Hahn Banach. separate $A:=\{f \in L^2(-1,1): f \mathrm{\ is\ continuous}, f(0)=0\}$, $B:=\{f \in L^2(-1,1): f \mathrm{\ is\ continuous}, f(0)=1\}$ Consider $L^2(-1,1)$ and the sets $A, B$ defined as 
$A:=\{f \in L^2(-1,1): f \mathrm{\ is\ continuous}, f(0)=0\}$, $B:=\{f \in L^2(-1,1): f \mathrm{\ is\ continuous}, f(0)=1\}$

Question: Does there exist an $\phi \in L^2(-1,1)'$  [that is the toplogical dual space of $L^2(-1,1)]$ such that $\phi(a)<\phi(b)\ \forall a \in A,\ b \in B$?

I want to use the following result:
$X...t.v.s., A, B \subseteq X$ disjoint, non empty, convex. $A$ open
$\Rightarrow \exists f \in X', \gamma \in \mathbb R: Re \ f(a) < \gamma \leq Re\ f(b)\ \forall a \in A, b \in B$
Now in our setting, $A, B$ are disjoint, non empty. Both are convex. 

How can it be shown that $A$ is open?

 A: No such $\phi$ can exist, since $A$ and $B$ are dense in $X = L^2(-1, 1)$. As I pointed out in the comments, we have
$$f_n(x) = \max\{1 - n|x|, 0\} \to 0$$
as $n \to \infty$, since
\begin{align*}
\|f_n - 0\|^2 &= \int_{-1}^1f_n^2 \\
&= \int_{-1}^{-1/n} f_n^2 + \int_{-1/n}^0 f_n^2 + \int_0^{1/n} f_n^2 + \int_{1/n}^1 f_n^2 \\
&= \int_{-1}^{-1/n} 0 \, \mathrm{d}x + \int_{-1/n}^0 (1 + nx)^2 \, \mathrm{d}x + \int_0^{1/n} (1 - nx)^2 \, \mathrm{d}x + \int_{1/n}^1 0^2 \, \mathrm{d}x \\
&= \int_{-1/n}^0 (1 + 2nx + x^2) \, \mathrm{d}x + \int_0^{1/n} (1 - 2nx + x^2) \\
&= \left[x + nx^2 + \frac{x^3}{3}\right]_{-1/n}^0 + \left[x - nx^2 + \frac{x^3}{3}\right]^{1/n}_0 \\
&= \frac{2}{3n^3} \to 0.
\end{align*}
Note, in particular, that $f_n$ is a sequence of functions such that $f_n(0) = 1$, but $\|f_n\|$ can be arbitrarily small. As such, if we have an arbitrary $g \in X$, then we can always consider the sequence of functions
$$g_n = g - g(0) \cdot f_n.$$
Note that $g_n(0) = 0$, hence $g_n \in A$. However, since $f_n \to 0$, we have $g_n \to g$. This proves $A$ is dense. You can try something similar for $B$.
Now, suppose there is a continuous linear functional from $X$ to $\Bbb{R}$ such that $\phi(a) < \phi(b)$ for all $a \in A$ and $b \in B$. Let $r = \sup \phi(B)$. Then,
$$\phi(A) \subseteq (-\infty, \sup \phi(B)] \implies A \subseteq \phi^{-1}(-\infty, \sup \phi(B)].$$
The latter set is a proper, closed (recall $\phi$ is continuous) subset of $X$ that contains $A$, which contradicts $A$ being dense.
A: $A$ is not open:


*

*The always vanishing map $f_0$ belongs to $A$.

*The piece-wise linear map $g$ defined by $g(x) = 0$ for $\vert x \vert \ge 1/2$ and $g(0) = 1$ doesn't belong to $A$ as well as all the $g(ax)$ for $a \neq 0$ as $g(0) \neq 0$.

*For any ball centered on $f_0$ of radius $r >0$, the map $g_r(x) = g(\frac{x}{r})$ satisfies $\int_{-1}^1 \left(\vert g_r - f_0 \vert^2\right)^{\frac{1}{2}} < r$
It is impossible to separate $A$ and $B$ with a continuous linear form. You can notice that $A,B$ are dense in $L^2(-1,1)$. Therefore if $\phi$ would separate $A$ and $B$, taking $a \in A$, we would have $\phi(f) \ge \phi(a)$ for all $f \in L^2(-1,1)$. That would imply that $\phi = 0$, a contradiction.
