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Let $X=C(S)$ where $S$ is compact. Suppose $T\subset S$ is a closed subset such that for every $g\in C(T),$ there is an $f\in C(S)$ such that: $f\mid_{T}=g$. Show that there exists a constant $C>0$ such that every $g\in C(T)$ can be continuously extended to $f\in C(S)$ such that: $\sup_{x\in S}\left|f(x)\right|\leq C\sup_{y\in T}\left|g(y)\right|$

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2 Answers 2

$C(S) \to C(T)$ is a surjective bounded linear map of Banach spaces (with sup norms), so there is a closed linear subspace $M \subset C(S)$ such that $C(S)/M \to T$ is bijective and bounded with the quotient norm. Inverse mapping theorem says that the inverse is a bounded linear map. The statement then follows.

By the way, how is this related to Banach-Steinhaus/Hahn-Banach?

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So $A:C(S)/M \rightarrow C(T)$, then $A^{-1}:C(T)\rightarrow C(S)/M$, apply the inverse mapping theorem, we obtain $\left\Vert A^{-1}g\right\Vert \leq\left\Vert A^{-1}\right\Vert \left\Vert g\right\Vert $, then $\inf_{f'\in M}\left\Vert f+f^{'}\right\Vert \leq C\left\Vert g\right\Vert $. We must show that there exists a function in $f+M$ whose restriction is $g$ and it's norm is less than C$\left\Vert g\right\Vert $

But How to show the existence of such function?

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Rewind the definition of the maps: what does all these $f+M$ restrict to? –  Soarer Feb 8 '11 at 1:52

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