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\begin{document}
\begin{center}
{\bf Math 6210 - Homework 5}\\
Try to finish before the Thanksgiving break.
\end{center}

Let $([0,1], \cM, m)$ be the standard Lebesgue measure space on $[0,1]$ and let $L^2[0,1] = L^2(m)$ (using the notation from Rudin).
\begin{enumerate}
\item Define $T:L^2[0,1] \to L^2[0,1]$ by $(Tf)(t) = tf(t)$. Show that:
\begin{enumerate}
\item $T$ is bounded and symmetric with $\|T\| = 1$;

\item has no eigenvalues;

\item and $T-\lambda I$ is surjective if and only if $\lambda \not\in [0,1]$.
\end{enumerate}
\end{enumerate}
Now let $([0,1]^n, \cM, m)$ be Lebesgue measure on the cube in $\reals^n$ and let $L^2([0,1]^n) = L^2(m)$. We want to find an orthonormal basis for $L^2([0,1]^n)$. Our strategy will be the same as it was for $L^1[0,1]$.

We can also define Fourier series for periodic functions on $\reals^n$. (If you like you can simplify things by assuming $n=2$.)

We need a bit of notation. If $\bxi \in \integers^n$ and $\bx \in \reals^n$ then $\bxi \cdot \bx  = \xi_1 x_1 + \cdots + \xi_n x_n$.  Let $|\bxi|_{\infty} = \max\{|\xi_1|, \dots, |\xi_n|\}$.

A function $f: \reals^n \to \cx$ is periodic if $f(\bx)  = f(\bx + \bxi)$ for all $\bx \in  \reals^n$ and $\bxi \in \integers^n$. Let $C(\mT^n)$ be continuous periodic functions on $\reals^n$. Of course, any function in $C(\mT^n)$ can be restricted to a function in $L^2([0,1]^n)$. We want to show that
the functions $e_\bxi(\bx) = e^{2\pi \imath \bxi \cdot \bx}$ are an orthonormal basis for $L^2([0,1]^n)$.
\begin{enumerate}
\setcounter{enumi}{1}
\item Show that $(e_{\bxi_0}, e_{\bxi_1}) = 1$ if $\bxi_0 = \bxi_1$ and $(e_{\bxi_0},e_{\bxi_1}) = 0$ if $\bxi_0 \neq \bxi_1$.

\item Show that $e_{\bxi}(\bx) e_{\bxi}(\by) = e_{\bxi}(\bx + \by)$.

\item (Optional) Let $$D_N(x) = \sum_{k = -N}^N e^{2\pi \imath k x}$$ and show that $$D_N(x)= \frac{\sin(\pi(2N+1)x)}{\sin(\pi x)}.$$

\item (Optional) Let $$K_N(x) = \frac{1}{N} \sum_{k=0}^{N-1} D_k(x)$$ and show that
$$K_N(x) = \frac{1}{N} \left(\frac{\sin(N\pi x)}{\sin(\pi x)}\right)^2.$$

\item Let 
$$ {\bf K}_N(\bx) = \frac{1}{N^n}\prod_{j=1}^n\sum_{M=0}^{N-1}  \sum_{k=-M}^M e^{2\pi \imath k x_j}.$$
Show that
$${\bf K}_N(\bx)
= \frac{1}{N^n}\prod_{j=1}^n \left(\frac{\sin (N\pi x_j)}{\sin (\pi x_j)}\right)^2$$
and that ${\bf K}_N(\bx)$ is a finite linear combination of $e_\bxi(\bx)$.
Conclude that
$$f_N(\bx) = \int_{[0,1]^n} f(\by){\bf K}_N(\bx - \by) dm(\by)$$
is a finite linear combination of $e_\bxi(\bx)$.

\item Periodic functions $\Phi_k \in C(\mT^n)$ form an {\em approximate identity} if
\begin{enumerate}
\item $\int_{[0,1]^n} \Phi_k(\bx) dm(\bx)=1$;

\item $\sup_n \int_{[0,1]^n} |\Phi_k(\bx)| dm(\bx) < \infty$;

\item For all $\delta>0$, $\int_{1/2>|\bx|> \delta} |\Phi_k(\bx)| dm(\bx) \to 0$.
\end{enumerate}
If $\Phi_k$ is an approximate identity and $f$ is continuous show that
$$ \int_{[0,1]^n} f(\by)\Phi_k(\bx - \by) dm(\by) \to f(\bx)$$
uniformly.
\item Show that ${\bf K}_N(\bx)$ is an approximate identity and therefore if $f$ is in $C(\mT^n)$ show that $f_N$ converges to $f$ uniformly. Conclude that the $e_\bxi$ are an orthonormal basis for $L^2([0,1]^n)$.

