C,\quad x\in\mathbb{R}.
\end{equation}
Combined with \eqref{e:mblgr.2} and definition of $w_0$ this implies that
\begin{multline*}
|v(x,y)| \leq C+|v_k(x,y)| = C + \left|\Phi_{k}*P_y\right|(x)\\ \leq C + w_0(2^{-k}) \leq Cw_0(y),\quad x\in\mathbb{R},\; 2^{-k+1}\geq\;y\geq 2^{-k},\; k\in\mathbb{N},
\end{multline*}
and, therefore, $v\in h^{\infty}_{w_0}$.\par
Now we prove that $v\in\mathcal{B}(\mathbb{R}^2_+)$. Fix any positive $y\leq 1$ and $m\in\mathbb{Z}$ such that $2^{-m+1}\geq y \geq 2^{-m}$. We have
\begin{equation*}
\left|\nabla v(x,y)\right| \leq \left|\nabla v(x,y) - \nabla v_m(x,y)\right| + \left|\nabla v_m(x,y)\right|,\quad x\in\mathbb{R}.
\end{equation*}
Repeating the estimate in \eqref{e:est} verbatim we get for any $x\in \mathbb{R}$,
\begin{multline}\label{e:estp}
\left|\nabla v(x,y) - \nabla v_m(x,y)\right|\\ \leq \left|\frac{\partial}{\partial y}v(x,y) - \frac{\partial}{\partial y}v_m(x,y)\right| + \left|\frac{\partial}{\partial x}v(x,y) - \frac{\partial}{\partial x}v_m(x,y)\right| \\=
\left|\sum_{j=m+1}^{\infty}\sum_{I\in \Delta_j}c_{I}\varphi_{I}*\left(\frac{\partial}{\partial y}P_y\right)\right|(x) + \left|\sum_{j=m+1}^{\infty}\sum_{I\in \Delta_j}c_{I}\varphi_{I}*\left(\frac{\partial}{\partial x}P_y\right)\right|(x)\\  \leq \frac{C2^{-m}}{y^2} \leq \frac{C}{y}.
\end{multline}
Recall that $\varphi \in C^{10}(\mathbb{R})$ and that for any two different intervals $I, J \in \Delta_j$ the supports of $\varphi_I$ and $\varphi_J$ are disjoint. A simple rescaling gives
\begin{equation}\notag
\left|\sum_{I\in \Delta_j}c_{I}\varphi_{I}*\left(\frac{\partial}{\partial x}P_y\right)\right|(x) + \left|\sum_{I\in \Delta_j}c_{I}\varphi_{I}*\left(\frac{\partial}{\partial y}P_y\right)\right|(x) \leq C2^j,\quad x\in\mathbb{R},\, y>0,\, j\in\mathbb{N}.
\end{equation}
It follows that
\begin{multline*}
\left|\frac{\partial}{\partial x}v_m(x,y)\right| + \left|\frac{\partial}{\partial y}v_m(x,y)\right| = \left|\Phi_{m}*\left(\frac{\partial}{\partial x}P_y\right)\right|(x) + \left|\Phi_{m}*\left(\frac{\partial}{\partial y}P_y\right)\right|(x)\\ \leq  C\sum_{j=0}^m2^j = C2^{m+1} \leq \frac{C}{y},\quad x\in\mathbb{R}.
\end{multline*}
This estimate and \eqref{e:estp} imply that $v\in \mathcal{B}(\mathbb{R}^2_+)$.\par
It remains to prove \eqref{e:ex}. Fix any $k\geq k_0$. Since $\|\Phi'_{\beta_k}\|_{\infty}\leq 2^{\beta_k+1}\|\varphi'\|_{\infty}$, we see that for any $x\in (0,1]$ such that $\left|\Phi_{\beta_k}(x)\right| \geq \frac{k-1}{2}$, there exists an interval $I_x = [x-\rho_k,x+\rho_k]$, $\rho_k = \frac{2^{-\beta_k-2}}{\|\varphi'\|_{\infty}}$, such that $\left|\Phi_{\beta_k}(t)\right| \geq \frac{k-3}{2},\; t\in I_x$. Clearly then
\begin{equation}\notag
\int_{\mathbb{R}\setminus I_x}P_{y}(x-t)\,dt \leq \frac{1}{4},
\end{equation}
for $0<y\leq \frac{\rho_k}{10}$.
Now if we fix such an $x$ and put $y_k = \frac{2^{-\beta_k-2}}{10\|\varphi'\|_{\infty}}$, we have
\begin{multline*}
|v_{\beta_k}(x,y)| = \left|\Phi_{\beta_k}*P_y\right|(x)\\ \geq \left|\int_{I_x}\Phi_{\beta_k}(t)P_y(x-t)\,dt\right| - \left|\int_{\mathbb{R}\setminus I_x}\Phi_{\beta_k}(t)P_y(x-t)\,dt\right|\\ \geq
