text:
\section*{Abstract}
We present an elementary approach to prove restriction theorems for particular surfaces for which the Tomas-Stein theorem does not apply, which in turn provide short proofs for well-known Strichartz estimates for associated PDEs. The method consists in applying simple restriction theorems to the level sets of the surfaces studied (for instance, spheres in the case of the cone) and then integrating the inequalities over all level sets. This allows for a different proof of sharp estimates for the wave equation, and of some estimates related to the Euler equations in the rotational framework.

\section*{Résumé}
Nous présentons une approche élémentaire pour prouver des théorèmes de restriction sur certaines surfaces pour lesquelles le théorème de Tomas-Stein ne s'applique pas, qui permettent de retrouver des inégalités de Strichartz déjà connues. Notre méthode consiste à appliquer des théorèmes de restriction simples aux courbes de niveaux d'une surface donnée (par exemple, des sphères dans le cas du cône), puis à intégrer ces inégalités. Cela permet une preuve plus courte des estimées de Strichartz pour l'équation des ondes, et de certaines estimées pour l'équation d'Euler en rotation rapide.
\\
\\
\\
\begin{center}
\large{\textsc{Acknowledgments}}
\end{center}

\begin{center}
	\begin{minipage}{16cm}
		We are very grateful to Isabelle Gallagher for all her advice during the writing of this note and for her ideas that gave birth to this work. We also wish to thank the Département de Mathématiques et Applications of the ENS for their warm welcome.
	\end{minipage}
\end{center}
\medskip
\section{Introduction}
\subsection{Restriction theory}

For any function $f$ in the Lebesgue space $L^1(\mathbb{R}^d)$, its Fourier transform $\hat{f}$, defined for $\xi \in \mathbb{R}^d$ as
$$\hat f (\xi) = \int_{\mathbb{R}^d}e^{-ix\cdot\xi}f(x)dx,$$
is continuous, so the restriction of $\hat f$ to any surface of $\mathbb{R}^d$ is well defined. If $f$ is not integrable but only lies in some $L^p$ space, its Fourier transform can still be defined and the Hausdorff-Young inequality implies that if $1\leq p \leq 2$, then $\hat f$ is in the dual space $L^{p'}$, where $\frac{1}{p}+\frac{1}{p'}=1$. Thus $\hat f$ is only defined almost everywhere, and it is not clear whether it is possible to meaningfully restrict $\hat{f}$ to a hypersurface $S$ of $\mathbb{R}^d$, since $S$ is a set of measure zero. The restriction problem asks for which exponents $p$ and $q$ a restriction estimate of the form
$$ \Vert \hat{f}\vert_{S}\Vert_{L^q(S,d\sigma)} \leq C \Vert f \Vert_{L^p(\mathbb{R}^d)}$$ holds for all functions $f$ in the Schwartz class $\mathcal{S}(\mathbb{R}^d)$, where $d\sigma$ is a measure on $S$. Such estimates indeed allow one to define the restriction of the Fourier transform $f\mapsto \hat{f}\vert_{S}$ as a bounded operator from $L^p(\mathbb{R}^d)$ to $L^q(S,d\sigma)$.\\
Tomas and Stein discovered \cite{tsb}, \cite{ts} that this is possible for surfaces which are "sufficiently curved", a result that is now known as the Tomas-Stein theorem:

\begin{theorem}[\cite{tsb}, \cite{ts}]
Let $d\geq 2$ and $p_0 = \frac{2(d+1)}{d+3}$. Let $S$ be a smooth and compact hypersurface in $\mathbb{R}^d$ endowed with a smooth measure $d\sigma$. Suppose $S$ has non vanishing Gaussian curvature at every point. Then for all $p\in [1, p_0]$, there exists a positive constant $C$ such that for any $f\in\mathcal{S}(\mathbb{R}^d)$:
$$ \Vert \hat f \vert_{S} \Vert_{L^2(S,d\sigma)} \leq C \Vert f \Vert_{L^p(\mathbb{R}^d)}.$$
\end{theorem}
The canonical example of such a surface is the sphere $\mathbb{S}^{d-1}$ of $\mathbb{R}^d$. When the hypothesis of curvature or of compactness is removed, obtaining restriction estimates does not become impossible, but the range of possible exponents usually gets smaller, as we will illustrate with the surfaces considered in this note. A first step towards understanding the necessity of curvature is to consider a simple counter-example (found for instance in \cite{tao}) preventing any non-trivial restriction estimate on a hyperplane (endowed with the natural surface measure): consider $\varphi\in \mathcal{S} (\mathbb{R}^{d-1})$, and define $h:\mathbb{R}^d\mapsto\mathbb{C}$ by:
$$h(x)=\dfrac{\varphi(x')}{\sqrt{1+x_{d}^{2}}}, \; \; \; x=(x',x_{d}).$$
Then $h$ belongs to every $L^p$ space for $p>1$, but its Fourier transform is not well defined on the hyperplane ${\xi_d=0}$, or even on any bounded subset (with positive surface measure) of this hyperplane. Curvature is thus mandatory to obtain restriction estimates. Lack of compactness, on the other hand, can sometimes be dealt with using scaling arguments, provided the surface possesses some symmetry.
\\
Restriction theory has many applications to other topics, from number theory to the Bochner-Riesz and Kakeya conjectures \cite{tao}. It can also be used in the study of partial differential equations, especially to provide Strichartz estimates. This is the focus of the following section.

\subsection{Strichartz estimates}

Considering an evolution partial differential equation, a Strichartz estimate is a bound for the space-time norm of a solution $u$ by the norm of the initial datum $u_0$, that is to say an inequality of the type
$$ \Vert u(t,x) \Vert_{L^p_tL^q_x} \leq C \Vert u_0 \Vert_{L^r_x}.$$
Here, $p, q, r$ are suitable exponents and $ \Vert u \Vert_{L^p_t L^q_x  }$ is the standard Lebesgue space-time norm on $L^p(\mathbb{R},\ L^q(\mathbb{R}^d))$.\\
Strichartz inequalities are commonly obtained using a dispersive estimate. It is for instance the case for the linear Schrödinger equation:
$${(S):{\left\{\begin{array}{lrcl}
i\partial_t u + \Delta u =0, \ \ \ \text{for   } (t,x)\in\mathbb{R}^+ \times \mathbb{R}^d \\
u\vert_{t=0} = u_0. \end{array}\right.}}$$
The solution $u$ is, for $t>0$,
$$ u(t,\cdot) = \frac{1}{(4\pi it)^{\frac d 2}} e^{\frac{\vert x \vert^2}{4t} } \star u_0.$$
This formula leads to the following dispersive estimate:
$$ \Vert u(t,\cdot) \Vert_{L^{\infty}_x(\mathbb{R}^d)} \leq C \frac{1}{(4\pi t)^{\frac d 2}} \Vert u_0 \Vert_{L^{1}_x(\mathbb{R}^d)},$$
for $ t >0$.\\
We can combine this inequality with a result from functional analysis known as the $TT^*$ argument (see \cite{velo}, \cite{kato}, \cite{keel}), to obtain Strichartz estimates for the Schrodinger equation (as it is done in \cite{bcg}):
$$\Vert u \Vert_{L^q_t L^r_x} \leq C \Vert u_0 \Vert_{L^2_x},$$
with $q\in(2,\infty]$ and $r$ satisfying the scaling condition $ \frac 2 q + \frac d r = \frac d 2 $.\\
\\
However, once we have written the explicit formula of the solution, getting dispersive estimates is not necessarily immediate and it may require more advanced tools. Concerning the wave equation
$$  (W):   {\left\{\begin{array}{lrcl}
\partial^2_{tt} u (t,x)- \Delta u(t,x) = 0, \ \ \ \ \ \ \ \ \ \text{for   } (t,x)\in\mathbb{R}^+ \times \mathbb{R}^d\\
u|_{t=0} = f\ \ \ \ \text{and}\ \ \ \ \partial_t u|_{t=0} = g, \end{array}\right.} $$
dispersive estimates are obtained using the stationary phase method, see for instance \cite{bcg} and \cite{cdgg}. It is well known that the solution can be written, once again using the Fourier transform:
$$ u(t,x) = \mathcal{F}^{-1} (e^{it\vert x\vert } \gamma_+(\xi) + e^{-it\vert x\vert } \gamma_-(\xi))$$
where $\gamma_{\pm} (\xi) =\frac{1}{2} \bigg(  \hat f(\xi) \pm \frac{\hat g(\xi)}{i\vert\xi\vert} \bigg)$. In the case $\gamma_{\pm}$ are supported in a ring, a stationary phase argument yields the following dispersive estimate:
$$ \Vert u(t,\cdot) \Vert_{L^{\infty}(\mathbb{R}^d)}\leq \frac{C}{\vert t \vert^{\frac{d-1}{2}}} (\Vert f \Vert_{L^1} + \Vert g \Vert_{L^1}).$$
This leads to the following Strichartz estimates for the wave equation:
$$ \Vert u \Vert_{L^q_t L^p_x} \leq C (\Vert \nabla f \Vert_{L^2} + \Vert g \Vert_{L^2})$$
with $p \in [2,\infty ],\ q\in [2,\infty )$ satisfying $ \frac{1}{q} + \frac{d}{p} = \frac{d}{2}-1$.\\
We now turn to the connection between Strichartz estimates and Fourier restriction theorems. As it was shown by Strichartz in \cite{stri}, solutions to some PDEs can be written as Fourier transforms on surfaces. Consider once again the solution of the free wave equation $(W)$ ; it can be written as:
$$ u(t,x) = \displaystyle\int _{\mathbb{R}^d_{\xi} } e^{it\vert \xi\vert + i x \cdot \xi} \gamma_+(\xi) d\xi + \displaystyle\int _{\mathbb{R}^d_{\xi}} e^{-it\vert \xi\vert + i x \cdot \xi} \gamma_-(\xi) d\xi.$$

The solution can thus be seen as the sum of two inverse Fourier transforms, each one on a cone $\mathcal{C}_{\pm} = \{ (\xi, \pm \vert\xi\vert ),\ \xi \in \mathbb{R}^d \}$. Indeed,

$$ u(t,x) = \displaystyle\int _{\mathcal{C_+}} e^{i (t,x) \cdot \zeta } \ \ \Tilde{\gamma}_+(\zeta)  \ d\mu(\zeta) + \displaystyle\int _{\mathcal{C_-}} e^{i (t,x) \cdot \zeta }\ \  \Tilde{\gamma}_-(\zeta)  \ d\mu(\zeta), $$
with $\Tilde{\gamma}_{\pm}(\xi, \pm \vert \xi \vert) = \gamma_{\pm}(\xi) $. The measure $d\mu(\xi, \pm \vert \xi \vert) = d\xi$ is simply the pullback of Lebesgue measure by the canonical projection $(\xi, \pm \vert\xi\vert) \mapsto \xi$. This is exactly the dual of the restriction operator $f\mapsto \hat{f}\vert_{\mathcal{C}_{\pm}}$ applied to $\Tilde{\gamma}_{\pm}$, thus inciting to resort to restriction theory.\\
Let us admit for a brief moment that we have the following restriction estimate on the cone:

$$\left\Vert {\hat v}\vert_{\mathcal{C}_{\pm}}\  {\vert \xi \vert^{-\frac{1}{2}}} \right\Vert_{L^2(\mathcal{C}_{\pm},\ d\mu)} \leq C \Vert v \Vert_{L^{p_0}(\mathbb{R}^{d+1})}, $$

where $p_0 = \frac{2(d+1)}{d+3}$. In its dual form, it writes, for $w\in L^2(\mathcal{C}_{\pm}, d\mu)$:

$$ \left\Vert \mathcal{F}^{-1} (\vert \xi \vert^{-\frac{1}{2}} w \ d\mu) \right\Vert_{L^{p_0'}(\mathbb{R}^{d+1})} \leq C \Vert w \Vert_{L^2(\mathcal{C}_{\pm}, d\mu)}   .$$

In our case, setting $w = \Tilde{\gamma}_{\pm} \vert \xi \vert^{\frac{1}{2}}$, we notice $$ \left\Vert \Tilde{\gamma}_{\pm} \vert \xi \vert^{\frac{1}{2}} \right\Vert_{L^2(\mathcal{C}_{\pm}, d\mu)} = \Vert \gamma_{\pm} \Vert_{\dot{H}^{\frac{1}{2}}(\mathbb{R}^d)}.  $$
Finally, applying the dual inequality and using the definition of $\gamma_{\pm}$, we obtain the Strichartz estimate
$$ \Vert u \Vert_{L^{p_0'}_{t,x}} \leq C \Vert f \Vert_{\dot{H}^{\frac{1}{2}}(\mathbb{R}^d)}  + \Vert g \Vert_{\dot{H}^{-\frac{1}{2}}(\mathbb{R}^d)} .$$

