text:
\section{Introduction}
Let $\Omega\subset \mathbb{R}^n$ be a bounded domain with Lipschitz boundary. To a positive real-valued function $\mbox{\fontfamily{pzc}\selectfont g}$ on $\Om$ with $0<c<\mbox{\fontfamily{pzc}\selectfont g}<c^{-1}$ we associate an elliptic operator $L_\mbox{\fontfamily{pzc}\selectfont g}$ in divergence form:
\[L_\gamma u := \div(\gamma \nabla u).\]
Given $f\in H^{1/2}(\partial \Omega)$, there exists a unique solution $u_f$ to the Dirichlet problem
\begin{align*}
L_\gamma u_f & = 0 \quad \text{in $\Omega$}\\
\eval{u_f}_{\partial\Omega} & = f,
\end{align*}
and  we define the Dirichlet-to-Neumann map $\Lambda_\gamma: H^{1/2}(\partial \Om) \to H^{-1/2}(\partial \Om)$ formally by
\[\Lambda_\gamma(f) := \eval{\gamma \frac{\partial u_f}{\partial \nu}}_{\partial \Omega},\]
where $\partial/\partial_\nu$ is the outward normal derivative at the boundary. By identifying $H^{1/2}(\partial\Om)$ with the quotient $H^1(\Om)/H^1_0(\Om)$, we can interpret this definition in a weak sense as follows: If $v \in H^1(\Om)$ satisfies $\eval{v}_{\partial \Om} = g$, then
\[\j{\Ld_\mbox{\fontfamily{pzc}\selectfont g}(f), g} := \int_\Om \mbox{\fontfamily{pzc}\selectfont g}\na u_f\cd \bar{\na v}\,dx,\]
where the notation $\j{\cd,\cd}$ indicates the pairing between $H^{-1/2}(\partial \Om)$ and $H^{1/2}(\partial \Om)$. Our main theorem is that the map $\mbox{\fontfamily{pzc}\selectfont g} \mapsto \Ld_\mbox{\fontfamily{pzc}\selectfont g}$ is injective for certain $\mbox{\fontfamily{pzc}\selectfont g}$:
\begin{theorem}
\label{maintheorem}
The map $\mbox{\fontfamily{pzc}\selectfont g} \mapsto \Ld_\mbox{\fontfamily{pzc}\selectfont g}$ is injective for $\mbox{\fontfamily{pzc}\selectfont g}\in W^{s,p}(\Om)$, where
\begin{align*}
(s,p)&=\begin{cases}
(1,n) & n=3,4\\
(1 + (1-\th)(\ff 1 2 - \ff 2 n), \ff{n}{1-\th}) & n=5,6
\end{cases}
\end{align*}
and $\th \in [0,1)$.
\end{theorem}

This problem was introduced by Calder\'on, who proved uniqueness in~\cite{Calderon1980} for the linearized problem. The basic approach to this problem in this paper is the method introduced by Sylvester and Uhlmann in~\cite{Sylvester1987}, where they proved uniqueness for $n\geq 3$ and $\mbox{\fontfamily{pzc}\selectfont g} \in C^2$ based on ideas in~\cite{Calderon1980}. It is of interest to determine how much this regularity condition can be relaxed. Uniqueness is known to fail (at least in the anistropic problem) for conductivities which are sufficiently singular, as was shown in~\cite{greenleaf2003,Kohn2008}.

For $n\geq 3$, the regularity assumption in~\cite{Sylvester1987} was relaxed to $\mbox{\fontfamily{pzc}\selectfont g} \in C^{3/2+}$ by Brown~\cite{Brown1996}, to $C^{3/2}$ by~\cite{Paivarinta2003}, to $W^{3/2,2n+}$ in~\cite{Brown2003}, and to $C^1$ conductivities or Lipschitz conductivities close to the identity in~\cite{Haberman2013}. Recently, the smallness condition for Lipschitz conductivities was removed in~\cite{Caro2014}. In~\cite{Nguyen2014}, uniqueness was shown in three dimensions for conductivities in $W^{3/2+,2}$.

In two dimensions, the low-regularity theory is fairly well-understood. There are essentially sharp results, even for anisotropic conductivities. In particular, uniqueness holds for $\mbox{\fontfamily{pzc}\selectfont g}$ in $L^\infty$, which is is invariant under the scaling associated to the Dirichlet problem. The methods in the plane are somewhat different, and we refer the reader to~\cite{Astala2011} and references therein.

There are some reasons to doubt that uniqueness holds in higher dimensions for conductivities with less than one derivative. Calder\'on's problem seems to be closely related to unique continuation; in particular, most of the progress in both of these problems involves Carleman estimates. Unique continuation in the plane holds for elliptic operators in divergence form when the coefficients are merely bounded~\cite{Alessandrini1992}; in higher dimensions, however, this is only known for Lipschitz coefficients~\cite{Aronszajn1962}. Furthermore, there are counterexamples to unique continuation for elliptic equations where the coefficients are $C^\al$ with any $\al < 1$~\cite{Plis1963, Miller1973, Mandache1998}.

The conductivity equation $\div (\mbox{\fontfamily{pzc}\selectfont g} \na u) =0$ is equivalent to
\[(\lp + A\cd \na) u = 0,\]
where $A = \na \log \mbox{\fontfamily{pzc}\selectfont g}$. Unique continuation holds for this equation as long as $A \in L^n$~\cite{Wolff1992,Koch2001}. Brown~\cite{Brown2003} conjectured uniqueness in the inverse conductivity problem for $\mbox{\fontfamily{pzc}\selectfont g} \in W^{1,n}$. We verify that this conjecture holds in dimensions three and four.

One can also study the closely-related problem of determining a Schr\"odinger potential $q$ from the Cauchy data associated with the operator $-\lp + q$. In this setting Lavine and Nachman used the $L^p$ Carleman estimates of~\cite{Kenig1987} to show that the Cauchy data determine $q\in L^{n/2}$ (see also~\cite{Chanillo1990,DosSantosFerreira2009,Nguyen2014} for similar results). These $L^p$ Carleman estimates are the starting point for our analysis.

It was shown in~\cite{Sylvester1987} that the inverse conductivity problem reduces to the inverse problem for $-\lp + q$, where $q = \mbox{\fontfamily{pzc}\selectfont g}^{-1/2} \lp \mbox{\fontfamily{pzc}\selectfont g}^{1/2}$. One step in this reduction is to show that the map $\Ld_\mbox{\fontfamily{pzc}\selectfont g}$ determines $\mbox{\fontfamily{pzc}\selectfont g}$ and its normal derivatives at the boundary. In~\cite{Kohn1984}, Kohn and Vogelius established that for smooth conductivities, the map $\gamma \to \Lambda_\gamma$ determines the values of $\gamma$ and all of its derivatives on $\partial \Omega$. This boundary determination result holds in much greater generality~\cite{Alessandrini1990,Sylvester1988}. In particular, Brown in~\cite{Brown2013} showed that the boundary values of a $W^{1,1}$ conductivity are determined by the Dirichlet-to-Neumann map. This improvement will be a crucial ingredient in this paper.

The key idea in~\cite{Sylvester1987} is that if $\mbox{\fontfamily{pzc}\selectfont g}_i$ are such that $\Ld_{\mbox{\fontfamily{pzc}\selectfont g}_1}=\Ld_{\mbox{\fontfamily{pzc}\selectfont g}_2}$, then
\[\int_{\Om} (q_1-q_2)\,u_1\,u_2\,dx = 0,\]
where the $u_i$ are arbitrary solutions to the Schr\"odinger equation $(-\lp + q)u_i = 0$ in $\Om$. It follows that one way to show that the potentials $q_1$ and $q_2$ coincide is to produce enough solutions to the corresponding Schr\"odinger equations that their products are dense in some sense. This idea goes back to the original paper of Calder\'on~\cite{Calderon1980}. In~\cite{Sylvester1987}, Sylvester and Uhlmann proved a uniqueness result for $C^2$ conductivities by constructing complex geometrical optics (CGO) solutions of the form $u_i = e^{x\cdot \zeta_i}(1 + \psi_i)$. Here the $\zeta_i\in \mathbb{C}^n$ are chosen so that $\zeta_i \cdot \zeta_i = 0$, so that $e^{x\cdot \zeta_i}$ is harmonic, and $e^{x\cdot \zeta_1} e^{x\cdot \zeta_2} = e^{ix\cdot k}$ for some fixed frequency $k\in \mathbb{R}^n$. In three or more dimensions, these conditions allow for an infinite family of pairs $\zeta_1,\zeta_2$ with $\abs{\zeta_i} \to \infty$. The remainders $\psi_i$ decay to zero in a suitable sense sense as $\abs{\zeta_i} \to \infty$, so that the product $u_1 u_2$ converges to $e^{ix\cdot k}$. Since $k$ is arbitrary, uniqueness follows from Fourier inversion.

To construct these CGO solutions, fix $\zeta \in \mathbb{C}^n$ such that $\zeta \cdot \zeta = 0$, and note that $e^{-x\cdot \zeta} \Delta( e^{x\cdot \zeta}\psi) = (\Delta + 2\zeta \cdot \nabla) \psi$. Then $u = e^{x\cdot \zeta}(1+\psi)$ solves $\Delta u = qu$ if
\begin{equation}
\label{cgoequation}
\Delta_\zeta \psi := \Delta \psi + 2 \zeta \cdot \nabla \psi = q (1+\psi).
\end{equation}
Let $m_q$ be the map sending $\psi$ to $q\psi$. We will treat this equation perturbatively, by viewing $\Delta_\zeta - m_q$ as a perturbation of $\Delta_\zeta$. The operator $\Delta_\zeta$ has a right inverse defined by
\[\widehat{\Delta_\zeta^{-1} f}(\xi) = p_\zeta(\xi)^{-1} \hat f(\xi),\]
where
\[p_\zeta(\xi) = -\abs{\xi}^2 + 2i\zeta \cdot \xi.\]
We take  $\z$ of the form $\tau (e_1 - i e_2)$, where $e_1,e_2$ are orthogonal unit vectors. Then $\lp_\z$ is characteristic on a codimension-2 sphere $\Si_\z$, with
\[\Si_\z = \{\xi: \xi\cd e_1 = 0, \a{\xi-\tau e_2}=\tau\}.\]

To construct a solution to~\eqref{cgoequation}, we show that $m_q$ is a perturbation of $\lp_\z$ with respect to an appropriate norm.

It was observed in~\cite{Paivarinta2003} that it is possible to construct CGO solutions for $C^1$ conductivities using Picard iteration in Sobolev spaces. However, because the inhomogeneity in~\eqref{cgoequation} is not bounded in the correct space, one cannot control these solutions.

A simplified explanation of the problem is as follows: in problems involving Carleman estimates, it is most natural to work with Sobolev spaces depending on a large parameter $\tau$ (or a small parameter in the semiclassical notation). In our problem $\tau$ is proportional to $\a{\z}$. The quantity $\tau$ is thought of as equivalent to a derivative $\na$. In view of this correspondence, we define the Sobolev space $H^s_\tau(\mathbb{R}^n)$ by
\[\n{u}_{H^s_\tau} := \n{(-\lp + \tau^2)^{s/2} u}_{L^2}.\]
For the operator $\lp_\zeta^{-1}$ the following Carleman estimate holds~\cite{Salo2009}:
\[\n{u}_{H^{1}_\tau} \ll \tau \n{\lp_\z u}_{H^{-1}_\tau},\]
where $u$ is supported in some fixed compact set. This means that once we account for the physical space localization in the problem, the operator $\lp_\z^{-1}$ (heuristically speaking) maps $H^{-1}_\tau$ to $H^{1}_\tau$ with constant $\tau$. On the other hand, we have $q = \lp \f 1 2 \log \mbox{\fontfamily{pzc}\selectfont g} + l.o.t.$
\begin{align}
\a{\j{m_q u, v}_{L^2}} &\ll \n{\na (\log \mbox{\fontfamily{pzc}\selectfont g}) \na(u \bar v)}_{L^1}+l.o.t. \\
&\ll \n{\mbox{\fontfamily{pzc}\selectfont g}}_{W^{1,\infty}} (\n{\na u}_{L^2} \n{v}_{L^2} + \n{u}_{L^2} \n{\na v}_{L^2}) + \dotsb  \\
&\ll \tau^{-1} \n{\mbox{\fontfamily{pzc}\selectfont g}}_{W^{1,\infty}} \n{u}_{H^1_\tau}\n{v}_{H^1_\tau}.
\label{sobolev}
\end{align}
By duality, this implies that $m_q$ maps $H^{1}_\tau$ back to $H^{-1}_\tau$ with constant $\tau^{-1}$. This means that the composition $m_q \lp_\zeta^{-1}$ is bounded, and there is some hope of construction CGO solutions perturbatively (see~\cite{Krupchyk2014} for this type of analysis). We are trying to solve $(\lp_\zeta - m_q) u = q$, and we certainly have $q \in H^{-1}_\tau$. However, because we lost a factor of $\tau$ in the Carleman estimate, our solution only satisfies an estimate of the form $\n{\psi}_{H^1_\tau} \leq \tau \n{q}_{H^{-1}_\tau}$. This is bad, because $\psi$ is supposed to be small in $H^1_\tau$.

In~\cite{Nguyen2014}, this problem with Sobolev spaces is circumvented by showing that the first iterate in the solution procedure is bounded on average. This avoids the use of specialized spaces at the expense of requiring slightly more differentiability.