\frac{k-3}{2} - \|\Phi_{\beta_k}\|_{\infty}\int_{\mathbb{R}\setminus I_x}P_y(x-t)\,dt \geq
 \frac{k-3}{2} - \frac{k+2}{4}\\ = \frac{k-4}{2} \geq \frac{w_0(2^{-\beta_k})-4}{2} ,
\end{multline*}
so that
\begin{equation}\notag
\lambda_1\left(\left\{x\in (0,1]: \left|v_{\beta_k}\right|(x,y) \geq \frac{w_0(2^{-\beta_k})-4}{2}\right\}\right) \geq \frac{1}{10},\quad 0<y\leq y_k.
\end{equation}
Again, following \eqref{e:est}, we obtain
\begin{equation}
|v_{\beta_k}(x,y_k) - v(x,y_k)| \leq C_0\|\varphi'\|_{\infty}.
\end{equation}
The doubling property of $w_0$ implies
\begin{equation}\notag
w_0(y_k) = w_0\left(\frac{2^{-\beta_k-2}}{10\|\varphi'\|_{\infty}}\right) \leq C_1\frac{w_0(2^{-\beta_k})-4}{2} - C_0\|\varphi'\|_{\infty} \leq C\frac{w_0(2^{-\beta_k})-4}{2}
\end{equation}
for $k$ large enough.
We therefore have
\begin{equation}\notag
\lambda_1\left(\left\{x\in (0,1]: \left|v\right|(x,y_k) \geq \frac{w_0(y_k)}{C}\right\}\right) \geq \frac{1}{10},\quad y_k = \frac{2^{-\beta_k-2}}{10\|\varphi'\|_{\infty}},
\end{equation}
which is \eqref{e:ex}.
\end{proof}
The way we did the construction of $v$ is, probably, not the most effective one. Unfortunately we could not use here the dyadic martingale methods, as described, for example, in \cite{GM:05}. Instead we decided to employ the wavelet-like series for Bloch functions (see \cite{M:01} for the description of $\mathcal{B}(\mathbb{R}^2_+)$ in terms of wavelet representation).\par
Note that the averaging process $u(x,\delta)\rightarrow H(x,\delta) = \int_{0}^1u(x,y+\delta)\,d\frac{1}{w(y)}$ can be viewed as an application of some multiplier $M$ to the boundary values of $u$,
\begin{equation}\notag
\widehat{Mf}(\tau) = \widehat{f}(\tau)\int_0^1e^{-2\pi y|\tau|}d\frac{1}{w(y)},\quad \tau\in\mathbb{R},
\end{equation}
where $f = u(\cdot,0)$ (these boundary values exist in some sense, at least as a distribution). The doubling condition \eqref{e:wdc} implies that
\begin{equation}\notag\
\int_0^1e^{-2\pi y|\tau|}d\frac{1}{w(y)}\sim \frac{1}{w(\frac{1}{|\tau|})},\quad |\tau| > 0,
\end{equation}
so we basically divide the Fourier transform of $u(\cdot,0)$ by $w$. It would be interesting to find out the image of $M$, we see at least that $Mu\in h^{\infty}_w(\mathbb{R}^{n+1}_+)\bigcap\mathcal{B}(\mathbb{R}^{n+1}_+)$ if $u\in h^{\infty}_w(\mathbb{R}^{n+1}_+)$. The example in Proposition \ref{p:ex} shows that the image of $M$ can (in the case of slowly growing weights) be a proper subset of $h^{\infty}_w(\mathbb{R}^{n+1}_+)\bigcap\mathcal{B}(\mathbb{R}^{n+1}_+)$. For more information about the multipliers on the growth spaces see \cite{EM:14}.
\section{Concluding remark}\label{s:cr}
Consider the function $\Phi\in L^1(\mathbb{R}^n)$, and let $\Phi_y(t) = \frac{1}{y^n}\Phi\left(\frac{t}{y}\right),\; t\in\mathbb{R}^n,\, y>0$. Assume now that $\Phi\in C^{k_0}(\mathbb{R}^n)$ for some $k_0\in\mathbb{N}$, and that $\Phi^{(k)}\in L^1(\mathbb{R}^n)$ for $k\leq k_0$. Denote by $h^{\infty}_{w,\Phi}$ the space of functions $u:\mathbb{R}^{n+1}_+$ of the form $u(x,y) = \left(f*\Phi_y\right)(x)$, such that
\begin{equation}\notag
|u(x,y)| \leq Kw(y),\quad x\in\mathbb{R}^n,
\end{equation}