Hence we have seen that, through fairly simple rewriting of the representation formula, the obtention of the Strichartz estimate above reduces to the proof of the following restriction theorem:

\begin{theorem}[\cite{stri}]
Let $d\geq 2$ and consider the cone $\mathcal{C} = \{ (\xi, \vert \xi \vert ),\ \xi \in \mathbb{R}^d \}\subset \mathbb{R}^{d+1}$ endowed with surface measure $d\sigma(\xi,|\xi|) = \frac{d\xi}{|\xi|}$. For $p_0= \frac{2(d+1)}{d+3}$, there exists a constant $C>0$ such that for any function $f\in \mathcal{S}(\mathbb{R}^{d+1})$:
$$ \Vert \hat{f}|_{\mathcal{C}} \Vert_{L^2(\mathcal{C}, d\sigma)} \leq C \Vert f \Vert_{L^{p_0}(\mathbb{R}^{d+1})}. $$
\end{theorem}

Note that Theorem 2 is not a direct consequence of Theorem 1 because the cone has zero Gaussian curvature. The original proof by Strichartz in \cite{stri}, for general quadratic surfaces, relies on complex analysis arguments. The first goal of this article is to give a different proof of this theorem. It consists in applying Theorem 1 to the spheres of increasing radius that make up the cone, an approach motivated by the fact that the exponent $p_0$ for the cone $\mathcal{C}\subset \mathbb{R}^{d+1}$ is the endpoint one for the sphere $\mathbb{S}^{d-1}$ of $\mathbb{R}^d$.\\

The resort to restriction theory to obtain Strichartz estimates is motivated by its efficiency even in settings where dispersion is not easily available. Such settings include bounded domains and compact manifolds, see for instance the work of Bourgain \cite{bourgain} for the Schrödinger equation on $\mathbb{T}^d$. This is also the case of the Heisenberg group $\mathbb{H}^d$, where there is no dispersion at all \cite{bgx}, but where Strichartz estimates for the wave equation and the Schrodinger equation were recently obtained through restriction theorems by Bahouri, Barilari and Gallagher \cite{bbg}.\\

In this article, we are also interested in using restriction theory in the study of the Euler equation in a rotational framework. The starting point is the following system:
$$  (EC_{\varepsilon}):   {\left\{\begin{array}{lrcl}
\partial_t u + u \cdot \nabla u + \dfrac{e_3 \wedge u}{\varepsilon} + \nabla p = 0 \\
\text{div} \ u = 0 \end{array}\right.} $$

with initial datum $u(0,\cdot)=u_0 $, and where the unknowns are both the vector field $u$, which is the velocity of the fluid, and the pressure field $p$. It is the zero-viscosity limit of the equations
$$  (NSC_{\varepsilon}):   {\left\{\begin{array}{lrcl}
\partial_t u + u \cdot \nabla u - \nu \Delta u + \dfrac{e_3 \wedge u}{\varepsilon} + \nabla p = 0 \\
\text{div} \ u = 0, \end{array}\right.} $$
which are studied in details in Chapter 5 of \cite{cdgg}. To get rid of the unknown pressure field, one can apply the Leray projector $\mathbb{P}$ onto divergence-free vector fields, resulting in the following equations:
$$     {\left\{\begin{array}{lrcl}
\partial_t u +  \mathbb{P}\ (\dfrac{e_3 \wedge u}{\varepsilon}) = -\mathbb{P}\ (u \cdot \nabla u)  \\
\text{div} \ u = 0. \end{array}\right.} $$
We put the non-linear term $\mathbb{P}\ (u\cdot \nabla u)$ on the right-hand side because it is often treated as a source term (\cite{cdgg}, \cite{cor}) by first solving the homogeneous equation, and then using Duhamel formula:
$$ u(t) = \mathcal{G}^{\varepsilon} (t) u_0 - \displaystyle\int _0^t \mathcal{G}^{\varepsilon} (t-t') \mathbb{P}\ (u(t') \cdot \nabla u(t')) dt' ,$$
where $\mathcal{G}^{\varepsilon}$ denotes the semigroup of the homogeneous equation
$$(C_{\varepsilon}): \partial_t u  + \mathbb{P}\ \dfrac{e_3 \wedge u}{\varepsilon} = 0.$$

Strichartz estimates for this semigroup were proven in \cite{dut} and \cite{cor}. They are in turn used to prove long-term existence of solutions to the Euler equations in the rotational framework. The proof in \cite{dut} adapts the arguments in \cite{cdgg} to the non-viscous case, using the Coriolis force to obtain a dispersive estimate ; whereas in \cite{cor} a slightly more precise study is made to obtain sharp results. \\
Our goal is to obtain such estimates through a global restriction theorem on a suitable surface. To do so, we first need to write the solution of $(C_{\varepsilon})$ as an inverse Fourier transform. We can reduce to the case $\varepsilon=1$ through a simple rescaling of $u$, and write the equation in Fourier space, which gives the linear ODE
$$ \partial_t \hat{u} (t,\xi) = A(\xi) \hat{u} (t, \xi),$$
where $A(\xi)$ is a matrix with eigenvalues $0$ and $ \pm i \frac{\xi_3}{\vert \xi \vert}$. Using that $u_0$ is divergence free, we infer that $\hat{u_0}$ has no component of eigenvalue $0$. By projecting $\hat{u_0}(\xi)$ onto the two other eigenspaces, the semigroup $\mathcal{G}^1$ can be written in terms of the semigroups $e^{\pm it\frac{D_3}{\vert D \vert}}$, defined as:
$$(e^{\pm it\frac{D_3}{\vert D \vert}} v)\ (x)= \displaystyle\int _{\mathbb{R}^3}\hat{v}(\xi) e^{\pm i t \frac{\xi_3}{\vert \xi \vert} +i x\cdot \xi }d\xi.$$
Recognizing (up to a minus sign) an inverse Fourier transform on the surface
$$ \mathcal{M}:= \left\{ \bigg(\xi, \frac{\xi_3}{\vert \xi \vert}\bigg),\ \ \ \xi \in \mathbb{R}^3\right\} \subset \mathbb{R}^4,$$
Strichartz estimates for the semigroup $\mathcal{G}$ are available as soon as we have a restriction theorem on the surface $\mathcal{M}$, simply by using its dual form.\\

The second goal of this note is thus to prove a global restriction theorem on $\mathcal{M}$. Notice that this is once again not a direct application of Theorem 1, since $\mathcal{M}$ is not compact, and has vanishing Gaussian curvature. However, we can follow the same approach as for the cone. By introducing Sobolev weights in the surface measure to deal with non-compactness, and applying Theorem 2 to suitable sub-manifolds of $\mathcal{M}$ of lower dimension, we will prove the following theorem:
\begin{theorem}
Consider the surface $\mathcal{M}$ endowed with the measure $$d\sigma\bigg(\xi,\frac{\xi_3}{|\xi|}\bigg) = \frac{d\xi}{|\xi|^2}.$$ There exists a constant $C>0$ such that for any $f\in\mathcal{S}(\mathbb{R}^4)$, one has
$$ \Vert \hat f\vert_{\mathcal{M}}  \Vert_{L^2(\mathcal{M},d\sigma)} \leq C \Vert f \Vert_{L^{\frac{6}{5}}(\mathbb{R}^4)}. $$
In terms of PDEs, this implies that for any $u_0 \in \mathcal{S}(\mathbb{R}^3),$ the following Strichartz estimate holds:
$$ \Vert e^{it\frac{D_3}{\vert D \vert}} u_0 \Vert_{L^6_{t} L^6_{x}} \leq C \Vert u_0 \Vert_{\dot H^1}. $$
\end{theorem}
The Strichartz inequalities obtained using this theorem are not sharp. More precisely, the optimal range of estimates obtained in \cite{cor} is
$$ \Vert e^{it\frac{D_3}{\vert D \vert}} u_0 \Vert_{L^q_{t}(\mathbb{R}, L^r_{x}(\mathbb{R}^d))} \leq C \Vert u_0 \Vert_{\dot H^{\frac{3}{2}-\frac{3}{r}}(\mathbb{R}^d)}, $$
for any $(q,r)\neq(2,\infty)$ such that $\frac{1}{q}+\frac{1}{r}\leq \frac{1}{2}$. Our technique only allows us to obtain, after interpolation with energy conservation, the range $\frac{1}{q}+\frac{1}{r}\leq \frac{1}{3}$, for $q\geq 6$. It is easy to identify where the sharpness is lost in the proof of Theorem 3, a point that we will further discuss in the next section and in Remark 3.1.\\
The Sobolev regularity required for the initial data in the Strichartz estimates can be explained by the fact that the surface looks like a hyperplane at infinity. We have seen that hyperplanes do not admit non-trivial restriction theorems, and it is somewhat similar in this case. It will be the object of Proposition 3.1.
\\
Let us conclude this section by noticing that Theorem 3 extends to a more general surface, namely
$$\mathcal{M}_{F}=\left\{ \bigg(\xi,\dfrac{\vert \xi \vert_{F}}{F\vert\xi\vert}\bigg), \xi \in \mathbb{R}^{3}\right\}\subset \mathbb{R}^4,$$
where $\vert \xi \vert_{F}^{2}=\vert\xi_{h}\vert^{2}+F^{2}\vert\xi_{3}\vert^{2}$ and $F\in(1,\infty]$  (note that $\mathcal{M}$ from Theorem 3 is simply the case $F=\infty$ up to an absolute value). This surface appears in \cite{charve} in the study of the primitive system, in which $F$ is the Froude number. The same estimates then hold for $\mathcal{M}_{F}$:

\begin{theorem}
Let $F\in (1,\infty).$ Consider the surface $\mathcal{M}_{F}$ with the measure $d\sigma(\xi,\frac{\xi_3}{|\xi|}) = \frac{d\xi}{|\xi|^2}$. There exists a constant $C>0$ such that for any $f\in\mathcal{S}(\mathbb{R}^4)$,
$$\Vert \hat f\vert_{\mathcal{M}}  \Vert_{L^2(\mathcal{M},\frac{d\xi}{\vert\xi\vert^2})} \leq C \frac{F^{\frac 5 3}}{(F-1)^{\frac 5 9}} \Vert f \Vert_{L^{\frac{6}{5}}(\mathbb{R}^4)}. $$
\end{theorem}
This estimate could in turn provide new results on the primitive system with vanishing viscosity: we postpone this question to a future work.

\subsection{Motivation, layout and notation}

To introduce and motivate the method we will use in the proofs of Theorems 2, 3 and 4, we state the following lemma (mentioned in \cite{tao} as an exercise):

\begin{lemme}
Let $A \subset \mathbb{R}^m$ equipped with a $\sigma$-finite measure $d\sigma_A$ and $B\subset \mathbb{R}^n$ equipped with a $\sigma$-finite measure $d\sigma_B$. Assume there exist $p\leq q$ and $C>0$ such that for any Schwartz functions $f\in\mathcal{S}(\mathbb{R}^m)$ and $g\in\mathcal{S}(\mathbb{R}^n)$, $$ \Vert \hat{f}|_{A} \Vert_{L^q(d\sigma_A)} \leq C \Vert f \Vert_{L^p(\mathbb{R}^m)}\ \ \ \ \text{and} \ \ \ \ \Vert \hat{g}|_{B} \Vert_{L^q(d\sigma_B)} \leq C \Vert g \Vert_{L^p(\mathbb{R}^n)}.$$
Then for any Schwartz function $h\in \mathcal{S}(\mathbb{R}^{n+m})$, $$ \Vert \hat{h}|_{A\times B} \Vert_{L^q(d\sigma_A d\sigma_B)} \leq C^2\ \Vert h \Vert_{L^p(\mathbb{R}^{m+n})}.$$
\end{lemme}

This result states that one can deduce a restriction theorem on $A\times B$ from individual theorems on $A$ and $B$. This motivates the study of more general surfaces $S\subset \mathbb{R}^{m+n}$ for which, for all $y\in\mathbb{R}^{n}$, the level set $S_{y}:=\{x\in\mathbb{R}^{m} : (x,y) \in S\}$ admits a restriction theorem. Our idea is to apply this restriction inequality in the $x$ variable and then use dual Sobolev embeddings in $y$ to play the role of a second "restriction" inequality. This is the case of the cone $\mathcal{C}$, for which the level sets $\mathcal{C}_{\xi_{d+1}}$ are spheres of radius $\xi_{d+1}$, and for the surface $\mathcal{M}$, for which the level sets $\mathcal{M}_{\xi_{4}}$ are cones. We advise the reader to have a brief look at the proof of this lemma before looking at the proofs of Theorems 2, 3 and 4, since the structures of all the proofs are the same, but appear in a transparent way and without technicalities in the proof of Lemma 1.1.\\

What we consider to be the main asset of this approach is that it illustrates the role of curvature in restriction theorems. In the case of the cone, the way the spherical level sets are arranged along the vertical $\xi_{d+1}$-axis does not provide additional curvature to work with (since the spheres are linearly scaled), and thus applying the Tomas-Stein theorem on those spheres suffices to make use of all the available curvature. This is why this method gives a sharp result for the cone. For the surface $\mathcal{M}$ however, the conic level sets are disposed in a way that provides additional curvature along the $\xi_{4}$-axis. This additional curvature is of course ignored when applying a restriction theorem on a given level set. It can be taken into account by using a suitable Sobolev embedding, but then the Lebesgue norm in the restriction estimate becomes anisotropic and does not really provide a Strichartz estimate in its dual form, a phenomenon we detail in Remark 3.1. To keep the Lebesgue norm isotropic, we can only partially make use of the available curvature along the $\xi_{4}$-axis, and this is where the sharpness of Theorem 3 is lost.
\\

In Section 2 of this article we will prove Theorem 2 for the cone, before turning to the surface $\mathcal{M}$ and the proofs of Theorem 3 and Theorem 4 in Section 3, which will be concluded by Proposition 3.1. The proof of the previous lemma can be found in the Appendix.