In~\cite{Haberman2013}, Tataru and the author dealt with this problem using specialised function spaces. These are inspired by the $X^{s,b}$ spaces of Bourgain~\cite{Bourgain1993}, and were used in the context of Carleman estimates by~\cite{Tataru1996a}. Define the space $\dot X^b_\zeta$ by the norm
\[\norm{u}_{\dot X^b_\zeta} = \norm{\abs{p_\zeta(\xi)}^{b} \hat u(\xi)}_{L^2},\]
where $p_\zeta(\xi) = -\abs{\xi}^2 + 2i\zeta \cdot \xi$ is the symbol of $\Delta_\zeta$. In this paper, we take $b=\pm 1/2$. It is easy to see that $\norm{\Delta^{-1}_\zeta}_{\dot X_\zeta^{-1/2} \to \dot X_\zeta^{1/2}} =1$. We will also make use of the inhomogeneous spaces $X_\zeta^{b}$ with norm
\[\norm{u}_{X_\zeta^{b}} = \norm{(\abs{\zeta} + \abs{p_\zeta(\xi)})^{b} \hat u(\xi)}_{L^2}.\]
The map $m_q$ satisfies
\begin{equation}
\label{lipschitzbound}
\n{m_q}_{\dot X^{1/2}_\z \to \dot X^{-1/2}_\z} \ll \n{\mbox{\fontfamily{pzc}\selectfont g}}_{\Lip},
\end{equation}
and we may solve~\eqref{cgoequation} perturbatively as before. This bound follows from~\eqref{sobolev}, the easy estimate
\begin{align}
\label{easyenergy}
\n{u}_{H^1_\tau} &\ll \tau^{1/2} \n{u}_{X^{1/2}_\z},
\end{align}
and the fact that for localized $u$, the $X^{1/2}_\z$ and $\dot X^{1/2}_\z$ norms are equivalent.

Solving in this way gives a CGO solution $\psi$ with $\n{\psi}_{\dot X^{1/2}_\z} \ll \n{q}_{\dot X^{-1/2}_\z}$. Unfortunately, the best bound on $\n{q}_{\dot X^{-1/2}_\z}$ is $\tau^{1/2} \n{\mbox{\fontfamily{pzc}\selectfont g}}_{H^1}$. This means that the $\dot X^{1/2}_\z$ norm of CGO solutions might grow like $\tau^{1/2}$ as $\tau \to \infty$.

By an averaging argument, however, $\n{q}_{\dot X_\z^{-1/2}}$ is bounded for a large set of $\z$ (which may depend on $q$) as long as $\mbox{\fontfamily{pzc}\selectfont g} \in H^1$. This is because the $\dot X^{-1/2}$ norm is only large when $\hat q$ concentrates near $\Si_\z$. As we vary $\z$, the characteristic set $\Si_\z$ varies through a family of growing codimension 2 spheres, and $\hat q$ cannot concentrate near all of them. In particular, the estimate $\n{q}_{\dot X^{-1/2}_\z} \ll \n{\mbox{\fontfamily{pzc}\selectfont g}}_{H^1}$ holds on average. Once this is established, uniqueness for $\mbox{\fontfamily{pzc}\selectfont g} \in C^1$ follows from the standard arguments.

If $\na \mbox{\fontfamily{pzc}\selectfont g}$ is unbounded, then we need to replace the $H^1_\tau$ norm on the left hand side of~\eqref{easyenergy} with an $L^p$ norm, where $p > 2$. We can obtain such an estimate using the methods of~\cite{Kenig1987}, which essentially give~\footnote{The author would like to thank Russell Brown for pointing this out.}
\begin{equation}
\label{halfkenig}
\n{u}_{2n/(n-2)} \ll \n{u}_{\dot X_\z^{1/2}}.
\end{equation}
This puts $u$ in a better $L^p$ space, but at the cost of a factor of $\tau$. Since there is no such room in~\eqref{lipschitzbound}, it seems that such a bound does not hold for $\mbox{\fontfamily{pzc}\selectfont g} \in W^{1,p}$ if $p < \infty$.

To do better, we need a refined version of~\eqref{halfkenig}. If $\hat u_\mu$ is supported in the region $\{\xi: d(\xi,\Si_\z) \sim \mu\}$, then we can replace~\eqref{halfkenig} by
\begin{equation}
\label{refinedhalfkenig}
\n{u_\mu}_{2n/(n-2)} \ll (\mu/\tau)^{1/n} \n{u_\mu}_{\dot X^{1/2}_\z}.
\end{equation}
and~\eqref{easyenergy} by
\begin{equation}
\label{refinedeasyenergy}
\n{u_\mu}_2 \ll (\mu\tau)^{-1/2} \n{u_\mu}_{X_\z^{1/2}}.
\end{equation}
Assume we are given $v_\nu$ satisfying a similar condition, with $\mu \leq \nu$. Define $f = \na \log \mbox{\fontfamily{pzc}\selectfont g}$. Then
\begin{align*}
\a{\int (\na f)  u_\mu \bar v_\nu}& \ll  \n{\na f}_n (\mu/\tau)^{1/n} (\nu \tau)^{-1/2} \n{u_\mu}_{X^{1/2}_\z} \n{v_\nu}_{X^{1/2}_\z}.
\end{align*}
Now we exploit the fact that the Fourier transform of $u_\mu v_\nu$ is supported in $\{\xi: \a{\xi} \ll \tau, \a{\xi\cd e_1} \ll \nu\}$. By orthogonality, we may restrict $f$ to this region, so that the above becomes
\[\a{\int( \na f)  u_\mu \bar v_\nu} \ll \n{D^{1/2-1/n} D_1^{1/n-1/2} f}_n \n{u_\mu}_{X^{1/2}_\z} \n{v_\nu}_{X_\z^{1/2}}.\]
where $D$ and $D_1$ are operators with symbols $\a{\xi}$ and $\a{\xi\cd e_1}$, respectively. An argument along these lines gives an estimate of the form
\[\n{m_{\na f}}_{X_\z^{1/2} \to X_\z^{-1/2}} \ll \n{D^{1/2-1/n} D_1^{1/n-1/2} f}_n.\]
Although we have lost $1/2-1/n$ derivatives in this estimate, this is counterbalanced by a gain of $1/2-1/n$ derivatives in the $e_1$ direction. This gain is useless if the Fourier support of $f$ concentrates near the plane perpendicular to $e_1$. However, we expect that this behavior does not occur on average, and we can take advantage of this by exploiting our freedom in choosing $\z$.

It is easiest to average over all choices of $e_1\in S^{n-1}$. In $L^2$ we have
\[\int_{e_1 \in S^{n-1}}\n{D^\al D_1^{-\al} f}_2^2 \,d\si(e_1) \ll \n{f}_2.\]
for $\al < 1/2$. Heuristically, we can interpolate this with the trivial observation that $\sup_{e_1 \in S^{n-1}} \n{f}_\infty \ll \n{f}_\infty$ to obtain
\[\sps{\int_{e_1 \in S^{n-1}} \n{D^\b D_1^{-\b} f}_p^p \,d\si(e_1)}^{1/p} \ll \n{f}_p,\]
when $\b < 1/p$. In three dimensions, we have $1/2-1/3 < 1/3$, and we find that $\n{m_{\na f}}_{X^{1/2}_\z\to X^{-1/2}_\z}$ is bounded on average. In four dimensions, we have $1/2-1/4=1/4$. This causes a logarithmic divergence, which turns out to be harmless. For $n\geq 5$, however, we do not have a way to avoid losing derivatives.

The averaging argument in this paper is somewhat different from the argument in~\cite{Haberman2013}. There, $k$ was taken to be fixed, and $\hat q(k)$ was determined by testing against CGO solutions with $\z \sim \tau(\eta_1-i\eta_2)$, where $\eta_1$ and $\eta_2$ are perpendicular to the frequency $k$. This approach does not give control of $D_1^\b D_1^{-\b} f$, since averaging only over $\eta_1$ perpendicular to $k$ is useless if $\hat f$ is concentrated along the $k$ direction.

Instead of fixing $k$, we vary the triple $\{k, \eta_1, \eta_2\}$ over an small open set of triples of orthonormal vectors. This set is essentially parameterized by $\{U e_1, Ue_2, Ue_3\}$, where the $e_i$ denote fixed orthonormal vectors, and $U$ is an orthogonal transformation. The idea is then to average the relevant quantities (which depend on $U$) with respect to the Haar measure on $O(n)$. Using all of the degrees of freedom in this way allows for an improvement in the estimate for $m_q$ and clarifies the estimate for $\n{q}_{X^{-1/2}_\z}$. This idea comes from~\cite{Nguyen2014}, where uniqueness is established in three dimensions for conductivities in $W^{3/2+,2}$. Remarkably, they showed that under this assumption the boundedness of $m_q$ on average can be proven without taking advantage of the curvature of $\Si_\z$. Using our framework, this corresponds (roughly speaking) to applying Bernstein's inequality to $u_\mu$ in the $e_1$ direction and Sobolev embedding to $f$ in the other directions, to obtain
\begin{align*}
\a{\int (\na f) u_\mu \bar v_\nu} & \ll \n{\na f}_{L^2_{e_1} L^\infty_{e_2,e_3}} \n{u_\mu}_{L^\infty_{e_1} L^2_{e_2,e_3}} \n{v_\nu}_{L^2}\\
&\ll \n{\j D^{2+} f}_2\, \mu^{1/2}\n{u_\mu}_2 \n{v_\nu}_2\\
&\ll \nu^{-1/2} \tau^{-1} \n{D^{2+} f}_2 \n{u_\mu}_{X^{1/2}_\z} \n{u_\nu}_{X^{1/2}_\z}.
\end{align*}
Taking $f$ supported in $\{\a{\xi} \ll \tau, \a{\xi_1} \ll \nu\}$, we obtain
\[\a{\int( \na f ) u_\mu \bar v_\nu} \ll \n{D^{1/2}D_1^{-1/2} D^{1/2+} f}_2 \n{u_\mu}_{X^{1/2}_\z} \n{v_\nu}_{X^{1/2}_\z}, \]
and we can estimate $\n{D^{1/2}D_1^{-1/2} D^{1/2+} f}_2$ on average as before.

When $n \geq 7$, the situation is essentially the same, but there is a new technical difficulty. Since our methods are global in space, we need to extend the conductivities $\mbox{\fontfamily{pzc}\selectfont g}_i \in W^{s,p}(\Om)$ to some $\mbox{\fontfamily{pzc}\selectfont g}_i \in W^{s,p}(\mathbb{R}^n)$ which agree outside of $\Om$. When $s\leq 1+ 1/p$, we can do this as long as $\mbox{\fontfamily{pzc}\selectfont g}_1=\mbox{\fontfamily{pzc}\selectfont g}_2$ on the boundary. However, when $s > 1+1/p$, we also need $\partial_\nu \mbox{\fontfamily{pzc}\selectfont g}_1 = \partial_\nu \mbox{\fontfamily{pzc}\selectfont g}_2$ on the boundary, and there does not seem to be such a boundary identification result in the literature.

It is possible that one can relax the uniform ellipticity condition on $\mbox{\fontfamily{pzc}\selectfont g}$. The natural condition to impose is then $\na \log \mbox{\fontfamily{pzc}\selectfont g} \in L^n$, in which case $\log \mbox{\fontfamily{pzc}\selectfont g}$ is only in BMO, which would correspond to the results of~\cite{Astala2011} in the plane. We will not address this issue, as it introduces numerous technical difficulties. We note, however, that the assumption in Theorem~\ref{estimatetheorem} is of this type.

\section{Notation}
Let $\zeta = \tau(e_1 - ie_2)$, where $e_1,e_2 \in \mathbb{R}^n$ are orthogonal unit vectors. Define the conjugated Laplacian
\[\lp_\z := e^{-x\cd \z} \lp e^{x\cd \z},\]
a differential operator whose symbol is
\[p_\z(\xi) := -\a \xi^2 + 2 i \z \cd \xi.\]
This symbol vanishes simply on the characteristic set
\[\Si_\zeta := \{\xi:  p_\zeta(\xi)=0\} = \{\xi: \xi_1 = 0, \a{\xi-\tau e_2} = \tau\},\]
which is a sphere of codimension two. In fact, it is not hard to check that
\begin{equation}
\label{pbehavior}
\a{p_\z(\xi)} \sim \begin{cases}
\tau d(\xi,\Si_\z) & d(\xi,\Si_\z) \leq \tau/8\\
\tau^2 + \a{\xi}^2 & d(\xi,\Si_\z) > \tau/8
\end{cases}
\end{equation}
where $d(\xi,\Si_\z) \sim \a{\xi_1} + \a{\a{\xi - \tau e_2}-\tau}$ is the distance from $\xi$ to $\Si_\z$. We will refer to this distance as the {\em modulation}.

Define the Banach spaces $\dot X^{b}_\z$ and $X^b_\z$ with norms
\begin{align*}
\n{u}_{\dot X^b_\z} & = \n{\a{p_\z(\xi)}^b \hat u}_{L^2} \\
\n{u}_{X^b_\z} & = \n{(\a{p_\z(\xi)}+\tau)^b \hat u}_{L^2} .
\end{align*}
We will use the Greek letters $\ld,\mu,\nu$ to represent dyadic integers of the form $2^{k}$, where $k \geq 0$. For $\ld > 1$ we define $E_\ld$ to be the set of $\xi$ with modulation comparable to $\ld$:
\[E_\ld := \{\xi: d(\xi,\Si_\z) \in (\ld/2,\ld]\}.\]
Similarly, for any $\ld$ we write
\[E_{\leq \ld} := \{\xi: d(\xi,\Si_\z) \leq \ld\}.\]
Since our problem is localized to a fixed compact set, the uncertainty principle implies that we need not distinguish frequencies which are separated on the unit scale. Therefore, by abuse of notation we will define $E_1 := E_{\leq 1}$. Let $m_\ld$ denote the characteristic function of $E_\ld$, and similarly for $m_{\leq \ld}$.

