proof of EX 4.13 Fourier analysis, stein

main.tex

\documentclass[a4paper]{article}
\usepackage{amsmath}
\usepackage{amssymb}
\newcommand{\abs}[1]{\lvert #1 \rvert}
\newcommand{\diff}{\mathrm{d}}
\newcommand{\idx}[1]{\mathbb{I}(#1)}
\usepackage{hyperref}

\begin{document}
I use a quite different method from yours.
For (1), we see that
\begin{equation}
    \begin{aligned}
        I=&\int_{\abs{x}\leq 1/2}\abs{H_t(x)}^2\diff x\\
        =&\int_{\abs{x}\leq 1/2}H_t(x)\overline{H_t(x)}\diff x\\
        =&\int_{\abs{x}\leq 1/2}\sum_{n}e^{-4\pi^2n^2t}e^{2\pi i n x}\sum_{k}e^{-4\pi^2k^2t}e^{-2\pi i k x}\\
        =& \sum_{n}\sum_{k}e^{-4\pi^2(n^2+k^2)t}\int_{\abs{x}\leq 1/2}e^{2\pi i (n-k)x}\diff x\\
        =& \sum_{n}\sum_{k}e^{-4\pi^2(n^2+k^2)t}\idx{n=k}\\
        =&\sum_{n}e^{-8\pi^2n^2t}
    \end{aligned}
\end{equation}
Note that
\begin{equation}
    \begin{aligned}
        I=&\sum_{n}e^{-8\pi^2n^2t}=1+2\sum_{n=1}^{\infty}e^{-8\pi^2n^2t}\\
        \leq& 1+2 \sum_{n=1}^{\infty}\int_{n-1}^n e^{-8\pi^2 x^2 t}\diff x=1+2\int_{0}^{\infty}e^{-8\pi^2 x^2 t}\diff x\\
        =&1+\int_{-\infty}^{+\infty}e^{-8\pi^2 x^2 t}\diff x
    \end{aligned}
\end{equation}
by the same trick we can show that
\begin{equation}
    \begin{aligned}
        I=&\sum_{n}e^{-8\pi^2n^2t}=-1+2\sum_{n=0}^{\infty}e^{-8\pi^2n^2t}\\
        \geq & -1+2 \sum_{n=0}^{\infty}\int_{n}^{n+1} e^{-8\pi^2 x^2 t}\diff x=-1+2\int_{0}^{\infty}e^{-8\pi^2 x^2 t}\diff x\\
        =&-1+\int_{-\infty}^{+\infty}e^{-8\pi^2 x^2 t}\diff x
    \end{aligned}
\end{equation}
and by the Normal distribution $N(0,\sigma^2)$, we see that (the constant might be wrong, but it does not matter.)
\begin{equation}
    \int_{-\infty}^{+\infty}e^{-8\pi^2 x^2 t}\diff x=\int\exp\left(-\frac{x^2}{2\frac{1}{16\pi^2t}}\right)\diff x=\sqrt{2\pi t }\cdot 4\pi=O(t^{-1/2}).
\end{equation}
Thus $I=O(t^{-1/2})$.

For (2),by the hint we have $c_1\sin^2(\pi x)\leq x^2\leq C_1 \sin^2(\pi x)$, for $\abs{x}\leq 1/2$, i.e., $x^2=O(\sin^2(\pi x))$, we write $x^2=C \sin^2(\pi x)$.
\begin{equation}
\begin{aligned}
    I_2=&\int_{\abs{x}\leq 1/2}x^2\abs{H_t(x)}^2\diff x=C\int\sin^2(\pi x)\abs{H_t(x)}^2\diff x\\
    =&C\int \frac{1}{2}(1-\cos(2\pi x))\sum_{n}\sum_{k}e^{-4\pi^2(n^2+k^2)t}e^{2\pi i (n-k)x} \diff x\\
    =&\frac{C}{2}\sum_{n}\sum_{k}e^{-4\pi^2(n^2+k^2)t}\int e^{2\pi i (n-k)x}\left(1-\frac{e^{2\pi i x}+e^{-2\pi i x}}{2}\right)\diff x\\
    =&\frac{C}{2}\sum_{n}\sum_{k}e^{-4\pi^2(n^2+k^2)t}\int e^{2\pi i (n-k)x}\diff x\\
    &-\frac{C}{4}\sum_{n}\sum_{k}e^{-4\pi^2(n^2+k^2)t}\int e^{2\pi i (n+1-k)x}\diff x\\
    &-\frac{C}{4}\sum_{n}\sum_{k}e^{-4\pi^2(n^2+k^2)t}\int e^{2\pi i (n-1-k)x}\diff x\\
    =&\frac{C}{2}\sum_{n}\sum_{k}e^{-4\pi^2(n^2+k^2)t}\idx{n=k}\\
    &-\frac{C}{4}\sum_{n}\sum_{k}e^{-4\pi^2(n^2+k^2)t}\idx{n=k-1}\\
    &-\frac{C}{4}\sum_{n}\sum_{k}e^{-4\pi^2(n^2+k^2)t}\idx {n=k+1}\\
    =&\frac{C}{2}\sum_{n}e^{-8\pi^2n^2t}-\frac{C}{4}\sum_{n}\left[e^{-4\pi^2(2n^2+2n+1)t}+e^{-4\pi^2(2n^2-2n+1)t}\right]\\
    =&\frac{C}{2}\sum_{n}e^{-8\pi^2n^2t}\left[1-\frac{e^{-4\pi^2(1+2n)t}+e^{-4\pi^2(1-2n)t}}{2}\right]\\
    =&\frac{C}{2}\sum_{n}e^{-8\pi^2n^2t}\left[1-e^{-4\pi^2t}\frac{e^{-8\pi^2nt}+e^{8\pi^2nt}}{2}\right]\\
\end{aligned}
\end{equation}
Note that $2\leq e^x +e^{-x}\leq 2 e^{\abs{x}}$

when $0\leq t \leq \delta$, $e^x \sim 1+x $, one can see
\begin{equation}
    1-e^{-4\pi^2t}\frac{e^{-8\pi^2nt}+e^{8\pi^2nt}}{2}\leq 1-e^{-4\pi^2t}\leq c 4\pi^2t .
\end{equation}
and
\begin{equation}
    1-e^{-4\pi^2t}\frac{e^{-8\pi^2nt}+e^{8\pi^2nt}}{2}\geq  1-e^{-4\pi^2t}e^{8\pi^2\abs{n}t}\geq c 4\pi^2(2\abs{n}-1)t .
\end{equation}
Finally we have
\begin{equation}
    \begin{aligned}
        I_2\leq K t \sum_{n}e^{-8\pi^2n^2t}=t O(t^{-1/2})=O(t^{1/2}),
    \end{aligned}
\end{equation}
and
\begin{equation}
    I_2\leq Kt \sum_{n}e^{-8\pi^2n^2t}(2\abs{n}-1)=2Kt \sum_{n}\abs{n}e^{-8\pi^2n^2t} -O(t^{1/2}).
\end{equation}
Note that
\begin{equation}
    \sum_{n}\abs{n}e^{-8\pi^2n^2t} \sim\int \abs{x}e^{-8\pi^2x^2t }\diff x=O(t^{-1/2}),
\end{equation}
which is $E(\abs{X})$ for $X\sim N(0,\sigma^2)$, and $E(\abs{X})=\sigma\sqrt{\frac{2}{\pi}}$.
For more details, see \href{https://math.stackexchange.com/questions/555831/the-expectation-of-absolute-value-of-random-variables}{this question}.

By far we have proved that $I_2=O(t^{1/2})$.
\end{document}