I am trying to solve the following exercise:
Let $V=L^2([0,1];\mathbb{R})$, $\hspace{3pt}$ $A=\{f\in C([0,1])\mid f(0)>0\}$ and $B=\{f\in C([0,1])\mid f(0)<0\}$.
Show that these two convex, disjoint sets cannot be separated with an hyperplane (we say $A$ and $B$ can be separated by an hyperplane if there exists $\phi \in V' \setminus \{0\}$ such that $\phi(a)\leq \phi (b) \quad \forall a \in A \quad \forall b \in B$)
What I did: every functional on $V$ is a $L^2$ function which acts by integration, so I tried to show that every function of that sort encounters some problems, but I couldn't manage to formalize it.
Lemma 1. For every $\varepsilon >0$, and every $M\in {\mathbb R}$, there exists some $f\in C([0, 1])$, such that $\|f\|_2<\varepsilon $, and $f(0)=M$.
Proof. Defining $$ f(x) = \left\{\matrix{ M(1-x/\delta ), & \text{if } 0\leq x<\delta ,\hfill\cr 0, & \text{if } \delta \leq x\leq 1, }\right. $$ and choosing $\delta $ small enough, gives the desired function. QED
Lemma 2. $A$ in dense in $L^2([0, 1])$.
Proof. Given any $\xi $ in $L^2([0, 1])$, and given $\varepsilon >0$, use the fact that $C([0,1])$ is dense to pick a continuous function $g$ with $\|\xi -g\|_2<\varepsilon /2$. Next use Lemma (1) to find some $f\in C([0, 1])$, such that $\|f\|_2<\varepsilon/2 $, and $f(0)=-g(0)+1$.
Setting $g'=g+f$, we then have that $$ g'(0) = g(0) + f(0) = 1, $$ so $g'$ lies in $A$. Furthermore $$ \|\xi -g'\|_2 = \|\xi -g-f\|_2 \leq \|\xi -g\|_2 + \|f\|_2 < \varepsilon , $$ proving that $A$ is dense in $L^2([0, 1])$. QED
OK, now back to the question.
Suppose by contradiction that there is a nonzero $\phi$ such that $\phi (a)\leq \phi (b)$, for every $a$ in $A$, and every $b$ in $B$. Then, fixing some $b$ in $B$, the above inequality implies that $$ \phi (\xi )\leq \phi (b), \quad\forall \xi \in L^2([0, 1]), $$ by Lemma (2) and the fact that $\phi $ is continuous.
However, a nonzero linear functional is never bounded above, so this says that $\phi =0$, a contradiction.
PS: If we are allowed to choose $\phi $ in the algebraic dual of $L^2([0, 1])$, that is, if we do not insist on the continuity of $\phi $, it is possible to separate $A$ and $B$. In fact, consider the linear functional $$ \phi _0:f\in C([0, 1])\mapsto f(0)\in {\mathbb R}, $$ (which is not continuous for the 2-norm) and let $\phi $ be any (necessarily discontinuous) extension of $\phi _0$ to $L^2([0, 1])$. Then clearly $\phi $ separates $A$ and $B$.