# Showing that smoothing operators are compact

Suppose I have a bounded, linear map $T: H^1(X) \to H^1(X)$ such that $T(H^1(X)) \subset C^\infty(X)$. Is $T$ a compact operator?

I'm guessing this depends on whether or not $X$ is (pre)compact, and in the case I am the most interested in, it is ($X$ is an open, bounded set in $\mathbb{R}^n$), but I am also curious about the general question.

My preliminary thoughts are Arzela-Ascoli: if we take a sequence of functions $f_n \in H^1(X), ||f_n||=1$, then $||Tf_n|| \leq ||T||$ so the sequence is uniformly bounded. Unfortunately, I'm not sure how I should translate the smoothness of $Tf_n$ (and compactness of $X$) into equicontinuity.

Or maybe this follows directly from something like Rellich's theorem or another Sobolev embedding result?

EDIT: Let us further refine the problem statement, so that $T$ is a bounded, linear map $H^1_0(X) \to H^1_0(X)$ that extends to a continuous map $\mathcal{E}'(X) \to C^\infty(X)$ (assumptions on $X$ to follow, $\mathcal{E}'$ are compactly supported distributions). Is $T$ compact?

Thoughts: if $X$ is an open, bounded set, then the Schwartz kernel theorem says that there exists $k \in C^\infty(X \times X)$ so that $$Tf(x) = \int k(x,y)f(y) \ dy.$$

If $k$ is uniformly continuous, then equicontinuity essentially follows immediately. But uniform continuity doesn't necessarily hold, as $X$ is only pre-compact.

If we replace $X$ by $\overline X$, then obviously a smooth $k$ on $\overline X \times \overline X$ is uniformly continuous, but does the Schwartz kernel theorem still hold in this case (every map $T:\mathcal{E}'(\overline X) \to C^\infty(\overline X)$ corresponds to a unique, smooth integral kernel)?

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Do we really need an Arzelà Ascoli tag? – Pedro Tamaroff May 14 '13 at 1:01
@PeterTamaroff - I don't know, I guess not? Do you have any insight into the original question? – BaronVT May 14 '13 at 2:57
Do you assume that $T$ is linear and bounded? – gerw May 15 '13 at 6:47