where $f$ is some distribution on $\mathbb{R}^n$ (we assume that this convolution exists).
The ideas and methods from \cite{EML:13} can be used to prove the following statement
\begin{theoremm}
Let $u$ be a function in $h^{\infty}_{w,\Phi}(\mathbb{R}^n_+)$. Put
\begin{equation}\notag
I_{\Phi}(x,\delta) = \int_{\delta}^1u(x,y)\,d\left(\frac{1}{w(y)}\right).
\end{equation}
If $w(y)y^{k_0-1}$ is bounded for all $y>0$, then the following inequality holds
\begin{equation}\label{e:phikern}
\limsup_{\delta\rightarrow0}\frac{I_{\Phi}(x,\delta)}{\sqrt{\log w(\delta)\log\log\log w(\delta)}} \leq C\|u\|_{w,\infty},\quad a.e.\, x\in\mathbb{R}^n
\end{equation}
for some absolute constant $C>0$.
\end{theoremm}
For $0<\alpha\leq 1$ let
\begin{equation}\notag
\Lip_{\alpha}(\mathbb{R}) = \{f:\mathbb{R}\mapsto\mathbb{R}: |f(x)-f(y)| \leq C|x-y|^{\alpha},\; x,y\in\mathbb{R}\},
\end{equation}
where $C$ is some constant that depends only on $f$, and denote by $\|f\|_{\alpha}$ the infimum of such constants.
We state the result (second part of Theorem \textbf{1}) from \cite{LN:13}
\begin{theoremm}
Fix $0<\alpha<1$. For $f \in \Lip_{\alpha}(\mathbb{R})$ and $0<\varepsilon<\frac12$ let
\begin{equation}\notag
\Theta_{\varepsilon}(f)(x) = \int_{\varepsilon}^1\frac{f(x+h)-f(x-h)}{h^{\alpha}}\frac{dh}{h},\quad x\in\mathbb{R}.
\end{equation}
Then there exists a constant $C(\alpha)$, independent of $f$, such that at almost every point $x\in\mathbb{R}$, one has
\begin{equation}\label{e:skern}
\limsup_{\varepsilon\rightarrow0}\frac{|\Theta_{\varepsilon}(f)(x)|}{\sqrt{\log \frac{1}{\varepsilon}\log\log\log \frac{1}{\varepsilon}}} \leq C(\alpha)\|f\|_{\alpha}.
\end{equation}
\end{theoremm}
These two results are actually very similar. Indeed, let $S(t) = \frac12\chi_{[-1,1]}$ be the box kernel, $w(h) = h^{\alpha-1},\; h>0,$ and $u(x,y) = (f'*S_y)(x),\; x\in\mathbb{R},\, y>0$ (here $f'$ is understood in the sense of distributions, so that the convolution is well defined). We see that the condition $f\in \Lip_{\alpha}$ is equivalent to $u \in h^{\infty}_{w,S}$. Then $\Theta_{\varepsilon}$ can be rewritten in the following way
\begin{multline*}
\frac{(1-\alpha)}{2}\Theta_{\varepsilon}(f)(x) = \int_{\varepsilon}^1\frac{f(x+h)-f(x-h)}{2h}\,(1-\alpha)h^{-\alpha}dh\\ =\int_{\varepsilon}^1\left(f'*S_{h}\right)(x)\,d\left(\frac{1}{w(h)}\right)\\=
\int_{\varepsilon}^1u(x,h)\,d\left(\frac{1}{w(h)}\right) = I_S(x,\varepsilon),\quad x\in\mathbb{R},\;\varepsilon >0.
\end{multline*}
If the box kernel $S$ were smooth enough, the inequality \eqref{e:skern} would follow immediately from \eqref{e:phikern}. Unfortunately that is not the case, and  we see that Theorem \textbf{C} cannot be deduced from Theorem \textbf{B}. To prove Theorem \textbf{C} Llorente and Nicolau used a different argument, which employed the fact that $f \in \Lip_{\alpha}$.\par
\textit{Acknowledgements}. We would like to thank A. Nicolau and E. Malinnikova for fruitful discussions of the problem. Part of the work was done while the author was visiting the Chebyshev laboratory at St. Petersburg State University. We are grateful to the laboratory for hospitality and great working conditions.