Let $Q_\ld, Q_{\leq \ld}$ be the Fourier multipliers with symbols $m_\ld, m_{\leq \ld}$. We will wish to distinguish the cases $\ld \leq \tau/8$ and $\ld \gg \tau/8$, so we define projections onto low and high modulation by
\begin{align*}
Q_l & = \sum_{1\leq \ld \leq \tau/8} Q_\ld \\
Q_h & = \sum_{\ld > \tau/8} Q_\ld,
\end{align*}
where the $\ld$ vary over dyadic integers. By~\eqref{pbehavior}, we have
\begin{align}
\label{hfestimate}
\norm{Q_h u}_{H^1_\tau} \ll \norm{u}_{\dot X^{1/2}_\zeta}
\end{align}
We will also need the standard Littlewood-Paley projections. For these we choose a smooth dyadic partition of unity, i.e. a function $\chi \in C_0^\infty(\mathbb{R})$ supported on $[1/2,2]$ such that
\[1 = \sum_{k=-\infty}^\infty \chi(2^{-k} \rho).\]
for any $\rho > 0$. For a dyadic integer $\ld> 1$, we set $\chi_\ld(\xi) = \chi(\a{\xi}/\ld)$, and by abuse of notation we again set $\chi_1 = \sum_{\ld \leq 1} \chi_\ld$. The Littlewood-Paley projections $P_\ld$ are defined as the Fourier multipliers with symbols $\chi_\ld$. Given a direction $\om \in S^{n-1}$, we can also define the Littlewood-Paley projections $P^\om_\ld$ in the $\om$ direction using the Fourier multipliers $\chi(\a{\xi \cd \om}/\ld)$.
\section{Strichartz estimates}
Our goal in this section is to prove $L^p$ estimates for functions in $\dot X^{1/2}_\z$. We follow~\cite{Kenig1987}. Since the symbol $p_\z(\xi)$ is characteristic on a sphere $\Si_\z$, we begin with the Stein-Tomas restriction theorem.
\begin{theorem}[\cite{Stein1993,Tomas1975}]
Suppose $p \geq (2d+2)/(d-1)$. Let $\si$ denote the surface measure on $S^{d-1}$. Then
\[\n{\widehat{f \,d\si}}_{L^{p}(\mathbb{R}^d)} \ll \n{f}_{L^2(S^{d-1})}.\]
\end{theorem}

Let $\tau S^{d-1}$ denote the sphere of radius $\tau$. Given a set $E$ we define its $\ld$-neighborhood
\[N_\ld(E) := \{\xi: d(\xi, E) \leq \ld\}.\]
We use the following rescaled and localized variant of the restriction theorem:
\begin{corollary}
\label{localrestriction}
Let $p$ be as above. Suppose that $\hat g$ is supported in $N_\ld(\tau S^{d-1})$, where $\ld \leq \tau/8$. Then
\[\n{g}_{L^{p}} \ll \ld^{1/2} \tau^{(d-1)/2-d/p}  \n{\hat g}_{L^2(N_\ld(\tau S^{d-1}))}\]
\end{corollary}
\begin{proof}
By Fourier inversion, we have
\begin{align*}
g(x)& =c_d\int \hat g(\xi) e^{ix\cd \xi}\,d\xi\\
&= c_d \int_{\tau-\ld}^{\tau+\ld} \int_{S^{d-1}} \hat g(\rho \om) e^{i\rho \j{x,\om}}\,\rho^{d-1}\,d\rho\,d\si\\
&= c_d\int_{\tau-\ld}^{\tau+\ld} \rho^{d-1} (\hat g(\rho \om) \,d\si)^\vee (\rho x)\,d\rho
\end{align*}
By Minkowski's inequality, the restriction theorem, Cauchy-Schwarz, and Plancherel this implies that
\begin{align*}
\n{g}_{L^{p}} & \ll \int_{\tau-\ld}^{\tau+\ld} \n{(\hat g(\rho \om)\,d\si)^\vee(\rho x)}_{L^{p}(\mathbb{R}^d)} \,\rho^{d-1} \,d\rho\\
&\ll \int_{\tau-\ld}^{\tau+\ld} \rho^{-d/p} \n{(\hat g(\rho \om)\,d\si)^\vee(x)}_{L^{p}(\mathbb{R}^d)} \,\rho^{d-1} \,d\rho\\
\\
&\ll \tau^{-d/p}\int_{\tau-\ld}^{\tau+\ld} \n{\hat g(\rho \om)}_{L^2(S^{d-1})}\,\rho^{d-1}\,d\rho\\
&\ll \tau^{-d/p} (\ld \tau^{d-1})^{1/2} \sps{\int_{\tau-\ld}^{\tau+\ld} \n{\hat g(\rho\om)}_{L^2(S^{d-1})}^2 \,\rho^{d-1} \,d\rho}^{1/2}\\
&=\ld^{1/2} \tau^{(d-1)/2 - d/p} \n{\hat g}_{L^2(N_\ld(\tau S^{d-1}))}.
\end{align*}
\end{proof}
We deduce the following Strichartz-type estimates
\begin{lemma}
Let $p=2n/(n-2)$, $\ld\leq \tau/8$. Then\footnote{Strictly speaking, the $X^{1/2}_\z$ norm should be replaced with $\dot X^{1/2}_\z$, but this will not be important.}
\begin{align}
\label{bandstrichartz}
\n{Q_\ld f}_{p} &\ll (\ld/\tau)^{1/n} \n{f}_{X^{1/2}_\zeta}.\\
\label{strichartz}
\n{f}_p& \ll \n{f}_{X_\zeta^{1/2}}.
\end{align}
\end{lemma}
\begin{proof}
By a change of coordinates, we may assume $e_1 = (1,0,\dotsc, 0)$. We use the notation $\xi = (\xi_1,\xi')$.

For~\eqref{bandstrichartz}, write $g = Q_\ld f$. Note that
\[E_\ld \subset \{\xi: \a{\xi_1} \leq c\ld, \a{\a{\xi' - \tau e_2'}-\tau} \leq c\ld\}.\]
We can write $g = \phi_\ld *_{x_1} g$, where $\phi_\ld(x_1) = \ld \phi(\ld x_1)$ for some Schwartz $\phi$ and the convolution is taken in the $x_1$ variable only. By Minkowski's inequality and Young's inequality, we have
\begin{align*}
\n{g}_{p} & = \ns{\int \phi_\ld(x_1-y_1) g(y_1,x')\,dy_1}_{p} \\
&\ll \ns{\int \a{ \phi_\ld (x_1-y_1) } \n{g(y_1)}_{L^{p}_{x'}}\,dy_1}_{L^{p}_{x_1}}\\
&\ll \ld^{1/2-1/p} \n{g}_{L^2_{x_1} L^p_{x'}}.
\end{align*}
If we regard $g$ as a function in the $x'$ variable, we see that its Fourier transform lies in $N_{c\ld}(\tau S^{n-2} + \tau e_2')$. By Corollary~\ref{localrestriction} and translation invariance, we have $\n{g(x_1)}_{L^p_{x'}} \ll \ld^{1/2} \tau^{(n-2)/(2n)}\n{\hat g(x_1)}_{L^2_{x'}}$ for each $x_1$. It follows that
\[\n{g}_p \ll \ld^{1/2} \ld^{1/2-1/p} \tau^{1/2-1/n} \n{\hat g}_{2} \ll \ld^{1/n} \tau^{-1/n} \n{f}_{X^{1/2}_\zeta}. \]

For~\eqref{strichartz}, we apply~\eqref{bandstrichartz} near $\Si_\z$ and Sobolev embedding away from $\Si_\z$. On $E$ we have
\begin{align*}
\n{Q_{l} f}_p & \ll \sum_{1 \leq \ld \leq \tau/8} \n{Q_\ld f}_p \\
&\ll \sum_{1 \leq \ld \leq \tau/8} (\ld/\tau)^{1/n}\n{Q_\ld f}_{X^{1/2}_\z}\\
&\ll \sps{\sum_{1\leq \ld \leq \tau/8} \n{Q_\ld f}_{X_\z^{1/2}}^2}^{1/2} \\
&\leq \n{f}_{X_\z^{1/2}}
\end{align*}
Away from $E$ we have
\[\n{Q_{h}f}_p \ll \n{Q_{h}f}_{H^1} \ll \n{f}_{X^{1/2}_\z}.\]
by~\eqref{hfestimate}. Combining these estimates gives the claimed inequality.
\end{proof}

\section{Bilinear estimates}
Given a tempered distribution $f\in \mc S'(\mathbb{R}^n)$, define the map $m_f: \mc S(\mathbb{R}^n) \to \mc S'(\mathbb{R}^n)$ by $m_f \phi := f \phi$. We would like to control $\n{m_f}_{X^{1/2}_\z\to X^{-1/2}_\z}$. By duality, this is equivalent to establishing a bilinear estimate of the form
\[\a{m_f(u,v)} \ll \n{u}_{X^{1/2}_\z} \n{v}_{X^{1/2}_\z},\]
where
\[m_f(u,v) = \j{m_f u, v}.\]
Suppose that $f \in L^{n/2}$. By~\eqref{strichartz}, we have
\begin{align}
\a{m_f(u,v)} & \ll \n{f}_{n/2} \n{u}_{2n/(n-2)} \n{v}_{2n/(n-2)}\\
&\ll \n{f}_{n/2} \n{u}_{X^{1/2}_\z} \n{v}_{X^{1/2}_\z}.
\label{lowerorder}
\end{align}
We also have
\begin{equation}
\a{m_f(u,v)} \ll \n{f}_\infty \n{u}_2 \n{v}_2 \ll \tau^{-1} \n{f}_\infty \n{u}_{X^{1/2}_\z} \n{v}_{X^{1/2}_\z}.
\label{linfinity}
\end{equation}
A more difficult task is to control $m_{\na f}$. We record the main computation in the following lemma:
\begin{lemma}
\label{calculation}
Let $s,p,\th$ be as in Theorem~\ref{maintheorem}. Let $1/q = 1/2-1/p$. There is some $\al > 0$ such for fixed $\ld \leq 100\tau$, we have
\begin{align}
\label{leq}
\sum_{\substack{\mu\leq \nu\leq \tau/8\\\nu <\ld}}  (\nu/\ld)^{(1-\th)/n}  \n{Q_\mu u}_q\n{Q_\nu v}_2 &\ll (\ld/\tau)^\al \ld^{s-2} \n{u}_{X^{1/2}_\z} \n{v}_{X^{1/2}_\z},
\end{align}
and
\begin{align}
\label{geq}
\sum_{\substack{\mu \leq \nu\leq \tau/8 \\ \ld \leq \nu < \tau/8} }\n{Q_\mu u}_q \n{Q_\nu v}_2 &\ll \ld^{-1}(\ld/\tau)^\al \n{u}_{X^{1/2}_\z} \n{v}_{X^{1/2}_\z}.
\end{align}
\end{lemma}
\begin{proof}
We may interpolate~\eqref{bandstrichartz} with the trivial estimate $\n{Q_\mu u}_2 \ll (\mu \tau)^{-1/2} \n{Q_\mu u}_{X^{1/2}_\z}$ to obtain
\[\n{Q_\mu u}_q \ll (\mu/\tau)^{(1-\th)/n} (\mu \tau)^{-\th/2} \n{Q_\mu u}_{X_\z^{1/2}}\]
where $1/q = 1/2-1/p$ and $\th$ is such that $p=n/(1-\th)$. Combining this with the trivial $L^2$ estimate for $v$, we obtain
\begin{align*}
\n{Q_\mu u}_q \n{Q_\nu v}_2& \ll B_{\mu,\nu}\n{Q_\mu u}_{X^{1/2}_\z} \n{Q_\nu v}_{X^{1/2}_\z}
\end{align*}
where
\[B_{\mu, \nu} := \tau^{-(1-\th)/n - \th/2-1/2} \mu^{(1-\th)/n-\th/2} \nu^{-1/2}.\]
Set
\[\b: = \frac{1-\th}{n} - \frac{\th}{2}.\]
Suppose first that $\b > 0$. When $\nu \geq \ld$ we have
\begin{align*}
B_{\mu,\nu}&\ll \tau^{-(1-\th)/n - \th/2-1/2} \ld^{(1-\th)/n - \th/2-1/2} (\mu/\nu)^{\b} \\
&=\ld^{-\th-1} (\ld/\tau)^{(1-\th)/n + \th/2 + 1/2} (\mu/\nu)^{\b},
\end{align*}
We take $\al \leq (1-\th)/n+(1+\th)/2$ and use the discrete Young's inequality to establish~\eqref{geq}.

Suppose now that $\nu < \ld$. When $n=3$, we set $\th = 0$,
\begin{align*}
(\nu/\ld)^{1/3}  B_{\mu,\nu} &= (\nu/\ld)^{1/3} \tau^{-5/6} \mu^{1/3} \nu^{-1/2}\\
&=(\mu/\nu)^{1/3} (\nu/\tau)^{1/6}   (\ld/\tau)^{2/3}\ld^{-1}.
\end{align*}
By Young's inequality we have~\eqref{leq} for $\al\leq 2/3$. When $n = 4$ we take $\th$ to be zero and obtain
\begin{align*}
(\nu/\ld)^{1/4} B_{\mu,\nu} & = (\nu/\ld)^{1/4} \tau^{-3/4} \mu^{1/4} \nu^{-1/2}\\
&= (\mu/\nu)^{1/4}(\ld/\tau)^{3/4} \ld^{-1}.
\end{align*}
Applying Young's inequality we have~\eqref{leq} for $\al\leq 3/4$.

When $n > 4$, we have
\begin{align*}
(\nu/\ld)^{(1-\th)/n} B_{\mu,\nu} & = (\mu/\nu)^\b \nu^{-1/2+2(1-\th)/n - \th/2} \ld^{s-2} (\ld/\tau)^{(1-\th)/n+\th/2+1/2}.
\end{align*}
In this case we have~\eqref{leq} for $\al \leq (1-\th)/n+\th/2+ 1/2$

In higher dimensions, we also want to consider the case $(1-\th)/n - \th/2 \leq 0$. For $\nu \geq \ld$ we have
\begin{align*}
B_{\mu,\nu} &\leq \mu^\b \ld^{-1/2} \tau^{-(1-\th)/n-\th/2-1/2}\\
&\ll \ld^{-1} \tau^{-(1-\th)/n-\th/2}.
\end{align*}
Then we have~\eqref{geq} for $\al < (1-\th)/n + \th/2$, since there are only $\sim \log \tau$ possible values of $\mu,\nu$.