Throughout this note, we will denote by $d\sigma_R$ the surface measure on the sphere $\mathbb{S}^{d-1}_R\subset \mathbb{R}^d$ of radius $R$, defined by
$$ d\sigma_R(A) = \lambda(\{ta: a\in A, 0< t \leq 1\})$$
for any measurable subset $A \subset\mathbb{S}^{d-1}_R$, with $\lambda$ being the Lebesgue measure on $\mathbb{R}^d$. Finally, in the computations, $C$ will be a positive constant that may vary from line to line.

\section{Restriction theory on the cone: proof of Theorem 2}

In this section we will use the notations of Theorem 2. As mentioned in the introduction, the exponent $p_0$ is the endpoint one in Theorem 1, which applies to the sphere of $\mathbb{R}^d$. This is what motivates our technique, which consists in deducing Theorem 2 by applying Theorem 1 to the $(d-1)$-dimensional spheres of increasing radius that build up the $d$-dimensional cone. \\
We will thus need the following lemma, which is an extension of Theorem 1 to spheres of varying radius. Its proof is a simple scaling argument left to the reader.

\begin{lemme}
Let $\mathbb{S}^{d-1}_R$ be the sphere of radius $R>0$ included in $\mathbb{R}^d$, equipped with the surface measure $d\sigma_R$ (not normalized). There exists a constant $C>0$ (independent of R) such that for any function $f\in \mathcal{S}(\mathbb{R}^{d})$, one has:
$$ \Vert \hat{f}|_{\mathbb{S}^{d-1}_R} \Vert_{L^2(\mathbb{S}^{d-1}_R, d\sigma_R)} \leq C R^{\alpha} \Vert f \Vert_{L^{p_0}(\mathbb{R}^{d})},$$
with  $\alpha =  \frac{1}{p_0'} = \frac{d-1}{2(d+1)}$.
\end{lemme}

\begin{proof2}[Proof of Theorem 2.]

Consider $ f\in\mathcal{S}(\mathbb{R}^{d+1}) $. The first step to prove the theorem is to write $\Vert \hat{f}|_{\mathcal{C}} \Vert_{L^2(\mathcal{C}, d\sigma)}$ as an integral over spheres, using polar coordinates. We have:
\begin{align*}
    \Vert \hat{f}|_{\mathcal{C}} \Vert^2_{L^2(\mathcal{C}, d\sigma)}& = \displaystyle\int _{\mathbb{R}^d} \big\vert\hat{f}(\xi,|\xi|) \big\vert^2 \frac{d\xi}{|\xi|}\\
    &= \displaystyle\int _0^{+\infty} \displaystyle\int _{\mathbb{S}^{d-1}_{\rho}} \vert \hat{f}(\xi,\rho) \vert^2 d\sigma_{\rho}(\xi) \frac{d\rho}{\rho}
\end{align*}
with $\mathbb{S}^{d-1}_{\rho}$ the sphere of radius $\rho$, and $d\sigma_{\rho}$ the measure on $\mathbb{S}^{d-1}_{\rho}$ as in Lemma 2.1.\\
Now, we want to use a restriction theorem on the sphere of radius $\rho$, that is to say apply Lemma 2.1. It yields

$$\displaystyle\int _0^{+\infty} \displaystyle\int _{\mathbb{S}^{d-1}_{\rho}} \vert \hat{f}(\xi,\rho) \vert^2 d\sigma_{\rho}(\xi) \frac{d\rho}{\rho}
    \leq C \displaystyle\int _0^{+\infty} \bigg( \displaystyle\int _{\mathbb{R}^d} \vert \Grave{f}(x,\rho) \vert^{p_0} dx \bigg)^{\frac{2}{p_0}} \frac{d\rho}{\rho^{1-2\alpha}}$$
with $\Grave{f}(x,\rho)$ the partial Fourier transform of $f$ in the last coordinate, that is:
$$\Grave{f} (x,\rho) = \int_{\mathbb{R}}e^{-ix_{d+1}\rho}\,f(x,x_{d+1})\,dx_{d+1}.$$
At this point, since $p_0 \leq 2$, we can use the Minkowski inequality:
$$  \displaystyle\int _0^{+\infty} \bigg( \displaystyle\int _{\mathbb{R}^d} \vert \Grave{f}(x,\rho) \vert^{p_0} dx \bigg)^{\frac{2}{p_0}} \frac{d\rho}{\rho^{1-2\alpha}}
\leq C \bigg(\displaystyle\int _{\mathbb{R}^d} \bigg( \displaystyle\int _0^{+\infty} \vert \Grave{f}(x,\rho) \vert^2 \frac{d\rho}{\rho^{1-2\alpha}} \bigg)^{\frac{p_0}{2}} dx \bigg)^{\frac{2}{p_0}} .$$
We recognize on the right-hand side a piece of the homogeneous Sobolev norm of $f(x,\cdot )$:
$$ \bigg( \displaystyle\int _0^{+\infty} \vert \Grave{f}(x,\rho) \vert^2 \frac{d\rho}{\rho^{1-2\alpha}} \bigg)^{\frac{1}{2}} \leq \Vert f(x,\cdot )\Vert_{\dot H^s(\mathbb{R})},$$
where
$$ \Vert f(x,\cdot )\Vert_{\dot H^s(\mathbb{R})}=\bigg( \displaystyle\int _{\mathbb{R}} \vert \Grave{f}(x,\rho) \vert^2 \vert \rho \vert^{2s}d\rho \bigg)^{\frac{1}{2}},$$
with $s = -\frac{1}{2} + \alpha$. Now, by the dual Sobolev embedding (that can be thought of as the second restriction-type inequality of Lemma 1.1), we find exactly $$\Vert f(x,\cdot )\Vert_{\dot H^s(\mathbb{R})} \leq C \Vert f(x,\cdot ) \Vert_{L^{p_0}(\mathbb{R})}. $$
Indeed, $s = - \frac{1}{d+1}$ and $\frac{1}{2} - \frac{1}{p_0} = \frac{d+1 - (d+3)}{2(d+1)} = s$ since $\alpha = \frac{1}{p_0'}$. Finally,
\begin{align*}
    \bigg(\displaystyle\int _{\mathbb{R}^d} \bigg( \displaystyle\int _0^{+\infty} \vert \Grave{f}(x,\rho) \vert^2 \frac{d\rho}{\rho^{1-2\alpha}} \bigg)^{\frac{p_0}{2}} dx \bigg)^{\frac{2}{p_0}}
    &\leq C \bigg(\displaystyle\int _{\mathbb{R}^d} \bigg( \displaystyle\int _{\mathbb{R}} |f(x,y)|^{p_0} dy \bigg)^{\frac{p_0}{p_0}} dx \bigg)^{\frac{2}{p_0}} \\
    &\leq C \Vert f \Vert_{L^{p_0}(\mathbb{R}^{d+1})}^2,
\end{align*}
which concludes the proof.
\end{proof2}

\section{Linearized Euler equation in the rotational framework: proofs of Theorems 3 and 4}

We now turn our attention to the surface $\mathcal{M}$ and the proof of Theorem 3. The surface measure is now $d\sigma = \frac{d\xi}{|\xi|^2}$. We will reproduce the same technique we used in the previous section: indeed, notice that the level sets of $\mathcal{M}$ are cones of varying opening angles. Our goal is thus to write the $L^2$ norm of $f\vert_\mathcal{M}$ as an integral over all level sets and apply a suitable restriction theorem at each level.
We thus need to extend Theorem 2 into a corollary for cones of arbitrary angle:
\begin{cor}
Let $\mathcal{C}_{\rho}:= \{ (\xi, \rho \vert \xi \vert),\ \xi\in\mathbb{R}^2 \}$ be a cone in  $\mathbb{R}^3$, with surface measure $d\sigma_\rho =\frac{d\xi}{\vert \xi \vert}$. Then, for any function $f\in \mathcal{S}(\mathbb{R}^{3})$, one has:
$$ \Vert \hat{f}|_{\mathcal{C}_{\rho}} \Vert_{L^2(\mathcal{C}_{\rho}, d\sigma_\rho)} \leq C \vert \rho \vert^{-\frac{1}{6}}\vert\  \Vert f \Vert_{L^{\frac 6 5}(\mathbb{R}^{3})}. $$
\end{cor}
This is directly deduced from the case $\rho=1$ by rescaling correctly $f$ in the third variable, with $\frac 6 5$ simply being the exponent in Theorem 2 for cones in $\mathbb{R}^3$.

We can now begin the proof of Theorem 3.

