text:
\section{Introduction}
A special case of the Strichartz inequality for the wave equation asserts that
\begin{equation} \label{classicalstrichartz} \|u\|_{q} \lesssim \left\|u\left( \cdot, 0\right)\right\|_{H^s\left(\mathbb{R}^d \right)} + \left\| u_t\left(\cdot, 0 \right) \right\|_{H^{s-1}\left( \mathbb{R}^d \right)} \quad \text{ for } \quad \Delta u - u_{tt} = 0, \end{equation}
where $d \geq 2$, $s=1/2$ and $q= \frac{2(d+1)}{d-1}$ \cite{strichartz,keel}. The purpose of this work is to prove a refinement of this inequality in certain cases, analogous to the refined Strichartz inequalities for the Schrödinger equation in \cite{li,du,du2}. The refined inequalities for the Schrödinger equation were used in \cite{li,zhang} to prove sharp results involving the almost everywhere pointwise convergence of Schrödinger solutions to the initial data. They were also used in \cite{du,zhang} to improve the known results on Falconer's distance set problem in dimensions $n \geq 3$. This was done through Mattila's approach \cite{mattila} via the spherical averages of Fourier transforms of measures. Here the refined Strichartz inequality for the wave equation will be used to improve results on the averages over the cone, rather than the sphere. As is well known \cite{wolff,erdogan2,cho}, the optimal decay of these conical averages is related to best possible $s$ in an inequality of the form \eqref{classicalstrichartz}, with the $L^q$ norm on the left replaced by an $L^2(\mu)$ norm for a fractal measure $\mu$.

The Fourier transform of a compactly supported Borel measure $\mu$ on $\mathbb{R}^n$ is given by
\[ \widehat{\mu}(\xi) := \int e^{-2\pi i \langle \xi, x \rangle } \, d\mu(x). \]
The measure $\mu$ is called $\alpha$-dimensional if
\[ c_{\alpha}(\mu) := \sup_{\substack{x \in \mathbb{R}^n \\ r>0 }} \frac{ \mu\left(B\left(x,r\right)\right)}{r^{\alpha}} < \infty. \]
Given a smooth compact surface $\Gamma \subseteq \mathbb{R}^n$, let $\beta(\alpha,\Gamma)$ be the supremum over all $\beta \geq 0$ which satisfy
\begin{equation} \label{coneaverages} \int \left| \widehat{\mu}(R \xi) \right|^2 \, d\sigma_{\Gamma}(\xi) \lesssim \|\mu\|c_{\alpha}(\mu) R^{-\beta} \quad \text{ for all } R >0, \end{equation}
uniformly over all Borel measures $\mu$ with support in the unit ball of $\mathbb{R}^{n}$. Here $\|\mu\|$ is the total variation norm, the implicit constant is allowed to depend on $\beta$, and $\sigma_{\Gamma}$ is the surface measure on $\Gamma$. The distance set problem is related to the case where $\Gamma = S^{n-1}$ \cite{mattila,liu}, see also \cite{luca} for the connection between $\beta\big(\alpha, S^{n-1}\big)$ and the pointwise convergence of solutions to the wave  and Schrödinger equations. Henceforth $\Gamma$ will be the truncated cone
\[ \Gamma := \left\{ \left(\xi, |\xi|\right) \in \mathbb{R}^{d+1} : 1 \leq |\xi| \leq 2 \right\}, \quad n =d+1, \]
and it may be assumed that $\sigma_{\Gamma}$ is actually the pushforward of the $d$-dimensional Lebesgue measure under the map $\xi \mapsto \left( \xi, \left|\xi\right| \right)$.

For $d \geq 3$, the best known lower bounds are
\begin{equation}  \label{cho} \beta\big(\alpha,\Gamma^d\big) \geq \left\lbrace
\begin{array}{lll}
\alpha &\text{ if } 0 < \alpha \leq \frac{d-1}{2} & \text{(Mattila \cite{mattila} and Rogers \cite{rogers})}\\
  \frac{d}{4} + \frac{\alpha}{2} - \frac{1}{4}  &\text{ if } \frac{d-1}{2} < \alpha \leq \frac{d+3}{2} & \text{(Cho, Ham and Lee \cite{cho})}\\
\alpha-1 &\text{ if } \frac{d+3}{2} < \alpha \leq d+1 & \text{(Sjölin \cite{sjolin}),} \end{array} \right. \end{equation}
and the best known upper bounds are
\begin{equation}  \beta\big(\alpha,\Gamma^d\big) \leq \left\lbrace
\begin{array}{lll}  \alpha &\text{ if }  0 < \alpha \leq d-2 &\\
  \frac{d}{2} + \frac{\alpha}{2} - 1  &\text{ if }  d-2 < \alpha \leq d  & \text{(Cho, Ham and Lee \cite{cho}).}\\
\alpha-1 &\text{ if } d \leq \alpha \leq d+1 & \end{array} \right. \end{equation}
These bounds are equal and give the exact value of $\beta\big(\alpha,\Gamma^d\big)$ when $d=3$. The exact value of $\beta\big(\alpha,\Gamma^2\big)$ was determined by Erdoğan in \cite{erdogan2} (Sjölin's lower bound $\beta\big(\alpha,\Gamma^d\big) \geq \alpha-1$ is a straightforward consequence of Parseval and duality). In Section \ref{conicalaverages}, it will be shown that
\begin{equation} \label{newdecay} \beta\big(\alpha,\Gamma^d\big) \geq \alpha -1 + \frac{d-\alpha}{d+1}, \quad \alpha \in \left(0,d+1\right], \end{equation}
improving \eqref{cho} in the range $\frac{d+1}{2}+\frac{2}{d-1} < \alpha < d$, which is nonempty only for $d \geq 4$. As in \cite{du}, the inequality \eqref{newdecay} will be shown through the linear refined Strichartz inequality; the proof of which is given in Section \ref{linearstrichartz}. In Section \ref{multilinearstrichartz}, a multilinear refined Strichartz inequality will be deduced from the linear one, as in \cite{du2}. By using the multilinear version, the bound \eqref{newdecay} could possibly be improved further through the methods in \cite{zhang}; the known values of $\beta\big(\alpha,\Gamma^d\big)$ for the $d$-dimensional cone tend to mimic the known values of $\beta\big(\alpha, S^{d-1}\big)$ for the sphere of one dimension less.

The upper bound
\begin{equation} \label{newdecay2} \beta\big(\alpha,\Gamma^d\big) \leq \alpha -1 + \frac{2(d+1-\alpha)}{d+1}, \quad \alpha \in \left(0,d+1\right), \quad d \geq 5. \end{equation}
will also be proved in Section \ref{conicalaverages}, this is the analogy to the upper bound for the spherical averages, due to Luca and Rogers \cite{luca}. Improvement holds in the range $\frac{d+1}{2} < \alpha < d-\frac{4}{d-3}$, and this interval is nonempty only for $d \geq 6$.

For $q \in [1,\infty]$, let $s_d(\alpha,q)$ be the infimum over all $s$ satisfying the inequality
\begin{equation} \label{strichartz} \|u\|_{L^q(\mu)} \lesssim C(\mu) \left(\left\|u\left( \cdot, 0\right)\right\|_{H^s} + \left\| u_t\left(\cdot, 0 \right) \right\|_{H^{s-1}} \right) \quad \text{ for } \quad \Delta u - u_{tt} = 0, \end{equation}
where
\[ \|f\|_{H^s} := \left( \int \left| \widehat{f}(\xi) \right|^2 \left( 1+ |\xi|^2 \right)^s \, d\xi \right)^{1/2}, \]
and
\[C(\mu) := \begin{cases} \left\|\mu\right\|^{\frac{1}{q}-\frac{1}{2}} c_{\alpha}(\mu)^{\frac{1}{2}} &\text{ if } 1 \leq q \leq 2 \\
c_{\alpha}(\mu)^{\frac{1}{q}} &\text{ if } 2 < q \leq \infty. \end{cases} \]
The quantity $s_d(\alpha,2)$ is related to $\beta\big(\alpha,\Gamma^d\big)$ through the equation
\begin{equation} \label{beta-s} \beta\big(\alpha,\Gamma^d\big)= d-2s_d(\alpha,2), \end{equation}
see \cite{wolff,erdogan2,cho,rogers}; a proof is given here in Proposition \ref{relationship}. The piecewise definition of $C(\mu)$ is what makes the proof of \eqref{beta-s} work, although for known counterexamples the choice of scaling does not make a difference (see e.g.~\cite{rogers}). For $d \geq 3$, the corresponding best known bounds for $s_d(\alpha,q)$ are
\begin{equation} \label{qsmall} s(\alpha,2,d) \leq s_d(\alpha,q) \leq \widetilde{s}(\alpha,2,d) \quad \text{for $q \in [1,2]$,} \end{equation}
and
\begin{equation} \label{qlarge} s(\alpha,q,d) \leq s_d(\alpha,q) \leq \widetilde{s}(\alpha,q,d) \quad \text{for $q \in [2,\infty]$,} \end{equation}
where (for $d \geq 2$),
\[ s(\alpha,q,d) := \begin{cases} \max\left( \frac{d}{2}-\frac{\alpha}{q},  \frac{d+1}{4} \right) &\text{ if } 0 < \alpha \leq 1 \\
\max\left( \frac{d}{2}-\frac{\alpha}{q}, \frac{d+1}{4} + \frac{1-\alpha}{2q},  \frac{d+2-\alpha}{4} \right) &\text{ if } 1 < \alpha \leq d  \\
\max\left( \frac{d}{2}-\frac{\alpha}{q}, \frac{d+1}{4} + \frac{d+1-2\alpha}{2q},  \frac{d+1-\alpha}{2} \right) &\text{ if } d < \alpha \leq d+1, \end{cases}  \]
and (for $d \geq 3$),
\[ \widetilde{s}(\alpha,2,d) :=  \begin{cases} \frac{d-\alpha}{2} &\text{ if } 0 < \alpha \leq \frac{d-1}{2} \\
 \frac{3d+1}{8} - \frac{\alpha}{4} &\text{ if } \frac{d-1}{2} < \alpha \leq \frac{d+3}{2} \\
\frac{d+1-\alpha}{2}  &\text{ if } \frac{d+3}{2} < \alpha \leq d+1, \end{cases} \]
and for $q>2$,
\[ \widetilde{s}(\alpha,q,d) :=  \begin{cases} \max\left( \frac{d}{2}-\frac{\alpha}{q},  \frac{d+1}{4}, \frac{3d+1}{8} - \frac{\alpha}{4} \right) &\text{ if } 0 < \alpha \leq 1 \\
\max\left( \frac{d}{2}-\frac{\alpha}{q}, \frac{d+1}{4} + \frac{1-\alpha}{2q}, \frac{3d+1}{8} - \frac{\alpha}{4} \right) &\text{ if } 1 < \alpha \leq d \\
\max\left( \frac{d}{2}-\frac{\alpha}{q}, \frac{d+1}{4} + \frac{d+1-2\alpha}{2q},  \frac{d+1-\alpha}{2} \right) &\text{ if } d < \alpha \leq d+1. \end{cases} \]
The piecewise intervals in $\widetilde{s}(\alpha,q,d)$ are only subdivided this way for simplicity, and could be made slightly more optimal as in $\widetilde{s}(\alpha,2,d)$. The lower bound in \eqref{qlarge} is from \cite{cho}, and the lower bound in \eqref{qsmall} is from \cite{rogers} (and works even if the constant $\|\mu\|^{\frac{1}{q}-\frac{1}{2}}c_{\alpha}(\mu)^{\frac{1}{2}}$ is relaxed to $c_{\alpha}(\mu)^{\frac{1}{q}}$). The upper bound in \eqref{qsmall} and \eqref{qlarge} is from \cite{cho}. Earlier results were obtained by Oberlin in \cite{oberlin}, see also \cite{oberlin2}. The inequalities \eqref{qsmall} and \eqref{qlarge} determine $s_3(\alpha,q) = s(\alpha,q,3)$ for all $q$. The equality $s_2(\alpha,q) = s(\alpha,q,2)$ holds for all $q$, this was determined for $\alpha \geq 1$ in \cite{erdogan2}, and for $\alpha \leq 1$ in \cite[Eq. 90]{wolff2} and \cite[Sect. 4]{rogers}.

The relation \eqref{beta-s} combined with the bound for $\beta\big(\alpha,\Gamma^d\big)$ in \eqref{newdecay} will give
\[ s_d(\alpha,2) \leq \frac{d}{2} - \frac{1}{2} \left(\alpha-1+ \frac{d-\alpha}{d+1} \right). \]
The range of improvement is the same as for $\beta\big(\alpha,\Gamma^d\big)$, given by $\frac{d+1}{2}+\frac{2}{d-1} < \alpha < d$. There would also be some small improvement for slightly larger $q$ by interpolation of $H^s$ spaces. For $q \geq 4$ the known inequalities are already optimal, since the lower and upper bounds in \eqref{qlarge} are the same.

The upper bound in \eqref{newdecay2} gives
\[ s_d(\alpha,2) \geq \frac{d}{2} - \frac{1}{2} \left(  \alpha -1 + \frac{2(d+1-\alpha)}{d+1} \right), \quad \alpha \in \left(0,d+1\right), \quad d \geq 5, \]
which improves the lower bound in \ref{qsmall} for $q=2$, $d \geq 6$ in the range $\frac{d+1}{2} < \alpha < d-\frac{4}{d-3}$. This shows that $s(\alpha,q,d)$ is not the optimal value for all $d$, which disproves a conjecture from \cite{cho}.

