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\begin{document}
\begin{enumerate}
\centerline{\sc Math 597.  Homework 11\\ Bryden Cais}
\medskip

\item Let $a,b >1/2$ and define
$$f(x)=\begin{cases}\frac{1}{\sqrt{x}|\log x|^a} & \text{if}\ 2\leq x <\infty \\ \frac{1}{\sqrt{x}|\log x|^b} & \text{if}\ 0 < x <\frac{1}{2}\\
0 & \text{otherwise} \end{cases}.$$
We claim that $f\in L^2(\R,\mu_{\rm Leb})$ but $f\not\in L^p(\R,\mu_{\rm Leb})$ for any $p\neq 2$.
Indeed, we have
\begin{align*}
    \int_{\R} |f|^2\,\mathrm{d}\mu &= \int_0^{1/2} \frac{\mathrm{d}x}{x|\log x|^{2b}}+ \int_2^{\infty} \frac{\mathrm{d}x}{x|\log x|^{2a}}\\
    &=-\frac{1}{1-2b}|\log x|^{1-2b}\Bigg]_0^{1/2}+  \frac{1}{1-2a}(\log x)^{1-2a}\Bigg]_2^{\infty} <\infty\\
\end{align*}
since $a,b >\frac{1}{2}$ so that both terms above tend to 0 as $x\rightarrow 0$ and $x\rightarrow \infty$ respectively.

Suppose that $p<2$.  Then we have
\begin{align*}
    \int_{\R} |f|^p\,\mathrm{d}\mu &\geq  \int_{2}^{\infty} \frac{\mathrm{d}x}{x^{p/2}(\log x)^{pa}}\\
    &=\int_{\log 2}^{\infty} \frac{\mathrm{d}t}{e^{(p/2-1)t}t^{pa}},
\end{align*}
which diverges since $p<2$ so the integrand behaves like $t^{-pa} e^{\alpha t}$ for some $\alpha >0$ and hence is unbounded as $t\rightarrow \infty$.

On the other hand, if $p>2$ then we have
\begin{align*}
    \int_{\R} |f|^p\,\mathrm{d}\mu &\geq  \int_{0}^{1/2} \frac{\mathrm{d}x}{x^{p/2}|\log x|^{pb}}\\
    &=\int_{-\infty}^{-\log 2} \frac{\mathrm{d}t}{e^{(p/2-1)t}|t|^{pb}}\\
        &=\int_{\log 2}^{\infty} \frac{\mathrm{d}t}{e^{(1-p/2)t}t^{pb}},
\end{align*}
which diverges since $p>2$ so the integrand behaves like $t^{-pb} e^{\alpha t}$ for some $\alpha >0$ and hence is unbounded as $t\rightarrow \infty$.

Thus, $f\in L^2$ but $f\not\in L^p$ for any $p\neq 2$.

\item
\begin{enumerate}
    \item Define
    $$f(x)=\begin{cases} n^2(x-n)+1 & \text{if}\ x\in [n-1/n^2,n]\ \text{for each integer}\ n\geq 2\\
    -n^2(x-n)+1 & \text{if}\ x\in [n,n+1/n^2]\ \text{for each integer}\ n\geq 2\\
     0 & \text{otherwise}\end{cases}.$$
    Observe that $f$ is continuous and that $f(n)=1$ for each integer $n\geq 2$.  Moreover, we have
    $$\int_{\R} |f|\,\mathrm{d}\mu= \sum_{n\geq 2} \frac{1}{2}\frac{2}{n^2} <\infty$$
    so that $f\in L^1(\R,\mu_{\rm Leb})$.  On the other hand, $f(x)\not\rightarrow 0$ as $x\rightarrow \infty$
    since $f(n)=1$ for every integer $n\geq 2$.

    \item Let $f$ be a function on $\R$ with continuous derivative $f^{'}$ and $f,f^{'}\in L^1(\R,\mu_{\rm Leb})$.
    Then by the Fundamental Theorem of Calculus we have
    $$f(x)-f(a)=\int_a^x f^{'}(t)\,\mathrm{d}t.$$  Since $f^{'}\in L^1(\R,\mu_{\rm Leb}),$ the limit
    $$\lim_{x\rightarrow \infty} \int_a^x |f^{'}(t)|\,\mathrm{d}t$$
    exists and is finite.  It follows that
    $$\lim_{x\rightarrow \infty} (f(x)-f(a))=\lim_{x\rightarrow\infty} \int_a^x f^{'}(t)\,\mathrm{d}t$$
    exists and is finite, so that $\lim_{x\rightarrow \infty} f(x)=c<\infty$ exists and is finite.
    Pick $\e>0$ and let $N_{\e}$ be such that for all $x>N_{\e}$ we have $|f(x)-c|<\e$.  Then for $x>N_{\e}$ we have
    $$\int_{N_{\e}}^{x}|f(t)|\,\mathrm{d}t \geq  \int_{N_{\e}}^{x} (|c|-\e)\,\mathrm{d}{t}=(x-N_{\e})(|c|-\e).$$
    If $c\neq 0$ we can find $\e>0$ so that $|c|-\e>0$ whence, taking the limit as $x\rightarrow \infty$ above,
    we find that $f\not\in L^1(\R,\mu_{\rm Leb})$, a contradiction.  It follows that $c=0$ so that
    $\lim_{x\rightarrow \infty} f(x)=0$.
\end{enumerate}