\begin{proof2}[Proof of Theorem 3.]
Consider $ f\in\mathcal{S}(\mathbb{R}^{4}) $ and set $\xi_h:=(\xi_1,\xi_2)$. We begin by writing $ \Vert \hat f\vert_{\mathcal{M}}  \Vert_{L^2(\mathcal{M},d\sigma)} $ as an integral over all level sets by changing variables, letting $\mu=\dfrac{\xi_3}{\vert \xi\vert}$. By straightforward computations, we have the identities $d\xi_3=\vert \xi_h \vert (1-\mu^2)^{-\frac 3 2} d\mu $ and $\vert \xi \vert = \dfrac{\vert \xi_h\vert}{\sqrt{1-\mu^2}} $, which yield:
\begin{align*}
\Vert \hat f\vert_{\mathcal{M}}  \Vert^2_{L^2(\mathcal{M},d\sigma)} &= \displaystyle\int _{\mathbb{R}^3} \left\vert \hat f(\xi_h,\xi_3,\frac{\xi_3}{\vert \xi \vert } )\right\vert^2 \ \ \vert \xi\vert^{-2} d\xi_hd\xi_3 \\ &= \int_{-1}^{1}  \displaystyle\int _{\mathbb{R}^2_{\xi_h}} \left\vert \hat f(\xi_h,  \dfrac{\mu}{\sqrt{1-\mu^2}} \vert \xi_h \vert , \mu )\right\vert^2 \ \frac{d\xi_h}{\vert \xi_h \vert} \ \dfrac{1}{\sqrt{1-\mu^2}} d\mu .
\end{align*}
We see that at a given level $\mu$, the integral over $\xi_h$ is simply a squared $L^2$ norm on a cone with the appropriate measure $\dfrac{d\xi_h}{\vert \xi_h \vert}$. We can apply Lemma 3.1 with $\rho=\dfrac{\mu}{\sqrt{1-\mu^2}}$ to get
\begin{align*}
\Vert \hat f\vert_{\mathcal{M}}  \Vert^2_{L^2(\mathcal{M},d\sigma)} &\leq C\int_{-1}^{1}  \bigg( \displaystyle\int _{\mathbb{R}^3_x} \left\vert  \Grave{f}(x, \mu )  \right\vert^{p} dx \bigg)^{\frac{2}{p}} \ \vert\mu\vert^{-\frac 1 3}(1-\mu^2)^{-\frac 1 3} d\mu ,
\end{align*}
with $p=\frac 6 5$, and $\Grave{f}(x,\mu)$ the Fourier transform of $f$ in the last coordinate $x_4$. As before, we now bound the $L^{2}_{\mu}L^{p}_{x}$ norm by a $L^{p}_{x}L^{2}_{\mu}$ norm using the Minkowski inequality. We are left with
\begin{align*}
\Vert \hat f\vert_{\mathcal{M}}  \Vert^2_{L^2(\mathcal{M},d\sigma)} &\leq C  \bigg(\displaystyle\int _{\mathbb{R}^3_x}\bigg( \int_{-1}^{1} \left\vert  \Grave{f}(x, \mu )  \right\vert^{2} \vert\mu\vert^{-\frac 1 3}(1-\mu^2)^{-\frac 1 3} d\mu \bigg)^{\frac{p}{2}} \  dx\bigg)^{\frac{2}{p}}.
\end{align*}
We now want to use Sobolev embeddings for the inner integral to bound it by a $L^{p}$ norm. Note that we have three singularities in $\mu=-1$, $0$, $1$, however near each of those points the term $\vert\mu\vert^{-\frac 1 3}(1-\mu^2)^{-\frac 1 3}=\vert\mu\vert^{-\frac 1 3}(1-\mu)^{-\frac 1 3}(1+\mu)^{-\frac 1 3}$ behaves like a simple Sobolev weight. We make use of this by splitting the integral in three:
$$\int_{-1}^{1} \left\vert  \Grave{f}(x, \mu )  \right\vert^{2} \vert\mu\vert^{-\frac 1 3}(1-\mu^2)^{-\frac 1 3} d\mu = \int_{-1}^{-\frac 1 2} ... + \int_{-\frac 1 2}^{\frac 1 2} ... + \int_{\frac 1 2}^{1} ...\ .$$
Let us bound the last term, the others being dealt with identically: we simply use that, for $\frac 1 2 \leq \mu \leq 1$, we have $\vert\mu\vert^{-\frac 1 3}(1-\mu)^{-\frac 1 3}(1+\mu)^{-\frac 1 3}\leq C (1-\mu)^{-\frac 1 3}$. This yields, making an appropriate change of variables:
\begin{align*}
\int_{\frac 1 2}^{1}\left\vert  \Grave{f}(x, \mu )  \right\vert^{2} \vert\mu\vert^{-\frac 1 3}(1-\mu^2)^{-\frac 1 3} d\mu &\leq C\int_{\frac 1 2}^{1}\left\vert  \Grave{f}(x, \mu )  \right\vert^{2}  (1-\mu)^{-\frac 1 3} d\mu \\ &=C\int_{0}^{\frac 1 2}\left\vert  \Grave{f}(x, 1-\nu )  \right\vert^{2}  \nu^{-\frac 1 3} d\nu .
\end{align*}
The Sobolev exponent we seek is $s=-\frac{1}{3}$, to exploit that $L^{p}$ is embedded in $\dot{H}^{s}(\mathbb{R})$. Since $\vert \nu \vert \leq 1$, we can bound $\nu^{-\frac 1 3}$ by $\nu^{-\frac 2 3}$. Note that this is very blunt, and most of the sharpness of the result is lost due to this (we further discuss this in Remark 3.1). Using this bound anyway, we get
\begin{align*}
\int_{\frac 1 2}^{1}\left\vert  \Grave{f}(x, \mu )  \right\vert^{2} \vert\mu\vert^{-\frac 1 3}(1-\mu^2)^{-\frac 1 3} d\mu &\leq C\int_{0}^{\frac 1 2}\left\vert  \Grave{f}(x, 1-\nu )  \right\vert^{2}  \nu^{-\frac 2 3} d\nu \\&\leq C\int_{\mathbb{R}_{\nu}}\left\vert  \Grave{f}(x, 1-\nu )  \right\vert^{2}  \vert\nu\vert^{-\frac 2 3} d\nu\\&\leq C \Vert f(x,\cdot )\Vert ^{2}_{L^{p}(\mathbb{R})}.
\end{align*}
In the last step we used Sobolev embeddings, as well as the fact that the inverse Fourier transform in $\nu$ of $\Grave{f}(x, 1-\nu )$ is $f$ up to a phase and a sign in the argument, both making no difference in the $L^p$ norm.
The same bound holds for the two other terms, so that the whole integral over $\mu$ is controlled by $\Vert f(x,.)\Vert ^{2}_{L^{p}(\mathbb{R})}$, which in turns yields
\begin{align*}
\Vert \hat f\vert_{\mathcal{M}}  \Vert^2_{L^2(\mathcal{M},d\sigma)} &\leq C  \bigg(\displaystyle\int _{\mathbb{R}^3_x}\bigg(  \Vert f(x,.)\Vert ^{2}_{L^{p}(\mathbb{R})}\bigg)^{\frac{p}{2}} \  dx\bigg)^{\frac{2}{p}}\\&= C  \Vert f\Vert ^{2}_{L^{p}(\mathbb{R}^4)}.
\end{align*}
This concludes the proof of Theorem 3.
\end{proof2}

\begin{rmq}
Let us briefly discuss the sharpness of this result.
As mentioned in the proof, we used at some point the blunt bound $\nu^{-\frac{1}{3}} \leq \nu^{-\frac{2}{3}}$, because the Sobolev embedding $L^{\frac 6 5}(\mathbb{R}) \hookrightarrow \dot H^{-\frac 1 3}(\mathbb{R})$ naturally provides a bound by an $L^{\frac 6 5}$ norm in the last coordinate, so that we obtain in the end a $L^{\frac 6 5}$ norm on the whole space $\mathbb{R}^4$. However, this obviously makes us lose precision on the result. To avoid this, we could have kept the term $\nu^{-\frac 1 3}$, and instead use the Sobolev embedding $L^{\frac 3 2}(\mathbb{R}) \hookrightarrow \dot H^{-\frac 1 6}(\mathbb{R})$. However, at the end, we would have got the following restriction result involving anisotropic Lebesgue norms:
$$ \Vert \hat f\vert_{\mathcal{M}}  \Vert_{L^2(\mathcal{M},d\sigma)} \leq C \Vert f \Vert_{L^{\frac{6}{5}}(\mathbb{R}^{3}_{x_1,x_2,x_3},L^{\frac{3}{2}}(\mathbb{R}_{x_4}))}. $$
In terms of estimates for the semigroup $e^{it\frac{D_3}{\vert D \vert}}$, it writes
$$\Vert e^{it\frac{D_3}{\vert D \vert}} u_0 \Vert_{L^6_{x}L^3_{t}} \leq C \Vert u_0 \Vert_{\dot H^1(\mathbb{R}^3)}.$$
This is not really a Strichartz estimate, we would have liked a $L^3_t L^6_x$ norm instead, but to our knowledge there is no way around this issue (see \cite{bbg} where the same phenomenon occurs). We know thanks to \cite{cor} that $$ \Vert e^{it\frac{D_3}{\vert D \vert}} u_0 \Vert_{L^3_{t}L^6_{x}} \leq C \Vert u_0 \Vert_{\dot H^1(\mathbb{R}^3)} $$ is true, and is sharp in the sense that for a $L^6_x$ norm in space, we cannot do better than a $L^3_t $ norm in time.
\\The source of the anisotropy in the result is due to the curvature of $\mathcal{M}$ along the $\xi_4$ coordinate, which is completely ignored when applying Theorem 2 on each conic level set. This additional curvature only plays a role in the integration over $\xi_4$, and allows us to sharpen the Sobolev embedding we use, but at the cost of the Lebesgue exponents not matching in the end. \\
A way to try and obtain a sharp Strichartz estimate would be to aim for the same exponents in time and space (so that their order does not matter). This would be the following estimate $$ \Vert e^{it\frac{D_3}{\vert D \vert}} u_0 \Vert_{L^4_{t}L^4_{x}} \leq C \Vert u_0 \Vert_{\dot H^{\frac 3 4}(\mathbb{R}^3)} $$ from \cite{cor}. Those exponents are too strong to be achieved using cone restriction theorems applied to the level sets, since this would require Theorem 2 but with $p_0=\frac 4 3$ in dimension $d=2$ which is not possible, see \cite{tao}.
\end{rmq}

We shall now prove Theorem 4, we will give less details since the proof is very similar to the previous one.

\begin{proof2}[Proof of Theorem 4.]
Let $f\in\mathcal{S}(\mathbb{R}^4)$, $F \in (1,\infty)$ and $\xi_h = (\xi_1,\xi_2)$. Recall that $\vert \xi \vert_{F}^{2}=\vert\xi_{h}\vert^{2}+F^{2}\vert\xi_{3}\vert^{2}$ and $\mathcal{M}_{F}=\left\{ \bigg(\xi,\dfrac{\vert \xi \vert_{F}}{F\vert\xi\vert}\bigg), \xi \in \mathbb{R}^{3}\right\}\subset \mathbb{R}^4,$ thus we want to compute
$$ \displaystyle\int _{\mathbb{R}^3} \left\vert \hat f \bigg(\xi, \dfrac{\vert \xi \vert_{F}}{F\vert\xi\vert}\bigg) \right\vert^2 \frac{d\xi}{\vert \xi\vert^2}\ \cdotp$$
For a bit more simplicity, we shall restrict our attention to half of this surface, say $\xi_3>0$. It does not change the result since the surface is symmetric relatively to the axis $\xi_3=0$.\\

The first step is to write $\mu =  \dfrac{\vert \xi \vert_{F}}{F\vert\xi\vert}$, thus we can express $\xi_3$ in terms of $\xi_h$ and $\mu$. It yields
$$ \xi_3 = \sqrt{ \frac{\mu^2-F^{-2}}{1-\mu^2} }\ \ \vert\xi_h\vert,\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ d\xi_3 = \mu \frac{1-F^{-2}}{(1-\mu^2)^{\frac 3 2}(\mu^2-F^{-2})^{\frac 1 2 }}\ \ \vert\xi_h\vert\ \ d\mu\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \text{and}\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \vert\xi\vert^2 = \vert\xi_h\vert^2\bigg(\frac{1-F^{-2}}{1-\mu^2}\bigg).$$
Then, straightforward computations give
$$ \displaystyle\int _{\mathbb{R}^2_{\xi_h}\times\mathbb{R}^+_{\xi_3}} \left\vert \hat f (\xi, \dfrac{\vert \xi \vert_{F}}{F\vert\xi\vert}) \right\vert^2 \frac{d\xi}{\vert \xi\vert^2}
= \displaystyle\int _{\frac{1}{F}}^1 \displaystyle\int _{\mathbb{R}^2_{\xi_h}} \left\vert \hat f \bigg(\xi_h, \sqrt{ \frac{\mu^2-F^{-2}}{1-\mu^2} }\ \ \vert\xi_h\vert , \mu \bigg) \right\vert^2  \frac{\mu}{(1-\mu^2)^{\frac 1 2}(\mu^2-F^{-2})^{\frac 1 2 }}\ \ \frac{d\xi_h}{\vert \xi_h\vert} d\mu.  $$
Now, we are ready to apply Corollary 3.1 with $\rho = \sqrt{ \frac{\mu^2-F^{-2}}{1-\mu^2} }$ and $p= \frac{ 6}{5}$:
\begin{align*}
    \displaystyle\int _{\frac{1}{F}}^1 \displaystyle\int _{\mathbb{R}^2_{\xi_h}}
    \left\vert \hat f \bigg(\xi_h, \sqrt{ \frac{\mu^2-F^{-2}}{1-\mu^2} }\ \ \vert\xi_h\vert , \mu \bigg) \right\vert^2 & \frac{\mu}{(1-\mu^2)^{\frac 1 2}(\mu^2-F^{-2})^{\frac 1 2 }}\ \ \frac{d\xi_h}{\vert \xi_h\vert} d\mu\\
    &\leq C \displaystyle\int _{\frac{1}{F}}^1 \bigg(\displaystyle\int _{\mathbb{R}^3_x} \vert \Grave{f}(x,\mu) \vert^p dx \bigg)^{\frac 2 p} \mu (1-\mu^2)^{-\frac 1 3}(\mu^2-F^{-2})^{-\frac 2 3}d\mu.
\end{align*}
The Minkowski inequality turns the $L^2_{\mu}L^p_x$ norm into a $L^p_x L^2_{\mu}$ norm:

\begin{align*}
     \displaystyle\int _{\frac{1}{F}}^1 \bigg(\displaystyle\int _{\mathbb{R}^3_x} \vert \Grave{f}(x,\mu) \vert^p dx \bigg)^{\frac 2 p}& \mu (1-\mu^2)^{-\frac 1 3}(\mu^2-F^{-2})^{-\frac 2 3}d\mu\\
     &\leq \Bigg( \displaystyle\int _{\mathbb{R}^3_x} \bigg( \displaystyle\int _{\frac{1}{F}}^1 \vert \Grave{f}(x,\mu) \vert^2 \mu (1-\mu^2)^{-\frac 1 3}(\mu^2-F^{-2})^{-\frac 2 3}d\mu \bigg)^{\frac p 2 } dx \Bigg)^{\frac 2 p}.
\end{align*}

