$\mathcal{L}_2$ continuous functions with $f(0)=\alpha$ are dense in $\mathcal{L}_2 [-1,1]$

Question

Let $X=\mathcal{L}_2 [-1,1]$ and for any scalar $\alpha$ we define $E_\alpha=\{f\in \mathcal{L}: f \text{ continuous in } [-1,1] \text{ and } f(0)=\alpha \}$.

1. Prove $E_\alpha$ is convex for any $\alpha$.
2. Prove $E_\alpha$ is dense in $\mathcal{L}_2$
3. Prove there is no $f\in X^*$ that separates $E_\alpha$ and $E_\beta$ for $\alpha \neq \beta$.

Part 1 is easy because for $f,g\in E_\alpha$ we have $\gamma f(0) +(1-\gamma)g(0)=\gamma \alpha + (1-\gamma)\alpha =\alpha$.

I'm having trouble with 2 and 3.

I don't know how to approach 2: I can't use the density of step functions with value $\alpha$ at zero as they are not continuous, and I can't use Luzin's theorem in $[-1,1]$ because I need the function to be continuous everywhere in the interval. I also don't see how polynomial approximations help here.

How do I attack 2 and 3? Edit: I was also hinted that in 3 i might want to describe $f(E_\alpha)$ for any $f\in X^*$, but I don't know how to follow through.

Thanks in advance.

Answer 1

This was getting too long for a comment. Analysis is not my thing, but I think you've reduced the problem to showing that for a continuous function $f\in \mathcal{C}^0(\left[0,1\right])$ and $\epsilon > 0$, you can find $g\in E_\alpha$ such that $\left|f-g\right|_2<\epsilon$. So, given a continuous $f$, let $g$ be equal to $f$ except on a super tiny closed interval around $0$, say $\frac{1}{10}\left[-\epsilon,\epsilon\right]$. This interval is compact, so $f$ is bounded there, say the max of $|f|$ is $L$, so, shrink the interval further to $\frac{1}{10(L+\alpha)}\left[-\epsilon,\epsilon\right]$. Then define $g$ on that interval so the graph of $g$ on that interval connects continuously at the endpoints and is a straight line $(0,\alpha)$. This makes the area between the graphs of $f$ and $g$ less than $\epsilon$. You can make the modifications so that the $L_2$ norm is less than $\epsilon$.

Answer 2

HINT: Once you have a continuous $g$ that is a good approximation to the generic $f\in L^2(-1, 1)$, take a $\delta>0$ and do a linear fit between the points $(-1, \alpha)$ and $(-1+\delta, g(-1+\delta))$. That is, consider the function $$ g_\delta(x)=\begin{cases} \text{linear}, & x\in[-1, -1+\delta) \\ g(x), & x\in [-1+\delta, 1] \end{cases} $$ You can choose $\delta$ as small as you wish. See if you can make the distance $\|g-g_\delta\|_2$ sufficiently small.