For $\ld \geq \nu$ we have
\begin{align*}
(\nu/\ld)^{(1-\th)/n} B_{\mu,\nu} & \ll \nu^{(1-\th)/n-1/2} \ld^{-2(1-\th)/n-\th/2-1/2} (\ld/\tau)^{(1-\th)/n+\th/2+1/2}.
\end{align*}
Thus we have~\eqref{leq} for $\al \leq  (1-\th)/n + \th/2 + 1/2$.
\end{proof}

Let $P_\ld$ denote the Littlewood-Paley projections, and let $P^1_\mu$ denote the Littlewood-Paley projections in the $e_1$ direction. Then
\begin{lemma}
\label{mqlemma}
Let $s,p$ be as in Theorem~\ref{maintheorem}. Then for any $f \in W^{s-1,p}(\mathbb{R}^n) \cap L^n(\mathbb{R}^n)$,
\begin{align*}
\n{m_{\na f}}_{X^{1/2}_\z \to X^{-1/2}_\z} & \ll \n{f}_n + \sup_{\nu\leq \ld \leq 100\tau} (\ld/\tau)^{\b} (\ld/\nu)^{1/p} \ld^{s-1} \n{P_\ld P^1_{\leq 8\nu}f}_p ,
\end{align*}
where $\b>0$
\end{lemma}
\begin{proof}
Write
\[m_{\na f}(u,v) = m_{\na f} (Q_h u, Q_h v) + m_{\na f}(Q_h u, Q_l v) + m_{\na f}(Q_l u, Q_h v) + m_{\na f}(Q_l u, Q_l v).\]
We can treat all but the last term using~\eqref{hfestimate},~\eqref{strichartz}. Integrating by parts,
\begin{align*}
\a{m_{\na f}(Q_h u,Q_h v)} & \ll \n{f}_n \n{Q_h \na u}_2 \n{Q_h v}_{2n/(n-2)} + \n{f}_n \n{Q_h u}_{2n/(n-2)} \n{Q_h \na v}_2 \\
&\ll \n{f}_n \n{u}_{X^{1/2}_\z} \n{v}_{X^{1/2}_\z}.
\end{align*}
Since $Q_l v$ is supported in $\a{\xi} \ll \tau$,
\begin{align*}
\a{m_{\na f}(Q_h u, Q_l v)} & \ll \n{f}_n \n{Q_h \na u}_2 \n{Q_l v}_{2n/(n-2)} + \n{f}_n \n{Q_h u}_{2} \n{Q_l \na v}_{2n/(n-2)} \\
&\ll \n{f}_n  \n{Q_h u}_{H^1_\tau} \n{Q_l v}_{2n/(n-2)} \\
&\ll \n{f}_n \n{u}_{X^{1/2}_\z} \n{v}_{X^{1/2}_\z}.
\end{align*}

It remains to estimate $m_{\na f}(Q_l u,Q_l v)$. We have
\begin{align}
\label{qsum}
m_{\na f}(Q_l u, Q_l v) & = \sum_{\mu,\nu,\ld}\int (\na P_\ld f)\, Q_\mu u \, \bar{Q_\nu v}\,dx.
\end{align}
Suppose $\mu \leq \nu$ (the case $\mu > \nu$ is identical). Because $Q_\mu u\, \bar{Q_\nu v}$ has Fourier support in $\{\xi: \a{\xi_1} \leq 2 \nu\}$, Plancherel's theorem and H\"older's inequality give
\begin{align*}
\as{\int (\na P_\ld f)\, Q_\mu u \, \bar{Q_\nu v}\,dx} & = \as{\int P^1_{\leq 8 \nu}( \na P_\ld f)\, Q_\mu u\,\bar{Q_\nu v}\,dx}\\
& \ll \n{P^1_{\leq 8 \nu} \na P_\ld f}_p \n{Q_\mu u}_q \n{Q_\nu v}_2.
\end{align*}
Furthermore, since $Q_\mu u \, \bar{Q_\nu v}$ has Fourier support in $\{\a{\xi} \ll 100\tau\}$, we can assume $\ld \leq 100\tau$ in this sum. Applying Lemma~\ref{calculation},
we get
\begin{align*}
\a{m_{\na f} (Q_l u, Q_l v)} & \ll \sum_{\substack{\nu\geq \ld\\\mu\leq \nu}}  \n{\na P_\ld f}_p \n{Q_\mu u}_q \n{Q_\nu v}_2 \\
&\quad\quad+ \sum_{\substack{\nu < \ld\leq 100\tau\\\mu\leq\nu}} (\ld/\nu)^{1/p} (\nu/\ld)^{1/p} \n{\na P_{\ld} P^1_{\leq 8\nu}f}_p \n{Q_\mu u}_q \n{Q_\nu v}_2\\
&\ll \sum_{\ld\leq 100\tau} \{(\ld/\tau)^\al \ld^{-1} \n{\na P_\ld f}_p \\
&\quad\quad\qq+ \sup_{\nu\leq \ld} (\ld/\tau)^{\al} (\ld/\nu)^{1/p} \ld^{s-2} \n{\na P_\ld P^1_{\leq 8\nu} f}_p\}\\
&\quad\quad\times\n{u}_{X^{1/2}_\z} \n{v}_{X^{1/2}_\z}\\
&\ll (\n{f}_p +  \sup_{\nu\leq \ld\leq 100\tau}(\ld/\tau)^{\al/2} (\ld/\nu)^{1/p} \ld^{s-1} \n{P_\ld P^1_{\leq 8\nu} f}_p)\\
&\quad\quad\times\n{u}_{X^{1/2}_\z} \n{v}_{X^{1/2}_\z}.
\end{align*}
\end{proof}

\section{Averaging}
Given any vector $\om\in S^{n-1}$, we define $P^\om_{\mu}$ to be Littlewood-Paley projection in $\om$ direction. Let $\mu$ denote Haar measure on $O(n)$, normalized so that if $\si$ is the usual spherical measure on $S^{n-1}$ and $f:S^{n-1} \to \mathbb{R}$ is integrable, then for any $\th\in S^{n-1}$ we have
\begin{equation}
\label{haarformula}
\int_{O(n)} f(U\th)\,d\mu(U) = \int_{S^{n-1}} f(\om) \,d\si(\om).
\end{equation}

\begin{lemma}
\label{planeavg}
Suppose $p \in [2,\infty]$. Let $f \in L^p(\mathbb{R}^n)$. For $U \in O(n)$ and $\nu \leq \ld$, define
\[A_{\ld,\nu}(U) = (\ld/\nu)^{1/p}\n{P_\ld P_{\leq \nu}^{Ue_1} f}_p.\]
Then
\[\n{A_{\ld,\nu}}_{L^p(O(n))} \ll \n{f}_p\]
\end{lemma}
\begin{proof}
We define an operator $T$ mapping functions on $\mathbb{R}^n$ to functions on $O(n) \times \mathbb{R}^n$ by
\[Tf(U,x) = P_\ld P_{\leq \nu}^{Ue_1} f(x).\]
The lemma asserts that this operator is bounded from $L^p(\mathbb{R}^n)$ to $L^p(O(n)\times \mathbb{R}^n)$. By interpolation, it suffices to establish this at the endpoints $p = 2$ and $p=\infty$.

When $p = \infty$ this is just the fact that the Littlewood-Paley projections are bounded on $L^\infty$.

When $p=2$ we use Plancherel's theorem and Fubini.
\begin{align*}
\n{Tf}_{L^2}^2 & \sim \int_{O(n)} \int_{\mathbb{R}^n} \a{\phi(\xi/\ld) \chi(\xi\cd (Ue_1)/\nu) \hat f(\xi) }^2\,d\xi \,d\mu(U)\\
&\leq \sps{ \sup_\xi \int_{O(n)} \a{\phi(\xi/\ld) \chi(\xi\cd (Ue_1)/\nu)}^2\,d\mu(U)} \n{f}_2^2.
\end{align*}
Here $\phi$ is supported on an annulus, and $\chi$ is supported on an interval. We estimate the last integral using~\eqref{haarformula} and spherical coordinates:
\begin{align*}
\int_{O(n)} \a{\phi(\xi/\ld) \chi(\xi\cd (Ue_1)/\nu)}^2 \,d\mu(U) &\ll \sup_{\a{\xi} \sim \ld} \int_{S^{n-1}} \a{\chi(\a{\xi} \om\cd e_1/\nu)}^2 \,d\si(\om)\\
&\ll\sup_{\a{\xi} \sim \ld}  \int_0^\pi \a{\chi(\a{\xi} \cos \th/\nu)}^2 \, \sin(\th)^{n-2}\,d\th\\
&\ll \sup_{\a{\xi} \sim \ld} \int_{-1}^1 \chi(\a{\xi} u/\nu) \,du\\
&\ll\sup_{\a{\xi} \sim \ld}  \frac{\nu}{\a \xi} \\
&\ll \frac{\nu}{\ld}.
\end{align*}
This shows that
\[\n{Tf}_2  \ll (\nu/\ld)^{1/2} \n{f}_2,\]
which completes the proof.
\end{proof}
Define $\z(\tau,U) = \tau U(e_1-ie_2)$. Our next lemma establishes that $\n{q}_{X^{-1/2}_{\z(\tau,U)}}$ is small on average. This is implied by~\cite[Lemma 3.1]{Haberman2013}, but we give a simpler proof here, based on~\cite{Nguyen2014}:
\begin{lemma}
\label{qavg}
If $f \in \dot H^{-1}$, then
\begin{align*}
M^{-1} \int_{M}^{2M} \int_{O(n)} \n{f}_{\dot X^{-1/2}_{\z(\tau,U)}}^2 \, d\mu(U) \,d\tau &\ll \n{P_{\geq 100M} f}_{\dot H^{-1}}^2 + M^{-1} \n{P_{< 100 M}f}_{\dot H^{-1/2}}^2.
\end{align*}
\end{lemma}
\begin{proof}
This is true if $f$ is supported at frequencies $\a{\xi} \geq 100M$, because there we have $\a{p_\z(\xi)} \geq \a \xi^2$. Thus we may assume that $f$ is supported at frequencies $\a{\xi} \ll M$, where we have $\a{p_\z(\xi)} \gg 2\tau\a{\xi\cd (U e_1)} + \a{-\a \xi^2 + 2\tau \xi \cd (Ue_2)}$. Here we use Plancherel and the identity $U^T = U^{-1}$ and estimate as in Lemma~\ref{planeavg} by
\begin{align*}
\n{\a{\na}^{-1/2}f}_2^2  \sup_{\a{\xi} \leq 100M} \frac{\a{\xi}}{M} \int_M^{2M}  \int_{O(n)} (2\tau \a {(U^{-1} \xi) \cd e_1} + \a{-\a{\xi}^2 + 2\tau (U^{-1} \xi)\cd e_2})^{-1} \,d\mu(U)\,d\tau.
\end{align*}
By~\eqref{haarformula}, the quantity inside the supremum is given by
\begin{align*}
\frac{1}{M} \int_M^{2M} \int_{S^{n-1}} (2\tau\a{\om \cd e_1} + \a{-\a\xi + 2\tau\om\cd e_2})^{-1} \,d\si(\om)\,d\tau.
\end{align*}
We view $(\tau,\om)$ as polar coordinates and change variables to $u = \tau\om$. Then in the region $\tau \in [M,2M]$ the volume element $du$ is bounded below by $M^{n-1} \, d\si(\om) \,d\tau$, so this integral is bounded by
\[\frac{1}{M^n} \int_{\a u \in [M,2M]} (2\a{u_1} + \a{-\a{\xi} + 2u_2})^{-1}\,du.\]
Writing $v = (u_1,u_2)$, and integrating over the remaining variables, we bound by
\begin{align*}
\frac{1 }{M^n} M^{n-2} \int_{B(0,2M)} (2\a {v_1} + \a{-\a\xi + 2v_2})^{-1} \,dv&\leq \frac{1}{M^2} \int_{B(0,2M)} \a{v}^{-1} \,dv\\
&\sim \frac{1}{M}.
\end{align*}
\end{proof}

We summarize our estimates so far in the following
\begin{theorem}
\label{estimatetheorem}
Let $s,p$ be as in Theorem~\ref{maintheorem}, and let $\mbox{\fontfamily{pzc}\selectfont g}$ be a positive real-valued function on $\mathbb{R}^n$ such that $\na \log \mbox{\fontfamily{pzc}\selectfont g} \in W^{s-1,p}$ and  $\mbox{\fontfamily{pzc}\selectfont g} = 1$ outside of a large ball $B$. For $q = \mbox{\fontfamily{pzc}\selectfont g}^{-1/2}\lp \mbox{\fontfamily{pzc}\selectfont g}^{1/2}$, we have
\begin{align}
\label{mqavgdecay}
M^{-1} \int_{M/2}^{2M} \int_{O(n)} \n{q}_{X^{-1/2}_{\z(\tau,U)}}^2 \,d\mu(U)\,d\tau  &\to 0.
\end{align}
Furthermore,
\begin{align}
\sup_{\tau\in[M/2,2M]}\n{m_q}_{X^{1/2}_{\z(\tau,U)} \to X^{-1/2}_{\z(\tau,U)}} & \leq C_M + A_M(U),
\label{qavgdecay}
\end{align}
where $C_M \to 0$ as $M \to \infty$ and
\begin{equation}
\label{liminf}
\sum_{k > 2} k^{-1} \n{A_{2^k}}_{L^p(O(n))}^p <\infty.
\end{equation}
\end{theorem}
\begin{proof}
First, we write
\[\mbox{\fontfamily{pzc}\selectfont g}^{-1/2} \lp \mbox{\fontfamily{pzc}\selectfont g}^{1/2} = \f 1 2 \lp \log \mbox{\fontfamily{pzc}\selectfont g} + \f 1 4 \a{\na \log \mbox{\fontfamily{pzc}\selectfont g}}^2 = \sum_i \na_i f_i + h,\]
where $f_i \in W^{s-1,p}$ and $h \in L^{p/2}$.

We decompose each term into a good part and a bad part. Let $\phi_\epsilon = \epsilon^{-n} \phi(x/\epsilon)$, where $\phi$ is a $C_0^\infty$ function supported on the unit ball and $\int \phi = 1$. Define $f_\epsilon = f * \phi_\epsilon$.

