$||f||_\infty$ bounded by $C||f||_2$

Question

Let $Y$ be a closed subspace of $L^2([0,1])$ each of whose elements may be represented by a continuous function on $[0,1]$.
Prove that there exists $C>0$ so that $||f||_{L^\infty}\leq C||f||_{L^2}$ for all $f\in Y$

Attempt:
If $f\in Y$, then we can identify $f$ with a continuous version, so $||f||_\infty = ||f||_u$
Suppose $\exists \{f_n\}\subset Y$ s.t. $||f_n||_u\geq n||f_n||_2$
I would like to use this sequence to contradict $Y$ being closed in $L^2$, but I'm not sure how to go about it.

Answer 1

Since $Y$ is closed in $L^2(0,1)$, $(Y,\|\cdot\|_2)$ is a Banach space. Now recall that $\|f\|_2\leq \|f \|_u$ for all $f\in C([0,1])$. Hence $Y$ is closed in the uniform norm as well, i.e. $(Y,\|\cdot\|_u)$ is a Banach space. The thesis then follows from the inverse mapping theorem.