This problem is out of Rudin's Functional analysis exercise 3.2. The problem is stated below. I'm really struggling with this chapter in general. It has a lot of new topics I have not seen before. Any help is appreciated. I am attaching some of my current thoughts on the problem.
Suppose $L^2 = L^2([-1,1])$, with respect to Lebesgue measure. For each scalar $\alpha$, let $E_\alpha$, be the set of all continuous functions $f$ on $[-1,1]$ such that $ f(0) = \alpha$. Show that $E_\alpha$ is convex and that each is dense in $L^2$. Thus $E_\alpha$ and $E_\beta$ are disjoint convex sets (if $\alpha \neq \beta$) which cannot be seperated by any continusous linear functionals $\Lambda$ on $L^2$. Hint: What is $\Lambda(E_\alpha)$.
proof:
Suppose the given hypothesis above.
To show convexity let functions $f, g \in E_\alpha$ be arbitrary. Then we observe that $h(x) = tf(x) + (1-t) g(x)$ for $0 \leq t \leq 1$ is a continuous function and that $h(0) = t f(0) + (1-t)g(0) = t\alpha + (1-t)\alpha = \alpha.$ Therefore $h(x) \in E_\alpha$ and we have $E_\alpha$ convex.
To show that $E_\alpha$ is dense in $L^2$. Let $A$ be a collection of continuous functions in $L^2$. Need to show that $E_\alpha \cap A \neq \emptyset$, but not sure how... I assume that given $\epsilon > 0$ we need to take an arbitrary function $f \in A$ and show there exists a function $g \in E_\alpha$ such that for $h(x) = f(x) - g(x)$, $$ \|h(x)\| = \left ( \int_{-1}^{1}|h(x)|^2 d\mu \right )^{1/2} < \epsilon $$
It makes sense that the sets $E_\alpha$ and $E_\beta$ are disjoint because every function in each of them will have a different value at $0$. But the last statement confuses me.
Also I am struggling with the hint, is $\Lambda(E_\alpha) = 0$?
I know that $L^2$ is its own dual space, since $p=2$ and its conjugate $q=2$ as well, so $\Lambda \in L^2$. It seems like symmetry of interval may have something to do with it, along with convexity or $E_\alpha$.
Thanks again for any help in advance.
$X=L^2[-1,1]$ is the completion of the continuous functions on $[-1,1]$ using the $L^2[-1,1]$ norm. Let $\epsilon>0$ $f \in X$ and choose $c$ such that $\|f-c\|_2 < {1 \over 2} \epsilon$. Let $M=\max (|\alpha|, \sup_x |c(x)|)$ and choose $\delta>0$ such that $2 \delta M^2 < {1 \over 2} \epsilon$. Define $c^*(x) = c(x)$ for $|x|> \delta$, and straight line interpolation between the three points $(-\delta, c(-\delta)), (0, \alpha), (\delta, c(\delta))$. It is straightforward to see that $\|c-c^*\|_2 \le 2 \delta M^2$, and so $\|f-c^*\|_2 < \epsilon$ and $c^*(0) = \alpha$. Hence $E_\alpha$ is dense in $X$ for all $\alpha$.
Choose a non zero $\Lambda \in X^*$. Then $\Lambda(f) \neq 0$ for some $f$ and since $\Lambda( c f) = c \Lambda (f)$ for all scalars $c$, we see that $\Lambda(X)$ is the entire set of scalars. Since $E_\alpha$ is dense in $X$, we see that $\Lambda(E_\alpha)$ is dense in the scalar field.
It follows that $E_\alpha \neq E_\beta$ (with $\alpha \neq \beta$) cannot be separated.