By~\eqref{linfinity}, we have
\begin{align*}
\n{m_{\na f_\e}}_{X^{1/2}_\z \to X^{-1/2}_\z} + \n{m_{h_\e}}_{X^{1/2}_\z \to X^{-1/2}_\z} & \ll \tau^{-1}( \n{\na f_\e}_\infty + \n{h_\e}_\infty)\\
&\ll \tau^{-1} \e^{-2}( \n{f}_n + \n{h}_{n/2}).
\end{align*}
We also have
\begin{align*}
\n{\na f_\e}_{X^{-1/2}_\z} & \ll \tau^{-1/2} \n{\na f_\e}_2 \\
&\ll \tau^{-1/2} \e^{-1} \n{f}_2 \\
&\ll \tau^{-1/2} \e^{-1} \n{f}_n,
\end{align*}
since $n > 2$ and $f$ is compactly supported. For $n\geq 4$ we have
\begin{equation}
\begin{aligned}
\n{h}_{X^{-1/2}_\z} &\ll \tau^{-1/2} \n{h}_2\\
&\ll \tau^{-1/2} \n{h}_{n/2},
\label{highhestimate}
\end{aligned}
\end{equation}
and for $n = 3$ we have
\begin{align*}
\n{h_\e}_{X^{-1/2}_\z} & \ll \tau^{-1/2} \n{h_\e}_2 \\
&\ll \tau^{-1/2} \e^{-1/2} \n{h}_{3/2}.
\end{align*}
Taking $\e = M^{-1/4}$, we find that if we replace $q$ with $q_\e$ then the left hand sides of~\eqref{mqavgdecay} and~\eqref{qavgdecay} vanish as $\tau \to \infty$.

It remains to treat the bad part $q-q_\e$. Let $g = f - f_\e$, and define
\[A(\tau,U) = \n{m_{\na g}}_{X^{1/2}_{\z(\tau,U)}\to X^{-1/2}_{\z(\tau,U)}}.\]
Using Lemma~\ref{mqlemma}, we have
\[\sup_{\tau\in [M/2,2M]} A(\tau,U) \ll \n{g}_{L^p} + \sps{\sum_{1\leq \nu\leq \ld \leq 4M} [(\ld/M)^{\b} \ld^{s-1} A_{\ld,\nu}(U)]^p}^{1/p}, \]
where $A_{\ld,\nu}(U) = (\ld/\nu)^{1/p} \n{P_\ld P_{\leq 8\nu}^{Ue_1} g}_{L^p}$. As $M \to \infty$, we have $\e = M^{-1/4} \to 0$, so $\n{g}_{L^p} \to 0$. We take $A_M(U)$ to be the second term on the right hand side of this inequality, which is clearly a measurable function on $O(n)$. Now, $P_\ld g = P_\ld P_{\sim \ld} g$, where $P_{\sim \ld} g = \sum_{\ld/16 \leq \mu\leq 16\ld} P_\mu g$. Applying Lemma~\ref{planeavg}, we have
\begin{align*}
\n{A_M(U)}_{L^p(O(n))}^p &\ll \sum_{1\leq \nu \leq \ld \leq M/4} [(\ld/M)^{\b} \ld^{s-1} \n{P_{\sim \ld} g}_{L^p}]^p\\
&\ll \log M \sum_{1\leq \ld \leq M/4} [(\ld/M)^\b \ld^{s-1} \n{P_{\sim \ld} f}_{L^p}]^p.
\end{align*}
We control this quantity by taking a weighted sum over dyadic integers $M$, as in~\cite{Nguyen2014}. Namely, we have
\begin{align*}
\sum_{M\geq 2} (\log M)^{-1} \n{A_{M}(U)}^p_{L^p(O(n))} & \ll \sum_\ld \sum_{M \geq 4\ld} (\ld/M)^{\b p} [\ld^{s-1} \n{P_{\sim \ld} f}_{L^p}]^p\\
&\ll \sum_\ld [\ld^{s-1} \n{P_{\sim \ld} f}_{L^p}]^p.
\end{align*}
The last term is controlled by $\n{f}_{W^{s-1,p}}$ as a consequence of the Littlewood-Paley square function estimate. Thus we obtain~\eqref{liminf}.

By Lemma~\ref{qavg}, we have
\[M^{-1}\int_{M/2}^{2M} \int_{O(n)} \n{\na g}^2\,d\mu(U)\,d\tau \ll \n{g}_{L^2}^2\ll \n{g}_{L^p}^2\to 0.\]

Next we treat $h-h_\e$. When $n \geq 4$ we have $\n{h-h_\e}_{X^{-1/2}_\z} \to 0$ by~\eqref{highhestimate}. When $n=3$, we have
\begin{align*}
\n{h-h_\e}_{X^{-1/2}_\z} & \ll \n{h-h_\e}_{H^{-1/2}}\\
&\ll \n{h-h_\e}_{3/2}\\
&\to 0
\end{align*}
by Sobolev embedding. Finally, by~\eqref{lowerorder} we have
\[\n{m_{h-h_\e}}_{X^{1/2}_\z\to X^{-1/2}_\z} \ll \n{h-h_\e}_{n/2} \to 0.\]
\end{proof}

\section{Localization}
Because our problem is localized to a compact set, the uncertainty principle implies that the $X^{1/2}_\z$ norm is equivalent to the $\dot X^{1/2}_\z$ norm. To make this precise, we state the following
\begin{lemma}[\cite{Haberman2013}]
Let $\phi$ be a fixed Schwartz function. Then
\begin{align}
\label{dualloc}
\n{\phi u}_{\dot X^{-1/2}_\z} &\ll_\phi \n{u}_{X_\z^{-1/2}}\\
\label{loc}
\n{\phi u}_{X^{1/2}_\z} &\ll_\phi \n{u}_{\dot X^{1/2}_\z},
\end{align}
where the constants depend on the seminorms $\n{x^\al \na^\b \phi}_\infty$.
\end{lemma}
In particular, we have
\begin{lemma}
Suppose that $q$ is compactly supported. Then
\begin{equation}
\n{m_q}_{\dot X_\z^{1/2}\to \dot X_\z^{-1/2}} \ll \n{m_q}_{X_\z^{1/2}\to X_\z^{-1/2}}.
\label{compactbilinear}
\end{equation}
\end{lemma}
\begin{proof}
Let $\phi$ be a Schwartz function that it equal to one on the support of $q$. Then
\begin{align*}
\a{\j{m_q u, v}} & = \a{\j{m_q \phi u, \phi v}} \\
&\ll \n{m_q}_{X^{1/2}_\z \to X^{-1/2}_\z} \n{\phi u}_{X^{1/2}_\z}\n{\phi v}_{X^{1/2}_\z}\\
&\ll \n{m_q}_{X^{1/2}_\z \to X^{-1/2}_\z} \n{u}_{\dot X^{1/2}_\z} \n{v}_{\dot X^{1/2}_\z}
\end{align*}
\end{proof}

We record the following useful fact:
\begin{lemma}
\label{fudgezeta}
Suppose $\z, \ti \z\in \mathbb{C}^n$ satisfy $\z \cd \z = \ti \z \cd \ti \z = 0$. Then
\[\n{u}_{X^{b}_\z} \ll (1+\a{\z-\ti \z})^{\a b} \n{u}_{X^b_{\ti \z}}.\]
\end{lemma}
\begin{proof}
We have
\begin{align*}
\a{p_\z}& \leq \a{p_{\ti \z}} + 2 \a{(\z-\ti \z) \cd \xi} \\
&\leq \a{p_{\ti \z}} + 2 \a{\z-\ti \z} \a{\xi}\\
&\ll ( 1 + \a{\z-\ti \z}) (\a{p_{\ti \z}} + \tau)
\end{align*}
by~\eqref{pbehavior}.
\end{proof}

\section{Proof of the main theorem}
We summarize some known results which allow us to extend the $\mbox{\fontfamily{pzc}\selectfont g}_i$ to all of $\mathbb{R}^n$. First we transfer the problem to the interior, as in~\cite{Sylvester1987}.
\begin{lemma}
\label{weakequality}
Suppose $n \geq 3$. Let $\mbox{\fontfamily{pzc}\selectfont g}_1,\mbox{\fontfamily{pzc}\selectfont g}_2 \in W^{1,n}(\mathbb{R}^n)$ be functions such that $0<c \leq \mbox{\fontfamily{pzc}\selectfont g}_i \leq c^{-1}$ for some $c$. If $\mbox{\fontfamily{pzc}\selectfont g}_1 = \mbox{\fontfamily{pzc}\selectfont g}_2$ outside $\Om$ and $\Ld_{\mbox{\fontfamily{pzc}\selectfont g}_1}=\Ld_{\mbox{\fontfamily{pzc}\selectfont g}_2}$, then for $q_j = \lp \sqrt {\mbox{\fontfamily{pzc}\selectfont g}_j}/\sqrt{\mbox{\fontfamily{pzc}\selectfont g}_j}$, we have
\[\j{q_1,v_1 v_2} = \j{q_2,v_1 v_2}\]
when each $v_j$ is a solution in $H^1_{\mathrm{loc}}(\mathbb{R}^n)$ to $\lp v_j - q_j v_j = 0$.
\end{lemma}
\begin{proof}
See~\cite{Brown2003}.
\end{proof}

The following argument is apparently due to Alessandrini. It amounts to the fact that $q_1 = q_2$ implies that the function $\log \mbox{\fontfamily{pzc}\selectfont g}_1 - \log \mbox{\fontfamily{pzc}\selectfont g}_2$ solves the Dirichlet problem $\div \sqrt{g_1 g_2}\na u = 0$ with $u = 0$ at infinity. See~\cite{Sylvester1987,Brown1996,Brown2003}.
\begin{lemma}
\label{gaidentity}
Let $\mbox{\fontfamily{pzc}\selectfont g}_i, q_i$ be as in Lemma~\ref{weakequality}, and suppose that $q_1 = q_2$ in the sense of distributions. Then $\mbox{\fontfamily{pzc}\selectfont g}_1 = \mbox{\fontfamily{pzc}\selectfont g}_2$.
\end{lemma}
\begin{proof}
First, have $q_i \in H^{-1}(\mathbb{R}^n)$ for each $i$. To see this we note that
\begin{align*}
\n{q}_{H^{-1}} & = \n{\f 1 2 \lp \log \mbox{\fontfamily{pzc}\selectfont g} + \f 1 2  \a{\na \log \mbox{\fontfamily{pzc}\selectfont g}}^2}_{H^{-1}}\\
&\ll \n{\na \log \mbox{\fontfamily{pzc}\selectfont g}}_2 + \n{\a{\na \log \mbox{\fontfamily{pzc}\selectfont g}}^2}_{H^{-1}}\\
&\ll \n{\na \log \mbox{\fontfamily{pzc}\selectfont g}}_n + \n{\na \log \mbox{\fontfamily{pzc}\selectfont g}}_{4n/(n+2)}^2 \\
&\ll \n{\na \log \mbox{\fontfamily{pzc}\selectfont g}}_n + \n{\na \log \mbox{\fontfamily{pzc}\selectfont g}}_n
\end{align*}
by Sobolev embedding and H\"older's inequality. It follows that we may test $q_1 - q_2$ against the function $g_1 g_2( \log g_1 - \log g_2) \in H^1(\mathbb{R}^n)$, where $g_i = \sqrt{\mbox{\fontfamily{pzc}\selectfont g}_i}$. This gives
\begin{align*}
0& = \int [\na g_1 \cd \na (g_2 (\log g_1 - \log g_2)) - \na g_2 \cd \na(g_1 (\log g_1 - \log g_2))]\,dx\\
&=\int (g_2 \na g_1 - g_1 \na g_2) \cd \na(\log g_1 - \log g_2)\,dx\\
&=\int g_1 g_2 \a{\na(\log g_1 - \log g_2)}^2 \,dx,
\end{align*}
which implies that $g_1 = g_2$.
\end{proof}

Now we apply the boundary determination result of~\cite{Brown2013}, which implies
\begin{theorem}
\label{boundarydetermination}
Suppose that $0<c<\mbox{\fontfamily{pzc}\selectfont g}_i <c^{-1}$. If $\mbox{\fontfamily{pzc}\selectfont g}_i \in W^{1,1}(\Om)$ and $\Ld_{\mbox{\fontfamily{pzc}\selectfont g}_1} = \Ld_{\mbox{\fontfamily{pzc}\selectfont g}_2}$, then $\mbox{\fontfamily{pzc}\selectfont g}_1 = \mbox{\fontfamily{pzc}\selectfont g}_2$ on $\partial \Om$.
\end{theorem}

\begin{proof}[Proof of Theorem~\ref{maintheorem}]
By Theorem~\ref{boundarydetermination}, we have $\mbox{\fontfamily{pzc}\selectfont g}_1 = \mbox{\fontfamily{pzc}\selectfont g}_2$ on $\partial \Om$. Our assumptions imply that $s -1/p \leq 1$. Thus by~\cite{Marschall1987} we may extend the $\mbox{\fontfamily{pzc}\selectfont g}_i$ to functions in $W^{s, p}$ such that $\mbox{\fontfamily{pzc}\selectfont g}_1 = \mbox{\fontfamily{pzc}\selectfont g}_2$ outside of $\Om$. By Lemma~\ref{weakequality}, this implies that
\begin{equation}
\label{weakglobal}
\j{q_1, v_1 v_2}  = \j{q_2, v_1 v_2}
\end{equation}
when each $v_j$ is a solution in $H^1_{\mathrm{loc}}(\mathbb{R}^n)$ to $\lp v_j - q_j v_j = 0$.

Fix $r> 0$ and three orthonormal vectors $\{e_1,e_2, e_3\}$, and define
\begin{align*}
\z_1(\tau,U) & = \tau U(e_1 - i e_2)\\
\z_2(\tau, U) & = -\z_1(\tau, U)\\
\ti \z_1 (\tau,U) &:= \tau Ue_1 + i (rUe_3 - \sqrt{\tau^2 - r^2}Ue_2)\\
\ti \z_2 (\tau,U) &:= -\tau U e_1 + i(rUe_3 + \sqrt{\tau^2 - r^2}Ue_2)
\end{align*}
In what follows, all of inequalities will implicitly depend on $r$. For example, we have $\a{\z_i - \ti \z_i} \ll 1$. In particular, by Lemma~\ref{fudgezeta}, the spaces $X^b_{\z_i}$ and $X^b_{\ti \z_i}$ have equivalent norms.

