Show that there exists $M \gt 0$ such that for all $f \in S,$ $\|f\|_{\infty} \leq M \|f\|_2.$ 
Let $S$ be a linear subspace of $C([0,1])$ which is closed  in $L^{2}([0,1]).$


(a) Show that $S$ is closed  in $(C([0,1]), \|\cdot\|_{\infty}).$


(b) Show that there exists $M \gt 0$ such that for all $f \in S,$ $$\|f\|_{\infty} \leq M \|f\|_2.$$

My attempt $:$
(a) Let $\overline {S}$ be the closure of $S$ in $(C([0,1]),\|\cdot\|_{\infty}).$ Let $f \in \overline {S}.$ Then there exists a sequence $\{f_n\}_{n \geq 1}$ in $S$ converging to $f$ w.r.t. $\|\cdot\|_{\infty}$ i.e. $\|f_n - f\|_{\infty} \to 0$ as $n \to \infty.$ Since $\|g\|_2 \leq \|g\|_{\infty},$ for any $g \in C([0,1])$ it follows that $\|f_n - f\|_2 \to 0$ as $n \to \infty.$ Since $S$ is closed in $L^2([0,1])$ it follows that $f \in S,$ proving that $S$ is a closed subset of $(C[0,1],\|\cdot\|_{\infty}).$
(b) For part (b) If we can prove that the inclusion map $T : (S,\|\cdot\|_2) \longrightarrow (C([0,1]),\|\cdot\|_{\infty})$ is bounded then we are through because then by setting $M = \|T\|_{\text {op}}$ we are through. Now we know that $(C[0,1],\|\cdot\|_{\infty})$ is a Banach space and since $S$ is a closed subspace of $L^2([0,1]),$ $(S,\|\cdot\|_2)$ is also a Banach space. Also inclusion map is always linear. So we have a linear map between two Banach spaces and hence by the virtue of closed graph theorem it follows that $T$ is bounded $\iff$ The graph of $T$ is closed.
How can I show that the graph of $T$ is closed? Any suggestion regarding this will be appreciated.
Thanks in advance.
 A: By the open mapping theorem (or its corollary), a bijective linear bounded map between Banach spaces has a bounded inverse.
In this case, the identity mapping $I:(S,\|\cdot\|_\infty)\to (S,\|\cdot\|_2)$ is clearly bijective, linear, and bounded by $\|f\|_2\le\|f\|_\infty$. The spaces are Banach as the asker rightly observes.
Hence the inverse mapping (which is also the identity) is bounded $$\|f\|_\infty\le M\|f\|_2$$
A: First of all note that the graph of $T$ is precisely $$\mathscr G(T) : = \{(f,f)\ :\ f \in S \} \subseteq \left (S,\|\cdot\|_2 \right) \times \left (S, \|\cdot\|_{\infty} \right ).$$ To prove that $\mathscr G (T)$ is closed we first take a convergent sequence $\{(f_n,f_n) \}_{n \geq 1}$ in $\mathscr G (T)$ converging to some $(f,g)$ with respect to the product topology. Note that since $S$ is a closed subspace of both $\left (L^2 ([0,1]), \|\cdot\|_2 \right )$ and $\left (C([0,1]), \|\cdot\|_{\infty} \right )$ (by (a) in the question) both $f,g \in S.$ If we can prove that $f = g,$ we are through.
Now since $f_n \to f$ w.r.t. $\|\cdot\|_2$ it follows from the proof of Riesz-Fischer theorem that there exists a subsequence $\{f_{n_k} \}_{k \geq 1}$ of $\{f_n \}_{n \geq 1}$ such that $f_{n_k} \to f$ pointwise almost everywhere on $[0,1].$ Since $f_n \to g$ w.r.t. $\|\cdot\|_{\infty}$ it follows that $f_n \to g$ uniformly and hence pointwise on $[0,1].$ But then by the uniqueness of limits it follows that $f = g$ almost everywhere on $[0,1].$ But $$f,g \in S \subseteq C([0,1]).$$ What can you conclude from here?