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text:
ASAD B. SĀMĀNḴODĀ
ASAD B. SĀMĀNḴODĀ, (Sāmānḵodāt in Naršaḵī), ancestor of the Samanid dynasty. Sāmānḵodā seems to have been a local landowner (dehqān) of the village of Sāmān in the district of Balḵ. Bīrunī gives a genealogy going back four generations from Sāmānḵodā to the Sasanian Bahrām Čūbīn (Āṯār al-bāqīa, p. 39; Chronology, p. 48); while Gardīzī traces the line back to Gayomarṯ, the first man (ed. Nazim, pp. 19-20; ed. Ḥabībī, p 145). This affiliation of the Samanids with the Sasanians, whatever its truth, was generally accepted by contemporaries (as Bīrūnī testifies). Sāmānḵodā became a Muslim at the hands of Asad b. ʿAbdallāh Qasrī (not Qošayrī), the Omayyad governor in Khorasan (106-09/724-27, 117-20/735-38; he was not under al-Maʾmūn, pace Gardīzī's information). Asad was named after this governor. Of his life nothing is known, but he must have flourished in the late 2nd/8th and early 3rd/9th centuries. His four sons Nūḥ, Aḥmad, Yaḥyā, and Elyās were appointed deputy governors in towns of Ṭoḵārestān and Transoxania by al-Maʾmūn's governor of Khorasan, Ḡassān b. ʿAbbād (202-05/817-20). With them began the growth of Samanid fortunes in Transoxania.
See also Barthold, Turkestan3, p. 209.
Naršaḵī, pp. 81-82, 104-05, 263; tr. R. N. Frye, The History of Bukhara, Cambridge, Mass., 1954, p. 76.
Ebn al-Aṯīr, VII, p. 279.
Frye in Camb. Hist. Iran IV, p. 136.
C. E. Bosworth, "The Heritage of Rulership in Early Islamic Iran and the Search for Dynastic Connections with the Past," Iran 11, 1973, pp. 58-59.
(C. E. Bosworth)
Vol. II, Fasc. 7, pp. 696-697
C. E. Bosworth, "ASAD B. SĀMĀNḴODĀ," Encyclopædia Iranica, II/7, pp. 696-697, available online at http://www.iranicaonline.org/articles/asad-b-1 (accessed on 30 December 2012).
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So, following up on Barb's post on reading goals, I thought I would put my own perspective in. I know it is considered declasse by people in the serious literary set, who, it turns out, are drunk a lot. I know, you should love reading and you shouldn't need goals and all that. Quality over quantity. I get it.
So, with that in mind, I just kind of kept it in my mind that I wasn't going to start reading short books just to make the goal. We read War and Peace and Infinite Jest the last two years, each of which was the equivalent of four smaller books, and which are not reading goal friendly, but I don't want to stop doing that, so I just keep it kind of loose and don't get wrapped up in the goal too much. I might mentally add it in, for my own purposes.
Also, if I read all of a literary journal, that's a book.
At one point, I was thinking, you know, there could be like a book equivalency of pages, like an FTE or some kind of sports stat, a truly evolved metric with a ∑ in it, and then I thought, maybe you are taking the book nerd thing a little too literally.
Previous postIt's that time of year: reading goals!
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I think its fun to look back and see all the books you have read! Thats my favorite part of setting the goal. I am also the person who read a bunch of graphic novels to get caught up!