The Strichartz inequality \eqref{strichartz} was also considered in \cite{rogers} for a restricted class of measures of the form $\nu = \mu \otimes \lambda$, where $\mu$ is a compactly supported $\alpha$-dimensional measure in the unit ball of $\mathbb{R}^d$ and $d\lambda = \chi_{[0,1]} \, dm$, where $m$ is the Lebesgue measure on $\mathbb{R}$. For this restricted class, the infimum over $s$ in \eqref{strichartz} was shown in \cite{rogers} to be related to Falconer's distance set problem. The optimal value of $s$ has no relationship to the optimal decay of conical averages in \eqref{coneaverages} however, since for this restricted class of measures the optimal decay
\[ \int \left| \widehat{\nu}(R \xi) \right|^2 \, d\sigma_{\Gamma}(\xi) \lesssim I_{\alpha}(\mu) R^{-(\alpha+2)} \quad \text{ for all } R >0, \]
follows directly from a straightforward computation (see Proposition \ref{trivial}) involving the characterisation of the energy $I_{\alpha}(\mu)$ as
\begin{equation} \label{energydef} I_{\alpha}(\mu) := \int \int \frac{1}{|x-y|^{\alpha}} \, d\mu(x) \, d\mu(y) =c_{\alpha,d} \int |\xi|^{\alpha-d} \left|\widehat{\mu}(\xi)\right|^2 \, d\xi. \end{equation}

\subsection{Notation} \label{notation}

Throughout, let $A: \mathbb{R}^{d+1} \to \mathbb{R}^{d+1}$ be the unitary operator defined through the standard basis by
\[ e_i \mapsto e_i \quad \text{for} \quad 1 \leq i \leq d-1, \quad \frac{ e_{d+1}+e_d}{\sqrt{2}} \mapsto e_d, \quad \frac{ e_{d+1}-e_d}{\sqrt{2}} \mapsto e_{d+1}, \]
and let $E$ be the extension operator for the subset $\Gamma_+ := \Gamma \cap \{\xi_d \geq 0\}$ of the cone, defined by
\[ Ef(x,t) = \int_{\{ \xi_d  \geq 0 \} \cap B(0,2) \setminus B(0,1)}  e^{2\pi i(\langle \xi, x \rangle + |\xi|t)} f(\xi) \, d\xi. \] The restriction to $\xi_d \geq 0$ is a minor technicality to ensure that the normal to the cone at any point $(\xi,|\xi|) \in \Gamma_+$ makes an angle $\gtrsim 1$ with the plane $A^*\left(\mathbb{R}^d \times \{0\}\right)$.

Let $m$ be the Lebesgue measure on Euclidean space. When used as a function the symbol $\pi$ will mean the natural projection from $\mathbb{R}^{d+1}$ to $\mathbb{R}^d$.

The statement that $f$ is essentially supported in $U$ will mean that $|f(x)| \lesssim_N R^{-N}\|f\|_2$ for $x$ outside $U$, where $N$ is arbitrarily large.

For non-negative $X$ and $Y$ the notation $X \lessapprox Y$ will mean $X \leq C_{\epsilon} R^{\epsilon} Y$ where $\epsilon$ is arbitrarily small. Similarly $X \approx Y$ means $X \lessapprox Y$ and $Y \lessapprox X$. For functions the statement $Ef(x) \approx g(x)$ will mean that $|Ef(x) - g(x)| \lesssim R^{-N}\|f\|_2$ for arbitrarily large $N$, where $E$ is an operator (usually the extension operator). For any box $X$ and constant $C \geq 0$, the set $CX$ will the box with the same centre, but with side lengths scaled by $C$.

\section{Modified wave packet decomposition for the cone}
The wave packet decomposition used here is based on those in \cite{guth,cho,li,shayya,ou}. In the decomposition, the truncated cone is partitioned into caps; to construct these, partition the $(d-1)$-dimensional sphere $S^{d-1}$ into spherical caps $C$ of diameter $\delta^{1/2}$, where $\delta \in (0,1)$, and define each $\tau=\tau_C$ by
\[ \tau = \left\{ \left(\xi, |\xi|\right) \in \Gamma: \frac{ \xi}{|\xi|} \in C \right\}. \]
The set $\tau$ is called a cap at scale $\delta^{1/2}$. It is contained in a box of dimensions
\[ \sim \delta^{1/2} \times \dotsm \times \delta^{1/2} \times 1 \times \delta, \]
where the second last coordinate refers to the flat direction in $\tau$, and the last direction is normal to the cone at $\tau$. The normal to $\tau$ is defined to be the normal to the cone at $x_{\tau}$, where $x$ is the barycentre of $\tau$ with respect to $\sigma_{\Gamma}$. The flat direction of $\tau$ is $y_{\tau}$, where $y_{\tau}$ is the barycentre of $C$ with respect to the surface measure on the sphere (the use of the barycentre is not important, any other point in $\tau$, $C$ would work).

The dimensions in the following wave packet decomposition are adapted to a rescaling argument to be used later, and are therefore different from the standard dimensions (in e.g. \cite{ou}).

\begin{proposition} \label{primitivewave} Fix $\delta>0$. Let $B$ be a box in $\mathbb{R}^d$ of dimensions
\[ R^{-1/4} \times \dotsm \times R^{-1/4} \times 1, \]
with sides parallel to the coordinate axes, and let $\mathcal{D}$ be a collection of boxes $D \subseteq \mathbb{R}^d$ of dimensions
\[ R^{3/4+ \delta} \times \dotsm \times R^{3/4+\delta} \times R^{1/2+\delta}, \]
also with sides parallel to the coordinate axes, such that the boxes $(1/2)D$ form a finitely overlapping cover of $\mathbb{R}^d$. Suppose that $\Phi \in C^{\infty}((4/3) B, \mathbb{R})$ satisfies
\begin{equation} \label{derivative} \text{$|\partial_i^k \Phi | \lesssim_k R^{3k/4}$ for $1 \leq i \leq d-1$}  \quad \text{and} \quad \text{$|\partial_d^k \Phi | \lesssim_k R^{k/2}$ for all $k \geq 1$.}  \end{equation}
Then for sufficiently large $R$, any $f \in L^2(\mathbb{R}^d)$ supported in $B$ can be decomposed as $f = \sum_D f_D$, such that each $f_D$ is supported in $(4/3)B$, $\widecheck{\Phi f_D}$ is essentially supported in $D$,
\begin{equation} \label{essorthogonality} \sum_D \left\| f_D \right\|_2^2 \lesssim \|f\|_2^2, \end{equation}
and
\[ \sum_{D: z \notin D} \left|\widecheck{ \Phi f_D}(z)\right| \lesssim R^{-N} \|f\|_2 \quad \text{for any $z \in \mathbb{R}^d$.} \]
\end{proposition}

\begin{proof} Let $\psi$ be smooth with $0 \leq \psi \leq 1$, such that $\psi$ is equal to 1 on $B$, vanishes outside $(5/4)B$ and satisfies
\begin{equation} \label{scaling} \text{$\left| \partial_i^k \psi \right| \lesssim_k R^{k/4}$ for $1 \leq i \leq d-1$} \quad \text{and} \quad \text{$\left| \partial_d^k \psi \right| \lesssim_k 1$ for $k \geq 1$.} \end{equation}
Let $\{\phi_D\}_D$ be a smooth partition of unity subordinate to the $(3/4)D$, so that
\[ f= \psi f = \sum_D \psi\left(\widehat{\phi_D} \ast f\right) =: \sum_D f_D. \]
The convergence holds in $L^2(\mathbb{R}^d)$ by Plancherel, and each $f_D$ is supported in $(4/3)B$ by the support condition on $\psi$. By the finite overlapping property and Plancherel, the $L^2$ norm of $f$ satisfies
\[ \|f\|_2^2 = \sum_{i,j}  \left\langle \phi_{D_i} \widecheck{f} ,\phi_{D_j} \widecheck{f} \right\rangle \\
\sim \sum_D \left\| \phi_{D} \widecheck{f} \right\|_2^2 = \sum_D \left\| \widehat{\phi_{D}} \ast f \right\|_2^2 \gtrsim \sum_D \left\| f_D \right\|_2^2, \]
which proves \eqref{essorthogonality}.

It remains to check the support conditions on $\widecheck{\Phi f_D}$. Let $w = \Phi \psi$, so that by \eqref{derivative} and \eqref{scaling},
\[ \text{$\left|\partial_i^k w \right| \lesssim_k R^{3k/4}$ for $1 \leq i \leq d-1$} \quad \text{and} \quad \text{$\left|\partial_d^k w\right| \lesssim_k R^{k/2}$ for $k \geq 1$.} \] After integrating by parts $L$ times, this implies that
\[ \text{$\widecheck{w}(z) \lesssim \frac{m(B)}{(|z_i| R^{-3/4} )^L}$ for $1 \leq i \leq d-1$} \quad \text{and} \quad \text{$\widecheck{w}(z) \lesssim \frac{m(B)}{(|z_d| R^{-1/2} )^L}$ for $z \in \mathbb{R}^d$.} \]
Hence for $z \notin D$,
\begin{align*} \left|\widecheck{ \Phi f_D}(z) \right| &= \left|\widecheck{ w} \ast \left(\phi_D \widecheck{f}\right)(z)\right|  \\
&\leq \|f\|_1 \int_{(3/4)D}  \left|\widecheck{w}(z-y)\right| \phi_D(y)  \, dy \\
&\lesssim R^{-N} \|f\|_2,\end{align*}
for any $N$, by taking $L$ large enough, depending on $\delta$. This shows that $\widecheck{ \Phi f_D}$ is essentially supported in $D$. Similarly, since the cover is finitely overlapping, applying similar working and summing a geometric series gives
\[ \sum_{D: z \notin D} \left|\widecheck{ \Phi f_D}(z)\right| \lesssim R^{-N} \|f\|_2 \quad \text{for any $z \in \mathbb{R}^d$,} \]
which proves the proposition. \end{proof}

\begin{proposition} \label{wavepacket} Let $\tau$ be a cap in the cone at scale $R^{-1/4}$. Then any $f \in L^2(\pi(\tau))$ can be decomposed in $L^2$ as $f= \sum_T f_T$ with
\begin{equation} \label{essorthogonality2} \sum_T \left\| f_T \right\|_2^2 \lesssim \|f\|_2^2, \end{equation}
such that each $f_T$ has support in $(4/3) \pi(\tau)$, where the $T$'s are boxes with dimensions
\[ \sim R^{3/4+\delta} \times \dotsm \times R^{3/4+ \delta} \times R^{1/2+\delta} \times R, \]
with long axis normal to $\tau$, short axis in the flat direction of $\tau$, and such that the restriction of $Ef_{T}$ to $B(0,R)$ is essentially supported in $T$, with
\begin{equation} \label{esssupport} \sum_{T: (x,t) \notin T} \left| Ef_T(x,t) \right| \lesssim R^{-N} \|f\|_2 \quad \text{for} \quad |(x,t)| \leq R. \end{equation}
\end{proposition}

\begin{proof} Since the condition $|(x,t)| \leq R$ is rotation invariant, after applying a rotation of $\mathbb{R}^{d+1}$ which fixes the cone, it may be assumed that the flat direction of $\tau$ is $\frac{e_{d+1}+e_d}{\sqrt{2}}$. After then applying the unitary $A$, it may be assumed that the normal to $\tau$ is completely in the $e_{d+1}$-direction, and that $(4/3)\tau$ is contained in the graph of the function
\[ h(\omega_1, \dotsc, \omega_{d}) := \frac{ \sum_{i=1}^{d-1}  \omega_i^2}{2 \omega_d}, \]
restricted to $(4/3)B$ where
\[ B:= \left[ -R^{-1/4}, R^{-1/4} \right] \times \dotsm \times \left[ -R^{-1/4}, R^{-1/4} \right] \times \left[ \sqrt{2}, 2 \sqrt{2} \right] \subseteq \mathbb{R}^d. \] Let $f_T = f_D$ from Proposition \ref{primitivewave} where each $T= D \times \mathbb{R}$. The inequality \eqref{essorthogonality2} and the support condition on the $f_T$'s both follow from Proposition \ref{primitivewave}. Moreover,
\[ Ef_T(x,t) =  \int e^{2\pi i \left( \langle \omega, x \rangle + h(\omega) t \right) } f_T(\omega) \, d\omega = \widecheck{ \Phi_t f_D }(x), \]
where $\Phi_t(\omega) := e^{2\pi ih(\omega)t}$ restricted to $(4/3) B$, and from the definition of $h$,
\[ |\partial_i^k \Phi_t |(\omega) \lesssim_k R^{3k/4} \quad \text{for} \quad 1 \leq i \leq d-1 \quad \text{and} \quad |\partial_d^k \Phi_t |(\omega) \lesssim_k R^{k/2}, \]
for all $k \geq 1$, uniformly for $|t| \leq R$ and $\omega \in (4/3)B$. By Proposition \ref{primitivewave} this means that for each fixed $t$ with $|t| \leq R$, $Ef_T( \cdot, t)$ is essentially supported in $D$. Therefore the restriction of $Ef_T$ to $B(0,R)$ is essentially supported in $T$, and the inequality \eqref{esssupport} follows from Proposition \ref{primitivewave}. This finishes the proof.
\end{proof}