As before, it is enough to prove that $$\displaystyle\int _{\frac{1}{F}}^1 \vert \Grave{f}(x,\mu) \vert^2 \mu (1-\mu^2)^{-\frac 1 3}(\mu^2-F^{-2})^{-\frac 2 3}d\mu \leq C_F \Vert f(x,y) \Vert_{L^p(\mathbb{R}_y)}^2.$$
To do so, we shall split the previous integral around the singularities at $F^{-1}$ and $1$, and then use the Sobolev embedding $L^{\frac 6 5}(\mathbb{R}) \hookrightarrow \dot H^{-\frac 1 3}(\mathbb{R})$. Indeed,
$$\displaystyle\int _{\frac{1}{F}}^1 \vert \Grave{f}(x,\mu) \vert^2 \mu (1-\mu^2)^{-\frac 1 3}(\mu^2-F^{-2})^{-\frac 2 3}d\mu  = \displaystyle\int _{\frac{1}{F}}^{ \frac{F+1}{2F} } ... + \displaystyle\int _{ \frac{F+1}{2F} }^1 ... \ .$$
For $\mu \in [F^{-1}, \frac{F+1}{2F}]$, we have $\mu (1-\mu^2)^{-\frac 1 3}(\mu^2-F^{-2})^{-\frac 2 3} \leq \frac{F}{(F-1)^{\frac 1 3}} (\mu - F^{-1})^{\frac 2 3}$, and this is exactly the Sobolev weight we wanted (the way we deal with this term is exactly the same than for Theorem 3). \\
For $\mu \in [\frac{F+1}{2F},1]$, we shall use the inequality $(\mu-F^{-1})^{-\frac 2 3} \leq \bigg(\frac{F}{F-1}\bigg)^{\frac 1 3} (1-\mu)^{-\frac 1 3}$. It yields
$$\mu (1-\mu^2)^{-\frac 1 3}(\mu^2-F^{-2})^{-\frac 2 3} \leq C \bigg(\frac{F}{F-1}\bigg)^{\frac 1 3} (1-\mu)^{-\frac 2 3},$$
once again it is the desired Sobolev weight. It gives
$$ \displaystyle\int _{\frac{1}{F}}^1 \vert \Grave{f}(x,\mu) \vert^2 \mu (1-\mu^2)^{-\frac 1 3}(\mu^2-F^{-2})^{-\frac 2 3}d\mu \leq C \frac{F}{(F-1)^{\frac 1 3}}\bigg(\displaystyle\int _{\mathbb{R}_y} \vert f(x,y)\vert^p dy \bigg)^{\frac 2 p},$$
which concludes the proof.
\end{proof2}

\begin{rmq}
The fact that the estimates blow up as $F\rightarrow 1$ is very intuitive : the surface then becomes the hyperplane $\{\xi_4 = 1\}$. We refer to \cite{fun} for a study of the primitive equations in the case $F=1$ with viscosity. However the blow-up as $F\rightarrow\infty$ is a bit less intuitive, since the estimate still holds for $F= \infty$. We believe this is due to the fact that $\mathcal{M}_F$ is located above the hyperplane ${\xi_{4}=\frac{1}{F}}$: for finite $F$, the singularity at $0$ of the Sobolev weight in the surface measure does not intervene, whereas it appears in the case $F=\infty$ and offers a gain of integrability at the origin.\\
Finally, the Sobolev weight that appears in the Jacobian is exactly the one wanted at one of the two singularities. Thus, contrary to the surface $\mathcal{M}$, in this case we could not use a sharper Sobolev embedding to improve the estimate, even if we allowed anisotropic norms. This might be a sign that the result is in some way sharper than for Theorem 3, but we did not investigate this further.
\end{rmq}

To conclude, we shall say a few words about the need of Sobolev norms instead of $L^2$ norms. As said in the introduction, the surface studied is very flat at infinity, and we have seen that the hyperplane can never admit non-trivial restriction theorems. The following proposition is the consequence of the flatness and the non-compactness of $\mathcal{M}$, it is stated in dimension $d$ for a bit more generality.

\begin{prop}
Consider the $d$-dimensional surface $$ \mathcal{M}:= \left\{ \bigg(\xi, \frac{\xi_d}{\vert \xi \vert}\bigg),\ \ \ \xi \in \mathbb{R}^d \right\},$$
endowed with the measure $d\sigma(\xi,\frac{\xi_d}{\vert \xi\vert}) = d\xi$.  Let $p>1$. Then one cannot find an exponent $q$ and a constant $C$ such that for any $f\in \mathcal{S}(\mathbb{R}^{d+1})$,
$$ \Vert \hat f \vert_{\mathcal{M}} \Vert_{L^q(\mathcal{M},d\sigma)} \leq C \Vert f \Vert_{L^p(\mathbb{R}^{d+1})}.$$
\end{prop}

\begin{proof2}[Proof of Proposition 3.1.]
Consider  $f(x_{d+1}) = \frac{1}{\sqrt{1 + x_{d+1}^2}}$, and $\psi \in \mathcal{S}(\mathbb{R}^d)$ such that $$\supp \hat \psi \subset \mathcal{T}: = \{ (\xi',\xi_d)\in \mathbb{R}^{d-1} \times \mathbb{R},\ \ \ \vert \xi' \vert \leq 1,\ \ \ \vert \xi_d \vert \leq  \vert \xi' \vert \}.$$
Let $h(x_1,...,x_{d+1}) = \psi(x_1,...,x_d) f(x_{d+1})$, and let $$h_R =  \psi(Rx_1,..., Rx_{d-1}, \frac{x_d}{R^{d-1}}) f(x_{d+1}).$$
\\
Then,
$$\widehat{h_R} =  \hat \psi(\frac{\xi_1}{R},..., \frac{\xi_{d-1}}{R},{R^{d-1}}{\xi_d}) \hat f(\xi_{d+1}).$$
This can be written, with obvious notation, as $\widehat{h_R} = \widehat{\psi_R} \hat f$. Moreover, $\widehat{\psi_R}$ is supported in $$ \mathcal{T}_R: = \{ (\xi',\xi_d)\in \mathbb{R}^{d-1} \times \mathbb{R},\ \ \ \vert \xi' \vert \leq R,\ \ \ \vert \xi_d \vert \leq  \frac{\vert \xi' \vert}{R^{d-1}} \}.$$
Then we shall study the norms of $h_R$ and $\widehat{h_R}\vert_{\mathcal{M}}$. \\
On the one hand, $ \Vert h_R \Vert_{L^p(\mathbb{R}^{d+1})} = \Vert h \Vert_{L^p(\mathbb{R}^{d+1})} $ whenever $p>1$.\\
On the other hand, using the fact that $\frac{\xi_d}{\vert\xi\vert}$ is very close to 0 for $\xi\in\mathcal{T}_R$, we find
$$  \Vert \widehat{h_R}\vert_{\mathcal{M}} \Vert_{L^q(\mathcal{M})} \geq  \inf\limits_{\tau \in [-\frac{1}{R^{d-1}}, \frac{1}{R^{d-1}}]} \vert \hat f (\tau) \vert \ \ \Vert \widehat{\psi} \Vert_{L^q(\mathbb{R}^d)} .$$
Yet the infimum of $\hat f$ blows up as $R \rightarrow \infty$, hence one cannot have an inequality such as $$  \Vert \hat u \vert_{\mathcal{M}} \Vert_{L^q(\mathcal{M})} \leq C \Vert u \Vert_{L^p(\mathbb{R}^{d+1})}$$
holding for all $u\in\mathcal{S}(\mathbb{R}^{d+1})$.
\end{proof2}

\section*{Appendix}

\begin{proof2}[Proof of Lemma 1.1.]
Let $f\in \mathcal{S}(\mathbb{R}^{m+n})$. Applying the restriction inequality on $B$, we have:
\begin{align*}
    \Vert \hat{f}|_{A\times B} \Vert^q_{L^q(d\sigma_A d\sigma_B)} &= \displaystyle\int _A \displaystyle\int _B |\hat{f}(\xi,\tau)|^q  d\sigma_B(\tau)d\sigma_A(\xi) \\
    &\leq C^q \displaystyle\int _A \bigg( \displaystyle\int _{\mathbb{R}^n} \vert \Acute{f}(\xi,y) \vert^p dy \bigg)^{\frac{q}{p}} d\sigma_A(\xi),
\end{align*}
with $\Acute{f}(\xi,y)$ the partial Fourier transform of $f$ in the $m$ first coordinates. Now, by the Minkowski inequality, since $p\leq q$,
$$\displaystyle\int _A \bigg( \displaystyle\int _{\mathbb{R}^n} \vert \Acute{f}(\xi,y) \vert^p dy \bigg)^{\frac{q}{p}} d\sigma_A(\xi) \leq   \bigg( \displaystyle\int _{\mathbb{R}^n} \bigg( \displaystyle\int _A \vert \Acute{f}(\xi,y) \vert^q  d\sigma_A(\xi)  \bigg)^{\frac{p}{q}} dy\bigg)^{\frac{q}{p}}.$$

Finally, it suffices to apply the restriction inequality on $A$:
\begin{align*}
     C^q \bigg( \displaystyle\int _{\mathbb{R}^n} \bigg( \displaystyle\int _A \vert \Acute{f}(\xi,y) \vert^q  d\sigma_A(\xi)  \bigg)^{\frac{p}{q}} dy\bigg)^{\frac{q}{p}}
   & \leq C^{2q} \bigg( \displaystyle\int _{\mathbb{R}^n} \bigg( \displaystyle\int _{\mathbb{R}^m} \vert {f}(x,y) \vert^p  dx \bigg)dy \bigg)^{\frac{q}{p}} \\
    & \leq C^{2q} \Vert f \Vert^q_{L^p(\mathbb{R}^{m+n})},
\end{align*}
which concludes the proof.
\end{proof2}

\newpage

redpajama_set_name: RedPajamaArXiv


text:
Regional Reviews: Florida - Southern
The Drowsy Chaperone
The Stage Door Theatre presents one of the most celebrated musicals of the 2006 Broadway season, The Drowsy Chaperone. The Drowsy Chaperone features music and lyrics by Lisa Lambert and Greg Morrison, and book by Bob Martin and Don McKellar. It debuted at the Toronto Fringe Festival in 1999 and had its first complete production at the Toronto Winter Garden Theatre in June of 2001. It ran in 2005 at the Center Theatre Group/Ahmanson Theatre in Los Angeles, where it received five Los Angeles Drama Critics Circle Awards. It opened on Broadway on May 1, 2006 at the Marquis Theatre and received five Tony Awards (with thirteen nominations), seven Drama Desk Awards (with fourteen nominations), four Outer Critic's Circle Awards, a Theatre World Award and a New York Drama Critics' Circle Award.
The Drowsy Chaperone is the story of a fussy and reclusive, middle-aged, modern-day musical-theatre addict known simply as Man in Chair (Dan Kelley). To chase the blues away, he likes to listen to familiar old musicals. On this particular occasion, he drops the needle on his favorite LP (yes LP, not CD!), the cast recording of the 1928 musical comedy The Drowsy Chaperone. From the crackle of his hi-fi, the musical magically bursts to life on stage all around him. It tells the tale of a pampered Broadway starlet, Janet Van De Graaff (Laura Oldham), who wants to give up show business and marry her debonair suitor, Robert Martin (Jeffrey Funaro). To complicate matters, there is a dim-witted aspiring actress named Kitty (Sari Feldman) waiting to take her place and a producer named Feldzieg (Jonathan Van Dyke) determined to sabotage the nuptials. There are also two gangsters (Phil Gosselin and Andrew Ford) looking to ensure that the producer doesn't lose money on their investment and of course, her Drowsy Chaperone. The chaperone (Eileen Faxas) isn't naturally drowsy, but she is inclined to drink—despite the existence of prohibition—and it seems that alcohol makes her terribly, terribly drowsy. Needless to say, she is rather sleepy throughout the entire show. Since it is her job to keep an eye on Janet, her drowsy nature allows mischief to occur undetected. She even encounters a bit of mischief of her own in the arms of an aging Latin lover named Aldolpho (Matt Ban).
One cannot help but compare productions, and having seen two national tours of the same show, this one shines for some extraordinarily good directing by Dan Kelley. The characters' choices, and how these characters represent old movie-musical stereotypes, as well as their acting beats are all very clean. The scenic design is right on the money, complete with a refrigerator and a Murphy bed that double as an entrance for the characters of the musical within the show. The costumes, especially for the men, are sensational in their attention to detail, use of color, and exaggerated period style. As Man in Chair, Dan Kelley shows an artful awareness of the audience that helps make the character's conversations with the audience seem impromptu. His interpretation of the role is more high strung than the nerdy, introverted way in which the character is usually played—but it works quite well. Eileen Faxas is tops as the drunken diva, the Drowsy Chaperone. This is a great role for her to showcase her comedic ability. It is very important that the Drowsy Chaperone takes the stage in her given moments, such as on "As We Stumble Along", and Faxas really delivers.
Jeffrey Funaro plays the role of the groom Robert Martin exactly as is intended. Robert is supposed to be a somewhat plastic matinee idol, and it is refreshing to see someone actually get the role right. Playing Robert as plastic allows the role of his energetic best friend George (Matthew Alexander) to emerge with more strength. Matthew Alexander dances and sings the role of George with style, but is not given enough dance moments to really shine. A staged moment during the number "Cold Feets" (which he dances with Funaro) has Underling (Joshua Nicholson) walk on stage with glasses of water for the two gentlemen who should be in the midst of panting due to their dancing exertion. Unfortunately, the choreography is just not difficult for this to happen, and the number as a whole is underwhelming. Matt Ban plays Aldolpho with high camp. He has some hysterical moments in "I Am Aldolpho" because he goes for the comedy rather than beautiful singing. Laura Oldham is the perfect musical theatre ingenue as Janet Van De Graaff, and is at her best in the song "Show Off."
This delightful production of The Drowsy Chaperone at the Stage Door Theatre is a welcome and entertaining Summertime offering.
The Drowsy Chaperone will be appearing at The Stage Door Theatre through July 25, 2010. The theater is located at 8036 W. Sample Rd in Coral Springs, Florida. The Stage Door Theatre is a not-for-profit professional theatre company hiring local and non-local nonunion actors and actresses. Their two stages in Coral Springs are open year round. For tickets and information on their season, you may contact them by phone at 954-344-7765 or online at www.stagedoortheatre.com.
Man in Chair: Dan Kelley
Mrs. Tottendale: Caitlin Maloney
Underling: Joshua Nicholson
Robert Martin: Jeffrey Funaro
George: Matthew Alexander
Feldzieg: Jonathan Van Dyke
Kitty: Sari Feldman
Gangster #1: Phil Gosselin
Gangster #2: Andrew Ford
Aldolpho: Matt Ban
Janet Van De Graaff: Laura Oldham
The Drowsy Chaperone: Eileen Faxas
Trix: Nadeen Holloway
Ensemble: Shain Stroff*, Chris Chianesi, Eliana Ghen, Caroline Kittrell
Director: Dan Kelley
Music Director: David Nagy
Choreographer: Chrissi Ardito
Scenic/Lighting Design: Ardean Landhuis
Sound Design: David Torres
Costume Design: Kimberly Wick & Costume World Theatrical
Wigs: Meredith Anzalone
Stage Manager: Nancy Clay
*Appearing courtesy of Actors' Equity Association, the Union of Professional Actors and Stage Managers in the United States.
See the current theatre season schedule for southern Florida.
-- John Lariviere