Doctor Who: Rakhee Thakrar in Talks for Season 10 Companion Role?
It's going to be a rough year for Doctor Who fans - with only the 2016 Christmas Special to sate viewers until the show's return in 2017. After a solid two years as the Time Lord, Peter Capaldi and the Doctor Who crew are taking a year off to regroup - and set the stage for a new showrunner (and likely new star) in the eleventh "modern series" season. However, before Steven Moffat bows as writer-producer of the series, handing the reins to Chris Chibnall, the BBC has another role to fill - a replacement for Jenna Coleman's companion, Clara Oswald, who departed the series in the season 9 finale.
While viewers have enjoyed fan-casting their own choices (we've got our own picks for who the network should cast as the new companion), no names had, up until now, been floated (officially or unofficially). Still, with a fan base as loyal and eager for news as Whovians, something was bound to slip out eventually - and, today, we may have discovered the identity of the next actor to journey with the Doctor through time and space.
According to a report from the Radio Times, actress Rakhee Thakrar is in talks with BBC to co-star in season 10 of Doctor Who alongside returning Twelfth Doctor Peter Capaldi. American viewers might not be familiar with Thakrar but the actress is well-known in the UK for her work as Shabnam Masood on the British drama, EastEnders.
The report has yet to be confirmed by BBC and Radio Times suggests the actress is one of several performers who will meet with producers; though, Mirror UK elaborated on the initial rumor suggesting that, unless something changes, Thakrar could be the front-runner for season 10's companion. It might sound like a wild rumor at first but the potential addition of Thakrar comes as the actress resigned her role on EastEnders. At this time, assuming the rumor is even true, it's possible Doctor Who producers came knocking when they heard the actress was about to become available.
Either way, the timing is suspect and certainly hints that Thakrar has a clear schedule to begin preparation for Doctor Who - which anyone familiar with the production knows is an extremely grueling and time-consuming role. In addition to press tours, convention appearances, and other secondary responsibilities, Doctor Who series regulars spend a significant amount of their time on set. Between the highly cinematic and special effects-heavy filming of Doctor Who as a TV production, the Doctor and his companion have the added challenge of carrying the show's main story - requiring a hefty amount of screen time, compared to similar 1-hour dramas with a much larger cast (and multiple plot arcs). That's all to say, while portraying the Doctor's companion can be one of the most rewarding jobs in TV and movies today, it's an enormous commitment that prevents performers from taking-on much else.
It may be awhile before we get any official word from BBC, especially since the series is prepping for one of its biggest overhauls to date; though, it would make sense that, with such a large and ambitious plan for Moffat's final season, the network would start lining up key pieces. It may be over a year before the show returns in full, aside from the 2016 Christmas Special, but episode writers and pre-production artists will need to get rolling on season 10 in the not-too-distant future - since, as previously mentioned, Doctor Who requires a lot of time to film and post-produce. As a result, even though the report may seem to come out of nowhere, Thakrar could, indeed, replace Jenna Coleman in the TARDIS.
For anyone familiar with Thakrar's work, the actress is an intriguing fit - and the exact kind of talented performer that BBC would need for a co-starring spot next to Capaldi. Bucking Moffat's trend toward twenty-something companion actresses, Thakrar is slightly older at 32 years - and would also represent the first non-white full-time companion since Freema Agyeman's one-season run as Martha Jones in series 3 (alongside David Tennant). Agyeman would return to the Doctor Who universe on multiple occasions after her exit, especially in the spin-off Torchwood, and other non-white performers appeared in smaller companion capacities over the years, as recently as Samuel Anderson's Danny Pink; still, Thakrar would be a unique shift away from the predominantly white central cast that had dominated Moffat's run.
As Doctor Who has grown into a global phenomena, fans (both longtime and new) have begun calling for greater diversity - with many going so far as to hope that the Doctor himself could soon regenerate into someone other than a caucasian male. For that reason, it's not surprising to see BBC meeting with actresses that could showcase greater diversity in the central cast - especially since Thakrar is first and foremost a very capable actress.
Until BBC offers official confirmation, treat this report as rumor for now. Different outlets are treating the rumor with varying levels of significance and, even if Thakrar is BBC's first choice, front-runners don't always make it to the finish line. There's plenty that could still go wrong that might prevent Thakrar from taking her spot as the season 10 companion. That said, she's an exciting choice - and one that, if selected, should have no trouble carrying the companion mantle.
Doctor Who will continue with a Christmas special in December 2016. Season 10 is slated to air in 2017.