\section{Linear refined Strichartz inequality} \label{linearstrichartz}
The proof of the linear refined Strichartz inequality will use the (sharp) decoupling theorem for the truncated cone from \cite{bourgain}.
\begin{theorem}[{\cite[Theorem 1.2]{bourgain}}]  \label{decoupling} Let $\delta \in (0,1)$, partition the truncated cone into caps $\tau$ at scale $\delta^{1/2}$, and let $f = \sum_{\tau} f_{\tau}$ be a function on $\mathbb{R}^{d+1}$ such that the support of $\widehat{f_{\tau}}$ is contained in the $\delta$-neighbourhood of $\tau$. Then for any $\epsilon >0$,
\[ \|f\|_q \leq C_{\epsilon} \delta^{-\epsilon} \left(\sum_{\tau } \|f_{\tau}\|_q^2 \right)^2, \quad \text{where $q = \frac{2(d+1)}{d-1}$.} \] \end{theorem}

 Most of the notation used throughout is similar to that in \cite{li}, to emphasise the analogy with the Schrödinger case (see also Subsection \ref{notation}).
\begin{theorem} \label{strichartzr} Fix $d \geq 2$. Suppose that $f \in L^2(\mathbb{R}^d)$ is supported in $B(0,2) \setminus B(0,1)$, and let $Y= \bigcup Q$ be a collection of $R^{1/2} A^*\mathbb{Z}^{d+1}$-lattice cubes inside a ball of radius $R$. If $\|Ef\|_{L^q(Q)}$ is dyadically constant as $Q$ varies over $Y$, and if the cubes are arranged in slabs of the form $A^*\left(\mathbb{R}^{d} \times [R^{1/2} j, R^{1/2} j + R^{1/2}]\right)$ with $j \in \mathbb{Z}$, such that each slab intersecting $Y$ contains $\sim \sigma$ cubes in $Y$, then for any $\epsilon >0$,
\[ \| Ef\|_{L^{q}(Y)} \leq C_{\epsilon}  R^{\epsilon} \sigma^{-\gamma}\|f\|_2, \]
where $\gamma = \frac{1}{2} - \frac{1}{q}$ and $q  = \frac{2(d+1)}{d-1}$.
\end{theorem}

\begin{remark} \label{scales} To apply induction, it will be easier to prove a slightly more general statement, where each $Q$ in the theorem is replaced by $CQ$, for a constant $C \in [1,2]$, but the slabs are unchanged. The inductive assumption will be that this slightly more general statement of the theorem holds whenever $R$ is replaced everywhere in the theorem by $\widetilde{R}$, for any $\widetilde{R} \leq R^{3/4}$ ($3/4$ is not important, any exponent in $(1/2,1)$ would work).

Given $\epsilon >0$, the inductive assumption is that
\[ \| Ef\|_{L^{q}(Y)} \leq C_{\epsilon} \widetilde{R}^{\epsilon}\sigma^{-\gamma} \|f\|_2, \]
 for any $f$ satisfying the (generalised) assumptions of the theorem, and any $\widetilde{R} \leq R^{3/4}$. Using this, it will be shown that for any $\delta>0$, the inequality
\[ \| Ef\|_{L^{q}(Y)} \leq C_{\epsilon} C_{\delta} R^{O(\delta)} R^{\epsilon/2} \sigma^{-\gamma}\|f\|_2, \]
holds for any $f$ satisfying the (generalised) assumptions of the theorem. By choosing $\delta>0$ small enough, depending only on $\epsilon$, and then taking $R$ large enough (depending only on $\epsilon$), this will give
\[ \| Ef\|_{L^{q}(Y)} \leq C_{\epsilon} R^{\epsilon} \sigma^{-\gamma}\|f\|_2, \]
with the same constant $C_{\epsilon}$, which will close the induction and prove the theorem. To simplify notation the $R^{\epsilon}$ and $R^{\delta}$ factors will be absorbed in the symbols $\approx$ and $\lessapprox$, and therefore this argument will not be carried out explicitly. \end{remark}

\begin{proof}[Proof of Theorem \ref{strichartzr}] After replacing $f$ by $e^{-2\pi i(\langle x_0, \xi \rangle + t_0 |\xi|)} f$ it may be assumed that the ball of radius $R$ containing $Y$ is centred at the origin. Fix $\delta>0$, and use Proposition \ref{wavepacket} to decompose
\[ f = \sum_{\tau,\Box_{\tau}} f_{\Box}, \quad Ef = \sum_{\tau,\Box_{\tau}} Ef_{\Box}, \]
where the caps $\tau$ partitioning the truncated half cone $\Gamma_+$ are at scale $\sim R^{-1/4}$, each contained in a corresponding box with dimensions
\[ \sim R^{-1/4} \times \dotsm \times R^{-1/4} \times 1 \times R^{-1/2}, \]
and each $Ef_{\Box}$ has distributional Fourier transform supported in the cap $\tau$ corresponding to $\Box$. When restricted to $B(0,R)$, each $Ef_{\Box}$ is essentially supported in a set $\Box$ of dimensions
\[ \sim R^{3/4+\delta} \times \dotsm \times R^{3/4+\delta} \times R^{1/2+\delta} \times R, \]
with long axis normal to $\tau$, and short axis in the flat direction of $\tau$. Although each set $\Box$ depends on a cap $\tau$, this will be suppressed in the notation. Each $Ef_{\Box}$ can be partitioned further
\begin{equation} \label{unity} Ef_{\Box} \approx \sum_{2S \cap \Box \neq \emptyset} \eta_S Ef_{\Box} \quad \text{in} \quad B(0,R), \end{equation}
where for each $\tau$, the sets $S$ partition physical space and have the same axis orientations as $\Box$, with dimensions
\[ R^{1/2+\delta} \times \dotsm \times R^{1/2+\delta} \times R^{1/4+\delta} \times R^{3/4+\delta}. \]
The functions $\eta_S$ form a smooth partition of unity with each $\eta_S \lesssim 1$ on $S$, non-negative, essentially supported on $2S \cap N_{R^{1/2+\delta}}(S)$, such that $\widehat{\eta_S}$ is supported in a box of dimensions
\begin{equation} \label{poisson} \sim R^{-1/2} \times \dotsm \times R^{-1/2} \times R^{-1/4} \times R^{-1/2}, \end{equation}
around the origin, with long axis in the flat direction of $\tau$. Such a partition can be constructed by using the Poisson summation formula at scale one on the integer lattice to get a smooth non-negative function $\eta$ with
\[ \sum_{k \in \mathbb{Z}^{d+1}} \eta(x-k) = 1 \quad \text{for all $x \in \mathbb{R}^{d+1}$,} \quad \text{and} \quad \supp \widehat{\eta} \subseteq B(0,c_d).\]
Rescaling by the dimensions in \eqref{poisson} and then grouping the functions together corresponding to scaled lattice points in $S$ gives the required functions $\eta_S$.

For each fixed set $\Box$, sort the boxes $S$ close to $\Box$ into sets $\mathbb{S}_{\lambda}$ according to the dyadic value $\lambda$ of $\left\|Ef_{\Box}\right\|_{L^q(2S)}$. By ignoring the very small values of $\lambda$ which do not contribute significantly, there are $\lesssim \log R$ relevant values of $\lambda$, which will be the only ones considered from now on. For each fixed $\lambda$ and dyadic number $\sigma_{\Box}$, define $\mathbb{S}_{\lambda,\sigma_{\Box}}$ by $S \in \mathbb{S}_{\lambda, \sigma_{\Box}}$ if and only if there are $\sim \sigma_{\Box}$ boxes in $\mathbb{S}_{\lambda}$ inside the slab of width $R^{3/4+\delta}$ parallel to the tangent plane at $\tau$. These slabs are different to those in the theorem statement, but will be rotated and scaled to use the inductive assumption. There are $\lesssim \log R$ values of $\sigma_{\Box}$, so there are $\lesssim \left(\log R\right)^2$ total relevant pairs $\lambda$ and $\sigma_{\Box}$. For any set $\Box$ let $Y_{\Box,\lambda,\sigma_{\Box}}$ be the union of boxes $S$ in $\mathbb{S}_{\lambda,\sigma_{\Box}}$. By Fubini (and the condition \eqref{esssupport} in Proposition \ref{wavepacket}),
\[ Ef \approx_{L^q} \sum_{\lambda, \sigma_{\Box}} \sum_{\Box} \eta_{Y_{\Box,\lambda,\sigma_{\Box}}}Ef_{\Box} \quad \text{in $B(0,R)$,} \quad \text{where} \quad \eta_{Y_{\Box,\lambda,\sigma_{\Box}}} := \sum_{S \subseteq Y_{\Box,\lambda,\sigma_{\Box}}} \eta_S, \]
where the symbol $\approx_{L^q}$ here means the difference  has $L^q(B(0,R))$ norm $\lesssim R^{-N} \|f\|_2$. By the triangle inequality and the standard pigeonhole principle, there is a fixed pair $(\lambda_0, \sigma_{\Box})$ such that
\begin{equation} \label{pigeon} \|Ef\|_{L^{q}(CQ)} \lessapprox \left\|  \sum_{\Box} \eta_{Y_{\Box,\lambda_0,\sigma_{\Box}}} Ef_{\Box} \right\|_{L^{q}(CQ)} + R^{-N} \|f\|_2,\end{equation}
for a fraction $\approx 1$ of the cubes $Q$. The quantity $R^{-N}\|f\|_2$ is a remainder term which absorbs the very small values of $\lambda$ and the error terms in the wave packet decomposition, and it may essentially be ignored in most of the inequalities that follow. Henceforth write $Y_{\Box} = Y_{\Box,\sigma_{\Box}}$. The sets $\Box$ can be sorted according to the dyadic value of $\|f_{\Box}\|_2$, and since there are only $\sim \log R$ relevant dyadic values, the triangle inequality and the standard pigeonhole principle applied again to the remaining cubes satisfying \eqref{pigeon} yield a subset $\mathbb{B}$ of sets $\Box$ such that $\|f_{\Box}\|_2$ is essentially constant over $\mathbb{B}$ and
\begin{equation} \label{pigeonhole} \|Ef\|_{L^{q}(CQ)} \lessapprox \left\|  \sum_{\Box \in \mathbb{B}} \eta_{Y_{\Box}}Ef_{\Box} \right\|_{L^{q}(CQ)} + R^{-N} \|f\|_2, \end{equation}
again for a fraction $\approx 1$ of the cubes $Q$. By pigeonholing the remaining cubes satisfying \eqref{pigeonhole} again, there is a dyadic number $\mu$ and a fraction $\approx 1$ of the cubes $CQ$, with union $Y'$, such that the bound in \eqref{pigeonhole} still holds and for each $CQ \subseteq Y'$ the cube $R^{2\delta}Q$ intersects $\sim \mu$ of the sets $Y_{\Box}$ with $\Box \in \mathbb{B}$.

 Let $\eta_Q$ be smooth non-negative functions such that $\eta_Q \sim 1$ on $2Q$, with each $\widehat{\eta_Q}$ compactly supported in a ball of radius $\sim R^{-1/2}$, and such that $\sum_Q \eta_Q \lesssim 1$ (this can be done, for example, by using Poisson summation again). To apply decoupling, the functions $\eta_{Y_{\Box}} \eta_{Q}Ef_{\Box}$ have Fourier transform supported in an $\sim R^{-1/2}$ neighbourhood of $\tau$, since the short directions in the sets $S\subseteq Y_{\Box}$ are in the flat direction of $\tau$. Moreover, for each fixed $\tau$, the sets $\Box$ corresponding to $\tau$ form a finitely overlapping cover of physical space and have side lengths at least as large as those of $Q$, which means that each cube $R^{2\delta}Q$ intersects $\lessapprox 1$ set $\Box$ associated to $\tau$, for each $\tau$. Hence the disjointness assumption in the decoupling theorem applies and so for each $CQ \subseteq Y'$,
\begin{align*} \|Ef\|_{L^{q}(CQ)} &\lessapprox \left\|  \sum_{\Box \in \mathbb{B}} \eta_{Y_{\Box}}Ef_{\Box}\right\|_{L^{q}(CQ)} + R^{-N} \|f\|_2  && \text{by \eqref{pigeonhole},} \\
&\lessapprox \Bigg\|  \sum_{\substack{\Box \in \mathbb{B} \\
Y_{\Box} \cap R^{2\delta}Q \neq \emptyset }} \eta_{Y_{\Box}}Ef_{\Box} \Bigg\|_{L^{q}(CQ)}+ R^{-N} \|f\|_2 \\
&\lesssim \Bigg\|  \sum_{\substack{\Box \in \mathbb{B}  \\
Y_{\Box} \cap R^{2\delta}Q \neq \emptyset }}\eta_{Y_{\Box}} \eta_{Q} Ef_{\Box} \Bigg\|_{q} + R^{-N} \|f\|_2  \\
&\lessapprox \Bigg( \sum_{\substack{\Box \in \mathbb{B} \\
Y_{\Box} \cap R^{2\delta}Q \neq \emptyset }}\left\|   \eta_{Y_{\Box}} \eta_{Q}Ef_{\Box} \right\|_{q}^2 \Bigg)^{1/2}  + R^{-N} \|f\|_2 && \text{by Theorem \ref{decoupling},}  \\
&\lesssim \mu^{ \frac{1}{2} - \frac{1}{q}}\left( \sum_{\Box \in \mathbb{B} }\left\| \eta_{Y_{\Box}}\eta_{Q} Ef_{\Box} \right\|_{q}^{q} \right)^{1/q} + R^{-N} \|f\|_2 .  \end{align*}