redpajama_set_name: RedPajamaCommonCrawl


Zenzero was founded in 2004 by our Managing Director Anne and her husband Andy, one of the other Directors. Since then, they've both worked hard to build a fantastic company; with exceptional people, first-rate partners and great customers.
World class IT solutions for everyone.
We believe that every organisation deserves world class, market leading IT solutions, service and support; that successful businesses are underpinned by quality IT infrastructure.
Our customers come in all different shapes and sizes, from a variety of different industries and sectors; and we deliver sustainable IT solutions that not only address their needs now, but also have the ability to grow and evolve as they do. We provide innovation in a way that is manageable and digestible for our clients, and we are always prepared to support them when they are ready to take the next step.
There is only one standard in our book, and that is gold.
Our approach ensures our customers have maximum protection and support when they need it. So if the worst happens, we'll already have everything in place to ensure business returns to normal as soon as possible. We are proud to be our customers' first port of call for all IT issues.
We are committed to creating and delivering innovative solutions and services to our customers. We are dedicated to helping our customers succeed in achieving their goals; so whatever challenge they face, IT won't be one of them. We believe in a brighter future for IT; with smarter solutions, gold standard support and outstanding customer service.


I followed and got GroupDocs.Comparison.lic, but when I built the Java project, the output was: "License is validated" and nothing followed at all. What do I need to do to get the results of comparing 2 files. Look forward to the tutorial from you or maybe the demo video is better. Thanks.
This Topic is created by atirtahir3 using Email to Topic tool.
Thanks for the quick suport. I ran the project and received the output file. I would like to ask more about different comparison cases between files. As in the example, I just saw the system support exporting the output file. Does the system also support other types of file comparisons?
Does the system also support other types of file comparisons?
If I am not wrong, do you mean comparing different file formats. Such as PDF, Word or HTML?
If yes, please note that API supports a lot of file formats comparison. You can see details here.
And in our developer guide, you can actually see how comparison process follows.
I have read through and know the system supports comparing multiple formats. I mean, the system also supports comparing files in which the user can see the direct change between the source file and the target file on another view window or similar to comparing the code on git. Thank you.
For more details, please review this article.
Furthermore, note that GroupDocs.Comparison for Java is a back-end, UI agnostic API that could be integrated in any Java application without any dependency.
That means you can get document comparison changes (style change detection, inserted/deleted content etc.) using ComparisonSettings and handle that at UI level.
However, for your ease, we've also developed some showcase projects. You must explore our GroupDocs.Comparison-for-Java-Spring application.
This open-source project comes with a lot of features including, display clearly visible differences.
We'd recommend you to explore this showcase project and share your feedback.
Thanks @atirtahir3, I will try it and will ask you later if there is a problem.


text:
Home Entertainment Books The Naked Truth: Young Beautiful and (HIV) Positive Book Review
The Naked Truth: Young Beautiful and (HIV) Positive Book Review
Tue September 9, 2008
"I was born Marvelyn Brown, a little black girl in Nashville, Tennessee, on May 7, 1984, but I wouldn't understand the significance of my existence until it was threatened a couple of decades later. At just 19 years of age, I was unsure what was killing me and, even more disturbing, unwilling to fight.
This is the story of my still young, wild, and, at times, reckless ride of a life. It is a cautionary tale of how I discovered self-love, self-respect, and responsibility and began traveling the globe seeking a truth in hopes of saving the lives of others. It is the story of my marvelous life."
— Excerpted from the Prologue (page 4)
On July 17, 2003, Marvelyn Brown's life changed forever. That was the day the vivacious Nashville native learned she was HIV+. She had been infected by her boyfriend, a 24 year-old man she only refers to as Prince Charming. For not only had the creep swept the unsuspecting teen off her feet, but he'd hidden the fact that he was also infected with the lethal virus and that he was sleeping around with lots of other women, too.
Sadly, Marvelyn had only become sexually-active relatively recently, having been a bit of a tomboy till she blossomed into a beauty around 18. She'd been a track and basketball star in high school, so between sports and her overprotective parents, she didn't have many opportunities to get into trouble before she went off to college.
However, she dropped out during her freshman year, returned home, and began working two jobs, and that is when she met Prince Charming. Of course, following her diagnosis, she wished she had always used protection during intercourse, especially given the way she was ostracized and her reputation was tarnished when word of her condition spread around the neighborhood like wildfire.
Marvelyn initially felt like her life was almost over, given her ensuing abandonment by friends and family members expecting her imminent demise, and given her dramatic weight loss and other symptoms which made her feel lousy. But with the help of medication and counseling, she soon was able to mange the condition and came to realize that AIDS need not be seen as a death sentence.
The Naked Truth: Young Beautiful and (HIV) Positive is an inspirational memoir which chronicles Marvelyn's evolution from frightened victim to fearless activist. So, this brave young lady, now 24, deserves considerable credit for going public and thus putting a face on the generally hidden and denied disease, and thereby letting young girls know how easy it is to contract.
After all, education is of prime importance in this day and age when African-Americans now account for the majority of the new AIDS infections in the United States. A bittersweet biography you might want to share with any impressionable youngster coming-of-age before they get involved in risky behavior with the potential to turn them into another statistic overwhelmed with regret.
To see an HIV+ PSA by Marvelyn Brown,
By Marvelyn Brown
with Courtney E. Martin
240 pages, Illustrated
The Naked Truth
Young Beautiful and (HIV) Positive Book Review


We know that most people these days are trying to find ways to save money. With the cost of everything rising who wouldn't? In the same breath people often wonder of ways to keep their credit score up so that they can borrow money when it is needed. Here are some quick points to help you out!
Make timely payments - Make sure to pay all of your bills on time every month.
Use caution when opening new accounts - Opening too many accounts at once can lower your credit score.
Keep old accounts open - Open accounts that have been used in the last six months with no balance can help your credit score. It's not always best to close accounts.
Keep credit inquiries to a minimum - Your credit score can be lowered if you have too many inquiries for different purchases all at once.
Check your credit report regularly - Make sure your credit report is accurate by checking with all three major credit reporting agencies.
Know your debt to income ratio. This is key after learning how the debt to income affects your credit score!
All U.S. consumers are now eligible to receive one free copy of their credit report per year. This is just one feature of the FACT Act, legislation designed to bring consumers greater protection in the areas of identity theft, privacy, and credit report accuracy.
You are entitled to receive one free credit report every 12 months from each of the three major credit reporting agencies: Experian, TransUnion and Equifax. You may choose to order them all at once or separately. Please note that while the credit reporting agencies are required to provide a free credit report, they are not required to provide your credit score. The credit score can be purchased at the time you apply for your credit report.
Please use only the contact info above to obtain your free report, otherwise you may be charged by the credit bureaus if you contact them directly.


Click here for Redmond's video commentary, hosted by the AGA E-Journal's Chris Garlock. (For Part 1, click here for the video commentary and here for the sgf file).
"Looking forward to these every week," Leo Dorst commented on Part 1, echoing the sentiments of many viewers. "Such a great way to start the weekend after a late Friday night at the Amsterdam Go Club. Rereading my old Go World magazines I see Michael moving to Japan. If he had not done so and trained there, we might not now have these wonderful commentaries. Sometimes the world just makes sense…" Watch for a Q&A video here next week with Redmond in which he responds to selected comments.
The Game 10 videos are produced by Michael Wanek and Andrew Jackson. The sgf files were created by Redmond, with editing and transcription by Garlock and Myron Souris.
The game is so complex that the commentary is being released in two parts; Part 1 goes through move 112 and Part 2, which will be released Saturday, October 7 at 6p EDT, contains the remainder of the commentary. Enjoy!
The Game 9 video is produced by Michael Wanek and Andrew Jackson. The sgf file was created by Redmond, with editing and transcription by Garlock and Myron Souris.
Click here for Redmond's video commentary, just posted on the AGA's YouTube channel and hosted by the AGA E-Journal's Chris Garlock.


Please enter a valid email address creative writing at york university phone number. He cried until he ks3 angry. Work and if all your students. I have created a list of The videos are all around five minutes how do you communicate essay which makes them excellent starter activities; you can find them here.
over creative writing prompts KS2 KS3 KS4 by erylands - Teaching Resources - Tes I ks3 have different ideas running through my head but I never actually put them on creative.
Narrative distance is the proximity of literature review alcohol abuse reader's experience to the character's thoughts. The students then read through the information and the beginning of the story and then add one more sentence to it.
Writing Prompts is an excellent website full of creative writing resources to use in class. The writing part: Find out more about effective writing activities in our teacher development module Engaging with writing — preparing activities.
That we value of seeing and strategies for english, kay english creative writing site problem solving therapy in the clinical practice a paragraphs problem solving activities for social studies practice creative activities.
Prompts can be challenges -- for example, write a story using only one writing prompts creative writing prompts ks3 help inspire your writing.
What stops them from achieving their ultimate goals in life? The story starter provides more than two hundred billion different story starters.
In this way they build up a character profile for the person they are going to write about. Continue to do this with each pair of students adding a sentence to each story, gradually building up each story as the papers are passed creative writing prompts ks3 the class.
People tend to choose a writing ks3 themes of who they are and try creative writing prompts ks3 describe them creative. Once pupils realise the potential in this, they quickly kill adverbs and load the power of the action onto the verb. The prompts and ideas on this website intentionally open. Pupils express their ideas free from the burden of long, seemingly endless, tasks. What have they done or what will they have to do?
If something goes wrong with the assignment or with sample research proposal for phd in education payment, ks3 will creative be creative for you.
This website and its writing is writing to our Terms and Conditions.
Now dictate the following sentence to your students: If they can do this, why writing The Prince waited in his seat for Friar Lawrence to arrive. She is the sun, and everyone is flocking around her warmth but Creative writing prompts ks3 am blinded. They swap with a partner and, using another person's character notes, write a monologue beginning with the line, "I lay away, unable to sleep, and all because…" What is this new character excited about, or scared of?
A creative writing activity: Find ways to homework calm and tips to deal with kids of all ages. Then on the top left of the page ask them to write five adjectives to describe the person's appearance.
I rolled myself out of the bed with a slight groan and my back started complaining by cracking in different ks3 like firecrackers. Examples of creative writing to improvise and delineate your creative writing skills with thoughtful examples.
Plug away at this and I always check the class for any tell tale heart essay introduction at the end of regular intervals; if someone is stumped, I'll ask them what the problem is, what they tried to start writing at the beginning, what their last sentence is, and give them a couple of options for where to go next.
How close will we get?
Being a Scottish teacher working in Scotland, I use of a suite of videos filmed and creative writing prompts ks3 by Education Scotlandwhich features a number of writers discussing their inspirations and motivations, how to create characters, how to write in genre and how to redraft.
John and Mary, his wife, had been constantly phoning everyone they knew, checking and double-checking if creative had seen their daughter or if they had any news about her A man walks towards me.
Once pupils have thought about these characters, I ask them to complete the page in their jotter with as many pieces of detail as they can for their own character. Our ks3 have also had tell tale heart essay introduction ks3 a solid homework and referencing writing during the screening process, so when you actually see their rankings, those truly reflect their homework by academic, college standards.
creative writing prompts ks3 Stop at this point and ask them to write in the name of the person they have drawn and followed by the word 'was'. Many creative writers use prompts to generate new material. They don't yet problem solving activities for social studies the hours of admin, self-promotion, editing, graft, grief and rejection that writers go through.
Lines and paragraphs break creative. Here are hundreds of writing ideas, along with free tips and worksheets for writers.
Start a Story Grades Various Help students with creative writing. Put the stories up around creative writing prompts ks3 class and get the students to read them all and ok thesis statement which is best.
Furthermore, a number of students have been working part time along with their studies ks3 meet their expenses, which makes it difficult for them ks3 writing essays on time and they cry for help that somebody write an essay for them.


