# How can I show that this operator is bounded on $L^2$?

Consider the integral operator

$$Tf(x) = {\int}_{-\infty}^{\infty} \frac{\sin(x - y)}{x - y}f(y)dy$$

How can I show that $T$ is bounded on $L^2$?

I know that bounded means there is a constant $c$ independent of $f$ such that ${\| Tf \| }_{L^2} \leq c \|f\|_{L^2}$ but I do not seem to succeed on solving this.

Any help would be very much appreciated.

## 2 Answers

\begin{align} Tf(x) &= \int_{-\infty}^{\infty}\frac{\sin(x-y)}{x-y}f(y)dy \\ &= \int_{-\infty}^{\infty}\frac{1}{2}\int_{-1}^{1}e^{is(x-y)}ds f(y)dy \\ &= \frac{1}{2}\lim_{R\rightarrow\infty}\int_{-R}^{R}\int_{-1}^{1}e^{is(x-y)}dsf(y)dy \\ &= \lim_{R\rightarrow\infty}\frac{1}{2}\int_{-1}^{1}\left(\int_{-R}^{R}e^{-isy}f(y)dy\right) e^{isx}ds \\ &= \frac{\sqrt{2\pi}}{2}\int_{-1}^{1}\hat{f}(s)e^{isx}ds. \end{align} The last equality holds because $\frac{1}{\sqrt{2\pi}}\int_{-R}^{R}e^{-isy}f(y)dy$ converges in $L^2(\mathbb{R})$ to $\hat{f}$ as $R\rightarrow\infty$. Therefore $Tf \in L^2$ and, by the Plancherel Theorem, $$\|Tf\|_2 = \pi\|\hat{f}\chi_{[-1,1]}\|_2 \le \pi \|\hat{f}\|_2 = \pi \|f\|_2.$$

It's been suggested that this follows from Young's inequality. It does not. One special case of Young is $||f*g||_\infty\le||f||_2||g||_2$ (of course that special case is just Cauchy-Schwarz.) That applies here, but it shows that $||Tf||_\infty\le c||f||_2$, not $||Tf||_2\le c||f||_2$. The case of Young's inequality that could show $T$ is bounded on $L^2$ is $||f*g||_2\le||f||_2||g||_1$, but that doesn't apply here, since the function $\sin(t)/t$ is not integrable.

The good news is that it's clear from Plancherel that $$||f*g||_2\le||f||_2||\hat g||_\infty,$$where $\hat g$ is the Fourier transform. So you only need to show that the Fourier transform of $\sin(t)/t$ is in $L^\infty$, which is true. (See "Detail" below...)

(Don't calculate the Fourier transform of $\sin(t)/t$ directly. Instead look in the back of the book: Calculate the Fourier transform of $\chi_{[-1,1]}$ and apply the $L^2$ inversion theorem.)

Note I'm not claiming that the Fourier transform of $\chi_{[-1,1]}$ is exactly $\sin(t)/t$. In a context like this it's impossible to say exactly what the Fourier transform of anything is, exactly. The reason is that various authors define the Fourier transform differently, putting the $2\pi$'s in different places. If you calculate the Fourier transform of $\chi_{[-1,1]}$ by whatever definition you're using you'll get something sort of like $\sin(t)/t$, close enough to let you figure out a $\phi\in L^\infty$ such that $\hat\phi(t)=\sin(t)/t$. So Plancherel implies that the Fourier transform of $\sin(t)/t$ is bounded.

$\newcommand{\ft}{\mathcal F}$ $\newcommand{\ift}{\mathcal F^{-1}}$ $\newcommand{\sp}{\mathcal S}$

Detail: Why is it that $||f*g||_2\le||f||_2||\hat g||_\infty$ for $f,g\in L^2$? The obvious argument would be this: $$||f*g||_2=||\widehat{f*g}||_2=||\hat f\hat g||_2\le||\hat f||_2||\hat g||_\infty=||f||_2||\hat g||_\infty.$$

But this raises the question of what $\widehat{f*g}$ is, since $f*g$ is not in $L^1$ or $L^2$. (I for one find it amusing that nobody's asked about this, given all the silly objections regarding elementary applications of Plancherel; this spot is at least somewhat problematic.) One could avoid the question by showing instead that $f*g=\ift(\hat f\hat g)$; there at least it's clear what everything means, because $\hat f\hat g\in L^1$.

But we get a simpler and more elegant argument by interpreting $\widehat{f*g}$ in the sense of tempered distributions. (Two fairly accesible references for the basic theory of tempered distributions would be the relevant chapters in Rudin Funtional Analysis and Folland Real Analysis.) Just to keep things clear, I'll be using the notation $\ft u$ for the Fourier transform of the tempered distribution $u$, reserving the notation $\hat f$ for $f\in L^1$ or $f\in L^2$. Wee need to show this:

$\ft(f*g)=\hat f\hat g$ for $f,g\in L^2$.

Proof: First, if $f,g\in L^1$ then $f*g\in L^1$, and a standard application of Fubini shows that $$\ft(f*g)=\widehat{f*g}=\hat f\hat g.$$

Now suppose $f,g\in L^2$. Choose $f_n,g_n\in L^1\cap L^2$ such that $f_n\to f$ and $g_n\to g$ in $L^2$. Then $f_n*g_n\to f*g$ uniformly. Hence $f_n*g_n\to f*g$ in $\sp'$ (that is, "in the sense of tempered distributions"). Since $\ft$ is continuous on $\sp'$ it follows that $\ft(f_n*g_n)\to\ft( f*g)$ in $\sp'$. Similarly, $\hat f_n\to\hat f$ and $\hat g_n\to\hat g$ in $L^2$, so $\hat f_n\hat g_n\to\hat f\hat g$ in $L^1$ and hence $\hat f_n\hat g_n\to \hat f\hat g$ in $\sp'$. And so we're done: $$\ft(f*g)=\lim\ft(f_n*g_n)=\lim\hat f_n\hat g_n=\hat f\hat g,$$(where $\lim$ denotes convergence in $\sp'$).

• The Fourier Transforms of $\sin t/t$ is not $\chi_{[-1,1]}$. it is not even bounded. the reason is $\sin t /t$ is not integrable so the Fourier transform is not invertible there. – Guy Fsone Dec 1 '17 at 14:52
• @GuyFsone For heaven's sake, have you heard of the Plancherel theorem? – David C. Ullrich Dec 1 '17 at 14:54
• @GuyFsone So tell me, what is the Fourier transform of $\chi_{[-1,1]}$? – David C. Ullrich Dec 1 '17 at 15:08
• @GuyFsone Maybe we shoulldn't call it the Fourier transform. Let's call it the Plancherel trnasform instead. That operator on $L^2$ that agrees with the Fourier transform on $L^2\cap L^1$. Take $\mathcal F$ to be the Plancherel transform; then it's absolutely and completely true that $\mathcal F^2 f(t) = f(-t)$ for every $f\in L^2$. Btw people "always" refer to the Plancherel transform as just the "Fourier transform". – David C. Ullrich Dec 1 '17 at 15:28
• @GuyFsone Proof, in case you haven't got it yet: It's clear from the $L^1$ inversion theorem that $\mathcal F^2f(t)=f(-t)$ for, say, $f$ in the Schwartz space. But $\mathcal F$ and the operator we might informally describe as $f(t)\mapsto f(-t)$ are both bounded on $L^2$, and the Schwartz space is dense in $L^2$. – David C. Ullrich Dec 1 '17 at 16:12