Suppose $M$ is a closed supbspace of $L^2[0,1]$ whic his contained in $C[0,1]$, where $C[0,1]$ has a topology provided by the supremum norm.
Use the open mapping theorem to prove that there exists a $C>0$ such that $||f||_{\infty} \leq C ||f||_2$ for all $f \in M$.
This is one of the problems I have been struggling on for awhile now, the open mapping theorem seems so hard to use! If somebody could help me with this I'd greatly appreciate it, thanks!
Consider the embedding $\pi: M \to C[0,1], f \mapsto f$. We want to show that $\pi$ is bounded and by the closed graph theorem that is equivalent to the graph of $\pi$ being closed. So let us show it is closed:
Let $(f_n)_{n \in \mathbb N}$ in $M$ such that $f_n \to f$ in $L^2[0,1]$ and $\pi f_n \to g$ in $C[0,1]$. We have to show that $g = \pi f$. Since $f_n \to f$ in $L^2[0,1]$, we infer that there is a subsequence $(f_{n_k})_{k \in \mathbb N}$ such that converges to $f$ almost everywhere. In particular, $\pi f_{n_k}$ converges pointwise to $\pi f$ almost everywhere. Hence, we have $\pi f = g$ almost everywhere. Since $g$ and $\pi f$ are continuous, it follows that $g(x) = (\pi f)(x)$ for all $x \in [0,1]$. So the graph of $\pi$ and closed.