A Proposition on the Banach space I want to prove the following proposition. It comes from the exercise of Conway, Functional Analysis.
Prop. Let $X$ be compact and suppose that $\mathfrak{X}$ 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 $f\in \mathfrak{X}$ such that $f|_E=g$, then there is a constant $c>0$ such that for each $g\in C(E)$, there is $f\in \mathfrak{X}$ with $f|_E=g$ such that
$$
\sup_{x\in X}|f(x)|\leq c\sup_{x\in E}|g(x)|
$$
By the open mapping theorem, I observed that $f\mapsto f|_E$ is an open map. Maybe it needs Baire Category Theorem, but I have no idea how to use it. Please give me any hints.
Thanks in advance.
 A: You are very nearly there. Let $R: \mathfrak{X} \to C(E)$ be the map $R(f) = f|_E$. You've shown that $R$ is an open map and hence there is an $r>0$ such that $B_{C(E)}(r) \subset R(B_{\mathfrak{X}})$ (where $B_Y(r)$ is the open ball of radius $r$ in $Y$ and $B_Y$ denotes the open ball of radius $1$).
Then, for $g \in C(E)$, we have that $\frac{rg}{2 \|g\|} \in B_{C(E)}(r)$. As a result, there is $f \in B_{\mathfrak{X}}$ such that $R(f) = \frac{rg}{2 \|g\|}$. In particular, $$R\bigg(\frac{2 \|g\|}{r} f \bigg) = g$$ and since $\|f\| \leq 1$, $\bigg \| \frac{2 \|g\|}{r} f \bigg \| \leq \frac{2}{r} \|g\|$ so you can take $c = \frac{2}{r}$ for the desired result.

More generally, if $X,Y$ are normed vector spaces and $T:X \to Y$ is bounded then $T$ is open if and only if there exists $C > 0$ such that for every $g \in Y$ there is $x \in X$ such that $Tx = y$ and $\|x\| \leq C \|y\|$. In turn, this is equivalent to the existence of an $r > 0$ such that $B_Y(r) \subset T(B_X)$. The proof of this is exactly the same as the proof in the special case given here.