Raising both sides to the power $q$ and summing over $CQ \subseteq Y'$ gives, since $\sum_{Q} \eta_{Q}  \lesssim 1$ and since $\eta_{Y_{\Box}}$ is essentially supported on $2Y_{\Box}:= \bigcup_{S \subseteq Y_{\Box}} 2S$,
\begin{align*} \|Ef\|_{L^{q}(Y')}^{q}  &\lessapprox \mu^{ q \left( \frac{1}{2}- \frac{1}{q}\right) } \sum_{\Box \in \mathbb{B}} \left\| Ef_{\Box}\right\|^{q}_{L^q(2Y_{\Box})} + R^{-Nq} \|f\|_2^q . \end{align*}
 Since $Y'$ contains a fraction $\approx 1$ of the cubes in $Y$, and since the cubes contribute equally, this gives
\begin{align} \notag \|Ef\|_{L^{q}(Y)}^{q} &\lessapprox \mu^{ q \left( \frac{1}{2}- \frac{1}{q}\right) } \sum_{\Box \in \mathbb{B}} \left\| Ef_{\Box}\right\|^{q}_{L^q(2Y_{\Box})} + R^{-Nq} \|f\|_2^q  \\
\label{rescaling} &\lessapprox \mu^{ q \left( \frac{1}{2}- \frac{1}{q}\right) } \sum_{\Box \in \mathbb{B}} \left(  \sigma_{\Box}^{-\gamma}  \|f_{\Box}\|_2 \right)^{q} \quad \text{(by Lorentz rescaling, see below)}\\
\notag &\lesssim \mu^{ q \left( \frac{1}{2}- \frac{1}{q}\right) } \sigma_{\Box}^{-\gamma q} \sum_{\Box \in \mathbb{B}}  \|f_{\Box}\|_2^{q} \\
\label{qthroot} &\lesssim \mu^{ q \left( \frac{1}{2}- \frac{1}{q}\right) }\sigma_{\Box}^{-\gamma q}  |\mathbb{B}|^{  1- \frac{q}{2}} \|f\|_2^{q}, \end{align}
since $\|f_{\Box}\|_2$ is essentially constant in $\Box$ and $\sum_{\Box \in \mathbb{B}} \left\|f_{\Box} \right\|_2^2 \lesssim \|f\|_2^2$. To justify the Lorentz rescaling step in \eqref{rescaling}, after a rotation it may be assumed that the flat direction of the cap $\tau$ corresponding to $\Box$ is $\frac{ e_{d+1}+e_d}{\sqrt{2}}$. Let
\[ Bx = \left( R^{1/4}x_1,\dotsc, R^{1/4}x_{d-1}, x_d, R^{1/2}x_{d+1}\right), \]
and define $\eta$ and $\left(\widetilde{x}, \widetilde{t}\right)$ by
\[ \left(\eta, \left|\eta\right|\right) = A^*BA (\xi,|\xi|) \quad \text{for $\xi \in \tau$,} \quad \left(\widetilde{x}, \widetilde{t}\, \right) = A^* B^{-1} A(x,t). \]
Under this change of variables, the boxes $2S$ are sent to cubes $2Q$ of side length $2R^{1/4+\delta}$, whose union is defined to be $\widetilde{Y}$. Moreover,
\[ \|Ef_{\Box}\|_{L^{q}\left(2Y_{\Box}\right)} \sim \|E'g\|_{L^{q}\left(\widetilde{Y}\right)} \quad \text{and} \quad \|f_{\Box}\|_2 \sim \|g\|_2, \]
where
\[ g\left( \eta \right) := R^{ \frac{d-1}{8}}   \left|\frac{d \xi}{d\eta}\right| ( f_{\Box} \circ \pi) \left( \xi, |\xi| \right), \]
and $E'$ is the extension operator for a compact subset $\Gamma_1$ of the cone, which is contained in
\[ \left\{ (\xi,|\xi|): \xi_d \geq 0\right\} \cap B\left(0,3 \sqrt{2}\right) \setminus B(0,1), \]
provided $\tau$ is at scale $cR^{-1/4}$ for some small constant $c$ depending only on $d$ (which may be assumed). The values $\|Ef_{\Box}\|_{L^q(2S)} \sim \|E'g\|_{L^{q}(2Q)}$ are essentially constant over $S,Q$ by definition of the dyadic values $\lambda$. The slabs are of the form $A^* \left( \mathbb{R}^d \times [t_0, t_0 + R^{1/4+\delta}]\right)$ and each contains $\sim \sigma_{\Box}$ cubes, all inside a ball of radius $R^{1/2+2\delta}$. Even though $\Gamma_1$ may be slightly outside the truncated half cone, by scaling $g$ slightly this only affects the inequality by a constant factor. Hence the generalised statement of the theorem at scale $R^{1/2+2\delta}$ can be applied to give
\[ \|Ef_{\Box}\|_{L^{q}(2Y_{\Box})} \lessapprox \sigma_{\Box}^{-\gamma}\|f_{\Box}\|_2, \]
which is \eqref{rescaling}. Taking the $q$-th root of \eqref{qthroot} gives
\begin{align} \label{bound} \|Ef\|_{L^{q}(Y)} &\lessapprox \mu^{ \gamma } |\mathbb{B}|^{ - \gamma} \sigma_{\Box}^{-\gamma}   \|f\|_2, \quad \text{where $\gamma = \frac{1}{2}-\frac{1}{q}$.} \end{align}

To finish the proof, it suffices to show that $\mu \lessapprox \frac{ \sigma_{\Box} |\mathbb{B}|}{\sigma}$. The multiplicity $\mu$ satisfies
\begin{align*} \mu m(Y) &\lessapprox \mu m(Y') \\
&\sim  \mu \sum_{CQ \subseteq Y'}  m(Q) \\
&\sim \sum_{CQ \subseteq Y'} \sum_{\substack{\Box \in \mathbb{B} \\
Y_{\Box} \cap R^{2\delta} Q \neq \emptyset }} m( Q) && \text{by definition of $\mu$,} \\
&\lessapprox \sum_{CQ \subseteq Y'} \sum_{\substack{\Box \in \mathbb{B} \\
Y_{\Box} \cap R^{2\delta} Q \neq \emptyset }} R^{1/4}m\left(Y_{\Box} \cap R^{3\delta} CQ\right) \\
&\lessapprox  \sum_{\Box \in \mathbb{B}} R^{1/4}m\left(Y_{\Box} \cap R^{3\delta}Y'\right)  && \text{by Fubini,} \\
 &\leq \sum_{\Box \in \mathbb{B}} R^{1/4}m\left(Y_{\Box} \cap R^{3\delta} Y\right). \end{align*}
Therefore it suffices to show that for each $\Box \in \mathbb{B}$,
\[m\left(Y_{\Box} \cap R^{3\delta} Y\right) \lessapprox \frac{ \sigma_{\Box} m(Y) R^{-1/4} }{\sigma}. \]
By breaking both sides of this inequality into slabs $A^*\left(\mathbb{R}^d \times \left[t, t+CR^{1/2 + 3\delta} \right] \right)$ covering $R^{3\delta}Y$ and containing $\approx \sigma$ cubes $Q$ in each slab, it suffices to show that
\begin{equation} \label{intermediate} m\left(Y_{\Box} \cap R^{3\delta} A^*\left(\mathbb{R}^d \times \left[t, t+CR^{1/2 + 3\delta} \right] \right)\right) \lessapprox \sigma_{\Box} m(Q) R^{-1/4}, \end{equation}
where $m(Q)$ only depends on $R$. The long axis of the set $\Box$ makes an acute angle $\gtrsim 1$ with the $CR^{1/2+3\delta}$ slab, and therefore intersects it in a set of diameter $\lessapprox R^{3/4}$. Hence there are $\lessapprox \sigma_{\Box}$ sets $S \subseteq Y_{\Box}$ in the intersection on the left hand side. Since each such set $S$ has long axis in the same direction as $\Box$, the intersection of $S$ with the $CR^{1/2+3\delta}$ slab is contains in a box of dimensions
\[  \sim R^{1/2+\delta}  \times \dotsm \times R^{1/2+\delta} \times R^{1/4+\delta} \times R^{1/2+3\delta}, \]
and therefore the intersection has measure $\lessapprox m(Q) R^{-1/4}$. Adding up the contributions of each $S$ gives \eqref{intermediate}, and this yields
\[ \mu \lessapprox \frac{ \sigma_{\Box} }{\sigma} |\mathbb{B}|. \]
Substituting this bound into \eqref{bound} gives
\begin{align*} \|Ef\|_{L^{q}(Y)} &\lessapprox \sigma^{-\gamma}   \|f\|_2. \end{align*}
Therefore, by the induction on scales argument explained in Remark \ref{scales}, this proves the theorem. \end{proof}

\begin{remark} A similar example to the Schrödinger case \cite{du2} shows that the inequality is sharp; $Y$ is a collection of $\sigma$ disjoint parallel tubes of radius $R^{1/2}$ and length $R$, $f$ is an essentially orthogonal sum $f= \sum_{\nu} f_{\nu}$ over the different tubes with $\left\|f_{\nu}\right\|_2$ constant in $\nu$, and each $Ef_{\nu}$ essentially constant and supported on a thinner tube inside the larger one.

To be more precise, let $\phi$ be a Schwartz function on $\mathbb{R}^d$, with $0 \leq \widehat{\phi} \leq 1$, $\widehat{\phi}$ equal to 1 on $B((3/2) e_d, \epsilon)$ and vanishing outside $B((3/2) e_d, 2\epsilon)$ for fixed small $\epsilon >0$. Then $|E\widehat{\phi}| \sim 1$ on a ball $B(0,C_{\epsilon})$ for some small $C_{\epsilon}>0$, and $E\widehat{\phi}$ is essentially supported on $B(0,R^{\delta})$. Let $\tau$ be a cap in the cone at scale $R^{-1/2}$ with centre line in the direction $e_{d+1}+e_d$. Let $\eta = \pi (A^*BA(\xi,|\xi|) \in \mathbb{R}^d$ where $B(\omega_1, \ldots, \omega_d, \omega_{d+1}) = (R^{1/2} \omega_1, \dotsc, R^{1/2} \omega_{d-1}, \omega_d, R\omega_{d+1})$ and define $\phi_{\tau}$ on the Fourier side by $\widehat{\phi_{\tau}}(\xi) = \left|\frac{d\eta}{d\xi}\right| \widehat{\phi}(\eta)$. Then $|E\widehat{\phi_{\tau}}| \sim 1$ on $(A^*BA)B(0,C_{\epsilon})$ by a change of variables, and the restriction of $E\widehat{\phi_{\tau}}$ to $B(0,R)$ is essentially supported on a larger box of similar dimensions $R^{1/2+\delta} \times \dotsm \times R^{1/2+\delta} \times R^{\delta} \times 2R$, by an integration by parts as in the wave packet decomposition. Let $\{T_{\nu}\}$ be a finitely overlapping cover of $B(0,R)$ with translates of this box by points in $A^*(\mathbb{Z}^d \times \{0\})$, each containing a corresponding translate $S_{\nu}$ of the smaller box. Let $c(\nu)$ be the centre of $S_{\nu}$. Define $\phi_{\tau,\nu}$ on the Fourier side by $\widehat{\phi_{\tau,\nu}}(\xi) = e^{-2 \pi i \left\langle c(\nu), (\xi,|\xi|) \right\rangle}  \widehat{\phi_{\tau}}(\xi)$, and let $f_{\nu} = \widehat{\phi_{\tau,\nu}}$. Let $\mathbb{T}$ be a subcollection of boxes $T_{\nu}$ which intersect $B(0,R/2)$ and are $R^{1/2+2\delta}$-separated from each other, such that $|\mathbb{T}| = \sigma$. Let $f = \sum_{T_\nu \in \mathbb{T}} f_\nu$. Cover each $S_{\nu}$ with $\sim R^{1/2-2\delta}$ disjoint $R^{1/2+2\delta}A^*\mathbb{Z}^{d+1}$-lattice cubes $Q$, and let $Y$ be the union of all such $Q$, over all $T_{\nu} \in \mathbb{T}$. By a change of variables, for any $Q$ intersecting $S_{\nu}$,
\[ \|Ef\|_{L^q(Q)}^q \approx \|Ef_{\nu} \|_{L^q(Q \cap S_{\nu})}^q \approx R^{-1/2} \|Ef_{\nu} \|_{L^q(S_{\nu})}^q \approx  R^{d/2}. \]
Similarly $\|Ef\|_{L^q(Y)}^q \approx \sigma R^{\frac{d+1}{2}}$ and $\|f\|_2^2 \approx \sigma R^{\frac{d-1}{2}}$, which gives $\|Ef\|_{L^q(Y)} \gtrapprox \sigma^{-\gamma} \|f\|_2$. This verifies that the theorem is sharp. \end{remark}

\section{Multilinear refined Strichartz inequality} \label{multilinearstrichartz}
As in the Schrödinger case \cite{du2}, the linear refined Strichartz inequality for the wave equation implies a multilinear version. The proof is similar to the one in \cite{du2}; it utilises the $k$-linear Kakeya inequality in $\mathbb{R}^n$ from \cite{bennett,guth3}. To state this, given $\nu >0$, a collection $\mathbb{T}_1, \dotsc, \mathbb{T}_k$ of $k$ sets consisting of tubes $T_i \in \mathbb{T}_i$, of infinite length and equal radius, is called $\nu$-transverse if
\[ \left| v_1 \wedge \cdots \wedge v_k \right| \geq \nu \quad \text{for all $T_i \in \mathbb{T}_i$, with $1 \leq i \leq k$,} \]
where $v_i$ is the infinite direction in $T_i$.
\begin{theorem} Suppose that $2 \leq k \leq n$ and that $\mathbb{T}_1, \dotsc, \mathbb{T}_k$ are $\nu$-transverse families of tubes in $\mathbb{R}^n$ of radius $\delta$. if $\frac{k}{k-1} < q \leq \infty$ then
\begin{equation} \label{kakeya} \left\| \prod_{i=1}^k  \left( \sum_{T_i \in \mathbb{T}_i} \chi_{T_i} \right) \right\|_{L^{q/k}\left(\mathbb{R}^n\right)} \leq C(\nu,q,n) \prod_{i=1}^k \left( \delta^{n/q} \left| \mathbb{T}_i \right| \right). \end{equation}
\end{theorem}

