How to prove that operator is not compact in $L_2 (\mathbb{R})$

Question

I have the operator

$(Af)(x) = \int _{\mathbb{R}} e^{{-(x-t)^2}/2} f(t) dt$. It seems to me that it isn't compact and I'm looking for some general <=> criterion for integral operators to be compact on $L_2 (\mathbb{R})$.

Answer 1

Recall that if $g$ is $L^1$, then $$ T_g:f\longmapsto f\ast g $$ is bounded from $L^p$ to $L^p$ since $\|f\ast g\|_p\leq \|f\|_p\|g\|_1$. So with the $L^1$ function $g(t)=e^{-\frac{t^2}{2}}$, in particular, $T_gf(x)=\int e^{-\frac{(x-t)^2}{2}}f(t)dt=(f\ast g)(x)$ is bounded from $L^2$ to $L^2$.

Note that $T_g$ is unitarily equivalent to the multiplication operator by $\hat{g}$ via the Fourier transform. So $T_g$ is compact if and only if $$ h\longmapsto \hat{g}h $$ is compact from $L^2$ to $L^2$. As is well-known, a multiplication operator by $\theta\in L^\infty$ on $L^2(\mathbb{R})$ is compact if and only $\theta=0$ a.e. Since $\hat{g}\neq 0$ a.e., it follows that $T_g$ is not compact.

Answer 2

Do you mean \begin{equation} A:L^2(\mathbb{R}) \rightarrow L^2(\mathbb{R}), \ \ \ (Af)(x) = \int_{\mathbb{R}} e^{-(x-t)^2/2} f(t) dt \ ? \end{equation} An operator is compact if it maps bounded sets to relatively compact sets, that is, to sets whose closure is compact. A set is compact if every bounded sequence in it contains converging subsequence. I have the feeling that $A$ is compact, because the Fourier transform is $L^2$-isometry up to a constant and $A$ is mapped to the form $\hat{A}f(\omega) = g(\omega)f(\omega)$, where $g(\omega) = \int_{-\infty}^\infty e^{-x^2/2} e^{-i\omega x} dx$. An orthogonal moving box sequence $f_n(\omega) = \chi_{[n,n+1)}(\omega)$ in the Fourier domain is mapped to a sequence that converges to $0$ in $L^2$.

---

However, there is another orthogonal moving impulse sequence $f_k(t) = \textrm{sinc}(t-k)$, where $\textrm{sinc}(x) = \frac{\sin(\pi x)}{\pi x}$ for $x \neq 0$ and $1$ for $x = 0$. This time we will show that for any subsequence $f_{k_n}$ there is lower limit estimate for $||Af_{k_n} - Af_{k_m}||_2$, when $k_n$ and $k_m$ are far from each other. Hence $Af_{k_n}$ cannot be a Cauchy-sequence and hence cannot converge. We have \begin{eqnarray} |(\mathcal{F}Af_{k_n}-\mathcal{F}Af_{k_m})(\omega)|^2 & = & |\mathcal{F}Af_{k_n}(\omega) - \mathcal{F}Af_{k_m}(\omega)|^2 = |g(\omega)\mathcal{F}f_{k_n}(\omega) - g(\omega) \mathcal{F}f_{k_m}(\omega)|^2 \\ & = & |g(\omega) e^{-i k_n \omega} \mathcal{F}f_0(\omega) - g(\omega) e^{-i k_m \omega} \mathcal{F}f_0(\omega)|^2 \\ & = & |e^{-i \frac{1}{2} (k_m + k_n) \omega} (e^{i \frac{1}{2} (k_m - k_n)} - e^{-i \frac{1}{2}(k_m - k_n)\omega})g(\omega) \mathcal{F}f_0(\omega)|^2 \\ & = & |2ie^{-i\frac{1}{2} (k_m + k_n)\omega}\sin(\frac{1}{2}(k_m-k_n)\omega)g(\omega)\mathcal{F}f_0(\omega)|^2 \\ & = & 2 \sin^2(\frac{1}{2}(k_m-k_n)\omega)|g(\omega)\mathcal{F}f_0(\omega)|^2 \\ & = & 2 \frac{1}{2} (1-\cos((k_m-k_n)\omega)|g(\omega)\mathcal{F}f_0(\omega)|^2 \\ & = & |g(\omega) \mathcal{F}f_0(\omega)|^2 - \cos((k_m-k_n)\omega)|g(\omega)\mathcal{F}f_0(\omega)|^2 \end{eqnarray} Hence \begin{eqnarray} ||Af_{k_n}-Af_{k_m}||_2^2 & = & 2\pi||\mathcal{F}Af_{k_n}-\mathcal{F}Af_{k_m}||_2^2 \\ & = & 2\pi||g\mathcal{F}f_0||_2^2 - 2\pi \int_{-\pi}^\pi |g\mathcal{F}f_0(\omega)|^2 \cos((k_m-k_n)\omega) d\omega \ . \end{eqnarray} Now for sufficiently large differences $k_n-k_m$ we have \begin{equation} ||Af_{k_n}-Af_{k_m}||_2^2 \geq \pi ||g\mathcal{F}f_0||_2^2 \end{equation} because coefficients of Fourier series tend to $0$. This follows from the Riemann-Lebesgue lemma. Now $g$ is everywhere nonzero and $\mathcal{F}f_0$ = $\frac{1}{2\pi} \chi_{[-\pi,\pi]}$ hence the product is nonzero on the interval $[-\pi,\pi]$ and we have a lower limit. The result can be generalized to other convolution operators as follows. There has to be an interval $[a,b]$ s.t. $\neg(g=0 \ \textrm{a.e.})$. Otherwise $g=0\in L^2(\mathbb{R})$. Choose $f_0$ s.t. $\mathcal{F}f_0 = \chi_{[a,b]}$. Then the product $g\mathcal{F}f_0 \neq 0 \in L^2(\mathbb{R})$. Estimate now implies divergence.