NSW Police Legacy attended the Police Association of NSW (PANSW) conference May 28-30, which was held this year at the Novotel in Wollongong. There were approximately 200 Police Association delegates in attendance, and we had many productive conversations about how they could get more involved with the Police Legacy family. There was a lot of support for who we are and what we do.
BACKUP for Life coordinator Esther Mckay gained some very useful information about industrial relations at one of the breakout sessions, and will be applying this new knowledge going forward as she continues to help Officers transition into careers outside the force.
The biggest news at the Conference, of course, was the retirement of Scott Weber after eight dedicated years as President of PANSW. Scott has been a loyal friend and supporter of Police Legacy and we wish him all the best in his new position. We also say a very warm welcome to new President Tony King, and look forward to our association going forward!


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Like he had a concussion last year, I mean he has concussions pretty much every know, we don talk about it but he does have concussions. After the carnage was analyzed that evening, at least a few mechanics must have gone hungry or worse, home for not properly choosing the right tire pressure prior to the stage.. At the end of the first quarter, Pittsburgh now leads Denver 7 6.. The menu also includes a sampling of wine and beer. In a letter sent to Elliott, which was, the league explained its findings and the punishment for violating the NFL's personal conduct policy. At San Diego State, he recruited running back Marshall Faulk you may have heard of him.. Done and D O N E!. So by playing that second year, and being encourage by the coach and the GM that, "Oh, it's a minor injury and we'll take care of it after the season and we'll just continue to shoot your shoulder up with pain medication, and by the end of the season, because we're so close to the Super youth nba jerseys Bowl and you're going to lead us there, and oh, by the way, we're going to take you from that position we were going to play you in last year". "There's greater awareness, which I think we're a large part of that because we brought that on, and I think we've promoted that. The Yankee Traveler is for campers who are 55 and older. Knights Inn The Knights Inn in Charlotte is located about three miles from Bank of America Stadium, home of the NFL Carolina Panthers. The veteran hasn't faced Seattle since his first season in Arizona in 2013, but neither of those contests was particularly memorable. These were like a half hour each. Mel Kiper, ESPN: Josh Doctson, WR, TCU. Ranks 4th in league annals with 456 career pass TDs 61,582 pass yards. Cutting edge technologies will be integrated throughout the building, including an Oculus video board. All but a handful of Raiders sat on their bench and seven Redskins took a knee while their teammates stood arm in arm along with owner Dan Snyder and president Bruce Allen.. The Colts complained that several footballs were under inflated and the NFL confirmed that 11 of the 12 footballs were under the limit. I'm not saying it wasn't inmy body before, but that doeskind of weigh on me. Is he one of the best 65 70 quarterbacks in the league? Yes. "Faggot Goodell" (also described as a "punk" and "dictator" by Harrison), Anderson ("another dummy who never played a down"), and Hanks, a former Pro Bowl safety with the Niners ("he needs to be ashamed because he played D before, though he never was what you call a real hitter"), conspired, he says, to target the Steelers, who have "too much force, too much swag, and are predominantly black." Says Harrison: "We sent them a tape of 27 hits from games that following week 27 hits like mine or worse but none of got flagged or fined. The day after meeting at the White House with South Carolina Sen. I just felt like I was prepared and ready enough to become a professional. I'm like, 'Wow, who's that kid?'". My pinkie's all jacked up; it gets in the way with handshakes and catching the ball sometimes. The odds are frequently also available even though odds can vary https://www.cheapjerseys.football/ in one site to another one; creates even better betting.. Il avait 43 ans et vingt ans de carrire NFL derrire lui.. Comments on this story are moderated according to our Submission Guidelines. Thompson, the attorney handling Hernandez's appeal in the 2013 murder case, also said Hernandez's death should be "carefully and thoroughly investigated."Hernandez's fiancee, Shayanna Jenkins, has not spoken publicly since cameras captured her sobbing in court, overwhelmed with emotions after the jury found him not guilty in the double murder case. Tried to keep it on the game, but it was very apparent though that people cheap basketball jerseys wanted to talk about the anthem. After an assistant coaching career at Purdue, Martin became the head coact at Missouri State and now is the head coach at Tennessee.. (Tom Brady, QB, New England). In response, the owners never justified their demands for a massive giveback which would have resulted in the worst economic deal for players in major league pro sports.. Among those kneeling was a white player, Seth DeValve. Step 6Determine how many days you will need to use the scaffolding units and multiply this number by the daily rate. Nashville, Tennessee based Bridgestone Americas, Inc. He still has that explosive nature. 28, 2016″ > >NFL draft preview: Big Ten prospects to watchChicago Tribune sports reporter Teddy Greenstein gives you the top Big Ten prospects as well as some local hopefuls.
The Australian Property Institute


