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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?

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    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, $\|Tg_n\|_X=1/(n\|f_n\|_{C^k}) \|Tf_n\|_X>1$.2012-11-12
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    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 $c$.2012-11-12
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    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 http://terrytao.wordpress.com/2009/04/19/245c-notes-3-distributions/ for more about this topology.2012-11-12
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    @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.2012-11-12
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    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.2012-11-12
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    @Cantor: to discuss continuity of a map, you need to know the topologies involved in the domain and codomain. What would be the topology of $C^\infty(K)$? Not that there isn't one, but you have to choose. Different choice of the topology change whether your map is continuous or not.2012-11-12
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    @MartinArgerami: The topology on $C^\infty_c(K)$, where $K \subset \mathbb{R}^d$, is generated by the family of norms $\|f\|_{C^k}:=\sup_{x\in \mathbb{R}^d} \sum_{j=0}^k|\nabla^j f(x)|$, for $k=0,1,\cdots$. This gives them the structure of Frech\'et spaces. A sequence $f_n \in C^\infty_c(K)$ converges to a limit $f\in C^\infty_c(K)$ iff $\nabla^jf_n$ converges uniformly on $K$ to $\nabla^j f$ for all $j=0,1\cdots$.2012-11-14

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