Is the convolution with Gaussian kernel is compact? Suppose $X = L^2(\mathbb{R},m)$ where $m$ is the lebesgue measure and considere $T : X \to X$ defined as
$$
(Tf)(x) = \int_\mathbb{R} e^{-(x-y)^2} f(y)dy
$$
The operator is bounded, just by using the definition. What I was wondering is whether or not this operator is also compact, and I don't have a clue on how to proceed to prove or disprove.
A results shows that

If $T : X \to Y$ is bounded and $\text{dim Range}(T) < \infty$ then $T$ is compact.

Which I've been trying see if this might be the case. However I didn't get far. I've also tried to use the definition of compact operator, still nothing. So I started to suspect it's not compact. There's this result

If $T : X \to Y$ is bounded, $T$ compact and $\text{Range}(T)$ is closed then $\text{dim Range}(T) < \infty$.

Which led me to think that if I can show that the range is closed and it's dimension is not finite then I can show that $T$ is not compact, but I also got stuck here.
Any suggestions?
 A: To see that this operator is not compact, we will construct a sequence of functions $f_j\in L^2$ such that $\|f_j\|=1$ and $\|Tf_j - Tf_k\|_{L^2}\geq \delta$ for all $j\not=k$ and some $\delta>0$.  In other words, we will show that one cannot cover the image of the unit ball by a finite union of balls of radius $\delta$.
The construction is simple.  Take $f_0$ to be the indicator function of the interval $[-1/2,1/2]$.  Since $e^{-x^2} > e^{-4}$ for $|x|\leq 2$, we have that $Tf_0(x) \geq e^{-4}$ for all $x\in[-1,1]$.  On the other hand, because the Gaussian decays rapidly, we also have that
$Tf_0(x) < e^{-100}$ for $x>20$.
Now set $f_j(x) = f_0(x-100j)$ be the indicator function for the interval $[100j-1/2,100j+1/2]$.  We can now estimate
\begin{align*}
\|Tf_j - Tf_k\|^2 
&= \int_{-\infty}^\infty |Tf_j(x) - Tf_k(x)|^2 \,dx \\
&\geq \int_{100j-1}^{100j+1} |Tf_j(x) - Tf_k(x)|^2 \,dx.
\end{align*}
On this interval $Tf_j(x) > e^{-4}$ and $Tf_k(x) < e^{-100}$, so the integral
evaluates to at least $e^{-10}$.  We conclude that for any $j\not=k$,
$\|Tf_j-Tf_k\|_{L^2} \geq e^{-5}$, which shows that there is no finite $e^{-5}$-net of the image of the unit ball.