The preceding theorem in dimension $n$ will be applied with $n=d+1$. The statement that $f_1, \dotsc, f_k \in L^2(\mathbb{R}^d)$ are transversally supported in $B(0,2) \setminus B(0,1)$ will mean that each $\supp f_i \subseteq B(0,2) \setminus B(0,1)$, and any $k$-tuple $(\xi_1, \dotsc, \xi_k)$ with $\xi_i \in \supp f_i$ satisfies
\[ \left| G(\xi_1) \wedge \cdots \wedge G(\xi_k) \right| \gtrsim 1, \]
where $G(\xi)$ is the unit normal to the cone at $(\xi,|\xi|)$.

\begin{theorem} \label{mstrichartzr} Fix $d \geq 2$. Suppose that $2 \leq k \leq d+1$ and that $f_1, \dotsc, f_k \in L^2(\mathbb{R}^d)$ are transversally supported in $B(0,2) \setminus B(0,1)$, and let $Y= \bigcup_{j=1}^N Q_j$ be a collection of $N$ $R^{1/2} A^*\mathbb{Z}^{d+1}$-lattice cubes in $B(0,R)$. If for each $i$, $\left\|Ef_i\right\|_{L^q(Q_j)}$ is essentially constant in $j$, then for any $\epsilon>0$,
\[ \left\| \prod_{i=1}^k \left|Ef_i \right|^{1/k} \right\|_{L^{q}(Y)} \leq C_{\epsilon}R^{\epsilon}N^{-\frac{(k-1)\gamma}{k}} \prod_{i=1}^{k} \|f_i\|_2^{1/k}. \]
where $\gamma = \frac{1}{2}-\frac{1}{q}$ and $q  = \frac{2(d+1)}{d-1}$.
\end{theorem}
\begin{proof} Using Proposition \ref{wavepacket}, decompose
\[ f_i = \sum f_{\Box,i}, \]
for each $i$, as in the proof of Theorem \ref{strichartzr}. Applying Hölder's inequality gives
\[ \left\| \prod_{i=1}^k \left|Ef_i \right|^{1/k} \right\|_{L^{q}(Y)} \leq \prod_{i=1}^k \left\| Ef_i \right\|_{L^{q}(Y)}^{1/k}. \]
To each $f_i$, apply the same steps as in the proof of Theorem \ref{strichartzr} up to \eqref{bound}, except without replacing the cubes $Q$ by $CQ$, and instead choose the set $Y'$ of cubes in $Y$ to be the same for each $i$ (by doing this first for $f_1$, and then $f_2$ on the good cubes from $f_1$, and so on). The sets $\Box_i$, $Y_{\Box_i}, \mathbb{B}_i$ and parameters $\mu_i$, $\sigma_{\Box,i}$ are all defined as before. Using \eqref{bound} for each $f_i$, and the generalised version of Theorem \ref{strichartzr} at \eqref{rescaling} (instead of the inductive assumption) yields
\begin{equation} \label{aim} \left\| \prod_{i=1}^k \left|Ef_i \right|^{1/k} \right\|_{L^{q}(Y)} \lessapprox \left(\prod_{i=1}^k  \frac{\mu_i}{\left|\mathbb{B}_i\right| \sigma_{\Box,i} } \right)^{\gamma/k}  \prod_{i=1}^{k} \|f_i\|_2^{1/k}. \end{equation}
Therefore, to finish the proof it suffices to show that
\begin{equation} \label{sufficient} N \prod_{i=1}^k  \mu_i^{\frac{1}{k-1}} \lessapprox \prod_{i=1}^k \left(\left|\mathbb{B}_i\right| \sigma_{\Box,i}\right)^{\frac{1}{k-1}}. \end{equation}
To this end, for each $Y_{\Box_i}$ let $Z_{\Box_i} = \bigcup_{S \subseteq Y_{\Box_i}} F(S)$, where for each box $S \subseteq Y_{\Box_i}$ of dimensions
\[ R^{1/2+ \delta} \times \dotsm \times R^{1/2+\delta} \times R^{1/4+\delta} \times R^{3/4+\delta}, \]
the set $F(S)$ is defined to be the box with dimensions
\[ R^{1/2+ 4\delta} \times \dotsm \times R^{1/2+4\delta} \times R^{1/2+4\delta} \times R^{3/4+4\delta}, \]
with the same centre and orientations as $S$, obtained by expanding $S$ by $\approx R^{1/4}$ in the second last direction and $\approx 1$ in all the other directions. This adjustment ensures that if the expanded cube $R^{2\delta}Q$ of side length $R^{1/2 + 2\delta}$ intersects $Y_{\Box_i}$, then $Q$ is completely contained in $Z_{\Box_i}$. Recall that each $\mu_i$ was chosen so that for each $Q \subseteq Y'$, there are $\sim \mu_i$ sets $\Box_i \in \mathbb{B}_i$ such that $Y_{\Box_i}$ intersects $R^{2\delta}Q$.

Let $\{B\}_B$ be a finitely overlapping cover of $B(0,R)$ by balls of radius $R^{3/4}$ such that each cube $Q \subseteq Y$ is completely contained in some $B$. Define
\[ \mathbb{B}_{i,B} = \left\{ \Box_i \in \mathbb{B}_i :  \Box_i \cap R^{2\delta} B \neq \emptyset \right\}. \] Invoking the definition of each $\mu_i$ in the left hand side of \eqref{sufficient} yields
\begin{align*}  N \prod_{i=1}^k  \mu_i^{\frac{1}{k-1}} &\lessapprox \frac{1}{R^{\frac{d+1}{2}}} \sum_{Q \subseteq Y'}  \int_Q \prod_{i=1}^k \mu_i^{\frac{1}{k-1}} \, dx\\
&\lesssim  \frac{1}{R^{\frac{d+1}{2}}} \sum_B \sum_{Q \subseteq B \cap Y} \int_Q  \prod_{i=1}^k\left( \sum_{\Box_i \in \mathbb{B}_{i,B}} \chi_{Z_{\Box_i}} \right)^{\frac{1}{k-1}} \, dx \\
&\leq  \frac{1}{R^{\frac{d+1}{2}}}\sum_B \int_B  \prod_{i=1}^k\left( \sum_{\Box_i \in \mathbb{B}_{i,B}} \chi_{Z_{\Box_i}} \right)^{\frac{1}{k-1}} \, dx.\end{align*}

 Each $\chi_{Z_{\Box_i}}$ in the product can be bounded by $\sum_{T \in \mathbb{T}_i} \chi_T$ where each $T$ is a tube of radius $R^{1/2+5\delta}$ and infinite length in the long direction of $\Box_i$ through the centre line of some $S \subseteq Y_{\Box_i}$ which intersects $R^{3\delta}B$. Since all the sets $S$ therefore come from $\lessapprox 1$ slab of width $\approx R^{3/4}$ in $\Box_i$, there are $\lessapprox \sigma_{\Box,i}$ such tubes $T \in \mathbb{T}_i$, for each $i$. Therefore, applying the $k$-linear Kakeya inequality \eqref{kakeya} with $n=d+1$ and $q$ close to $\frac{k}{k-1}$ gives

\begin{align} \label{halfway}  N \prod_{i=1}^k  \mu_i^{\frac{1}{k-1}} &\lessapprox \left(\prod_{i=1}^k \sigma_{\Box,i}^{\frac{1}{k-1}} \right)  \cdot \sum_B  \left(\prod_{i=1}^k \left| \mathbb{B}_{i,B} \right|^{\frac{1}{k-1}}\right). \end{align}

For each $\Box_i \in \mathbb{B}_{i}$, let $T_{\Box_i}$ be a tube of infinite length and radius $R^{3/4+5\delta}$ in the long direction of $\Box_i$, with the same centre line as $\Box_i$. The sum on the right hand side of \eqref{halfway} satisfies
\begin{align*} \sum_B \left(\prod_{i=1}^k\left| \mathbb{B}_{i,B} \right|^{\frac{1}{k-1}}\right) &\leq \frac{1}{R^{\frac{3(d+1)}{4}}}  \sum_B \int_B \prod_{i=1}^k \left( \sum_{\Box_i \in \mathbb{B}_{i,B}} \chi_{T_{\Box_i}} \right)^{\frac{1}{k-1}} \, dx \\
&\lesssim \frac{1}{R^{\frac{3(d+1)}{4}}}   \int_{B(0,2R)} \prod_{i=1}^k \left( \sum_{\Box_i \in \mathbb{B}_{i}} \chi_{T_{\Box_i}} \right)^{\frac{1}{k-1}} \, dx \\
&\lessapprox \prod_{i=1}^k \left| \mathbb{B}_i \right|^{\frac{1}{k-1}}, \end{align*}
by the $k$-linear Kakeya inequality. Combining this with \eqref{halfway} and \eqref{aim} gives
\[ \left\| \prod_{i=1}^k \left|Ef_i \right|^{1/k} \right\|_{L^{q}(Y)} \lessapprox  N^{-\frac{(k-1)\gamma}{k}} \prod_{i=1}^{k} \|f_i\|_2^{1/k}, \]
which finishes the proof.  \end{proof}

\section{Decay of conical averages} \label{conicalaverages}
This section contains a lower bound on the decay rate $\beta\big(\alpha,\Gamma^d\big)$ of the conical averages, followed by a proof of the relationship between $\beta\big(\alpha,\Gamma^d\big)$ and $s_d(\alpha, 2,d)$. The section concludes with an upper bound on $\beta\big(\alpha,\Gamma^d\big)$.

Similarly to Corollary 3.3 of \cite{du}, the lower bound follows from Theorem \ref{strichartzr}; the proof is included for completeness.
\begin{corollary} \label{averages} For any $\alpha \in (0,d+1]$,
\[ \beta\big(\alpha,\Gamma^d\big) \geq \alpha -1 + \frac{d-\alpha}{d+1}. \]
\end{corollary}
\begin{proof} Without loss of generality, it may be assumed that $\Gamma$ is restricted to $\xi_d \geq 0$. Let $\pi: \mathbb{R}^{d+1} \to \mathbb{R}^d$ be the natural projection. If $\mu$ is a Borel measure supported in the unit ball with $c_{\alpha}(\mu) < \infty$, then for any $R>0$,
\begin{equation} \label{duality} \left(\int \left|\widehat{\mu}(R\xi) \right|^2 \, d\sigma_{\Gamma}(\xi) \right)^{1/2} = \sup_{\|f\|_2 = 1} \left| \int \widehat{\mu}(R \xi) \left(f \circ \pi\right)(\xi) \, d\sigma_{\Gamma}(\xi) \right|. \end{equation}
For any fixed $f$ with $\|f\|_2=1$,
\begin{align} \notag \left| \int \widehat{\mu}(R \xi) \left(f \circ \pi\right)(\xi) \, d\sigma_{\Gamma}(\xi) \right|&= \left| \int \widehat{\mu}(\xi) \left(f \circ \pi\right)(\xi/R) \, dR_{\#}\sigma_{\Gamma}(\xi) \right| \\
\notag &= \left| \int Ef(-Rx,-Rt) \, d\mu(x,t) \right| \\
\label{duality2} &\leq R^{-\alpha} \|Ef\|_{L^1(\mu_R)},  \end{align}
where $\mu_R(E) := R^{\alpha}\mu(-R^{-1}E)$ for any Borel set $E$. The measure $\mu_R$ satisfies $c_{\alpha}(\mu_R) \leq c_{\alpha}(\mu)$. If $\psi$ is a fixed non-negative Schwartz function equal to 1 on the ball $B( 0, 3\sqrt{2}) \supseteq \Gamma$, and vanishing outside a slightly larger ball, then $\widehat{Ef} = \widehat{Ef}  \psi$, which implies that $|Ef| \leq |Ef| \ast \left| \widecheck{\psi} \right|$. Therefore using Fubini and the Schwartz decay of $\widecheck{\psi}$ gives
\[ \|Ef\|_{L^1(\mu_R)} \lesssim \int_{B(0,2R)} \left| Ef\right| H \, dx + R^{-N} \|f\|_2, \quad \text{where} \quad H := \left|\widecheck{\psi}\right| \ast \mu_R. \]
By the uncertainty principle (see e.g. the scaled version of Lemma 2.5 in \cite{erdogan} with $s=1$), the function $H$ satisfies
\[ \|H\|_{\infty} \lesssim c_{\alpha}(\mu), \quad \|H\|_1 \lesssim \|\mu\|R^{\alpha} , \quad \text{and} \quad \int_{B((x,t),r)} H \, dy \lesssim c_{\alpha}(\mu) r^{\alpha}, \]
for every $(x,t) \in \mathbb{R}^{d+1}$ and $r>0$ (the third inequality holds for $r<1$ by the first one). Partition $B(0,2R)$ into cubes $Q$ of side length $\sqrt{2R}$, and sort each cube $Q$ into sets $\mathbb{S}_{\lambda}$ according to the dyadic value $\lambda$ of $\|Ef\|_{L^q(Q)}$. For each $\lambda$, sort the cubes $Q$ in $\mathbb{S}_{\lambda}$ further according to the number of cubes $\sigma$ in $\mathbb{S}_{\lambda}$ lying in the $(2R)^{1/2}$ slab parallel to $A^*\left( \mathbb{R}^d \times \{0\} \right)$ which contains $Q$. Since there are $\lesssim \left( \log R \right)^2$ significant pairs $(\lambda, \sigma)$, by standard pigeonholing and the triangle inequality there exists a pair $(\lambda, \sigma)$ with associated cubes $Y = \bigcup Q$ such that (ignoring the term $R^{-N} \|f\|_2$)
\[ \left\| Ef H^{1/2} \right\|_{L^2(B(0,2R)} \lessapprox \left\| Ef H^{1/2} \right\|_{L^2(Y)}. \]
Therefore, with $\gamma= \frac{1}{2}- \frac{1}{q}$, applying Theorem \ref{strichartzr} gives
\begin{align*} \int_{B(0,2R)} \left| Ef\right| H \, dx &\lesssim \|\mu\|^{1/2} R^{\alpha/2} \left\| Ef H^{1/2} \right\|_{L^2(B(0,2R))} \quad \text{by Cauchy-Schwarz,} \\
&\lessapprox \|\mu\|^{1/2}  R^{\alpha/2} \left\| Ef H^{1/2} \right\|_{L^2(Y)} \\
&\leq \|\mu\|^{1/2}  R^{\alpha/2}  \left\| Ef \right\|_{L^q(Y)} \left\| H^{1/2} \right\|_{L^{q'}(Y)} \quad \text{where $\frac{1}{q'}= \frac{1}{2}-\frac{1}{q}$,}\\
&\lessapprox \|\mu\|^{1/2}c_{\alpha}(\mu)^{1/2}  R^{\alpha/2}  \sigma^{-\gamma} \|f\|_2 \left( R^{1/2} \sigma  R^{\alpha/2} \right)^{\gamma}\\
&=  \|\mu\|^{1/2}c_{\alpha}(\mu)^{1/2} R^{ \frac{1}{2} \left( \alpha + \gamma + \alpha \gamma\right) }\|f\|_2. \end{align*}
Substituting this into \eqref{duality} and \eqref{duality2} yields
\[ \int \left|\widehat{\mu}(R\xi) \right|^2 \, d\sigma_{\Gamma}(\xi) \lessapprox \|\mu\|c_{\alpha}(\mu) R^{- \alpha+ \gamma + \alpha \gamma } = \|\mu\|c_{\alpha}(\mu) R^{ -\left(\alpha-1 + \frac{d-\alpha}{d+1}\right)}, \]
which proves the lower bound. \end{proof}

