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\begin{document}
\begin{enumerate}
\centerline{\sc Math 597.  Homework 13\\ Bryden Cais}
\medskip

\item
\begin{enumerate}
    \item  It is clear that for $f,g\in L^1(T)$ and $2\pi$ periodic the convolution $f*g$ is also $2\pi$ periodic:
    indeed, we have
    \begin{align*}
        f*g(x+2\pi)&=\int_T f(y)g(x+2\pi-y)\,dy\\
        &=\int_T f(y)g(x-y)\,dy\\
        \intertext{since $g$ is $2\pi$ periodic,}
        &=f*g(x).
    \end{align*}
    Moreover, observe that $|f(y)g(x-y)|$ is nonnegative and measurable.  Since Lebesgue measure is complete and $\sigma$-finite,
    it follows from the Tonelli Theorem that
    \begin{align*}\
        \int_T |f*g(x)|\,dx &\leq \int_T\int_T |f(y)||g(x-y)|\,dy\,dx\\
        &=\int_T\int_T |f(y)||g(x-y)|\,dx\,dy\\
        &=\int_T\int_T |f(y)||g(x)|\,dx\,dy\\
        \intertext{since $g$ is $2\pi$ periodic,}
        &=\|f\|_{L^1(T)} \|g\|_{L^1(T)},
    \end{align*}
    so that
    $$\|f*g\|_{L^1(T)}\leq \|f\|_{L^1(T)} \|g\|_{L^1(T)}.$$

    \item Let $\e >0$, and let $g$ be a continuous $2\pi$ periodic function with $\|f-g\| < \e/3$.
    Moreover, since $g*K_n\rightarrow g$ uniformly (since $g$ is continuous), let $N$ be such that for all
    $n>N$ we have $|g*K_n(x)-g(x)|<\e/6\pi$ for all $x\in T$.  Then we have
    \begin{align*}
        \|f*K_n-f\|&=\|f*K_n-g*K_n+g*K_n-g+g-f\|\\
        &\leq \|(f-g)*K_n\|+\|g*K_n-g\|+\|g-f\|\\
        &\leq \|f-g\|\|K_n\| +\|g*K_n-g\|+\|g-f\|\\
        \intertext{by part (i),}
        &\leq \frac{\e}{3} + \int_T \frac{\e}{6\pi}\,dx+ \frac{\e}{3}\\
        \intertext{where we have used the fact that $\|K_n\|=1$ since $K_n>0$ and $\int_T K_n=1$,}
        &\leq \e.
    \end{align*}
    It follows that the $n^{\text{th}}$ Cas\`{a}ro partial sum of the Fourier Series of $f$ tends to $f$
    in $L^1$ as $n\rightarrow\infty$.

\end{enumerate}

\item
\begin{enumerate}
    \item Suppose that $f\in C^k(T)$ has Fourier Series $\sum_{j\in\Z} c_je^{ijx}$.
    Then we have $c_j=\frac{1}{2\pi}\int_T f(x)e^{-ijx}\,dx$.
    Now, if $l <k$ then $f\in C^l(T)$ and $f^{(l+1)}$ exists and is continuous.  It follows
    that
    \begin{align*}
        \int_T f^{(l)}(x)e^{-ijx}\,dx - \int_T \frac{1}{ij}f^{(l+1)}(x)e^{-ijx}\,dx &= -\frac{1}{ij}f^{(l)}(x)e^{-ijx}\Bigg]_0^{2\pi}\\
        &=0
    \end{align*}
    since $f^{(l)}$ is $2\pi$ periodic.  Thus we have for $l < k$
    $$\int_T f^{(l+1)}(x)e^{-ijx}\,dx=(ij)\int_T f^{(l)}(x)e^{-ijx}\,dx.$$
    It then follows by induction that $f^{(k)}$ has Fourier Series
    $$\sum_{j\in\Z} (ij)^k c_j e^{ijx}.$$

    \item Suppose that $f\in C^k(T)$.  Then since $T$ is compact and $f,f^{(k)}$ are continuous, they are both bounded in
    absolute value on $T$; it follows that $\|f\|^2_2$ and $\|f^{(k)}\|^2_2$ are finite.  On the other hand, by part (i)
    and Parseval's identity, we have
    \begin{align*}
        \|f\|_2^2 &= \sum_{j\in\Z} |c_j|^2,\\
        \intertext{and}
        \|f^{(k)}\|_2^2 &= \sum_{j\in\Z} |j^k c_j|^2.
    \end{align*}
    Since $\|f\|_2^2,\ \|f^{(k)}\|_2^2 <\infty$ we see that
    $$\|f\|_2^2+\ \|f^{(k)}\|_2^2=\sum_{j\in\Z} (1+|j|^{2k})|c_j|^2=\|f\|^2_{H^k}<\infty$$
    where we are able to rearrange the sum however we like since all the terms are positive.