Now let
\[F(\tau,U) = \sum_i \n{m_{q_i}}_{X^{1/2}_{\z_i(\tau,U)} \to X^{-1/2}_{\z_i(\tau,U)}}^p + \sum_{i,j} \n{q_i}_{X^{-1/2}_{\z_j(\tau, U)}}^2.\]
By Theorem~\ref{estimatetheorem} and the fact that $\sum_{k>1} k^{-1} = \infty$, we have
\begin{equation}
\label{globalestimate}
\liminf_{M\to\infty} M^{-1} \int_{M}^{2M}\int_{O(n)} F(\tau,U) \,d\mu(U)\,d\tau = 0.
\end{equation}
Now we use an argument from~\cite[Section 3.3]{Nguyen2014} to select $\tau$ and $U$.
Their first observation is that if $\e > 0$ and $B_\e = \{U \in O(n): \n{U - I} < \e\}$, then by simply restricting~\eqref{globalestimate} we have
\[\liminf_{M\to\infty} M^{-1} \mu(B_\e)^{-1} \int_{M}^{2M}\int_{B_\e} F(\tau,U) \,d\mu(U)\,d\tau =0.\]
Thus we may choose, for a sequence of $M=M_l$ such that $M_l \to \infty$, some $\tau = \tau_{\e,l} \in [M_l, 2M_l]$, $U = U_{\e,l} \in B_\e$ and $\d=\d_{\e,l} > 0$ such that
\begin{equation}
\label{smallness}
\sum_i \n{m_{q_i}}_{X^{1/2}_{\z_i(\tau,U)} \to X^{-1/2}_{\z_i(\tau,U)}} + \sum_{i,j} \n{q_i}_{X^{-1/2}_{\z_j(\tau, U)}} \leq \d
\end{equation}
where $\d_{\e,l} \to 0$ as $l\to \infty$.

By~\eqref{compactbilinear}, we have
\[\n{m_{q_i}}_{\dot X^{1/2}_{\ti\z_i(\tau,U)} \to \dot X^{-1/2}_{\ti \z_i(\tau,U)}} \ll \n{m_{q_i}}_{X^{1/2}_{\ti\z_i(\tau,U)} \to X^{-1/2}_{\ti \z_i(\tau,U)}}\]
It follows that,
\[\n{m_{q_i}}_{\dot X^{1/2}_{\ti\z_i(\tau,U)} \to \dot X^{-1/2}_{\ti \z_i(\tau,U)}} \ll \d_{\e,l}\]
Since $\d_{\e,l} \to 0$ as $l \to \infty$, we can choose $l$ large enough that the left hand side is less than $1/2$. Since $\n{\lp_\z^{-1}}_{\dot X^{-1/2}_\z \to \dot X^{1/2}_\z} = 1$ for any $\z$, we can use the contraction mapping principle to construct solutions $\psi_i \in \dot X^{1/2}_{\ti \z_i(\tau,U)}$ to the equations $(\lp_{\ti\z_i(\tau,U)} - m_{q_i})\psi_i = q_i$, satisfying
\[\n{\psi_i}_{\dot X^{1/2}_{\ti \z_i(\tau,U)}} \ll \n{q}_{\dot X^{-1/2}_{\ti \z_i(\tau,U)}}.\]
Note that by~\eqref{easyenergy}, such a solution lies in $H^1_{\mathrm{loc}}(\mathbb{R}^n)$. This implies that the corresponding solution $v_i = e^{x\cd \ti\z_i(\tau,U)}(1+\psi_i)$ to the Schr\"odinger equation $(\lp-q_i)v_i$ lies in $H^1_{\mathrm{loc}}(\mathbb{R}^n)$ as well.

Let $k =2r U e_3$. By~\eqref{weakglobal},
\begin{align*}
0&=\j{q_1 - q_2, e^{ik\cd x} (1 + \psi_1)(1 + \psi_2)} \\
&=\j{q_1 - q_2, e^{ik\cd x}} + \j{q_1-q_2, e^{ik\cd x} \psi_1 \psi_2} + \j{q_1-q_2, e^{ik\cd x} (\psi_1 + \psi_2)}.
\end{align*}
We need to show that the second and third terms are small. Let $\phi$ be a Schwartz function that is equal to one on the support of $q$. Then
\begin{align*}
\a{\j{q_1, e^{ik\cd x}\psi_1 \psi_2}} & =\a{\j{m_{q_1} e^{-ik\cd x} \bar \psi_2, \psi_1}} \\
&\ll \n{e^{-ik\cd x} \phi \bar \psi_2}_{ X^{1/2}_{\z_1(\tau,U)}} \n{\phi \psi_1}_{X^{1/2}_{\z_1(\tau,U)}}\\
&= \n{e^{ik\cd x} \phi \psi_2}_{X^{1/2}_{\z_2(\tau,U)}} \n{\phi \psi_1}_{X^{1/2}_{\z_1(\tau,U)}}\\
&\ll \n{\psi_2}_{\dot X^{1/2}_{\ti \z_2(\tau,U)}} \n{\psi_1}_{\dot X^{1/2}_{\ti \z_1(\tau,U)}}\\
&\ll \n{q_2}_{ X^{-1/2}_{\z_2(\tau, U)}}\n{q_1}_{X^{-1/2}_{\z_1(\tau, U)}},
\end{align*}
since the seminorms of $e^{-ik\cd x} \phi$ are bounded with a bound depending only on $r$. We can bound the $q_2$ term in the same way. On the other hand, we have
\begin{align*}
\a{\j{q_i, e^{ik\cd x} \psi_1}} & \ll \n{q_i}_{X^{-1/2}_{\z_1(\tau,U)}} \n{\psi_1}_{\dot X^{1/2}_{\ti \z_1(\tau,U)} }\\
&\ll \n{q_i}_{X_{\z_1(\tau,U)}^{-1/2}} \n{q_1}_{X^{-1/2}_{\z_1(\tau,U)}}
\end{align*}
by duality of $\dot X^{1/2}_{\z_1(\tau,U)}$ and $\dot X^{-1/2}_{\z_1(\tau,U)}$. The terms with $\psi_2$ are the same. In summary, we obtain
\begin{align}
\label{tozero}
\a{(\hat q_1 - \hat q_2)(2rUe_3)}& \ll \sum_{1\leq i,j,k,l\leq 2} \n{q_i}_{X^{-1/2}_{\z_j(\tau,U)}} \n{q_k}_{X^{-1/2}_{\z_l(\tau,U)}} \ll \d^2
\end{align}
by~\eqref{smallness}.

To finish the proof, we again follow~\cite{Nguyen2014}. Since $B_\e$ is compact, we may pass to a subsequence such that $U_{\e,l} \to U_\e$ for some $U_\e \in B_\e$. Since the $\hat q_i$ are continuous, we may pass to the limit in~\eqref{tozero} to obtain
\[\a{\hat q_1 - \hat q_2}(2r U_\e e_3) \ll \lim_{l\to \infty} \d_{\e,l}^2 = 0.\]
Note that by construction, we have $U_\e \to I$ as $\e \to 0$. Thus, by taking limits again, we obtain $(\hat q_1 - \hat q_2)(2 r e_3) = 0$. Since $e_3 \in S^{n-1}$ and $r$ were arbitrary, this means that $\hat q_1 - \hat q_2=0$.
\end{proof}
\section*{Acknowledgments}
The author would like to thank his advisor, Daniel Tataru, for his patient guidance and encouragement. He would also like to thank Gunther Uhlmann, Russell Brown, Alberto Ruiz and Mikko Salo for many helpful conversations. Finally, the author would like to thank the anonymous referee for carefully reading the manuscript and suggesting many corrections and improvements.

redpajama_set_name: RedPajamaArXiv


High Fibre Breakfast Recipes, Fibre Rich Indian breakfast recipes.
High Fibre Breakfast Recipes, Fibre Rich Indian breakfast recipes. Breakfast, as the name suggests is breaking your fast after waking up after long hours of sleep. It is often said that this meal is the most important meal of the day as it boosts our metabolism and provides us with the necessary energy to stay satisfied for longer hours in the later part of the day.
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Lehsuni Bajra aur Kaddu ki Roti uses bajra which is fiber rich and can be paired with curds to make a perfect breakfast combo. Spinach and Paneer paratha or jowar bajra spring onion roti are good options. If you want the classic, then opt for jowar roti or bajra roti and have it with some healthy subzi.
And on days when you want to include sprouts in your breakfast, turn to Sprouts Pancake and accompany it with Palak, Kale and Apple Juice.
Jowar palak appe makes a great option for fiber rich south Indian breakfast. Or there is oats dosa or oats mutter dosa which is far healthier than regular dosa as there is no usage of rice in this recipe. The healthiest dosa got to be buckwheat dosa which is very high in fiber. What you should definitely avoid is the tasty rava dosa as rava is a refined carb.
Check out the nutrients you would gain for each of the breakfast recipes in the Nutritional Table shared with each recipe to be assured of the exact amount of fiber intake for breakfast. Accompany these fiber rich breakfast with a fruit of your choice to add in some more fiber and complete this first meal of the day.
Avoid tea and coffee for breakfast, see why?
Try and avoid having tea or coffee for breakfast as hinders the absorption of nutrients. So have your coffee or tea one hour before your breakfast or one hour after your breakfast. The objective is to let the digestion take place and the nutrients getting absorbed.
Enjoy our High Fibre Breakfast Recipes, Fibre Rich Indian breakfast recipes and other fiber recipe articles below.


text:
The Metaphysics of Scripture
05 Feb 2014 Posted by Andrew Fulford
Thinkers have sometimes stated that scripture is not metaphysical, or does not do metaphysics. This is obviously correct in one sense: the writings of the Bible are not generally characterized by philosophical jargon, but rather for the most part appear in everyday language. However, this statement is not true in another sense: the Bible does make statements that rule out some metaphysical ideas, and imply others. I want to briefly show how this is the case with a few Aristotelian concepts.
One reason why this elementary exercise is important is that it shows just how basic certain philosophical ideas and concepts are. Modern critics of using "philosophy"– or at least specific kinds of philosophy– in exegesis or theology almost always load the term with specific connotations. It is assumed that "philosophy" must be complicated, founded on many different and complex assumptions, and overly systematic. Yet when it comes to the basic elements of something like Aristotelian philosophy, this is not the case at all. It would be more accurate to say that "philosophy" is the intentional activity of defining terms and reflecting upon how concepts work, something that can be done by anyone at any time. As will be demonstrated, this can be quite simple.
Material Cause
Aristotle is famous for his so-called "four causes," i.e., material, formal, efficient, and final. These four causes are affirmed as real within the first book of scripture, and continuously throughout the rest of scripture. The material cause of an object is the stuff out of which it is made, which makes a concrete object concrete. In Genesis 1:2 we see the statement: "The earth was without form and void… ." By saying that "the earth was," Moses conveys that there was a concretely existing thing, the earth. There was not simply an abstract idea of an earth, but an actual material earth.
Formal Cause
The formal cause of an object is the pattern that makes it the type of thing it is. By virtue of having this same pattern of features, multiple individuals are considered "the same kind of thing". The idea of a form appears in Genesis 1:21: "So God created the great sea creatures and every living creature that moves, with which the waters swarm, according to their kinds, and every winged bird according to its kind." There are multiple individual creatures of the same "kind," or form.
Essential and Accidental Properties of Substances
Within this type of cause there's another important distinction. In Genesis 4:15b, we read: "And the Lord put a mark on Cain, lest any who found him should attack him." Genesis tells us Cain received a mark that he didn't have previously. This implies that Cain could receive a "mark" of some kind without ceasing to be the same individual he was before. He could either have the mark or not have it, and still be Cain.
On the other hand, Genesis 6:17 says: "For behold, I will bring a flood of waters upon the earth to destroy all flesh in which is the breath of life under heaven." Insofar as the flood is the agent of destruction in this story, the natural understanding of the text is that God kills (besides human beings) all animals by drowning them. This kind of death, however, does not immediately vapourize the body. What it does is cut off the capacity of the body to breathe, and thereby eventually stop the body from functioning as a living organism altogether. It is this cessation that Genesis describes as "destroying", which is another way of saying "bringing an end of existence to something". An animal can be wounded and continue to be the same animal; but an animal that ceases to be animated no longer exists.
What these two stories require us to say is that there are different kinds of features (a.k.a. "properties") an individual thing (a.k.a. "substance") can have, kinds that have been conventionally labeled "accidental" and "essential." There are types of features a thing can have or not have and still be the same thing, and then there are types of features without which a thing would not exist at all. A person can have or not have "a mark"; but an animal cannot lose its life without as such being destroyed. For that organic functioning to come to an end, either by violence or otherwise, is to destroy it as the kind of thing it is.
It is important to note that the identity of a person or substance through a process of changes implies the existence of an essence, that is, of a substantial form. (For Aristotle, a substantial form that actually exists is a substance.) If there were no such distinction, and the identity of a thing required that it never changed properties or relations of any kind, then any change whatsoever would imply that one substance ceased to exist and another began to be in its place. It would not be possible to speak truly as human beings normally do, of people being born, growing up, doing various kinds of activities, acquiring and losing properties and features, or any such thing. But the Bible manifestly does speak in this common-sense manner.
One further point on this matter needs explicit emphasis. This is the distinction between substance and property. For Aristotle, a substance is a being that is capable of possessing properties, whereas a property is something that exists only in a substance. These categories basically correspond to the linguistic distinction between subject and predicate, and as such correspond to obvious truths of reality. Some things exist in themselves, others only exist as features of other things. For example, "whiteness," a property possessed by many things (e.g., snow, robes), does not exist on its own, whereas an individual (e.g., a specific tree, a specific sword, King David) exists on its own rather than as a property of another thing.
Efficient Cause
The next cause is the "efficient cause," which denotes what causes a change to occur. This appears in Genesis 1:1: God creates the heavens and the earth. It appears again in Genesis 5:14, where God commands Noah to make an ark. Later in Exodus 15:6, Moses speaks in his song "Your right hand, O Lord, glorious in power, your right hand, O Lord, shatters the enemy." In all three texts, the author asserts that a being produced (or should produce) an effect of some kind.
Final Cause
The last cause is, appropriately, the "final cause," which is the purpose or aim of something. According to Aristotle, this "aimed" aspect of reality is evident everywhere in the world, not simply in conscious human beings. Rather, even unconscious beings reflect being aimed at a certain end, when they continually behave the same way. They show they have an intrinsic tendency to act a certain way by regularly doing so. This phenomenon appears, for example, in Genesis 1:12: "The earth brought forth vegetation, plants yielding seed according to their own kinds, and trees bearing fruit in which is their seed, each according to its kind." Of course, the narrator is giving an etiology of plants and trees, assuming our knowledge that plants and trees continue to behave this way. But that repeated reproductive behavior of vegetation displays "directedness," that is, it displays an intrinsic tendency these things have to behave in a certain manner.
In the gospels, Jesus actually argued according to precisely this logic:
Do men gather grapes from thornbushes or figs from thistles? Even so, every good tree bears good fruit, but a bad tree bears bad fruit. A good tree cannot bear bad fruit, nor can a bad tree bear good fruit. Every tree that does not bear good fruit is cut down and thrown into the fire. Therefore by their fruits you will know them. (Matt. 7:16b-20)
Act and Potency
Two more concepts are known for their Aristotelian provenance: act and potency. Act, or actuality, denotes the existence of things. Genesis affirms this in the first verse, when it tells us that God created the world. At that point in history, then, the world existed, or was actual. Potency is Aristotle's term for the idea of possibility. That is, potency refers to the idea that some things (whether substances or properties) could exist that presently do not exist. Such an idea is implied in Genesis 1:28: "Be fruitful and multiply and fill the earth and subdue it, and have dominion over the fish of the sea and over the birds of the heavens and over every living thing that moves on the earth." Here God commands human beings to multiply. This denotes bringing further human beings into existence which did not exist at that point. It entails that such a state of affairs would be possible, though it wasn't at that point actual. Which is (for Aristotle) just another way of saying, the first human couple had the potency of multiplying, though that potency had not yet been actualized.
The Principle of Causality
One last important philosophical principle deserves mention. Aristotle affirmed the basic idea that is now known as the principle of causality. This principle states that no potency can be made actual except by something already actual. This principle is assumed in various places in scripture, but two examples are present in the Gospel of John. In John 5:36 Jesus says, "But the testimony that I have is greater than that of John. For the works that the Father has given me to accomplish, the very works that I am doing, bear witness about me that the Father has sent me." Jesus is teaching here that his works "bear witness," and the way that they do this is by being effects that demand a divine cause as their only sufficient explanation. As Nicodemus says earlier in the Gospel (3:2), "Rabbi, we know that you are a teacher come from God, for no one can do these signs that you do unless God is with him."
This kind of reasoning would not be natural unless we assumed that every effect must have a cause, that every possibility that is made real (like a miraculous effect) must be made real by something already actual. Otherwise, e.g., people could come back to life from the dead for no reason at all, and no inference from any effect to any cause could be reliable, for it would always be equally possible that an effect was uncaused.
A reader may now reply: "What was the point of showing that scripture assumes the obvious?" But that is the point. In large measure, Aristotle's metaphysics is just that, the obvious truth about the world. It's also common sense, because, in general, human beings grasp the obvious. 1 Indeed, Aristotle intended to describe common sense, and to defend it against arguments like those of the Greek thinkers Parmenides and Heraclitus who had sought to argue that common sense beliefs about the world were mistaken. And it is as common sense, rather than as a doctrine originally thought up by Aristotle, that the writers of scripture teach the same things that the Philosopher did.
Interestingly, Dr. Susan Gelman has argued that children have an innate tendency to think like little Aristotelians about metaphysics in her book The Essential Child: Origins of Essentialism in Everyday Thought. ↩
Written by Andrew Fulford
Andrew Fulford is currently studying for a PhD in Reformation history.