\begin{corollary} \label{fractalstrichartz}For any $\alpha \in (0,d+1]$,
 \[ s_d(\alpha,2) \leq \frac{d}{2} - \frac{1}{2} \left(\alpha-1+ \frac{d-\alpha}{d+1} \right). \]
\end{corollary}
Corollary \ref{fractalstrichartz} is a consequence of Corollary \ref{averages} combined with the following relation between $s_d(\alpha,2)$ and $\beta\big(\alpha,\Gamma^d\big)$ from \cite{wolff}, the proof is included here for convenience, but is essentially the same as the one in \cite{rogers}.

\begin{proposition} \label{relationship} For any $\alpha \in (0,d+1]$,
\[ \beta\big(\alpha,\Gamma^d\big) = d-2s_d(\alpha,2). \]

\end{proposition}
\begin{proof} Assume now that $E$ denotes the extension operator for the unrestricted cone, and let $\mu$ be a finite Borel measure with support in the unit ball. For any fixed $f$ with $\supp f \subseteq B(0,2) \setminus B(0,1)$ and $\|f\|_2=1$,
\begin{align} \notag \left| \int \widehat{\mu}(R \xi) \left(f \circ \pi\right)(\xi) \, d\sigma_{\Gamma}(\xi) \right|&= \left| \int \widehat{\mu}(\xi) \left(f \circ \pi\right)(\xi/R) \, dR_{\#}\sigma_{\Gamma}(\xi) \right| \\
\notag &= R^{-d}\left| \int Ef_R \, d\mu \right| \\
&\leq R^{-d} \left\|Ef_R \right\|_{L^1(\mu)},  \end{align}
where $f_R(\xi) := f(-\xi/R)$. Since $\supp f_R \subseteq B(0,2R) \setminus B(0,R)$, applying the definition of $s_d(\alpha,1)$ to the wave equation with solution $Ef_R$ gives
\[  R^{-d}\left\|Ef_R \right\|_{L^1(\mu)} \lessapprox R^{- \left( \frac{d}{2} -s_d(\alpha,1) \right) } \left\|\mu\right\|^{1/2}c_{\alpha}(\mu)^{1/2}. \]
Taking the supremum over $\|f\|_2 =1$ with $\supp f \subseteq B(0,2) \setminus B(0,1)$ yields
\[ \int \left|\widehat{\mu}(R \xi)\right|^2 \, d\sigma_{\Gamma}(\xi) \lessapprox R^{- \left(d-2s_d(\alpha,1) \right) } \left\|\mu\right\| c_{\alpha}(\mu). \]
This proves the inequality $\beta\big(\alpha,\Gamma^d\big) \geq d-2s_d(\alpha,1)$. But $s_d(\alpha,1) \leq s_d(\alpha,2)$ by the Cauchy-Schwarz inequality, so this implies that
\[ \beta\big(\alpha,\Gamma^d\big) \geq d-2s_d(\alpha,2). \]

For the other direction, let $f$ be given and let $g= \widehat{f} \chi_{B(0,2R) \setminus B(0,R) }$. The characterisation of the $L^2(\mu)$ norm through the distribution function is
\begin{align*} \int \left| Eg \right|^2 \, d\mu &= 2\int_0^{\infty} \lambda \mu\left\{ |Eg| > \lambda \right\} \, d\lambda. \end{align*}
For each $\lambda>0$, define the probability measure $\mu_{\lambda}$ by
\[ \mu_{\lambda}(F) = \frac{ \mu\left( F \cap \left\{ |Eg| > \lambda \right\}\right)}{\mu\left\{ |Eg| > \lambda \right\}}, \]
for any Borel set $F$.

For any finite Borel measure $\nu$ supported in the unit ball,
\begin{multline*} \left| \int Eg \, d\nu \right| = \left| \int (g \circ \pi)(\xi)  \widehat{\nu}(\xi)  \, d\sigma_{\Gamma}(\xi) \right| \\
\lessapprox \|g\|_2 \left( R^{d-\beta(\alpha,\Gamma^d)} \left\|\nu \right\| c_{\alpha}\left(\nu\right) \right)^{1/2} = \|g\|_2  R^{\frac{d-\beta(\alpha,\Gamma^d)}{2}} \left\|\nu \right\|^{1/2} c_{\alpha}\left(\nu\right)^{1/2}. \end{multline*}
By repeating this with $\nu$ replaced by the restriction of $\nu$ to a set where $\arg Eg$ is essentially constant, the absolute value can be moved inside the integral to get
\[ \left\|Eg\right\|_{L^1\left(\nu \right)} \lessapprox \|g\|_2  R^{\frac{d-\beta(\alpha,\Gamma^d)}{2}} \left\|\nu \right\|^{1/2} c_{\alpha}\left(\nu\right)^{1/2}. \]

Hence, taking $\nu = \mu_{\lambda}$ yields
\begin{align}\notag \lambda &\leq \left\| Eg\right\|_{L^1\left(\mu_{\lambda}\right)} \\
\notag &\lessapprox \left\|\mu_{\lambda}\right\|^{1/2} c_{\alpha}\left(\mu_{\lambda}\right)^{1/2} R^{\frac{d-\beta(\alpha,\Gamma^d)}{2}} \|g\|_2  \\
\label{weak} &\leq \left( \frac{ c_{\alpha}(\mu)}{ \mu\left\{ |Ef| > \lambda \right\} } \right)^{1/2} R^{\frac{d-\beta(\alpha,\Gamma^d)}{2}} \|g\|_2. \end{align}
Assume for the moment that $\|g\|_2 =1$. By the trivial bound $\|Eg\|_{\infty} \lesssim R^{d/2}$,
\begin{align*} \int \left| Eg \right|^2 \, d\mu &= 2\int_0^{R^{\frac{d-\beta(\alpha,\Gamma^d)}{2}}} \lambda \mu\left\{ |Eg| > \lambda \right\} \, d\lambda +  2\int_{R^{\frac{d-\beta(\alpha,\Gamma^d)}{2}}}^{CR^{d/2}} \lambda \mu\left\{ |Eg| > \lambda \right\} \, d\lambda \\
&\lessapprox R^{d-\beta(\alpha,\Gamma^d)} c_{\alpha}(\mu)\|g\|_2^2, \end{align*}
by using \eqref{weak} for both terms. By scaling this holds even if $\|g\|_2 \neq 1$. Fix $\epsilon >0$. Applying the triangle inequality results in
\begin{multline*} \left\| E \widehat{f} \right\|_{L^2\left(\mu\right)} \lesssim_{\epsilon} c_{\alpha}(\mu)^{1/2} \left\| \widehat{f} \chi_{B(0,1) } \right\|_2  \\
+ \sum_{\substack{j=0 \\ R = 2^j}}^{\infty} R^{\frac{d-\beta(\alpha,\Gamma^d)+\epsilon}{2}} c_{\alpha}(\mu)^{1/2} \left\| \widehat{f} \chi_{B(0,2R) \setminus B(0,R) }\right\|_2. \end{multline*}
The $\ell^2$ Cauchy-Schwarz inequality then yields
\begin{align*} \left\| E \widehat{f} \right\|_{L^2\left(\mu\right)} \lesssim_{\epsilon} c_{\alpha}(\mu)^{1/2} \|f\|_{H^{\frac{d-\beta(\alpha,\Gamma^d)}{2}+\epsilon}}, \end{align*}
which shows that
\[ \beta\big(\alpha,\Gamma^d\big) \leq d - 2s_d(\alpha, 2), \]
and therefore proves the equality. \end{proof}

The following calculation shows that if the class of measures is restricted, there is no relation between the conical averages and fractal Strichartz inequalities.
\begin{proposition} \label{trivial} If $\nu = \mu \otimes \lambda$, where $\mu$ is a Borel measure with compact support in the unit ball of $\mathbb{R}^d$ and $d\lambda = \chi_{[0,1]} \, dm$, where $m$ is the Lebesgue measure on $\mathbb{R}$, then
\begin{equation} \label{optimal} \int \left| \widehat{\nu}(R \xi) \right|^2 \, d\sigma_{\Gamma}(\xi) \lesssim I_{\alpha}(\mu) R^{-(\alpha+2)} \quad \text{for all $R >0$,} \end{equation}
where $I_{\alpha}(\mu)$ is defined by \eqref{energydef}.
\end{proposition}

\begin{proof} The left hand side of \eqref{optimal} is

\begin{align*} \int_{\Gamma} |\widehat{\nu}(R \xi)|^2 \,d\sigma_{\Gamma} &= \int_{B(0,2) \setminus B(0,1)} |\widehat{\nu}(R \xi, R|\xi|)|^2 \, d\xi \\
&= R^{-d} \int_{B(0,2R) \setminus B(0,R)} |\widehat{\nu}(\xi,|\xi|)|^2 \, d\xi \\
&= R^{-d} \int_{B(0,2R) \setminus B(0,R)} |\widehat{\mu}(\xi)|^2\left|\widehat{ \chi_{[0,1]}}(|\xi|)\right|^2 \, d\xi \\
&\lesssim R^{-d} \int_{B(0,2R) \setminus B(0,R)} |\widehat{\mu}(\xi)|^2  |\xi|^{-2} \, d\xi \\
&\lesssim R^{-(\alpha+2)} \int_{B(0,2R) \setminus B(0,R)} |\xi|^{\alpha-d} |\widehat{\mu}(\xi)|^2 \, d\xi  \\
&\lesssim R^{-(\alpha+2)} I_{\alpha}(\mu). \end{align*}
This proves the proposition. \end{proof}

Returning to the case of a general measure, the following upper bound for the decay of the conical averages is based on the counterexample for the spherical case from \cite{luca}.
\begin{proposition} Let $d \geq 5$. Then
\[ \beta\big(\alpha,\Gamma^d\big) \leq \alpha-1 +  \frac{2(d+1-\alpha)}{d+1} \quad \text{for} \quad \alpha \in (0,d+1). \]
\end{proposition}

\begin{proof} Let $\alpha \in (0, d+1)$, let $\epsilon, \kappa \in (0,1)$ and let $R>1$. Define
\[ \Lambda =  \left( R^{\kappa-1} \mathbb{Z}^{d+1} + B(0,\epsilon R^{-1}) \right) \cap B(0,1), \]
and let $\mu$ be the Lebesgue measure on $\mathbb{R}^{d+1}$ restricted to $\Lambda$.