Sasha was "born" during the making of her 2003 hit single "Crazy in Love." In an interview with Entertainment Weekly magazine, Knowles noted that her alter-ego is "strictly for the stage." The magazine has described it as "the singer's sensual, aggressive alter ego." Knowles wears a "roboglove" in public to compliment Sasha Fierce. It was Knowles' concept, and New York-based jeweler Lorraine Schwartz made the titanium glove. It consists of several pieces, including a ring, a glove, and a separate component that covers Knowles' upper arm. The glove first appeared in the video for "Single Ladies (Put a Ring on It)".
Knowles later wore it on the red carpet during the 2008 MTV Europe Music Awards, on her performance at the Saturday Night Live show, on the cover of Gotham magazine, and in the music video to "Diva". Read more on Last.fm. User-contributed text is available under the Creative Commons By-SA License; additional terms may apply..


text:
// Copyright (c) 2012 The Chromium Authors. All rights reserved.
// Use of this source code is governed by a BSD-style license that can be
// found in the LICENSE file.

// Multiply-included message file, so no include guard.

#include <stdint.h>

#include <string>
#include <vector>

#include "build/build_config.h"
#include "content/common/content_export.h"
#include "content/public/common/common_param_traits.h"
#include "content/public/common/webplugininfo.h"
#include "ipc/ipc_message_macros.h"

#undef IPC_MESSAGE_EXPORT
#define IPC_MESSAGE_EXPORT CONTENT_EXPORT
#define IPC_MESSAGE_START UtilityMsgStart

//------------------------------------------------------------------------------
// Utility process messages:
// These are messages from the browser to the utility process.