redpajama_set_name: RedPajamaCommonCrawl


All iPads Gamers, how much space are your games taking up?
Discussion in 'iPad' started by SnackeyG, Dec 6, 2013.
I'm trying to decide on a 32gb or 64gb iPad. Storage wise it will hold my music(6gb right now) and games. Regular apps use under 3gb total.
I don't want to undercut my storage because I can't stand not having room. I'm an all or nothing guy. I of course also don't want to waste $100 if I'll never fill it before it's time to upgrade in 2-3 years.
Also how many games do you have?
Optional - How many over 100 mb? Over 500mb? 1gb plus?
Now, I have a 32GB iPad and have used all the storage. But that's because I have 10GB of music and I need to delete about half or more of those 102 apps.
I have exactly 60 games. I only play 65% of them. I should delete the ones I don't play, but since I have the room, I hang on to them. 32GB is the perfect size for me.
Out of 32GB, I have 12.5 of space left and will be deleting apps that I haven't used in ages or ever used.


THAT predators often hunt in packs is a commonplace. Wolves do it. Killer whales do it. Even Velociraptor, a species of dinosaur made famous by "Jurassic Park", is believed to have done it. These are, or were, all intelligent species, capable of exchanging and interpreting information. But the logic of pack hunting, that many may achieve what one alone cannot, and that individual pack members may perform different roles, does not depend on intelligence. Indeed, evidence has now emerged that this logic applies to viruses, the simplest biological entities of all. It was published this week in Cell, by Edze Westra and Stineke van Houte at the University of Exeter, in England.
The viruses in question are bacteriophages, which "hunt" bacteria. They do not eat their prey. Rather, they take over its genetic apparatus to create replicas of themselves, killing the host as a consequence. To do so they have to penetrate a bacterium's cell wall and then subvert its internal defences, of which there are several. One of the best known, because it is the basis of an emerging gene-editing technology (see article), is called CRISPR. The CRISPR system detects and cuts up alien DNA. In the wild, such DNA will almost always have come from a virus. To counter this, some bacteriophages have evolved ways of gumming up CRISPR's cellular machinery. Dr Westra and Dr van Houte have shown that, in essence, such phages collaborate. Some do the gumming. Others hijack the genetic apparatus.
Dr Westra and Dr van Houte were able to deduce what was going on by watching oddities in the rise and fall of bacterial and phage numbers in cultures. On the face of things, a population of CRISPR-armed bacteria would be expected to plummet in the presence of phages counter-armed with anti-CRISPR mechanisms. But this does not always happen. Instead, bacteriologists studying the matter have noticed that phages with anti-CRISPR traits are sometimes unsuccessful in attacking bacteria with CRISPR defences, and die out. Perplexed by this, the two researchers decided to take a closer look.
To do so, they and their colleagues generated a population of CRISPR-armed bacteria and another of phages with anti-CRISPR traits, and monitored exactly what happened when they introduced the one to the other. To start with, the density of viruses always declined. In other words, most of the early anti-CRISPR attacks were unsuccessful. These failed attacks did not leave the bacteria unscathed though. They resulted in the CRISPR defensive systems being weakened, a process the researchers were able to track by stopping an attack in midstream, washing away the phages, and testing the ability of the remaining bacteria to chop up alien DNA.
After this initial fall in viral numbers, if the culture was left long enough—and if there were enough phages in the first place—things eventually turned round. As the number of bacteria with weakened defences increased, more and more of them were subject to subsequent, lethal attacks, resulting in the creation of more phages. Ultimately, as the phages multiplied, the bacteria were overwhelmed and wiped out. Whether bacteria or viruses prevailed thus depended on the initial ratios of the two. Below a certain threshold of phage abundance at the beginning, the bacteria prevailed; above it, the viruses did.
Intriguingly, the evolutionary success of the phages' approach depends on a second phenomenon—also first studied in social animals—as well. This is kin selection. It relies on the fact that genetically determined behaviour that harms an individual can nevertheless spread if it disproportionately helps kin that carry the same genetic trait. In the case of the phages, the anti-CRISPR mechanism is exactly such a trait. Some viruses carrying it sacrifice themselves so that others may multiply.
Understanding this interaction between phages and bacteria is important, though, for reasons beyond its evolutionary elegance. One such is that phages are under consideration as alternatives to chemical antibiotics, particularly in situations where bugs are immune to those antibiotics. A second is that phages are a crucial, though ill-understood, part of the gut microbiome, the importance of which to human life is becoming clearer by the day. A third is that the interactions of phages and their hosts may be analogous to those of other viruses and other hosts, including human beings. Though animals do not employ CRISPR as part of their defence against viruses, they have a host of other antiviral mechanisms.
Dr Westra and Dr van Houte argue that theories about the spread of disease do not sufficiently take into account the possibility of these defences being damaged and weakened by failed attacks when determining the threats posed by specific sorts of viral pathogen. Monitoring such damage, and the degree to which it makes organisms vulnerable to later attacks, might improve control of the transmission of such diseases, and also the treatment of those who catch them.


Lead has been known to be a poison long before it was added to gasoline. When it was finally decided to remove lead from gasoline it took another 25 years to accomplish this task. The main reason was not due to health concerns but the lead caused clogging of the catalytic converters.
If you live in a much older home, you may have lead pipes that are increasing lead exposure through the water. Lead was in the paint in older homes.
Chronic exposure: central nervous system damage, gastrointestinal disturbances, anemia, reduced hemoglobin production , fatigue, headaches, tremors and hypertension. Very high lead exposure can result in lead encephalopathy with symptoms of hallucinations, convulsions, and delirium. Kidney dysfunction and possible injury has also been associated with chronic lead poisoning. Chronic over-exposure to lead has been implicated as a causative agent for the impairment of male and female reproductive capacity. Lead can cross the placental barrier and unborn children may suffer neurological damage or developmental problems due to excessive lead exposure. Lead and lead compounds are listed as an A3 carcinogen. IARC has listed lead compounds as Group 2A Carcinogens (Probably Carcinogenic to Humans) while lead metal is listed as Group 2B (Possibly Carcinogenic to Humans). The NTP lists lead and lead compounds as Reasonably Anticipated to be a Human Carcinogen. OSHA and the EU do not currently list lead as a human carcinogen.
The reasons for purposeful allowed exposure to lead is baffling. This is just another one that makes one question does the industry care about health or money or is it about bad health equals money?
Lead poisoning: A history of lead in pipes, makeup, cups, wine, paint, and gasoline.


Murad Abdu is a journalist based in Aden, Yemen.
Obtained a bachelor degree from Faculty of Languages - Aden University.
Worked as a reporter for international news agencies in the southern port city of Aden and neighboring provinces several years.
A news analysis about the ongoing armed confrontations in the Red Sea coast city of Hodeidah in western Yemen.


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Equip your printer with the Samsung Color Toner Cartridge Value Pack for professional quality document copies. The Samsung color laser printer toner is perfectly compatible with most of your Samsung printers in the CLT, CLX and Xpress range. The Samsung Color Toner Cartridge Value Pack comes with a collection of color series Yellow, Magenta, Cyan, and Black. This toner offers impressive prints for your home and office. The Samsung toner yields assured crisp output. This multi-color toner series makes your life comfortable by offering smear-free and lasting prints.


is a poem I wrote for this Flag Day/Father's Day/Trinity Sunday! Since all these things happened in one weekend this year, I figure I would try to tie together the "three-in-one" aspects of each. And the connection with all this? Well, my father is the one who inculcated with a deep reverence for our national flag and puts "Old Glory" out on our front porch every flag day. Also, my parents were married on Trinity Sunday many moons ago! So all my best wishes and blessings to all the fathers (especially and particularly my own, "Mr. B., the Entertainer's Entertainer!"), and flag wavers, and Christians seeking to grow in a deeper appreciation of the Mystery of the Holy Trinity.
"The Stripes They Bore on the Back of Christ....."


