I am working through problems in Conway studying for an exam, and I have become stumped on one. The problem is in the section on Open Mapping Theorem (OMT) and Closed Graph Theorem (CGT).

III.13.6: Let $X$ be compact and suppose that $\mathscr{X}$ is a Banach subspace of $C(X)$. If $E$ is a closed [hence compact] subset of $X$ such that for every $g \in C(E)$ there is an $f\in \mathscr{X}$ with $f_E = g$, show that there is a constant $c > 0$ such that for each $g \in C(E)$ there is an $f \in \mathscr{X}$ with $f_E$ and $\max\{|f(x)| : x \in X\} \leq c \max\{|g(x)| : x \in E\}$.

My current solution: Of course the final statement is equivalent to stating that $\|f\|_{L^\infty(C(X))} \leq c \|g\|_{L^\infty(C(E))}$. I note that $A : \mathscr{X} \to C(E)$ via $Af = f|_E$ is a bounded linear operator (in fact $\|A\| \leq 1$). Since we have the precondition that for every $g \in C(E)$ we can find an $f \in \mathscr{X}$ such that $Af = g$, then $A$ is a surjective map. So the OMT applies. Note that $\|Af\| = \|f|_E\|_{L^\infty(C(E))}$, so we really want to show that $1/c\|Af\| \geq \|f\|$. In the previous problem in the book, we in fact have that this holds if and only if $\text{ran}(A)$ is closed and $\text{ker}(A) = \{0\}$ (we needed the OMT, in fact the inverse mapping theorem, to prove the backwards direction). Now the first condition most certainly holds because a subset of a Banach space is Banach if and only if it is closed and we have surjectivity, but the second I suspect is not necessarily true.

My question: is there a way I can adjust $A$ so as to make it injective, am I missing something, or should I be going about this in a different manner? Thanks.

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    $\begingroup$ I might be missing something but consider $X = [0,1]$, $E = \left[0, \frac12\right]$ and $\mathscr{X} = C[0,1]$. Every $f \in C\left[0, \frac12\right]$ can be extended to a function in $C[0,1]$ but $\|\cdot\|_{\infty, [0,1]}$ is not dominated by a constant times $\|\cdot\|_{\infty, \left[0,\frac12\right]}$. $\endgroup$ – mechanodroid Jul 18 '18 at 23:22
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    $\begingroup$ Look at $\hat{A}:C(X)/Ker(A)\to C(E)$. It is onto, continuous, and injective. The open mapping theorem gives you that the inverse is bounded. Now, the finiteness of the norm of the inverse gives you $\inf_{m\in Ker(A)}\|f+m\|\leq \|\hat{A}^{-1}\|\|f|_{E}\|$. So, you can take $c$ to be $> \|\hat{A}^{-1}\|$, and since the infimum before is strictly smaller you can pick an $m$ such that the inequality holds. $\endgroup$ – user574889 Jul 18 '18 at 23:22
  • $\begingroup$ @mechanodroid "there is an $f \in \mathscr{X}$ ..." $\endgroup$ – user574889 Jul 18 '18 at 23:23
  • $\begingroup$ @cactus Sorry I don't get it. $\endgroup$ – mechanodroid Jul 18 '18 at 23:24
  • $\begingroup$ @mechanodroid The conclusion is that there is a $c$ such that for every $g$ there is an extension $f$ such that the norm of that particular extension is dominated by the norm of $g$ times $c$. $\endgroup$ – user574889 Jul 18 '18 at 23:25

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