Fourier-like operator on $L^2$ is compact

Question

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!

Answer 1

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 $$\|Tx\|^2+\|Tx\|^2- \langle Tx,Tx\rangle - \langle Tx, Tx\rangle =0.$$

Answer 2

Hints: let $\{f_n\}$ be a uniformly bounded sequence in $L^2(\mathbb{R}^n)$. Let $g_n = \mathcal{F}\chi_{[-a,a]}f_n$. Note that each $g_n$ is smooth.

1. Show that the $g_n$ are uniformly bounded in $L^{\infty}$. (Hint: get an $L^1$ estimate on $\chi_{[-a,a]}f_n$.)
2. Show that the derivatives $g_n'$ are uniformly bounded. (This should not be that different from step 1.) Conclude that the sequence $\{g_n\}$ is uniformly equicontinuous.
3. The Arzela-Ascoli theorem says there is a subsequence $\{g_{n_k}\}$ such that $\chi_{[-b,b]}g_{n_k}$ converges uniformly in $L^{\infty}$. Use this to conclude that $\mathcal{F}^{-1}\chi_{[-b,b]}\mathcal{F}\chi_{[-a,a]}f_{n_k}$ converges in $L^2$.

Answer 3

In general the following is true: if $\Omega$ is an open subset of $\mathbb{R}^n$ (for the sake of simplicity in this proof we just take $\Omega=\mathbb{R}^n$, which is the case of the problem's statement), and $K\in L^2(\Omega\times\Omega)$, then $T: L^2(\Omega)\to L^2(\Omega)$ defined via $(Tf)(x) = \int_{\Omega} K(x,y)f(y) dy$, is a compact continuous linear operator. First, notice it's well-defined (i.e. $(Tf)(x)$ converges for ae $x$) because $|(Tf)(x)| \le ||K(x,\bullet)f(\bullet)||_{L^1}\le ||K(x,\bullet)||_{L^2}||f||_{L^2}$ and $||K(x,\bullet)||_{L^2}<\infty$ for ae $x$, otherwise $\int |K(x,y)|^2 dy =\infty$ for all $x$ in some set $S$ of positive measure, so $\int\int |K(x,y)|^2 dy dx \ge \int_{x\in S}\int |K(x,y)|^2 dy dx=\infty$, contradicting $K\in L^2$. Clearly $T$ is linear, and

$$||Tf||^2_{L^2} \le \int \left( \int |K(x,y)f(y)| dy\right)^2 dx \le \int \left(\int |K(x,y)|^2 dy\right)\left(\int |f(y)|^2 dy\right) dx \le ||K||^2_{L^2}||f||^2_{L^2},$$

i.e. $||T||\le ||K||_{L^2}$ so $T$ is continuous. Finally we prove the hard part: $T$ it compact. Let $\{\phi_n(x)\}_{n=1}^\infty$ be an orthonormal basis for $L^2(\Omega)$, then $\{\phi_n(x)\overline{\phi_m(x)}\}_{n,m=1}^\infty$ is an orthonormal basis for $L^2(\Omega\times\Omega)$, so we may write $K(x,y) = \lim_{N\to\infty}\sum_{n,m\le N}\alpha_{n,m}\phi_n(x)\overline{\phi_m(y)}$, with the limit in $L^2$. Let $T_N$ denote the operator with kernel $\sum_{n,m\le N}\alpha_{n,m}\phi_n(x)\overline{\phi_m(y)}$, we prove that $T_N\to T$ in norm. Since each $T_N$ is of finite rank (since its image is contained in the span of $\phi_1,\ldots,\phi_N$), this will prove $T$ is compact, as desired. We have $$ ||T_N-T||\le ||K(x,y) - \sum_{n,m\le N}\alpha_{n,m}\phi_n(x)\overline{\phi_m(y)}||_{L^2}$$ by the same computation as above. The RHS goes to 0 as $N\to\infty$ so the LHS also goes to $0$, QED.

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This reduces the problem to proving the integral at the end of the problem indeed converges, which is equivalent to proving $$\int_{(x,y)\in[-R,R]\times [-a,a]} \bigg|\frac{\sin(2\pi b(x-y))}{x-y}\bigg|^2 dy dx < C,$$ for some $C$ independent of $R$ and $R\to\infty$. Extend the region of integration to the parallelogram with vertices $(-R-2a, -a), (-R, a), (R+2a, a), (R, -a)$. With $z=x-y$, the integral is bounded by $$ 2\sqrt{2}a\int_{-R-a}^{R+a}\bigg|\frac{\sin(2\pi bz)}{z}\bigg|^2 dz.$$ After re-scaling $z$ and letting $R\to\infty$, we just have to prove that $$\int_{-\infty}^\infty \bigg|\frac{\sin z}{z}\bigg|^2 dz <\infty.$$ Away from $z=0$ this converges since it's dominated by $1/z^2$, and near $z=0$ we use the taylor series to see that we're integrating $O(z^4)$ so it obviously converges. Alternatively near $z=0$ the function is continuous so it must be locally integrable at 0, QED.

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Finally, the other operator isn't compact because if it were, then the addition of two compact operators is compact so $\chi_{[-a,a]}$ would be compact. But this acts like the identity on $L^2([-a,a])$ which is an infinite-dimensional Hilbert space, and thus the identity operator on this space cannot be compact, QED.