I'm Tosh! I'm a professional singer and songwriter out of Los Angeles, Ca. I've worked with Gold artist Quietdrive and many talented up and coming solo artists and bands. I specialize in Pop, EDM, and Rock.
I provide high quality vocals with thought out top line melodies. I'm a very versatile writer and can work with multiple genres. I work out of a home studio(Blue Bottle Rocket Stage 2, UA710, Apogee Duet) with high quality vocals that are hard to distinguish from a professional studio. i also have access to professional studios if needed be. I am well-versed in Logic and take care of a bit of post-editing myself. I would love the opportunity to work with you on your next project!
Excellent work! Professional and provides a high quality product!
Tosh is supremely talented, incredibly gifted a great communicator and entirely professional, had an amazing time collaborating, highly recommend! gifted !!
Tosh is a true creative! Her work was amazing and she was more than happy to experiment with her voice and writing. She brought what was missing to this track and made it truly something special.
Did a great job wrighting lyrics on our song! She nailed the topic we had in mind, and delivered fast. Totally worth the money.
Had another great experience working with Tosh. I feel privileged to mix such cool songs that she sends me. She is an amazing songwriter and I can't wait to see where her career takes her.
Tosh was great to work with! Very talented and a good sounding voice!
She is very eager to write what you are looking for, therefore she works as long as possible to deliver the right thing! This includes multiple versions!
I can definitely recommend working with her!
I had a great experience working with Tosh. She's a fantastic songwriter. I'm looking forward to working with her again!
It was a pleasure to work with Tosh. She is a top pro. Very efficient with time, and makes sure she get's the best from the song. Fully recommend working with Tosh, can't wait to do more with her in the near future!
will work with her again!
Fantastic ideas, and very easy to work with. Very professional and talented writer!
Tosh really knows how to deliver exactly what you're looking for. She's a very talented songwriter. Her lyrics and vocal melodies are really catchy and creative. It's always a pleasure to work with her. Highly recommend Tosh!
A: Acceptance, Coldplay, One Republic, Halsey, Paramore, Katy Perry, Sia, Twenty-one Pilots, and Jessie J.
A: Absolutely. My friend Mickey Shiloh! She's a killer vocalist and writer.
A: I promise to give a final product that's the best of my ability.
A: I LOVE what i do and I can be creative for a living.
A: I would love to work with Sia. She's a music genius in my book. She has a very unique vocal approach and style.
A: I work on music in the pop realm, whether is be Pop/Pock, Dance/Pop..


text:
Home/technology/Amazon announces next-generation Echo Dot and Echo Studio speakers
Amazon announces next-generation Echo Dot and Echo Studio speakers
flashcardtreeSeptember 28, 2022
The online presentation of Amazon Devices was rich in new features unveiled, and among these, some Echo speakers which benefit from a brand new version. This is particularly the case with the manufacturer's high-end speaker, the Echo Studio. For its second iteration, the eldest of the Echo family largely retains its initial design, still with the famous light ring, the mechanical buttons and the vent present on the lower part of the speaker. The biggest changes are internal.
First, the Amazon Echo Studio would be equipped with a brand new sound spatialization algorithm capable, according to the manufacturer, of "reproducing the performance of a stereo Hi-Fi system, in order to obtain a better feeling of vocal presence in the center of the stage and a better definition of the instruments positioned on the sides of the stage for a more immersive listening". Nothing is said about the system's capabilities on multi-channel content, but the speaker still supports Dolby Atmos and Sony 360 Reality Audio content. Amazon specifies that this algorithm has been added to its Echo Show 15 smart display and will be rolled out to other Echo speakers in the future via firmware updates.
The second improvement also affects signal processing, this time with an improvement in the speaker's frequency response for deeper bass and a gain in midrange clarity. No technical information has been provided to support this promise.
The fifth generation of Echo Dot and Echo Dot with Clock also does not change its design, but has the luxury of a new acoustic architecture. Among other things, the speaker benefits from a new long-excursion full-range speaker designed specifically for it, thus promising "twice as much bass", "50% less distortion" and "clear vocals". Suffice to say that you will have to wait to hear it to really know the sound performance.
The Echo Dot and Dot with Clock now have an accelerometer and a temperature sensor to expand their range of features: it will be possible to tap directly on the speaker to manage the playback of a song or stop an alarm, for example. Alexa will also be able to indicate the indoor temperature and activate certain connected objects according to the measured temperature. Promised to be brighter, the small display of the Dot with Clock will now be able to scroll through the name of a musical title and its artist or even the name of an event set in its personal agenda.
Last novelty distilled by Amazon during this presentation for the Echo Dot: the integration of a Wi-Fi eero repeater function. This functionality will also be unlocked by a firmware update on the fourth generation Echo speakers on October 20 and in the following months for the Echo Dot/with Clock.
Amazon has unfortunately "torpedoed" the presentation of its products and has, at the time of writing, not yet shared further details, in particular on the arrival and the price of these new products in France. We are obviously watching the arrival of information closely, which we will add as soon as possible. In any case, the Amazon Echo Studio and Echo Dot/Echo Dot with Clock have been announced at respective prices of $200 and $50. A "child" version of the Echo Dot, at $60, was also presented with a slightly revisited design and a one-year subscription to Amazon Kids+ included.
Netflix drops 1899 series after just one season
Acer unveils two electric scooters to bite into the urban mobility cake
Scientists invent a robot capable of escaping from a cage by liquefying
Opel relaunches its muscular label GSe under the claw of the electric
DxO PhotoLab 6: ever more powerful and accessible


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/ Maggie Koerth-Baker / 7:39 am Mon Dec 26, 2011
How long have we known that dinosaurs were birds?
I spent most of my childhood with books about dinosaurs that played up the ancient beasties as overgrown lizards. The connection between dinosaurs and birds, while kind of flipping obvious once somebody points it out, was not much discussed among laypeople until I was in my teens. (That would be the 1990s, FYI.)
But, among scientists, the idea of a dinosaur-bird relationship is nothing new. In fact, Thomas Henry Huxley was making that connection back in the 1860s. On the Dinosaur Tracking blog, Brian Switek tells the fascinating story of how Huxley started to realize that dinosaurs and birds were related—a discovery that's all the more impressive because he figured it out without the help of some of the key transitional fossils we have access to today.
Huxley did not suggest that birds were the direct descendants of dinosaurs. So much geologic time was unaccounted for, and so few dinosaurs were known, that Huxley could not point to any known fossil creature as the forerunner of birds. Instead he made his argument on anatomical grounds and removed the issue of time. Dinosaurs were proxies for what the actual bird ancestor would have been like, and flightless birds (such as the ostrich and emu) stood in for what Huxley thought was the most archaic bird type. (We now know that Huxley got this backwards—the earliest birds could fly, and flightless birds represent a secondary loss of that ability.) As Huxley went about collecting evidence for his case, though, he also gave dinosaurs an overhaul. They were not the bloated, plodding, rhinoceros-like creatures that Richard Owen had envisioned. Dinosaurs were more bird-like than anyone had imagined.
In October of 1867, Huxley met with John Philips, an English geologist and a curator of Oxford's museum. As Huxley related in his 1870 paper "Further Evidence of the Affinity Between the Dinosaurian Reptiles and Birds," Philips wanted to discuss details of marine reptiles called ichthyosaurs in his museum's collection, but as he and Huxley made their way over toward the displays they stopped to look at the bones of the carnivorous dinosaur Megalosaurus. Then Huxley spotted something peculiar:
As Prof. Phillips directed my attention to one after the other of the precious relics, my eye was suddenly caught by what I had never before seen, namely, the complete pectoral arch of the great reptile, consisting of a scapula and a coracoid ankylosed together. Here was a tangle at once unravelled. The coracoid was totally different from the bone described by Cuvier, and by all subsequent anatomists, under that name. What then was the latter bone? Clearly, if it did not belong to the shoulder-girdle it must form a part of the pelvis; and, in the pelvis, the ilium at once suggested itself as the only possible homologue. Comparison with skeletons of reptiles and of birds, close at hand, showed it to be not only an ilium, but an ilium which, though peculiar in its form and proportions, was eminently ornithic in its chief peculiarities.
Earlier naturalists had made a mistake. They had misidentified the shoulder girdle, and one part of what was thought to be part of the shoulder was actually part of the hip. Another strange piece, previously thought to be a clavicle, also turned out to belong to the pelvis. This rearrangement immediately gave the dinosaur a more bird-like character.
If you look at the bottom of the image at the top of this post, you can see how much the re-arrangement of megalosaurus' parts changed our conception of what the whole creature looked like. Where other scientists saw a lumpy, obese crocodile, Thomas Henry Huxley saw a saber-toothed chicken.
Image: Ballista via CC
birds / dinosaurs / History / Huxleys / reptiles / Science / wrong
The new £50 notes will feature Alan Turing (whilst HMG proposes bans on Turing complete computers AND working crypto)
The Bank of England has unveiled its new £50 notes, which had been earmarked to honour a distinguished British scientist, and which will feature Alan Turing, the WWII hero who discovered many of the foundational insights to both modern computing and cryptography, and whose work with the codebreakers of Bletchley Park are widely believed to […]
Science offices throughout U.S. government closing under Trump at alarming rate
The great science purge, they'll call it one day. Donald Trump is closing science offices throughout the federal government. 'As of June, around 85 percent of all scientific posts in the federal government, including an official scientific advisor to the President, were left unfilled,' write the editors of I F***ing Love Science blog in an […]
Show Your Stripes: visualizing climate change in your location by displaying 100 years of average temperatures in color bars
Ed Hawkins, a climate scientist, created Show Your Stripes as a way to easily visualize the past century's climate change: give it a location and it will render a series of stripes representing a century's worth of average annual temperatures (above: global average temperature); as Kottke notes: "The warming patterns for particular regions are not […]


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Edition #13: February 2019
Mary Sophou (left) from PRONIA with team members from the project 'Gambling – Secret no More', photo: Paul Jeffers
Cultural nuance is key to preventing gambling harm
by Lisa Clausen
Step inside any one of the Greek coffee shops found on many of Melbourne's shopping strips and you're likely to see tables of men exchanging cards, counters and coins through a swirl of conversation. Social gambling on games like cards and backgammon, or tavli, has been woven through Greek life for generations, and brought to Australia by migrants seeking to build new lives. For many older Greek–Australians, in particular, such pastimes are precious links to their traditions and culture, and strengthen vital social connections.
But, as in the broader Australian population, for some in the Greek community gambling can take a darker turn, affecting relationships, finances and health. To open up a culturally sensitive conversation about this rarely discussed problem, particularly among older Greek–Australians, the Melbourne-based Australian Greek Welfare Society, PRONIA, conceived the project Gambling – Secret No More. The year-long project was one of 12 the Victorian Responsible Gambling Foundation funded in 2018 for culturally and linguistically diverse (CALD) communities across the state.
Crossing the divide
Gambling – Secret No More involved a series of gambling awareness information sessions and resources highlighting the risks of gambling harm and the culturally relevant help at hand, including PRONIA's range of family, financial and health support services. PRONIA's family and community services manager, Mary Sophou, says gambling has always been a divisive topic in her community because of people's 'love–hate relationship' with it. 'We know gambling harm is an issue that needs a lot of community education and awareness, and that we, as an organisation, need to advocate and take a stand against it,' she says.
Social gambling on games like cards and backgammon, or tavli, has been woven through Greek life for generations.
While gambling has traditionally been the domain of older Greek men, Mary says services like PRONIA are facing new and difficult challenges. Younger people are flocking to online betting. Some of the thousands of migrants who have fled the economic crisis in Greece since 2014 are turning to gambling to dull the stress of resettlement. PRONIA's busy counselling and support staff also encounter increasing numbers of older Greek women facing financial difficulties because of pokies. Their traditional responsibilities in running households mean 'women gamblers are even more of a hidden topic – it's even more shameful,' says Mary.
The difficulty in talking about gambling harm was starkly revealed when Mary and her colleagues tried to tap into the influential network of Greek senior citizens clubs in the project's two target municipalities – Whittlesea and Monash; areas with both large populations of Greek–Australian residents and heavy annual gambling losses. Not one of the more than a dozen clubs PRONIA contacted was willing to host education sessions. 'They said they'd call back, or that their members weren't ready,' says Mary. 'We were surprised by the number of knock-backs we got.'
'Women gamblers are even more of a hidden topic – it's even more shameful.'
Mary Sophou, PRONIA
Undeterred, PRONIA approached two women-only senior citizens clubs, which hosted two major forums that were attended by the mayors of both municipalities, and where more than 100 women heard from a range of Greek-speaking experts. The bilingual setting and cultural nuances of those discussions, and of other information sessions, community activities and coverage in the local Greek media, was critical, says Mary. 'Unless you have information that is culturally and linguistically relevant, people just tune out.'
Signs of hope
Part of the impetus for the project was the community's poor awareness about the potential harm gambling can cause and how to tackle it. But feedback following the forums confirmed their positive effect, with more than 90 per cent of participants reporting they consequently better understood the signs of gambling harm. At one forum, a member even publicly challenged her club on why it regularly organised group outings to the casino. Mary says such conversations are hopeful signs that need long-term, multi-faceted encouragement. 'People are starting to think about this issue so we need to capitalise on that. People need long-term consistent messaging to change their behaviour; that is the challenge we face.'
'Unless you have information that is culturally and linguistically relevant, people just tune out.'
Gambling – Secret No More was made possible by the Foundation's prevention grants program for CALD communities. Last year, the program assisted projects in a range of emerging and established communities, including Cambodian, Greek, Serbian and Somali groups. Research shows that although people in CALD communities in Australia tend to gamble less than the general population, those who do are more likely to experience gambling harm and barriers to accessing help.
Janet Dore, the Foundation's interim CEO, says the program benefited from the rich partnerships existing in CALD communities, and also saw new connections forged. 'Being able to provide funding for projects that are run by CALD communities for CALD communities helps to break down cultural and linguistic boundaries. Many of these organisations are keen to continue to collaborate with their new partners on future prevention initiatives.'
If you have concerns about your gambling, or are affected by someone else's gambling, call Gambler's Help on 1800 858 858. We can arrange an interpreter for free if you need one.
Find out more about getting support.
#Edition 13
#Community education
#Counselling and treatment
#Gender and gambling
#Health and welfare professionals
#Mental health and gambling
#Public health
Lisa Clausen
Lisa Clausen is a writer and author based in Melbourne. A former Time magazine journalist, her writing has appeared in many publications, including The Monthly and Good Weekend.
Helping people realise they aren't the only ones
by Elizabeth Papettas
Supporting communities to know the risks
by Karen Harbutt
Gambling in culturally and linguistically diverse communities in Australia
by Marissa Dickins and Anna Thomas


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text:
Santa Anita to Install Equine MRI Machine
Karen Klawitter, CEO of the Southern California Equine Foundation, said Tuesday the machine which allows for horses to be scanned while standing should be fully operational "by early in the upcoming Santa Anita season."
By Associated Press	• Published on November 19, 2019 at 9:50 pm
Icon Sportswire via Getty Images
A fundraising campaign to install an equine MRI at Santa Anita has been successful, and it's expected to be in place during the troubled track's winter-spring meet that begins on Dec. 26.
Karen Klawitter, CEO of the Southern California Equine Foundation, said Tuesday the machine which allows for horses to be scanned while standing should be fully operational "by early in the upcoming Santa Anita season."
Funding came from a variety of groups and individuals, including prominent breeder and owner John Harris, veterinarians Vince Baker, Melinda Blue and Ryan Carpenter, Del Mar, Los Alamitos, the California Thoroughbred Trainers and California Thoroughbred Horsemen's Foundation.
The equine imaging center at Santa Anita already has nuclear scan technology and will add an equine PET scan, with track owner The Stronach Group kicking in $500,000 for it and another $200,000 coming from a private foundation dedicated to research on equine health issues.
There had been a Nov. 15 deadline to raise money for the equine MRI so it could be used this winter.
A total of 37 horses have died in racing and training at Santa Anita since last December, including Mongolian Groom in the $6 million Breeders' Cup Classic at the track earlier this month.