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\begin{document}
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\large



\title{Math 533, Spring 2020, HW9} 
\author{Due in class, April 20}
\maketitle

\begin{problem}
 Let $G_t(x) = (4\pi t)^{-n/2} e^{-|x|^2/4t}$, and if $f \in \sS'(\bR^n)$, let $u(x,t) = f * G_t(x)$. Prove 
 \begin{enumerate}
  \item $u$ satisfies $(\partial_t -\Delta)u = 0$ on $\bR^n \times (0, \infty)$ and $u(\cdot, t) \to f$ in $\sS'$ as $t \to 0$.
  \item If $f$ is a tempered function, then $u(x,t) \to f(x)$ a.e. as $t \to 0$.
 \end{enumerate}

\end{problem}

\begin{problem}
 Let $W_t$ be the inverse Fourier transform of $(2\pi |\xi|)^{-1} \sin (2\pi |\xi|t)$. Prove the following.
 \begin{enumerate}
  \item If $n=1$, $W_t = \frac{1}{2} \chi{(-t,t)}$
  \item If $n=3$, let $\sigma_R$ denote surface measure on the sphere $|x| = R$.  Then $\hat{\sigma}_R(\xi) = 2R|\xi|^{-1} \sin(2\pi R |\xi|)$, and hence $W_t = (4\pi t)^{-1} \sigma_t$.
  \item If $n = 2$, think of $\xi \in \bR^2$ as an element of $\bR^2 \times \{0\} \subset \bR^3$.  Transform the integral
  \[
   \frac{2 R \sin(2\pi R|\xi|)}{|\xi|} = \int_{|x| = R} e^{-2\pi i x \cdot \xi} d\sigma_R(x)
  \]
  as an integral over the disc $D_R = \{y: |y| \leq R\}$ in $\bR^2$.  Conclude that for $n=2$,
  \[
   W_t(x) = (2\pi)^{-1}(t^2-|x|^2)^{-\frac{1}{2}} \chi_{D_t}(x).
  \]

 \end{enumerate}

\end{problem}

\begin{problem}
 Solve $(\partial_t - \partial_x^2)u = 0$ on $(a,b) \times (0, \infty)$ with boundary conditions $u(x,0) = f(x)$ on $(a,b)$, $u(a,t) = u(b,t) = 0$ for $t>0$.  Solve this again, but with boundary condition $u(a,t) = u(b,t) = 0$ replaced by $\partial_x u(a,t) = \partial_x u(b,t)=0$.
\end{problem}



\begin{problem}
 If $\mu$ is a positive Borel measure on $\bT^1$ with $\mu(\bT^1)=1$, show that $|\hat{\mu}(k)|<1$ for all $k \neq 0$ unless $\mu$ is a convex combination of the point masses at $0, m^{-1}, ..., (m-1)m^{-1}$ for some $m \in \bN$, in which case $\hat{\mu}(km) =1$ for all $k \in \zed$.
\end{problem}



\begin{problem} Show that if $\Delta(\bR^n)$ is the set of finite linear combinations of point masses on $\bR^n$, then $\Delta(\bR^n)$ is vaguely dense in $M(\bR^n)$.
\end{problem}


\end{document}

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