Assume that $R$ is large enough to ensure that $R^{ \kappa} > 2$, which makes the balls disjoint. If $0 < r \leq \epsilon R^{-1}$ then
\[ \frac{ \mu(B(x,r)) }{r^{\alpha} } \leq  r^{d+1-\alpha} \lesssim R^{\alpha-(d+1)}. \]

If $\epsilon R^{-1} \leq r \leq R^{ \kappa-1}$ then
\[ \frac{ \mu(B(x,r)) }{r^{\alpha} } \lesssim  (\epsilon R^{-1})^{(d+1)-\alpha} \sim R^{\alpha-(d+1)}. \]
If $R^{\kappa-1} \leq r \leq 1$, then choose a positive integer $N$ so that $r \sim NR^{\kappa -1}$, so that $N \lesssim R^{1- \kappa}$. Then
\[ \frac{ \mu(B(x,r)) }{r^{\alpha} } \lesssim  \frac{ N^{d+1} (\epsilon R^{-1} )^{d+1} }{N^{\alpha} R^{ (\kappa-1) \alpha}} \lesssim R^{-\kappa (d+1)}. \]
Since $\mu$ is supported in the unit ball, this gives
\[ c_{\alpha}(\mu) \lesssim \max\left( R^{-\kappa (d+1)}, R^{\alpha-(d+1)} \right). \]
Define $\kappa \in (0,1)$ by $\kappa (d+1) = (d+1)-\alpha$.  Then
\[ \|\mu\| \lesssim R^{\alpha-(d+1)}, \quad c_{\alpha}(\mu) \lesssim R^{\alpha-(d+1)}. \]
By Plancherel, Cauchy-Schwarz and the definition of $\beta\big(\alpha,\Gamma^d\big)$, every function $f$ on the cone with $\|f\|_{L^2\left(\sigma_{\Gamma} \right)} \leq 1$ satisfies
\begin{equation} \label{decaydefn} \left| \int \widehat{ f \sigma_{\Gamma}}(Rx) \, d\mu(x) \right| \lessapprox R^{\alpha-(d+1)} R^{\frac{-\beta(\alpha,\Gamma^d)}{2}}. \end{equation}
Let
\[ E = \left\{ (\xi, |\xi|) \in \Gamma : R^{\kappa} (\xi, |\xi| )  \in \mathbb{Z}^{d+1} \right\}. \]
By adding up the lattice points on each of the $\sim R^{\kappa}$ relevant spheres of integer radius $\sim R^{\kappa}$, and applying the number theoretic estimate which says that for $d\geq 5$ there are $\sim R^{\kappa(d-2)}$ lattice points on each sphere \cite[Theorem 20.2]{iwaniec}, the cardinality of $E$ is
\begin{equation} \label{numbertheory} |E| \sim R^{\kappa(d-1)}. \end{equation}

For a fixed small $\rho>0$, to be chosen later (not depending on $R$), let
\[ \Omega = \Omega(\rho) = \left\{ \xi \in \Gamma : \dist(\xi, E) \leq \rho R^{-1} \right\}, \quad f = \frac{ \chi_{\Omega} }{ \|\chi_{\Omega}\|_{L^2(\sigma_{\Gamma})}}. \]
Then for large $R$,
\[ \sigma_{\Gamma}( \Omega ) \sim  |E| R^{-d}. \]
Let $F = \mathbb{Z}^{d+1} \cap B(0, R^{1-\kappa})$, so that
\begin{multline*} \|\chi_{\Omega}\|_{L^2(\sigma_{\Gamma})} \int \widehat{ f \sigma_{\Gamma}}(Rx) \, d\mu(x) = \int_{B(0,1)} \int_{\Gamma} e^{ -2\pi i \langle \xi , Rx \rangle } \chi_{\Omega}( \xi) \, d\sigma_{\Gamma}(\xi) \, d\mu(x) \\
= \sum_{n \in F}\sum_{\omega \in E}  \int_{B(R^{\kappa-1}n ,\epsilon R^{-1})} \int_{\Gamma \cap B(\omega,\rho R^{-1})} e^{ -2\pi i \langle \xi , Rx \rangle }  \, d\sigma_{\Gamma}(\xi) \, d\mu(x). \end{multline*}
For $(\xi,x)$ in the domain of integration, and the corresponding $(\omega,n)$, the Cauchy-Schwarz inequality gives
\begin{align*} \left| \left\langle \xi, Rx \right\rangle - \left\langle \omega, R^{\kappa} n  \right\rangle \right| &\leq \left| \langle \xi - \omega, Rx \rangle \right| + \left| \left\langle \omega, R( x- R^{\kappa-1} n )  \right\rangle \right| \\
&\leq 3(\rho + \epsilon). \end{align*}
Hence
\begin{align*} &\|\chi_{\Omega}\|_{L^2(\sigma_{\Gamma})} \left|\int \widehat{ f \sigma_{\Gamma}}(Rx) \, d\mu(x) \right|\\
&= \sum_{n \in F} \sum_{\omega \in E} \left|\int_{B(R^{ \kappa -1} n,\epsilon R^{-1})} \int_{\Gamma \cap B(\omega,\rho R^{-1})} 1  + O(\rho+\epsilon) \, d\sigma_{\Gamma}(\xi)\, d\mu(x)\right| \\
&\sim |E||F| \left(\epsilon R^{-1}\right)^{d+1} \left(\rho R^{-1} \right)^d\left(1+ O(\rho +\epsilon) \right). \end{align*}
Therefore if $\epsilon,\rho$ are taken small enough (not depending on $R$), then
\[ \left|\int \widehat{ f \sigma_{\Gamma}}(Rx) \, d\mu(x)\right|  \sim |E|^{1/2} |F| R^{ -\frac{3d}{2} -1}. \]
Combining this with \eqref{decaydefn} and \eqref{numbertheory} gives
\[ \beta\big(\alpha,\Gamma^d\big) \leq  \alpha-1 + \frac{2(d+1-\alpha)}{d+1}, \]
by the definition of $\kappa$. \end{proof}


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Home Around The Block Accomplished Teenage Songwriter & Broadway Performer Ava Della Pietra Discusses New Single...
Accomplished Teenage Songwriter & Broadway Performer Ava Della Pietra Discusses New Single 'My Boyfriend'
Naomi Rodriguez
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The almost-17-year-old singer-songwriter is well beyond her years, as she's written over 120 songs, and her new single, "my boyfriend," shows off her blossoming musical abilities and devotion to her craft.
Released on May 20th, "my boyfriend" displays her gentle floating melodies, and emphasizes her youthful poetry, which she isn't hesitant to convey. Pietra's vivid outlook and unfettered lyrics allow us to experience her reality. The fluffy clichés transport us back to our adolescent years and make us reminisce on our first love.
At age 7, Pietra performed in 27 cities as Little Cosette with the 25th Anniversary National touring production of Les Misérables. By 10, she established herself as she began a two-year run on Broadway as a member of the original cast of School of Rock. After a powerful run on the theatrical stage, she chose to further pursue singing and songwriting.
In a short time, Pietra has garnered an audience with her 1M+ YouTube views and 700K+ Spotify streams. She is a talented pop musician on the rise with a bright future ahead, especially with her new EP, truth or truth, looming on the horizon.
We got the chance to chat with the young performer about the new single, her theater days, and what lies ahead.
I see you've been writing songs since you were 10-years-old, and at just 16-years-old, you've already established yourself as an artist. Who or what influenced you to start making music and performing?
I've always been in love with performing ever since I was really little. I loved singing, stomping my feet around the house trying to tap dance, and coming up with melodies to tell my parents about my day! When I booked my first professional role as Little Cosette in the National Tour of Les Miserables when I was seven, my career as a performer began.
In addition to musical theater, I became really interested in songwriting because of my love for storytelling. I recorded my first song when I was 10 in a recording studio in New York City while I was in School of Rock on Broadway. My music when I was little was influenced by Broadway music or Disney music, but now people like Dua Lipa, Alec Benjamin, Lorde, Maroon 5, Sam Smith, Conan Gray, and so many more artists inspire my style.
What has your schooling life been like, and does music and performing take priority?
Since I've been performing from a very young age, I've gotten pretty good at balancing school with performing. I do really well in all my classes while still finding time to write music or play my instruments. Songwriting is a form of relaxation for me, so it's a nice way to wind down at the end of the day.
So you recently dropped your new single, "my boyfriend." Can you talk about your vision behind this song and perhaps the feelings behind it?
"my boyfriend" was inspired by an experience I had while I was hanging out with my friends who were all talking about their crushes or boyfriends. All of a sudden, I realized that up until right then I hadn't been super interested in dating anyone. I had crushes here and there, but all of a sudden I really wanted a boyfriend! Then I picked up my ukulele and tried to think about what I want my first boyfriend to be like, and how I want to fall in love with him.
Who helped produce it and bring it to life?
"my boyfriend" was produced by Steven Martinez. I really love the production and the harmonies, because I think it reflects the dreamy lyrics and sentiment of the song.
What can you tell us about the upcoming EP, truth or truth?
truth or truth is a five song EP of music I wrote over the past year. The title was inspired by one of my favorite lines in "my boyfriend". I think the name also reflects how honest I am in my lyrics, and how I'm discovering my own truth in telling my story. I can't wait to see what people think!
So tell us more about your time performing on School of Rock and your other theater appearances. What were these experiences like, and what might you have learned so far?
School of Rock was an incredible experience because I was a swing where I understudied six roles. Being a swing was SO fun because I always had to be ready to perform at any moment. Some of my castmates would call out a few minutes before the show started, or a few girls even called out in the middle of the show! Performing on Broadway with so many talented people really solidified my love for musical theater. I also learned a lot about having confidence and energy on stage.
What has been one of your favorite/most memorable moments in your young career?
Performing as Summer Hathaway in School of Rock 15 minutes into the show is definitely one of my most memorable moments in musical theater! I've also performed at the Tony's which was extremely exciting. In terms of my songwriting career, I had this experience performing at Sundance Film Festival when I was about 13 where I had just finished my first ever 30-minute set and someone in the audience yelled "encore!" This was really special to me, and I think that was the first moment I realized that people truly enjoyed my music!
Do you have a dream performance or collaboration?
I would love to collaborate or perform with Conan Gray, Alec Benjamin, or Olivia Rodrigo! They are all so inspiring.
Where do you hope to see yourself in five years?
In the next five years, I hope to release a few albums and maybe even appear in a TV show or movie! Maybe I will be on a music tour! I can't wait to continue to share my music with the world.
What else might fans expect from Ava Della Pietra this summer, and frankly the rest of the year?
In addition to my EP, I will be releasing an LP at the end of the year along with a few music videos. I really hope people can relate to my songs and enjoy them. I have so many stories to share!
Photo by Shervin Lainez
Ava Della Pietra
Broadway performer
indie music blog
truth or truth
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City Information & Guides
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Magnificent Places of North India
By Sara Hasan on January 30, 2012 0
The Northern India holds a great history and glory which connects its people's present to their past. The Northern part of India includes gorgeous locations which are liked by national and international tourists. The Northern part of India includes hill stations, pilgrimages and places of historical and cultural importance. These places make India the cherished destination among tourists even from outside borders. There are different places which are considered as the jewel of Northern India. Kashmir, Nainital, Shimla, Lucknow, Vrindavan, Mathura, Amritsar etc. are few places which lure the Northern India. Apart from these beautiful destinations there is Golden Triangle Trip. Golden Triangle Trip is recognised for its key attractions and cultural value. The Golden Triangle Trip of Northern India involves major cities of the country like Delhi, Jaipur and Agra. These three cities are considered to be the most attractive destination among the tourists. It provides with the best locations of the country, its delicious cuisine and of course its heart whelming culture and tradition that attracts tourists from all over to globe to the mysterious country like India. The key attractions of the famous Golden Triangle Tour cities include various monuments and architectural structures along with the enriching value of culture of every city it covers.
Delhi the first place of the Golden Triangle Tour involves great historical monuments and other eye-catching places. Situated on the banks of river Yamuna, Delhi houses many important building of the country India. The important buildings include Parliament House, Supreme Court, Rashtrapati Bhavan etc. There are various other places of historical importance like the Red Fort(also known as Laal Quila in Hindi), Quitab Minar, Humayun's Tomb, National Museum, Birla Mandir, Akshardham temple, India Gate, Lotus Temple, Jama Masjid, Safdarjung Tomb, Firoz Shah Kotla Tomb etc are places which are of great historical importance and attractions to the tourist visiting India. Visiting Delhi through Gloden Triangle Tour is a eye treat for tourists.
Jaipur is the next tour of Golden Triangle Tour. The royal city is known as the Pink city and is the capital of Rajasthan which is known as the India's royal state. The city of Jaipur is very well known for its Forts and palaces. Some of the famous forts and palaces involve Jaigarh Fort, City Palace, Naharagarh Fort, Sheesh Mahal, Amber Fort, Hawa Mahal, Jal Mahal etc. Other places include Birla Temple, Jantar Mantar, Lakshmi Nrayan Temple and Chokhi Dhani which is a cultural village resort.
Golden Triangle Tour and Agra is the favoured destination among tourists both national and international as it houses the most beautiful monument of the world. Taj Mahal was built during the rule of Mughals. It was made by Shah Jahan who in the love and memory of his third wife Mumtaj Mahal built the astonishing model of love. The famous Taj Mahal is a great instance of Indian, Persian and Turkish architecture done by Mughal Emperor. It took many years for Taj Mahal to be built, this is probably the reason it is the most admired mausoleum in the entire globe. The dedication of Emperor and the hard work of the workers and architects involve in the making of Taj Mahal today makes this the beautiful masterpiece of architecture in the entire world. There are other places in Agra which are of great historical significance like Red Fort and Fatehpur Sikri.
Sara has deep interest in writing informative articles on Travel and Tourism in India. Also given her words to a portal for Rajasthan Travel where one can find information on Golden Triangle Tours, Golden Triangle Tour Packages and other destinations of India. Explore: travel-to-rajasthan.net
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Rajasthan Travel Golden Triangle Tours Golden Triangle Tour Packages Golden Triangle Tour
Rajasthan Travel - The City of Affluent History
Rajasthan is the largest state of India in the context of area. It is one of the most favored traveler destinations in India and the Rajasthan tour package is a rage among the travelers across the world.
Written by: Mike Anderson
Tour to North India Explore The Popular Attractions
North India is popular tourist zone of the country. There is high concentration of tourist attractions in this part of the country. Tourist visit North India to explore the popular attractions of the place.
Written by: Andy Rodick
North India Tours Know About Some Popular Destinations
Are planning to go on North India tours? Well, you have taken a right decision to enjoy the incredible charm of India tourism. There are many fascinating travel destinations in northern region of India which are full of tourists' delights.
Written by: Upendra Chansoria
Holidays in North India- An Enduring Experience
Holidays in North India is on every vacationer's bucket list. North India makes all its tourists in sheer awe due to its scenic beauty, the much-awaited sight-seeing, the lip-smacking food and a considerable amount of fun element.
Written by: Christina Wunch


Wife and a mother to two sweet little boys. Addicted to laughing, traveling and a good glass of wine.
Neutrogena wet kids is really great. I use this product on my kids when I need to get around there face and neck. I think it works really well in those areas.
These Medela breast shields worked great when I would pump breast milk. I love all of Medelas product's.
My boys love the honest apple juice boxs! They have great taste and I love the fact that they are organic!
I really like these monster energy zero ultra drinks. They taste good and they give me that extra boost of energy when I need it. There is zero calories in these and they are low in sugar. You can usually get these for a good price.
I love Dole pineapple chunks. They are a perfect little snack for if you want something fruity. Just put them in a bowl and eat them. My kids love them too! They taste so good. We are always buying them in our household. A great affordable option for a snack.
I love this body wash it works great. It smells good, green tea and aloe is the scent. It's pretty affordable.
I buy this brand for my husband. I think he likes it. It smells great. He has had no complaints about this at all.
I love Tim Hortons! They have great coffee, awesome egg sandwiches and there timbits are sooo good! It's also very affordable!


