$\exists c>0$ so that $f$ can be chosen to satisfy $||f|| \le c||g||$

Question

Let $K$ be compact and $X$ be Banach subspace of $C(K)$. Let $E \subseteq _{closed} K$ so that $\forall g \in C(E), \exists f \in X$ with $f|_{E} =g$. To show: $\exists c>0$ such that $f$ can be chosen to satisfy $||f|| \le c||g||$.

I think we have to apply bounded inverse theorem somewhere but I'm not sure how to approach.

Answer 1

This is part of the proof of Open Mapping Theorem. Apply the theorem to the bounded linear operator $T: C(K) \to C(E)$ defined by $T(f)=f|_E$ (the restriction of $f$ to $E$).