TITLE: How can I show that this operator is bounded on $L^2$?
QUESTION [5 upvotes]: 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.

REPLY [3 votes]: \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,
$$
               \|Tf\|_2 = \pi\|\hat{f}\chi_{[-1,1]}\|_2 \le \pi \|\hat{f}\|_2 = \pi \|f\|_2.
$$