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Let $X$ be compact and suppose that $Y$ is a Banach subspace of $C(X)$. If E is a closed subset of $X$ such that for every $g\in C(E)$ there is an $f\in Y$ with $f_{|_{E}}=g$. Show that there is a constant $c>0$ such that for each $g\in C(E)$ there is an $f\in Y$ with $f_{|_{E}}=g$ and $||f||\leq c||g||$.

For this I think about using the inverse mapping theorem for the functional $\phi:Y\to C(E)$ but it's not injective.

Please help me.

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marked as duplicate by Norbert, Davide Giraudo, Hakim, 91500, amWhy May 30 '14 at 11:07

This question was marked as an exact duplicate of an existing question.

the answer could be found here – Norbert May 30 '14 at 9:07
It is quite non-obvious in what way this is a duplicate of the older question. At the very least, it should be quite clearmthat that questions are very different... – Mariano Suárez-Alvarez May 31 '14 at 1:55

Your $\phi$ is continuous, so $\ker\phi$ is closed and $Y/\ker\phi$ is a Banach space. We have an induced map $\bar\phi:Y/\ker\phi\to C(E)$ which is continuous and bijective, so its inverse $\bar\phi^{-1}:C(E)\to Y/\ker\phi$ is continuous; let $K$ be its norm. If $g\in C(E)$, then $\bar\phi^{-1}(g)$ is the class $h+\ker\phi$ of some $h\in Y$. Now $\inf_{r\in\ker\phi}\Vert h+r\Vert=\Vert\bar\phi^{-1}(g)\Vert\leq K\Vert g\Vert$, so there are $f$s in $Y$ which restrict to $g$ on $E$ and with norm arbtrarily close to being below $K\Vert Y\Vert$. Can you continue?

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