    \item Suppose that $f\in H^s(T)$ has Fourier Series $\sum_{j\in\Z} c_je^{ijx}$.
    Then we have
    \begin{align*}
        \sum_{j\in \Z}|c_j| &= \sum_{j\in \Z}|c_j|\sqrt{1+|j|^{2s}}\frac{1}{\sqrt{1+|j|^{2s}}}\\
        &\leq \left(\sum_{j\in\Z}(1+|j|^{2s})|c_j|^2\right)^{1/2} \left(\sum_{j\in\Z}\frac{1}{1+|j|^{2s}}\right)^{1/2}\\
        \intertext{by Minkowski's identity,}
        &\leq \|f\|_{H^s}\left(\sum_{j\in\Z}\frac{1}{1+|j|^{2s}}\right)^{1/2}\\
        & <\infty,
    \end{align*}
    since $f\in H^s(T)$ and we have used the fact the $s>1/2$ so the sum $\sum_{j\in\Z}\frac{1}{1+|j|^{2s}}$ converges.

    \item Let $f\in H^s(T)$ with $t>1/2$.  Then by part (c) the Fourier Series for $f$ converges {\em uniformly}
    to $f$ on $T$; this follows from the Weirstrass $M$-test and the fact that $\sum |c_j|=\sum |c_je^{ijx}|<\infty$.
    Thus, the Fourier Series of $f$ converges to a continuous function that agrees with $f$ a.e., as required.

    \item
    Since $f\in H^{l+s}(T)$ for any $l\leq k$, we have
    \begin{align*}
        \sum_{j\in \Z}|j^l c_j| &= \sum_{j\in \Z}|j^l c_j|\sqrt{1/|j|^{2l}+|j|^{2s}}\frac{1}{\sqrt{1/|j|^{2l}+|j|^{2s}}}\\
        &\leq \left(\sum_{j\in\Z}(1+|j|^{2(s+l)})|c_j|^2\right)^{1/2} \left(\sum_{j\in\Z}\frac{1}{1/|j|^{2l}+|j|^{2s}}\right)^{1/2}\\
        \intertext{by Minkowski's identity,}
        &\leq \|f\|_{H^{l+s}}\left(\sum_{j\in\Z}\frac{1}{1/|j|^{2l}+|j|^{2s}}\right)^{1/2}\\
        & <\infty,
    \end{align*}
    as in part (c). We now proceed by induction.  Suppose that for $l<k$ $f\in C^l(T)$ and set $f_n=\sum_{-n}^n c_je^{ijx}$.
    Then our calculations above (via the Weierstrass $M$-test) show that $f_n^{(l)}$ converges uniformly to $f^{(l)}$
    and that $f^{(l+1)}=\lim_{n\rightarrow\infty} f_n^{(l+1)}$ exists a.e.  (See, for example, \cite{rud}[p. 152]).
    In fact, it is not difficult to see, as before, that $f_n^{(l+1)}$ converges uniformly, so that $f\in C^{l+1}(T)$.
    The base case is part (d) so that $f\in C^k(T)$ by induction.

    \item Let $f\in C^{\infty}$ have Fourier Series $\sum_{j\in\Z} c_je^{ijx}$.  Then by part (a), we find that
    $Sf$ has Fourier series $\sum_{j\in\Z} d_je^{ijx}$ where
    $$d_j=c_j \sum_{l=0}^k (ij)^l a_l,$$
    where for convenience we set $a_k=1$.  Now set $p(z)=\sum_{l=0}^k a_k (iz)^k$.
    Let $g\in C^{\infty}$ have Fourier Series $\sum_{j\in\Z} b_je^{ijx}$.
    As long as $p(j)\neq 0$
    for any integer $j$, the Fourier Series $\sum_{j\in\Z} \frac{b_j}{p(j)}e^{ijx}$ defines a $C^{\infty}$
    function $f$ with $Sf=g$ so that $S$ in this case is surjective (the fact that $f\in C^{\infty}(T)$
    follows from the fact that for large $j$ we have $|p(j)|>1$ so that $\sum_{j\in\Z} |j^k b_j/p(j)|$
    converges for all $k>0$ since the tail end tends to 0 since $g\in C^{\infty}(T)$).  Conversely, if
    $p(j)=0$ for some integer $j$, then $e^{ijx}\in C^{\infty}(T)$ is {\em not} in the image of $S$
    (since the coefficient of $e^{ijx}$ in the Fourier Series of $Sf$ for any $f$ is 0 as we calculated above).
    Thus, $S$ is surjective if and only if $p(z)$ has no integer roots.
\end{enumerate}



\end{enumerate}

\begin{thebibliography}{10}
    \bibitem{rud} Rudin, W. {\em Principles of Mathematical Analysis}.  McGraw-Hill, Inc. 1976.
\end{thebibliography}
\end{document}
