Exercise 2, Chapter 3 Rudin Functional Analysis (first edition) I am trying to tackle the following exercise.

Suppose $L^2=L^2([-1,1])$ with respect to the lebesgue measure. For each scalar $\alpha$ let $E_{\alpha}$ the set of $f$ continuous functions on $[-1,1]$ such that $f(0) = \alpha$.  Prove that each $E_{\alpha}$ is convex and dense in $L^2$.

Convexity is easy to show. For the density not sure, I've been trying to show that every continuous linear functional that vanishes on a subspace containing $E_{\alpha}$ also vanishes for any $f$ in $L^2$. However i don't know if this is the right approach. Any help is appreciated.
I've seen an approach that works out an approximation of arbitrary precision, I wonder however if separation theorems can somewhat be used in this case as well.
 A: I hope the following approach brings more functional analytic favor instead of heavy analysis and so fulfills what you expected.
As you have noted that, one can prove that $(\left<E_{\alpha}\right>)^{\perp}=\{0\}$. In other words, one shall prove that
\begin{align*}
\int_{-1}^{1}f(x)\varphi(x)dx=0,\quad\varphi\in E_{\alpha}
\end{align*}
implying that $f=0$ a.e.
Let $g\in C[-1,1]$ be arbitrary. Fix an $N\in\mathbb{N}$ and consider
\begin{align*}
\varphi_{N}(x)=g(x)+\dfrac{\alpha-g(0)}{(1+|x|)^{N}},
\end{align*}
so $\varphi_{N}(0)=\alpha$ and $\varphi_{N}\in E_{\alpha}$. By assumption, it follows that
\begin{align*}
\int_{-1}^{1}f(x)\varphi_{N}(x)dx&=0\\
\int_{-1}^{1}f(x)g(x)dx&=(g(0)-\alpha)\int_{-1}^{1}\dfrac{1}{(1+|x|)^{N}}dx\\
\int_{-1}^{1}f(x)g(x)dx&=2(g(0)-\alpha)\int_{0}^{1}\dfrac{1}{(1+x)^{N}}dx=2(g(0)-\alpha)\dfrac{1}{N-1}\left(1-\dfrac{1}{2^{N-1}}\right).
\end{align*}
Taking $N\rightarrow\infty$ we have
\begin{align*}
\int_{-1}^{1}f(x)g(x)dx=0.
\end{align*}
Since $C[-1,1]$ is $L^{2}$ dense in $L^{2}[-1,1]$, we conclude by Riesz representation theorem (Hilbert theory version) that $f=0$ in $L^{2}$.
EDIT:
The above answer is not complete. It only shows that $\overline{\left<E_{\alpha}\right>}=L^{2}[-1,1]$. But we can use the same trick again:
For $g\in\left<E_{\alpha}\right>$, set again
\begin{align*}
\varphi_{N}(x)=g(x)+\dfrac{\alpha-g(0)}{(1+|x|)^{N}},
\end{align*}
so $\varphi_{N}\in E_{\alpha}$ and
\begin{align*}
\|g-\varphi_{N}\|_{L^{2}[-1,1]}=|\alpha-g(0)|\left(\int_{-1}^{1}\dfrac{1}{(1+|x|)^{2N}}dx\right)^{1/2}\rightarrow 0
\end{align*}
for large $N$.
A: To prove the density of $E_\alpha$, let
$$\phi_\delta(x) = \begin{cases} 
      1 - \frac{|x|}{\delta} &  |x| \le \delta \\
      0      &\lvert x\rvert > \delta 
\end{cases}
$$
Then $\lVert \phi_\delta\rVert_2^2 = \frac{2}{3}\delta$. Let $f\in L^2$, and for $\epsilon > 0$, choose continuous $g$ such that $\lVert g - f \rVert_2 < \epsilon/2$. This is possible because the continuous functions of compact support are dense in $L^p \left(1\le p < \infty\right)$. Suppose $g(0) = \beta$, and let $$\tilde{g} = g + (\alpha - \beta)\phi_\delta,$$ where $\delta$ is chosen small enough that $\lVert \tilde g - g\rVert< \epsilon/2$. Then $\tilde{g}\in E_\alpha$, $\lVert \tilde g - f \rVert < \epsilon$, and we have shown that $E_\alpha$ is dense in $L^2$.
See also Disjoint convex sets which cannot be separated by any continuous linear functional, which contains a similar density proof, and also a proof that $E_\alpha$ and $E_\beta$ ($\alpha\ne\beta$) cannot be separated by any continuous linear functional $\Lambda$ on $L^2$, as requested in Rudin's exercise.
