I come across this exercise to use the closed graph theorem, if anyone can give and idea about it, thanks
Let $M$ be a closed subspace of $L^2([0,1])$ that is included on $C([0,1])$(with the supremum norm $\|.\|$) I have to show that exist a constant $C$ such that $$\|f\|\leq C\|f\|_2$$ for all $f\in M$.
Edit: Thanks for the answer, now i get other part of the problem, I have to prove that
$$\dim(M)\leq C^2.$$
And I have no idea how... So if you guys can help me with that too.
On
The inclusion $$ (C([0,1]), || \cdot ||_{\infty}) \longrightarrow (L^2([0,1]),|| \cdot ||_2) $$ is continuous since for all $f \in C([0,1])$
$$|| f ||_2^2 = \int_0^1 f^2 \leq \int_0^1 ||f||_{\infty}^2 = ||f||_{\infty}^2 $$
If you restrict the inclusion to $M$
$$i: (M, || \cdot ||_{\infty}) \longrightarrow (M,|| \cdot ||_2) $$
is continuous and bijective.
Since $M$ is closed, it is a Banach space, hence $i$ is open (by the open mapping theorem), bijective and continuous. This means that the inverse $i^{-1}$ is continuos as well, i.e. there exists $C>0$ such that for all $f \in M$ holds $$|| f ||_{\infty} \leq C|| f ||_2$$
Define $\iota\colon (M,\lVert \cdot\rVert_2)\to (C[0,1],\lVert\cdot\rVert_\infty)$ by $\iota(f)=f$. We have to show that $\iota$ is continuous.
Since $M$ and $C[0,1]$ endowed with their respective norms are complete, we have to show that if $\lVert f_n-f\rVert_2\to 0$ and $\lVert f_n-g\rVert_\infty=0$ then $f=g$. To see this, notice that $f_n\to g$ a.e. and $$\int |g-f|=\int\liminf_n|f_n-f|\leqslant \liminf_n\int|f_n-f|=0,$$ hence $f=g$ a.e.
For the dimension problem, assume that $f_1,\dots,f_{N+1}$ are linearly independent elements of $M$. Then using the orthogonalization process we can assume that this is an orthonormal family (for the usual inner product of $L^2$). We thus have for each $(c_1,\dots,c_{N+1})$ in the unit ball of $\mathbf R^n$ and $x\in [0,1]$ that $$\left|\sum_{j=1}^{N+1}c_jf_j(x)\right|\leqslant C.$$ Choosing $c_j(x):= f_j(x)/\sqrt{\sum_{l=1}^{N+1}f_l(x)^2}$, we obtain $N+1\leqslant C^2$.