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. 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.
For each $x \in [0,1]$, prove that there exists a function $k_x \in M$ such that:
$f(x) = \int_0^1 \overline{k_x(y)}f(y)dy$ for all $f \in M$ where $\overline{k_x(y)}$ is the conjugate of $k_x(y)$.
The function $K(x,y)=k_x(y)$ defined on $[0,1]\times[0,1]$ is called the reproducing kernel for $M$.

So I feel like I should get the Riesz representation theorem involved but I'm struggling to figure out how exactly... I've been stuck on this problem for at least a week and the gears arn't really  turning at all.. I could really use some help on this one!! Thanks I appreciate it so much!
 A: Hints:


*

*By the Riesz representation theorem, it suffices to show that the evaluation functional $\pi_x : M \to \mathbb{R}$, where $\pi_x(f) = f(x)$, is continuous with respect to the $L^2$ norm on $M$.

*The evaluation functional is definitely continuous on $C([0,1])$, so it would be enough to show that the inclusion map $i : M \to C([0,1])$ is continuous (where $M$ is equipped with the $L^2$ norm and $C([0,1])$ with the sup norm).

*To prove that $i$ is continuous, try using the closed graph theorem.
A: For the sake of my own review and to benefit the OP, I'm going to fill in some of the subtle details in the set up.
First of all, denote $X$ and $Y$ as the subspaces $M$ with the $\sup$ and $L^2$ norms, respectively. We know $Y$ is closed so it is a Hilbert space. The motivation is to say the linear functional $T_x:Y\to \mathbb{C}:f\mapsto f(x)$ is bounded. The existence of $k_x\in Y$ then follows from Riesz Representation.
Note that the mapping $T_x$ is well-defined since for each equivalence class of continuous functions in $L^2$, there is a unique representative that is continuous pointwise everywhere. Put differently, $T_x$ can be understood as $T_x: Y \stackrel{i_Y}\to X \to \mathbb{C}:y\mapsto y\mapsto y(x)$, where $i_Y$ is the identity mapping. The mapping $i_Y$ is well-defined by the comment above.
Obviously, we want to show $|T_x f|\leq C ||f||_2$ for all $f\in Y$. An intuitive thing to try is 
$$
|T_x f|=|f(x)|\leq ||f||_\infty\leq C ||f||_2.
$$
But without more information, it's not obvious that $C$ exists.
What is clear is that by Holders,
$$
||f||_2^2=\int_0^1 |f(t)|^2 \,dt \leq |||f|^2||_\infty\, ||1||_1=||f||_\infty^2
$$
so $||f||_2 \leq ||f||_\infty$. We can show $X$ is closed; therefore, $X$ is a Banach space. Suppose $f_n \stackrel{X}\to  f$, where $f_n \in X$. Then
$$
||f_n-f||_2\leq ||f_n-f||_\infty \to 0.
$$
Hence, $f_n \stackrel{Y}\to f$. Since $Y$ is closed, we must have $f \in M$. Therefore, $f \in X$.
The open mapping theorem says that since $i_X: X \to Y: f\mapsto f$ is surjective (bijective actually) and $X$ is a Banach space, this implies $i_X$ is an open mapping. In other words, a continuous inverse $i_Y: Y \to X: f \mapsto f$ exists. By definition, there exists $C>0$ such that $||f||_\infty=||i_Y(f)||_\infty \leq C ||f||_2$.
From here, we know $T_x$ is bounded so we just apply Riesz representation theorem on the Hilbert space $Y$.