Founded in 1941 by the Solie family based on family care and family service, operations were turned over to the Lorentzen family in the 1960s to continue the same philosophy.
We are the last family owned funeral home in Everett, WA and one of very few in the eastern Puget Sound area.
We are a full service funeral home with two chapels and viewing rooms. A reception room is also available on site with off street parking.
Our lower chapel has a viewing room with a viewing window to the crematorium for witness cremations.
In case you are interested in preplanning your memorial, you can be sure your legacy will be secure and that you may have peace of mind. This location has proudly served the local community with exceptional care for years and will help guide your household through funeral etiquette, individualize your tribute, funeral costs, directions to cemeteries, guestbook, online obituary creation, and telling your life story.
This business is situated 2.5 miles coming from downtown Everett, Washington which is in Snohomish County and serves the surrounding areas of: Naval Station Everett, Lowell, Pinehurst, Eastmont, Silver Lake, Hartford, Tulalip, Maxwelton, Possession, Mill Creek, Larimers Corner, Clearview, Machias, Maltby, Brier, Queensborough, Kennard Corner, Thrashers Corner, Mountlake Ter, and N Lakewood.
Solie Funeral Home & Crematory in Everett, WA -- Map, Phone Number, Reviews, Photos and Video Profile for Everett Solie Funeral Home & Crematory.
No Solie Funeral Home & Crematory Obits are listed at this time but if you need to send memorial flowers or funeral flowers, you can do so by clicking this link.


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ASM to Ship 'PRON: The XXX Parody' from Goodnight Media
David Maxwell Mar 30, 2011 2:00 PM PDT
LOS ANGELES — Adult Source Media has announced it will begin shipping "PRON: The XXX Parody," the first release from new studio Goodnight Media, on Thursday.
Lee Roy Myers and Sam Hain co-directed the sci-fi fantasy spoof that will hit the streets about the same time as the Disney release of the original "TRON" film will reach stores.
April O'Neil, Andy San Dimas, Lexi Belle, Zoe Voss and Diamond Foxx star with Anthony Rosano as the male lead.
Myers is the 2011 XBIZ Award-winning Director of the Year - Body of Work. He also directed the 2011 XBIZ Parody Movie of the Year, a spoof of the "Big Lebowski" for New Sensations. Hain, meanwhile, has previously helmed the "Sex Files I & II," a pair of acclaimed parodies for Revolution X/Digital Sin.
Visit GoodNightMovies.com to download the "PRON" trailer.


Co-Founder of CFTE – The Center for Finance, Technology and Entrepreneurship, Industry Fellow at Imperial College Business School. She is part of the Top 100 Women in Fintech and Women in Fintech Powerlist.
Tram Anh firmly believes that in a world dominated by technology, people must remain at the centre of society and organisations, and that education and training are key to helping individuals adapt and thrive within our fast paced world, in particular the rapidly developing AI world.
This motivation drove her to co-establish CFTE, an education platform supported by senior leaders from the largest institutions, startups and universities around the world – to help bridge the worlds of finance and technology, so individuals acquire the new skills of Finance 2.0.
She strongly believes in lifelong learning. She fights for diversity and is an Ambassador for EPIC Foundation, a new model of disruption in the philanthropy sector supporting children and youth globally.
Starting her career at Standard Chartered, Tram Anh was a trader and market maker on the emerging market desk in New York. She joined Dresdner Kleinwort Benson in London, then moved to UBS Wealth Management, where she worked with UHNW families. She has navigated extensively between the corporate and entrepreneurial world, and understands the excitement and challenges that it brings.
She is a frequent speaker at Money2020 Europe, Money2020 Asia, AI Congress, LendIt Conference, Citi Symposium, Europlace Paris, European Women in Payment, Oxford Fintech & SmartLaw.
Of Vietnamese origin, she was born in France and lived in Paris, New York and now London. She is married with 3 wonderful daughters.


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Q: Как запретить доступ к файловой системе TeX live компилятору Есть сервер на Node Js, который с помощью командной строки запускает TeX Live компилятор. Так как код для компиляции задают пользователи, существует возможность читать и редактировать файлы на сервере с помощью возможностей самого LaTeX-а, а это не очень приятно. Но, к сожалению, полностью ограничивать доступ нельзя, хотя бы потому что компилятор должен сгенерировать финальный файл. Какие способы ограничить возможности на один каталог/пространство существуют? Желательно способ попроще.

A: Техник для Sandboxing-а несколько, основные
Контейнеры: chroot, docker, lxc(в режиме виртуализации приложения)
Неймспэйсы: через unshare, freejail
Контроль доступа через: AppArmor или SeLinux
Способ попроще - настроить freejail c noroot и overlay и удалить из overlay все что не требуется для tex.

redpajama_set_name: RedPajamaStackExchange


Whether you're newly diagnosed with high blood pressure or it's something you've been managing for a while, you likely already know that your diet needs to change.
You may not realise that your diet doesn't mean having to limit yourself to bland, unpalatable foods.
In the days of time restraints due to family & work commitments changing your lifestyle might seem hard.
Your health is something you should not take lightly but your new food routine can be fast, nutritious and tasty.
Do You Know That Watching Your Intake Of This 1 Item Can Reduce Your Blood Pressure?
This ranges from adding salt to your food or the sodium intake by eating processed & packaged food.
We like to think of watching what you eat when you have high blood pressure as not a diet but more as a way of eating.
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It is important to remember if you don't change & take action now your life may be at risk.
The product is a list of special recipes specifically designed to dramatically lower your blood pressure quickly and effectively. Once you incorporate these special dishes into your daily diet, you will significantly lower your blood pressure for good. It consists of a picture of the end dish, short list of ingredients and written instructions step by step on how to put it together quickly. All these are designed by our professional naturopath advisor.
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(a) Perform deliveries of children in this state.
(b) Read mammograms and perform breast-imaging-guided procedures in this state.
(c) Perform emergency care on an on-call basis for a hospital emergency department.
(d) Plan to reduce or increase emergency on-call hours in a hospital emergency department.
(e) Plan to relocate outside the state.
(f) Practice medicine in this state.
(g) Plan to reduce or modify the scope of their practice.
(2) The Department of Health must report its findings to the Governor, the President of the Senate, and the Speaker of the House of Representatives by November 1 each year. The department shall also include in its report findings, recommendations, and strategic planning activities as provided in s. 381.4018. The department may also include other information requested by the Physician Workforce Advisory Council.
History.—s. 3, ch. 2007-172; s. 31, ch. 2010-161.
1Note.—Also published at s. 458.3192.


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Top Plant-Based Proteins and How To Add Them to Your Diet
Of all the health messages floating around today, two big ones are particularly tricky to reconcile: we should eat less meat and we should eat more protein. Eating too much meat can contribute to weight problems, heart disease, strokes and diabetes. Eating more protein helps us get stronger, lose weight, lower blood pressure and more. The tricky part? Meat is such a generous source of protein!
Luckily, though, meat is not the only way to get our protein levels up. Plant-based sources of protein are abundant and varied and versatile, and they generally have additional benefits that we could all use.
The recommended daily intake for protein is around 50 grams — about 46 grams for women and 56 for men. (This New York Times article takes an interesting look around the high-protein trend and why sticking to the allowances — rather than going overboard — is wise.) We've pulled together a list of top plant-based proteins that will help you hit your protein goals in a meatless, delicious and healthy way.
It's hard to beat edamame — which are soybeans harvested while they're still young and soft — in terms of plant-based protein content. We're talking 17-20 grams per cup. Eat them on their own (steamed with a pinch of salt, if you're dining with me) or mixed into a stir-fry, soup or salad — but educate yourself on the controversy surrounding the effect of soy on women's bodies before you go too crazy with them.
Quinoa hit celebrity status awhile back as an alternative to pasta and rice; it's a solid source of fiber and minerals, and has 8 grams of protein in one cup. Plus, it's gluten free. You see why people love it? Use quinoa to bulk up salads and wraps, to stuff roasted vegetables or to mix with spinach, feta and cherry tomatoes. And, in case you're wondering, it's pronounced "keen-wah."
Source: Spinach for Breakfast
Ah, green peas . . . as a child you attached them individually to the end of each prong on your fork. Now, you can treat them with far more respect. A cup of green peas — which have been credited with preventing all sorts of problems including stomach cancer, Alzheimer's and arthritis — can have over 7 grams of protein! If you're not a fan of green peas on their own, try this fresh pea hummus recipe to spread on some tasty crackers.
If you're vegetarian or vegan, beans probably already play a big role in your diet. They're filling, adaptable, tasty and nutritious. But they're also a great source of protein. Outside of soybeans, most beans — such as white, pinto, navy, black, kidney and even lima beans — have around 15 grams of protein per cup. Bring on the vegetarian chili!
SEEDS AND NUTS
It's stunning how much protein can be squeezed into tiny little seeds and nuts, which makes them super-convenient and effective sources of protein. For example, an ounce of chia or sesame seeds has about 4.7 grams of protein; an ounce of almonds or pistachios has nearly 6 grams of protein; and peanuts (which are actually legumes, but who's counting?) have 6.6 grams in an ounce! So take a handful of almonds on your next hike, sprinkle some peanuts in your veggie stir-fry and start adding seeds to your smoothies.
Dark green vegetables are supremely healthy and relatively low calorie — so you can eat loads guilt-free and just watch your protein numbers rise all day. These leafy options are particularly protein-dense: collard greens have 5.15 grams per cup, spinach has 5.35 grams per cup, and kale has 2.47 grams per cup. Use spinach as a base for your next salad, toss kale into your green smoothies, steam yourself some collards, or keep those leaves raw and just nibble away.
Nutrition, Plant-based, protein
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Best case scenario is a xbox gaming rig converted via adaptor to Wii 2.
What are your personal thoughts, Steve?
like the wii-mote and perhaps (not sure) a classic controller.
Agreed all round. The Wii U looks like a nightmare for broad accessibility.
another avenue of gaming for the disabled.
Keep up the good work with the coverage!
I hope you've all been following our E3 coverage. If not, it's not too late.
keep up with our tweets, posts, video and live blogging from the E3 floor.
first time E3 has had representation for disabled gamers live.


typically after a photo session, i post a sneak peek for the family. but when it's the first week back to pre-K and your parents are in town visiting (i.e. babysitting the kids), you are able to put a bit more hustle into all of your work that next thing you notice, you've finished editing and delivering a gallery in record time.
so instead of a traditional sneak peek, here are some of my favorite photos from this family fun session that i'm absolutely in love with! this family? they are the coolest family ever! they live in new jersey, was spending some time with their family in the bay area and trekked down to south san jose to have me photograph their family which i think was pretty cool. their oldest son loves rock climbing so the location was a bit of a bonus as he got to climb, jump and have fun on most of the boulders. their littlest daughter was an amazing trooper as she was probably dealing with a bit of jetlag (our session was at 7p which probably felt like 10p) but she loved hanging out with her dads, maybe showing a small smile when it came to bubbles, big smiles while swinging but saving the best smiles and laughs for her brother who could make her easily laugh.


LIBERTY CITY, USA. WELCOME TO AMERICA'S WORST CITY.
Featuring a fully 3-D living city, a combination of narrative driven and non-linear gameplay and a completely open environment, Grand Theft Auto III represents a huge leap forward in interactive entertainment. For the first time, players are put at the heart of their very own gangster movie, and let loose in a fully-realised 3 dimensional city, in which anything can happen and probably will.
Data and credits for this game contributed by anarcho_selmiak , tardounderdog , odino , R351D3NT3V1L4 , gamesadictsback , bobby190 , Mookiethebold , Xcarvenger , Cosmic! , Blueberry Buttface , LordAndrew , _Genesis , bcks , BGoldTLE , oliist , Faxwell23 , BartSmith , MuddyMaestro , and Krystal109 .
One type of pedestrian, which appears randomly along with all other pedestrians, wear a construction worker outfit and say lines such as "Young man", "There's no need to feel down", "In the YMCA" and "In the navy".
This all references the disco band Village People - one of them performs dressed as a construction worker, and the lines are parts of different song lyrics.