Let K be a compact set. How does one show the following?

If a linear map $T:C^\infty_c(K) \to X$ into a normed vector space X is continuous then there exists $k \geq 0$ and $C>0$ such that $\|Tf\|_X \leq C\|f\|_{C^k}$ for all $f \in C^\infty_c(K)$.

$C^\infty_c(K)$ is not a normed vector space, correct? What notions of continuity hold for a map $C^\infty_c(K) \to X$, just the notion for a map between two topological spaces?

  • $\begingroup$ If I can use the sequential notion of continuity, then I have a proof. By way of contradiction there exists functions $f_n$ in $C^k$ such that $\|Tf_n\|_X > n\|f_n\|_{C^k}$. Let $g_n=f_n/(n\|f_n\|_{C^k})$. Then $\|g_n\|_{C^k}=1/n$, hence $g_n \rightarrow 0$. However, $\|Tg_n\|_X=1/(n\|f_n\|_{C^k}) \|Tf_n\|_X>1$. $\endgroup$ – Cantor Nov 12 '12 at 10:02
  • $\begingroup$ Do you mean $C_c^{\infty}(K)$ or $C_c^{\infty}(X)$? Because if $K$ is already compact, there is no need to use the subscript $c$. $\endgroup$ – Christopher A. Wong Nov 12 '12 at 10:03
  • $\begingroup$ You can use the sequential notion of continuity. The so-called 'smooth topology' on $C_c^\infty(K)$ is first countable when $K$ is compact, which is a sufficient condition for equivalence of 'topological' continuity and 'sequential' continuity. Check out terrytao.wordpress.com/2009/04/19/245c-notes-3-distributions for more about this topology. $\endgroup$ – icurays1 Nov 12 '12 at 10:06
  • $\begingroup$ @ChristopherA.Wong, what can I say, you have a point. But what do you write when you want to restrict $C^\infty_c(X)$ to compact set $K$? you just replace $X$ with $K$ or you also drop the lowerscript $c$? I did the former. $\endgroup$ – Cantor Nov 12 '12 at 10:11
  • $\begingroup$ Actually, continuity in the topological sense implies that for every convergent sequence $f_n\rightarrow f$, the sequence $Tf_n\rightarrow Tf$, it is the converse of this statement that requires first countability. Since I am only using the first statement (and not it's converse) I don't need to use $C^\infty_c(K)$ is first countable. $\endgroup$ – Cantor Nov 12 '12 at 10:34

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