Given a dense vector subspace there exists a discontinuous functional?

Question

I know that given an infinite dimensional Banach Space $E$ there exists a discontinuous functional $f$ (thanks to uncle Hamel) then its kernel will be a dense subspace of $E$. But what I can't solve is that given a dense vector subspace $X$ of $E$ if there exists a discontinuous functional $g:E\rightarrow \mathbb{R}$ such that $ker(g)=X$ ??

Answer 1

You need an additional assumption: $X$ has codimension $1$. That is, there exists some $u \notin X$ such that $X \oplus \operatorname{span}(u) = E$. Then, you can define a discontinuous linear functional $f$ by setting $f(x) = 0$ for all $x \in X$, and $f(\lambda u) = \lambda$ for $\lambda u \in \operatorname{span}(u)$.

EDIT: Here's an explicit counterexample: take $E = C[0, 1]$ with the supremum norm, and let $X$ be the space of polynomial functions (which is dense by the Stone-Weierstrass theorem). I claim that there is no functional $f : E \to \mathbb{R}$ such that $\operatorname{ker} f = X$.

Suppose there was, and consider the values of $f(\sin)$ and $f(\cos)$. Note that these numbers must not be $0$. But then we have $$f\left(f(\sin)\cos - f(\cos)\sin\right) = 0,$$ hence $f(\sin)\cos - f(\cos)\sin \in X$, which is to say, it is a polynomial. But, this would imply that some $n$th derivative of $f(\sin)\cos - f(\cos)\sin$ is $0$, which would in turn imply that $\sin$ and $\cos$ are linearly dependent, which they are clearly not.

This is a contradiction; no such $f$ exists.