Dimension of a closed subspace of $C[0,1]$ Let $M$ be a closed subspace of $L^2([0, 1]; m)$ that is contained in $C([0, 1])$, where $m$ denotes Lebesgue measure. I have already proved that there exists some positive number $K$ such that $\|f\|_{\sup} \leq K \|f\|_2$ for all $f \in M.$ This part was an application of Closed Graph theorem.
Now, I want to show that for each $x \in [0, 1]$ there exists $g_x \in M$ such that $f(x) =\langle f, g_x\rangle $ for all $f \in M,$ and $\|g_x\|_2 \leq K$. It looks like an application of Riesz representaion theorem, but I could not prove it.
Moreover, my textbook says this information implies that $M$ has dimension at most $K^2$. There is a hint that says Bessel Inequality might help. Can anyone please give me some hints? I am completely stuck. Thanks so much. Any suggestions would be highly appreciated.
 A: Note that from $\lVert f \rVert_{\infty} \le K\lVert f \rVert_2$ we have $|\delta_x(f)| = |f(x)| \le \lVert f \rVert_{\infty} \le K\lVert f \rVert_2$. Therefore, by Riesz representation theorem applied to the Hilbert space $M$ (closed subspace of Hilbert space $(L^2[0,1],m)$) there is $g_x \in M$ s.t., $\delta_x = \left<g_x, \cdot\right>$ and in particular $\lVert g_x \rVert_2 \le K$.
Now, to see that $M$ is finite dimensional, let $\left\{g_j\right\}_{j=1}^n$ be an orthonormal set of functions in $M$. Then, for every $a = (a_1, \cdots , a_n) \in \mathbb{S}^n$ consider $\displaystyle g_a := \sum\limits_{j=1}^n a_jg_j$. Then, by orthogonality of the functions $g_j$ we know $\lVert g_a \lVert_2^2 = \sum\limits_{j=1}^n |a_j|^2 = 1$. Also, for every $x \in [0,1]$ (fixed) we have $$\left|\sum\limits_{j=1}^n a_jg_j(x)\right| = |g_a(x)| = |\delta_x(g_a)| \le K\lVert g_a \rVert_2 = K, \, \forall a \in \mathbb{S}^n.$$ 
Therefore, $\sum\limits_{j=1}^n |g_j(x)|^2 \le K^2, \forall x \in [0,1]$. Integrating both sides of the inequality we have $n \le K^2$. I.e., $\operatorname{dim}(M) \le K^2$. 