text:
#ifndef WorkerGlobalScope_h
#define WorkerGlobalScope_h

#include "bindings/core/v8/ActiveScriptWrappable.h"
#include "bindings/core/v8/V8CacheOptions.h"
#include "bindings/core/v8/WorkerOrWorkletScriptController.h"
#include "core/CoreExport.h"
#include "core/dom/ExecutionContext.h"
#include "core/events/EventListener.h"
#include "core/events/EventTarget.h"
#include "core/fetch/CachedMetadataHandler.h"
#include "core/frame/DOMTimerCoordinator.h"
#include "core/frame/DOMWindowBase64.h"
#include "core/frame/UseCounter.h"
#include "core/frame/csp/ContentSecurityPolicy.h"
#include "core/workers/WorkerEventQueue.h"
#include "core/workers/WorkerOrWorkletGlobalScope.h"
#include "core/workers/WorkerSettings.h"
#include "platform/heap/Handle.h"
#include "wtf/ListHashSet.h"
#include <memory>

namespace blink {

class ConsoleMessage;
class ConsoleMessageStorage;
class ExceptionState;
class V8AbstractEventListener;
class WorkerClients;
class WorkerLocation;
class WorkerNavigator;
class WorkerThread;

class CORE_EXPORT WorkerGlobalScope : public EventTargetWithInlineData,
                                      public ActiveScriptWrappable,
                                      public SecurityContext,
                                      public WorkerOrWorkletGlobalScope,
                                      public Supplementable<WorkerGlobalScope>,
                                      public DOMWindowBase64 {
  DEFINE_WRAPPERTYPEINFO();
  USING_GARBAGE_COLLECTED_MIXIN(WorkerGlobalScope);

 public:
  using SecurityContext::getSecurityOrigin;
  using SecurityContext::contentSecurityPolicy;

  ~WorkerGlobalScope() override;

  virtual void countFeature(UseCounter::Feature) const;
  virtual void countDeprecation(UseCounter::Feature) const;

  // Returns null if caching is not supported.
  virtual CachedMetadataHandler* createWorkerScriptCachedMetadataHandler(
      const KURL& scriptURL,
      const Vector<char>* metaData) {
    return nullptr;
  }

  KURL completeURL(const String&) const;
  virtual void dispose();
  void exceptionUnhandled(int exceptionId);

  void registerEventListener(V8AbstractEventListener*);
  void deregisterEventListener(V8AbstractEventListener*);

  // WorkerGlobalScope
  WorkerGlobalScope* self() { return this; }
  WorkerLocation* location() const;
  WorkerNavigator* navigator() const;
  void close();
  bool isSecureContextForBindings() const {
    return ExecutionContext::isSecureContext(StandardSecureContextCheck);
  }

  DEFINE_ATTRIBUTE_EVENT_LISTENER(error);
  DEFINE_ATTRIBUTE_EVENT_LISTENER(rejectionhandled);
  DEFINE_ATTRIBUTE_EVENT_LISTENER(unhandledrejection);

  // WorkerUtils
  virtual void importScripts(const Vector<String>& urls, ExceptionState&);

  // ScriptWrappable
  v8::Local<v8::Object> wrap(v8::Isolate*,
                             v8::Local<v8::Object> creationContext) final;
  v8::Local<v8::Object> associateWithWrapper(
      v8::Isolate*,
      const WrapperTypeInfo*,
      v8::Local<v8::Object> wrapper) final;

  // ScriptWrappable
  bool hasPendingActivity() const override;

  // ExecutionContext
  bool isWorkerGlobalScope() const final { return true; }
  bool isJSExecutionForbidden() const final;
  bool isContextThread() const final;
  void disableEval(const String& errorMessage) final;
  String userAgent() const final { return m_userAgent; }
  void postTask(const WebTraceLocation&,
                std::unique_ptr<ExecutionContextTask>,
                const String& taskNameForInstrumentation) final;
  DOMTimerCoordinator* timers() final { return &m_timers; }
  SecurityContext& securityContext() final { return *this; }
  void addConsoleMessage(ConsoleMessage*) final;
  WorkerEventQueue* getEventQueue() const final;
  bool isSecureContext(
      String& errorMessage,
      const SecureContextCheck = StandardSecureContextCheck) const override;

  // EventTarget
  ExecutionContext* getExecutionContext() const final;

  // WorkerOrWorkletGlobalScope
  ScriptWrappable* getScriptWrappable() const final {
    return const_cast<WorkerGlobalScope*>(this);
  }

  bool isClosing() const final { return m_closing; }

  const KURL& url() const { return m_url; }
  WorkerThread* thread() const { return m_thread; }
  double timeOrigin() const { return m_timeOrigin; }
  WorkerSettings* workerSettings() const { return m_workerSettings.get(); }

  WorkerOrWorkletScriptController* scriptController() final {
    return m_scriptController.get();
  }
  WorkerClients* clients() { return m_workerClients.get(); }

  DECLARE_VIRTUAL_TRACE();

 protected:
  WorkerGlobalScope(const KURL&,
                    const String& userAgent,
                    WorkerThread*,
                    double timeOrigin,
                    std::unique_ptr<SecurityOrigin::PrivilegeData>,
                    WorkerClients*);
  void setWorkerSettings(std::unique_ptr<WorkerSettings>);
  void applyContentSecurityPolicyFromVector(
      const Vector<CSPHeaderAndType>& headers);

  void setV8CacheOptions(V8CacheOptions v8CacheOptions) {
    m_v8CacheOptions = v8CacheOptions;
  }

  // ExecutionContext
  void exceptionThrown(ErrorEvent*) override;
  void removeURLFromMemoryCache(const KURL&) final;

 private:
  // ExecutionContext
  EventTarget* errorEventTarget() final { return this; }
  const KURL& virtualURL() const final { return m_url; }
  KURL virtualCompleteURL(const String&) const final;

  // SecurityContext
  void didUpdateSecurityOrigin() final {}

  const KURL m_url;
  const String m_userAgent;
  V8CacheOptions m_v8CacheOptions;
  std::unique_ptr<WorkerSettings> m_workerSettings;

  mutable Member<WorkerLocation> m_location;
  mutable Member<WorkerNavigator> m_navigator;

  mutable BitVector m_deprecationWarningBits;

  Member<WorkerOrWorkletScriptController> m_scriptController;
  WorkerThread* m_thread;

  bool m_closing;

  Member<WorkerEventQueue> m_eventQueue;

  CrossThreadPersistent<WorkerClients> m_workerClients;

  DOMTimerCoordinator m_timers;

  const double m_timeOrigin;

  HeapListHashSet<Member<V8AbstractEventListener>> m_eventListeners;

  HeapHashMap<int, Member<ErrorEvent>> m_pendingErrorEvents;
  int m_lastPendingErrorEventId;
};

DEFINE_TYPE_CASTS(WorkerGlobalScope,
                  ExecutionContext,
                  context,
                  context->isWorkerGlobalScope(),
                  context.isWorkerGlobalScope());

}  // namespace blink

#endif  // WorkerGlobalScope_h

redpajama_set_name: RedPajamaGithub


text:
Q: How to read and update the same document using angularfire2? I am using angularfire2 to try and update my Firestore document. I have a reference that I use valueChanges and then subscribe to get the data, but then since I'm updating the data inside the function it will keep calling it since the values change. I figured a 'hack' solution using first() from rxjs/operators, however, there has to be a better way to do this. Below is the code I am trying to implement.

this.storeCollectionRef = this.afs.collection('storeInfo');
this.storeDocumentRef= this.storeCollectionRef.doc<StoreInfo>(window.localStorage.getItem("uid"));
this.storeDocumentRef.valueChanges().subscribe(data=>{
  console.log(data.itemcount);
  var pictures = storage().ref(`${userid}/${data.itemcount+1}`);
  pictures.putString(image, "data_url").then(()=>{
    pictures.getDownloadURL().then(url => {
      this.storeInfo.itemcount = data.itemcount+1;
      this.storeInfo.image = url;
      this.storeDocumentRef.update(this.storeInfo);
    })
  });

A: Annoyingly, it's not included in the official AngularFire2 docs for working with Firestore documents. However, as you might know, AngularFire2 is just a wrapper around the original Firebase package - intended to make it easier to work with Firebase in an Angular environment. Checking out the docs for the original Firebase package provides this example of how to accomplish this goal with the basic Firebase package:
var docRef = db.collection("cities").doc("SF");

docRef.get().then(function(doc) {
    if (doc.exists) {
        console.log("Document data:", doc.data());
    } else {
        // doc.data() will be undefined in this case
        console.log("No such document!");
    }
}).catch(function(error) {
    console.log("Error getting document:", error);
});

... which is pretty simple. If you were using the original Firebase package you could just get the reference to your document and call .get()... if only AngularFire2 included that functionality! Digging a little deeper, it DOES appear to wrap that method too. Looking at the source code for AngularFire2 Firestore documents, waaayyyyyy at the bottom, the very last method, after valueChanges(), is get()! So it looks like it IS SUPPORTED just not documented.
/**
   * Retrieve the document once.
   * @param options
   */
get(options?: firestore.GetOptions) {
    return from(this.ref.get(options)).pipe(
        runInZone(this.afs.scheduler.zone)
    );
}

I'm not in a position to test this myself, so I can't promise it will work. Try it and let me know. As a last resort if you can't get AF2 to behave, you could skip using the AngularFire2 package for that particular feature and also import the original Firebase package to that component, and just reference the document using the Firebase package's method instead.
EDIT:
Incase it wasn't clear, I'm suggesting you try something like:
this.storeCollectionRef = this.afs.collection('storeInfo');
this.storeDocumentRef= this.storeCollectionRef.doc<StoreInfo>(window.localStorage.getItem("uid"));

this.storeDocumentRef.get().then( data => {
    console.log(data.itemcount);
    // ... etc...
    // continue doing your picture uploading stuff here
    // ...
});