// Tells the utility process that it's running in batch mode.
IPC_MESSAGE_CONTROL0(UtilityMsg_BatchMode_Started)

// Tells the utility process that it can shutdown.
IPC_MESSAGE_CONTROL0(UtilityMsg_BatchMode_Finished)

#if defined(OS_POSIX) && defined(ENABLE_PLUGINS)
// Tells the utility process to load each plugin in the order specified by the
// vector. It will respond after each load with the WebPluginInfo.
IPC_MESSAGE_CONTROL1(UtilityMsg_LoadPlugins,
                     std::vector<base::FilePath> /* plugin paths */)
#endif

//------------------------------------------------------------------------------
// Utility process host messages:
// These are messages from the utility process to the browser.

#if defined(OS_POSIX) && defined(ENABLE_PLUGINS)
// Notifies the browser when a plugin failed to load so the two processes can
// keep the canonical list in sync.
IPC_SYNC_MESSAGE_CONTROL2_0(UtilityHostMsg_LoadPluginFailed,
                            uint32_t /* index in the vector */,
                            base::FilePath /* path of plugin */)

// Notifies the browser that a plugin in the vector sent by it has been loaded.
IPC_SYNC_MESSAGE_CONTROL2_0(UtilityHostMsg_LoadedPlugin,
                            uint32_t /* index in the vector */,
                            content::WebPluginInfo /* plugin info */)
#endif

redpajama_set_name: RedPajamaGithub


Please send us an email or contact us by phone : 0975 101 212 (MR HÙNG). We have a bigger selection of houses to show than on the website.
Lakeview villas for rent in District 9, Ho Chi Minh City (near Maguerite Duras French school), safe and quiet. Near the river, clean, fresh and green environment, beautiful design and luxury facilities like swimming pool, three tennis courts, a gymnasium,team bath, sauna, a children's playground where you can enjoy your life after a hard working day.
It takes 30-40 mins to city center and 15mins to Industrial Parks (HiTech-Parks, Song Than, Bien Hoa, VSIP, Linh Trung, Amata,..). Lakeview includes 113 villas from 2-5 bedrooms (fully furnished or unfurnished) in the area of 10ha with the floor area ranges from 164 to 600 sqm.
The rental is from $1600 to $3500/month. The rent includes VAT, pest control, fire protection, cable TV, shuttle bus service (school, downtown), use of clubhouse (Gym, swimming pool, tennis court, sauna & steam room, indoor children playground).


Web & Graphic Design - Reveries Co.
Does your blog need a redesign?
Are you looking to set up a brand-new website?
Do you need graphics for social media?
This is an appearance-only redesign of your website. It includes fonts, colors, a header, a logo, and other small tweaks.
Websites will be created using a variety of images, fonts, and graphics from sites of the designer's choice. If you want a specific element outside the designer's resources, you will have to cover the cost.
Rush Orders: additional $20/30 depending on the time limit.
This is an appearance-only redesign of your website. It includes fonts, colors, a header, a logo, and up to five sidebar images.
We'd love to create a brand-new personalized website for you from scratch. We can do this on Wordpress, Weebly, or Wix! Includes fonts, colors, a header, a logo, and up to 3 pages.
Price does not include hosting, domains, and additional graphics costs. Those will be added to the total. Websites will be created using a variety of images, fonts, and graphics from sites of the designer's choice. If you want a specific element outside the designer's resources, you will have to cover the cost.
Rush Orders: additional $30/50 depending on the time limit.
Additional pages are $20 each.
Basic logos cost $25. This includes 2 sets of revisions.
(If you want a website designed!) What do you want to accomplish with it? What sort of elements will it need to have?
You can be as specific or as vague here as you want.
This is for legal reasons. If you are under 18, you will be required to have a cosigner when we create a contract.
Feel free to be as specific as you want about your expectations for your cover design!
We do not accept erotica, horror, novels with extreme sexual content or language, or novels with obvious or prevalent anti-Christian/anti-God messages. We reserve the right to refuse any novel at our discretion. Our service providers may also refuse a project at any time without explanation. Our service providers are independent contractors and are responsible for their own taxes.
Site design and elements copyright 2018, Reveries Co. Founded by Kellyn Roth. Design Consultant: Estetico Designs (esteticodesigns.com).
Current managers: Kellyn Roth and Angela R. Watts.