For real numbers $a,b>0$, consider the following continuous operator:
$$ \mathcal{G}:=\mathcal{F}^{-1}\chi_{[-b,b]}\mathcal{F}\chi_{[-a,a]}: L^2(\mathbb{R})\to L^2(\mathbb{R}),$$
where the $\chi$'s represent the multiplication operator that multiplies by the corresponding indicator function, and $\mathcal{F}$ is the Fourier transform. How could one prove that it is compact? Moreover, houw could we prove that the following similar-looking operator
$$ \mathcal{F}^{-1}(1-\chi_{[-b,b]})\mathcal{F}\chi_{[-a,a]}: L^2(\mathbb{R})\to L^2(\mathbb{R})$$ isn't compact? For the latter I guess we must find a sequence of functions with bounded $L^2$-norms such that no subsequence of their images converges in $L^2$, and maybe $e^{\pi inx/a}$ would work (since they form an orthonormal basis for $L^2([-a,a])$), where $n\in\mathbb{Z}$, but I'm having some trouble finishing the computations. For the former, a straightforward computation gives:
$$\mathcal{G}f(x) = \frac{1}{\pi}\int_{-a}^a f(y)\frac{\sin(2\pi b(x-y))}{x-y} dy $$
(unless I'm making a mistake). So it is (up to the $1/\pi$ constant) an integral operator with kernel
$$ K(x,y) = \frac{\sin(2\pi b(x-y))}{x-y} \chi_{|y|\le a}, $$
and I read somewhere that such an integral operator is compact if an only if its kernel is in $L^2$. In this case, this would reduce the problem to proving that
$$\int_{\mathbb{R}}\int_{-a}^a \bigg|\frac{\sin(2\pi b(x-y))}{x-y}\bigg|^2 dy dx <\infty,$$
but is this theorem I cited true? If so, how could one prove it? I couldn't find a proof online.
Thank you so much for your help!
Best Answer
To see compactness what you can do is start with an arbitrary sequence $x_n$ whose norm is bounded and then see that $Tx_n$ admits a limit point, where $T$ is the operator in question.
Well, if I should do this I don't like the form your operator is in. I will massage it a bit to get something nicer, the compactness of the massaged thing will be equivalent to the compactness of the original operator. The operator I want to look at is:
$$T:L^2(\Bbb R)\to L^2(\Bbb R), \quad x \mapsto\left( t\mapsto \chi_{[-a,a]}(t) \int_{\Bbb R}\frac{\sin(b(y-t))}{y-t} x(y)\,dy\right).$$
Why is this more or less the same operator? Well this operator is nothing other than (up to a constant $2\pi$ or something like that) $x\mapsto \chi_{[-a,a]}\ \cdot (\mathcal F(\chi_{[-b,b]}) * x)$ where $*$ is the convolution. Now we can use a calculational rule of the convolution with the Fourier transform to get that $$T(x)= \chi_{[-a,a]} \cdot [\mathcal F(\chi_{[-b,b]}) * \mathcal F(\mathcal F^{-1}(x))] = \chi_{[-a,a]}\cdot \mathcal F[ \chi_{[-b,b]}\ \cdot \mathcal F^{-1}(x)]$$ well this is your original map except the order is reversed. Or you could also say that your original map is $\mathcal F^{-1}\circ T \circ \mathcal F$ (with the role of the constants $a,b$ swapped), ie it is unitarily equivalent to $T$. Hence compactness of $T$ is the same as compactness of your original map.
Why is $T$ compact? Let $x_n$ be some norm bounded sequence in $L^2(\Bbb R)$. Since Hilbert spaces are reflexive there is some weak limit point of this sequence, call it $x$ and pass to a subsequence so that $x_n$ converges weakly to $x$. We will show that $Tx_n$ converges in norm to $Tx$. The first step is to show that $\|Tx_n\|^2\to \|Tx\|^2$. Here: $$\|Tx_n\|^2 = \int_b^b \left|\int_{\Bbb R}\frac{\sin(b(y-t))}{y-t} x_n(y)\ dy\right|^2\ dt$$ Well the inner integral is a scalar product of an $L^2$ function with $x_n$ and hence converges to $\int_{\Bbb R} \frac{\sin(b(y-t)}{y-t} x(y)\ dy$ as $n\to\infty$ for all $t$ (remember that $x_n$ converges weakly to $x$). Furthermore the inner integral is bounded by $\left|\int_{\Bbb R} \frac{\sin(by)}y\ dy\right| \cdot \|x_n\|$ for all $t$, hence bounded by $C\cdot \chi_{[-b,b]}(t)$ for an appropriate constant $C$. By the Lebesgue dominated convergence theorem you then get that the entire expression converges to $\|Tx\|^2$.
Now:
$$\|Tx-Tx_n\|^2 = \|Tx\|^2 + \|Tx_n\|^2 - \langle Tx, Tx_n\rangle - \langle Tx_n, T x\rangle$$ Since $T$ is norm continuous it is weak continuous and $Tx_n$ converges weakly to $Tx$. Hence this entire expression converges to $$\|Tx\|^2+\|Tx\|^2- \langle Tx,Tx\rangle - \langle Tx, Tx\rangle =0.$$