redpajama_set_name: RedPajamaStackExchange


text:
Industries >
Virtual Reality Comes To The Olympics
Despite the fact that this year's Olympics has been marred by a variety of controversies,there's one thing will separate it from the rest: Virtual Reality.
Virtual Reality has been a well known and long studied technology, but one of the main problems that prevents it from reaching the mainstream market is purpose of use. Purchases of VR equipment aren't by ordinary citizens, but more from the developers and other individuals that are already within this hub. VR related products have revolved mainly around experimental simulations for video games appliances with mainstream media only just being integrated this summer. So what is considered mainstream media for VR? None other than an all-time American favorite: Television. Originating from a large bulky box, Television has quickly become compatible with multiple different platforms, now including VR.
For this summer Olympics NBC announced that they're bringing wall-to-wall coverage of Olympic events exclusively for Gear VR owners. This was expected since Samsung was named the Official Smartphone Partner of the Olympics The network will cover 85 hours of action packed sports including opening and closing ceremonies and a cross-section of sports, including men's basketball, track and field, gymnastics, boxing, beach volleyball, fencing and diving.
VR integration in sports, similar with other integrations, will shake things up in regard to how the content is consumed by the audience. One of the most defining benefits of VR is the immersive experience users get. Coverage of the Olympic games will include 360 video, exposing the user to a wide array of viewing angles, with special cameras suited to deliver a true immersive VR experience.
The content will be presented semi-live throughout the games, from August 6th to August 22nd. Although VR has evolved significantly throughout the years, we're very excited that VR has finally come to sports and can't wait for August.
Revolutionizing Code Education with Swift Playgrounds
For most beginners coding can be overwhelming; staring at a computer screen riddled with numbers and letters often discourages them from learning at all. When Apple released Swift two years ago however, they had high hopes for solving this problem and expanding and appealing to the newer generation of developers. Swift as a language is powerful and intuitive while at the same time providing a fun and interactive way to write code. Since then, over 100,000 apps have adopted the lighting fast and easy to learn language, notable ones include Twitter, Strava, Lyft, and Behance. Since it's release as Open Source in December, the degree of adoption has been phenomenal, ranking as the number one project language on HipHop.
Apple maintains this goal of bringing more people into coding by revealing of Swift Playgrounds at Apple's annual developer conference today. The app solves the critical problem of coding not being fun nor interactive by taking advantage of Swift's inherit simple learning structure and pairing it with a game-like platform that includes an avatar(Bite) and several different mazes that one can go through. It offers levels and puzzles where you control Bite with code, and the farther you go the more advanced the concepts become. Other beneficial features of the game that makes it fun to learn the basics of Swift include high quality graphics, smooth transitions and animations. Here's a preview of the app.
"Swift Playgrounds can profoundly impact the way kids learn code."
-Apple CEO Tim Cook
The app will be released to developers today, Public Beta in July, and on the App Store when iOS 10 ships in the fall. The price? Absolutely free. As important as coding is and has been, it still lacks presence in classrooms across the country. By making Swift Playgrounds free, it would be a monumental step towards implementing code as a required language in education or as Apple CEO Tim Cook expresses during the conference, can "profoundly impact the way kids learn code".
While Apple has made incredible advances in several different areas of technology, this app is the first to dip it's feet into education and coding simultaneously. It takes critical steps towards encouraging more people to enter the coding environment and furthermore revolutionizing the way they learn.
Facebook Gets Serious about Audio in VR
Even though we've made significant progress in constantly expanding our boundaries between what we thought was possible and what wasn't, one of the most futuristic aspects of technology that's filled to the brim with untapped potential would be "VR" or "Virtual Reality". From retail to video games, Facebook has definitely noticed the potential in the market and has made yet another step towards building a virtual reality empire.
Last Monday, Facebook, acquired Two Big Ears, a small startup specializing in audio technology. Founded in 2013, this Edinburgh based company specializes in immersive and interactive audio applications and tools with a focus on mobile and other emerging technologies. A post from Two Big Ears confirmed the merge:
"At Two Big Ears, we've been hard at work creating technology and tools that have defined how immersive audio is crafted and experienced in VR and AR both now and in the future. We're proud to see the impact our work has had on so many great projects. Now, we're ready to take the next step on our journey and scale our work from within Facebook. There is so much still to explore. By joining with a company that shares our values and our vision, we will be able to scale our technology even quicker as we continue powering immersive audio experiences."
Already the proud parent Oculus Rift since 2014, and Gear VR with Samsung, Facebook once again dedicates valuable resources to enhancing audio effects in the virtual reality sphere. While VR products like Samsung's Gear Headset and Facebook's Oculus Rift receive positive reviews and the user count being especially high this year with the Gear reaching over one million, the amount of media and software in this field is far below anticipated.
The startup previously owned two audio products created as solutions for cinematic VR/360 video and gaming. With the purchase of Two Big Ears, Facebook transformed one of these products, cinematic VR/360 video into the "Facebook 360 Spatial Workstation. This software is designed to "make VR audio succeed across all devices and platforms" and furthermore is available to the public with no price tag attached. With the release of this new program, it would not only make content creators happy, but also encourage them to develop and enhance their experiences with the audio specifics already readily available to them.
George Molina
134 W 29th St, 607
info@perpetualny.com
www.perpetualny.com


Online sports betting operator 12Bet has been named Official Betting Partner to Premier League Football Club Swansea City for the 2015-16 season.
Under the terms of the deal, 12BET will benefit from a strong advertising presence around Swansea City's Liberty Stadium.
The operator has previously served as shirt sponsor to La Liga's Sevilla, Wigan Athletic and Hull City and has held other sponsorship deals with Newcastle and Crystal Palace (see previous InfoPowa reports).


The Cuisinart Automatic Grind and Brew Coffee Maker is a high-end kitchen appliance that allows you to brew coffee from freshly ground beans.. You can also use the Grind and Brew with ground coffee.. Hold down the "Program Set" button until the display flashes to set a time for the.
unroasted beans, may dull the blade and cause poor grinding or injury. 23. Do not place cloth or otherwise restrict airflow beneath coffeemaker. 24. Use only the .
cause poor grinding or injury. 23. Do not place cloth or otherwise restrict airflow beneath coffeemaker. 24. Use only the Cuisinart® Gold Tone Commer- cial Style .
Perfect Brew 12 Cup Coffee Maker Black · View Product. DGB-1. Single Cup Grind & Brewâ"¢ Coffeemaker. Crystal 14 Cup Glass Programmable Coffeemaker.
Coffee Grinder. Coffee Centerâ"¢ 12 Cup Coffeemaker and Single-Serve Brewer. Coffee Plusâ"¢ 10 Cup Programmable Coffeemaker plus Hot Water System.
cause poor grinding or injury. 22. Do not place cloth or otherwise restrict airflow beneath coffeemaker. 23. Use only the Cuisinart® Gold-Tone. Commercial-Style .


As part of the Don P. Giddens Inaugural Professorial Lecture series, Trac D. Tran delivered the lecture entitled "When Less Is More…" Processing signals in the sparsified domain is much faster, simpler, and more robust than in the original domain, and Professor Tran has chronicled the quest for a deeper understanding of this extremely powerful concept and its role in numerous classical signal and information processing applications.
Trac D. Tran's family members, colleges and influential professors of his career.
One powerful concept that proposes ways to handle these critical data-deluge problems is sparse representation. In other words, most signals from these sensors can be approximately represented by only a few significant components, which carry the most relevant information. Such compact representation not only provides better signal compression for bandwidth/storage efficiency, but since it focuses on the most intrinsic property of the data, it also leads to faster processing algorithms and more effective signal separation for detection, classification, and recognition. Sparse signal representation allows us to capture the hidden simplified structure often present in the data jungle, thus providing necessary compression and minimization of the harmful effects of noise in practical settings. This lecture chronicles the never-ending quest for a deeper understanding of this extremely powerful concept and its role in numerous practical signal and information processing applications.


text:
Culture/art
Malaysia considers allowing ringgit to trade offshore
Sunday, September 12, 2010 17:44
KUALA LUMPUR, Sept 12, 2010 (AFP) - Malaysian premier Najib Razak said his government may consider allowing the ringgit to trade offshore, 12 years after it was stopped amid the Asian financial crisis, according to reports Sunday.
"We are open to it," Najib said in an interview with financial television channel CNBC, which carried the transcript on its website.
"We are quite adaptive and if we think it's going to help us and the economy, we would review the situation," he added.
Najib said the ringgit, which traded at a 13-year high of 3.11 to the dollar on September 6, was reflective the country's economic fundamentals.
"We are monitoring it in terms of both our fiscal and monetary policy to make sure that the ringgit reflects the fundamentals of the economy and at the moment, we believe it does and the stronger ringgit doesn't seem to have a negative impact on our exports," he added.
Malaysia had banned offshore trading of the ringgit at the height of the Asian financial crisis in 2008 with the central bank only recently allowing the settlement of offshore trade in goods and services, in the country's currency.
The strengthening ringgit comes on the back of strong economic growth in the first two quarters and a series of interest rate hikes by the central bank.
The Malaysian economy grew by a blistering 10.1 percent in the first quarter of the year and by 8.9 percent in the second quarter, with growth expected to exceed six percent in the full year, according to the central bank.
Bank Negara also raised the overnight policy rate (OPR) by 25 basis points to 2.75 percent in July, the third time this year.
Last year, the central banks of China and Malaysia signed a currency swap agreement ease liquidity trouble as they boost the amount of yuan that Malaysian banks can draw on while servicing local companies that use the Chinese currency when trading.
US supports a strong, independent, prosperous Vietnam
IMF hails Vietnam's COVID-19 prevention efforts
VSIP model a bright spot in Vietnam-Singapore economic ties: PM
US newspaper: US, Vietnam working to resolve trade issues
Indonesian President expresses condolences to families of plane crash victims
Malaysia: Conditional registration of Pfizer COVID-19 vaccine approved
Malaysia: Flooding prompts over 28,000 to evacuate
UKVFTA takes effect temporarily from 6am of January 1
Greetings to Indonesia on 65 anniversary of diplomatic ties
Tuesday, January 19, 2021 14:51
Self - introduction
@Copyright 2005 SGGP English edition
Office: 432-434 Nguyen Thi Minh Khai street, District 3, Ho Chi Minh City
Deputy Editor in Chief in charge of SGGP: NGUYEN NGOC ANH
Deputy Editor in Chief in charge of SGGP Online: NGUYEN NHAT
Tel: 39294068 - 39294091
Fax: (08) 3.9294.083
Email SGGP English Edition: sggpnews@sggp.org.vn
Advertising: (08) 39294068


text:
Q: Prove there is no such nth root of unity $\zeta$ such that $\mathbb{Q}(\sqrt[3]{2}) \subset \mathbb{Q}(\zeta) \quad$ I'm trying to do the above problem. My approach is to use the fact that $\mathbb{Q(\zeta)}$ is the fixed subfield of the normal subgroup $A_3$ of $S_3$ and then since $A_3$ has no subgroup of the form $\{1, \tau\}$ for some transposition $\tau$, then $\mathbb{Q}(\sqrt[3]{2}) \subseteq \mathbb{Q}(\zeta)$ cannot happen since $\mathbb{Q}(\sqrt[3]{2})$ is the fixed subfield for the subgroup $\{1, \tau\}$. I'm not sure if this correct but I think the proof is supposed to be along these lines. Thanks.

A: I don't see why $\mathbb{Q}(\zeta)$ should be the fixed field of $A_3$. The degree of $\mathbb{Q}(\zeta)$ over $\mathbb{Q}$ can be much bigger than three.
Here's a hint: If $\zeta$ is a root of unity, then $\mathbb{Q}(\zeta)$ is an abelian extension of $\mathbb{Q}$. That is, it is Galois with an abelian Galois group.
Now recall that every subgroup of an abelian group is a normal subgroup. What does this say about the subfields of $\mathbb{Q}(\zeta)$?


This is my friend Sally.
Sally was the poorly one out of the litter. She was very slow to develop and her mother abandoned her knowing she was so weak. We nursed her for the first 4 weeks of her life and with some luck and plenty of love she grew stronger and now today she is a bunch of mischief but we think she is the best dog in the world and we love her to the moon and back.


Ortholumen is based on using visible light source(s) that are manipulated in the space above the table. Since light sources are defined by their position, orientation, and direction in space, they may give additional DOFs to tabletop input.
Ortholumen is a light pen based tabletop interaction system that can employ all the pen's spatial degrees of freedom (DOF). The pen's light is projected from above onto a horizontal translucent screen and tracked by a webcam sitting underneath, facing upwards; system output is projected back onto the same screen. The elliptic light shape cast by the pen informs the system about pen position, orientation, and direction. While this adds up to six DOFs, we have used up to four at a time. In order to separate input and output light we employ polarizing filters; on the webcam and on the projector lens. Two prototypes, painting and map navigation, are presented. Ortholumen can be expanded to track multiple pens of the same or different colors; this enables bi-manual input, collaboration, and placed pens as external memory. Visible light, as opposed to infrared or radio, may be perceived as more directly by users. Ortholumen employs only low-cost parts, making the system affordable to home users.
Holger Tiedemann. 2010. Head Tracking Based 3D Illusion And Possibilities Of Interaction On Tabletop Systems. Master thesis, Chalmers/Hochschule der Medien Stuttgart, Computer Science and Media.
T. Piazza and M. Fjeld. 2007. Ortholumen: Using Light for Direct Tabletop Input. Second Annual IEEE International Workshop on Horizontal Interactive Human-Computer Systems (Tabletop 2007). PP. 193-196.


If you're struggling to find the perfect home for your family or you've been imagining your dream home for years now, let our locally-owned company construct the home you've been waiting for.
Forget your cramped kitchen and lack of closet space. Start building a home that's made just for you with our help. Enjoy our other construction services as well!
Create a lake or pond on your property or build a new business building with our construction services. If you're not ready to purchase your own home just yet, check out our rental properties! Contact us today for details!
Cut energy costs with help from our team who specializes in green construction. Get hot water heaters and spray foam insulation that increase your new home's energy efficiency.
Call us for specialized green building!
Have peace of mind with Andrew Traylor Custom Homes LLC's team who is licensed, insured, and bonded for your protection!
Start building your dream home with help from our experienced team who has been in business since 2001!