{\bf Hint:} If $[a,b]$ is an interval show that
$$ \int_{[a,b] \times [-1/2,1/2]^{n-1}} {\bf K}_N(\bx) dm(\bx) = \int_a^b K_N(x) dm(x)$$
and therefore for all $\delta> 0$ we have
$$\underset{N\to \infty}{\lim} \int_{[\delta,1/2] \times [-1/2,1/2]^{n-1}} {\bf K}_N(\bx) dm(\bx) \to 0$$
with the same statement holding if we replace $[\delta, 1/2]$ with $[-1/2, -\delta]$.
Use the fact that the set of $\bx$ with $1/2 > |\bx|> \delta$ can be covered by $2n$ sets of the form $[\delta', 1/2] \times [-1/2, 1/2]^{n-1}$ or $[-1/2, -\delta'] \times [-1/2,1/2]^{n-1}$ for some $\delta'>0$ to show that ${\bf K}_N(\bx)$ satisfies (c).

\item Let $H^n_s$ be sequences $u_\bxi$ indexed by $\bxi \in \integers^n$ such that
$$\sum_{\bxi \in \integers^n} (1 + |\bxi|^2)^{s }|u_\bxi|^2 < \infty.$$
Show that if $s \ge \lfloor n/2 \rfloor + 1$ and $u \in H^n_s$ then the series $\sum |u_\bxi|$ converges. Conclude that $\sum u_\bxi e_\bxi(\bx)$ converges to a continuous function. (Hint: Write $|u_\bxi| = (1 +|\bxi|^2)^{-s/2} (1 + |\bxi|^2)^{s/2} |u_\bxi|$ and apply the Cauchy-Schwarz inequality. You will use the assumption that $s \ge \lfloor n/2 \rfloor +1$ to show that $\sum (1+|\bxi|^2)^{-s}$ converges.)

If $s\ge \lfloor n/2 \rfloor + m + 1$ show $\sum u_\bxi e_\bxi(\bx)$ has partial derivatives of all order $\le m$.

\item We can make $H^n_s$ into a Hilbert space by defining the inner product
$$(u, v)_s  = \sum_{\bxi \in \integers^n} (1+|\bxi|^2)^s u_\bxi \overline{v}_\bxi.$$
If $s>t$ show that the inclusion of $H^n_s$ in $H^n_t$ is a compact operator.

%In class we saw that if $M \subset \cH$ is a closed linear subspace then every $x \in \cH$ has a unique representation $x = Px + Qx$ with $Px \in M$ and $Qx \in M^\bot$ where $P$ and $Q$ are linear maps. The map $P$ is the {\em orthogonal projection} from $\cH$ to $M$.
%
%\item Let $A \subset [0,1]$ be Lebesgue measurable. Show that $L^2(A)$ is a closed subspace of $L^2[0,1]$ and that the map $f \mapsto f\chi_A$ is the orthogonal projection of $L^2[0,1]$ onto $L^2(A)$.
%
%The spectral theorem that we proved in class doesn't apply to $T$ (why not?). However, there is another version of the spectral theorem that applies to all bounded, symmetric operators. To state it we need the definition of a {\em spectral resolution} of a Hilbert space.
%
%Let $\cH$ be a Hilbert space and $P(\cH)$ orthogonal projections. Then $E: \reals \to P(\cH)$ is a spectral resolution of $\cH$ if:
%\begin{enumerate}
%\item $\|E(\lambda)f\|$ is an increasing function in $\lambda$ for every $f \in \cH$;
%
%\item there is a compact interval $[a,b]$ such that $E(\lambda) = 0$ if $\lambda<a$ and $E(\lambda) = I$ if $t>b$;
%
%\item $E(\lambda)$ is right continuous in the sense that $\underset{\mu \rightarrow \lambda^+}{\lim} E(\mu) = E(\lambda)$.
%\end{enumerate}
%
%The spectral theorem for bounded symmetric operators is then:
%\begin{theorem}\label{spectral}
%If $T$ is a bounded, symmetric operator then there exists a spectral resolution $E$ such that
%$$(Tf,g) = \int_a^b \lambda(E(\lambda)f,g) d\lambda.$$
%\end{theorem}
%
%\item If $(f,g) = (f', g)$ for all $g \in \cH$ show that $f = f'$.
%
%\item If $T$ is a compact, symmetric operator deduce Theorem \ref{spectral} from the spectral theorem that we proved in class.

\end{enumerate}


\end{document}