\item
\begin{enumerate}
    \item Let $\phi=f_1\chi_{I_1}+\cdots +f_N\chi_{I_N}$ be a step function and set $I=\bigcup I_j$ and $F=|f_1|+\cdots +|f_N|$.
    Since each $I_j$ is a bounded interval, we have $\mu(I)<\infty$.  Let the closure of $I_j$ be $[a_j,b_j]$.  We have
    \begin{align*}
        \left|\int_{\R} \phi(x)e^{-ixt}\,\mathrm{d}x\right| &= \left|\sum_{j=1}^N f_j\frac{e^{-itb_j}-e^{-ita_j}}{-it}\right|\\
        &\left|\leq \sum_{j=1}^N f_j\frac{2}{t} \right|\leq \frac{2F}{|t|},
    \end{align*}
    since $|e^{-ixt}|\leq 1$ for any $x,t\in \R$.
    It follows that $|\hat{\phi}(t)|\rightarrow 0$ as $t\rightarrow \infty$ or $-\infty$, as required.

    \item Now let $f\in L^1(\R,\mu_{\rm Leb})$.  Then for any $\e>0$ there exists a step function $\phi$
    with $\|\phi-f\|_1 <\e$.  Thus we have
    \begin{align*}
        |\hat{f}(t)-\hat{\phi}(t)|&= \left|\int_{\R} (f(x)-\phi(x))e^{-ixt}\,\mathrm{d}{x}\right|\\
        &\leq \int_{\R} |f(x)-\phi(x)||e^{-ixt}|\,\mathrm{d}{x}\\
        &=\int_{\R} |f(x)-\phi(x)|\,\mathrm{d}{x}\\
        &=\|f-\phi\|_1<\e,
    \end{align*}
    by assumption.  Now since $\hat{\phi}(t)\rightarrow 0$ as $t\rightarrow \pm\infty,$
    for any $\e >0$ and any sequence $\{x_n\}$ with $x_n\rightarrow \pm\infty$, there exists $N$
    such that for all $n>N$ we have $|\hat{\phi}(x_n)|<\e$ so that
    $|\hat{f}(x_n)|<2\e$.  It follows that $\hat{f}(t)\rightarrow 0$ as $t\rightarrow \pm\infty.$
\end{enumerate}

\item Let $f,g\in L^1(\R,\mu_{\rm Leb})$.
\begin{enumerate}
    \item We have
    \begin{align*}
        \widehat{f*g}(t)&=\int_{\R}\int_{\R} g(y)f(x-y)e^{-ixt}\,\mathrm{d}y\,\mathrm{d}x\\
        &=\int_{\R}\int_{\R} g(y)e^{-iyt}\,f(x-y)e^{-i(x-y)t}\,\mathrm{d}y\,\mathrm{d}x\\
        &=\int_{\R}\int_{\R} g(y)e^{-iyt}\,f(x-y)e^{-i(x-y)t}\,\mathrm{d}x\,\mathrm{d}y\\
        \intertext{by Fubini's Theorem}
        &=\int_{\R}\int_{\R} g(y)e^{-iyt}\,f(u)e^{-iut}\,\mathrm{d}u\,\mathrm{d}y\\
        \intertext{by the linear change of variables $x-y=u$ (since $y$ is fixed in the $x$ integral)}
        &=\hat{f}(t)\hat{g}(t),
    \end{align*}
    as required.

    \item Suppose that there exists some $g\in L^1(\R,\mu_{\rm Leb})$ satisfying $f*g=f$ for all $f\in L^1(\R,\mu_{\rm Leb})$.
    Then by part (a), we have
    $$\widehat{f*g}=\hat{f}\hat{g}=\hat{f}.$$
    Now there exists some $f\in L^1(\R,\mu_{\rm Leb})$ with $\hat{f}(t)\neq 0$ for all $t\in \R$.  Indeed, we may take
    $$f(x)=\begin{cases}1 & \text{if}\ x\in [-\pi/2,\pi/2]\\ 0 & \text{otherwise}\end{cases}.$$
    Then it is not difficult to check that $\hat{f}(2)=2/t\neq 0$ for any $t\in \R$.  Thus, in order for the above
    equality to hold for all $f\in L^1$ we must have $\hat{g}\equiv 1$, which by 3(b) is impossible as
    we must have $\hat{g}(t)\rightarrow 0$ as $t\rightarrow \pm\infty$.
\end{enumerate}


\end{enumerate}
\end{document}
