Let $X$ be an infinite dimensional locally convex space that separates points and $X^*$ its dual. I would like to prove that no $\sigma(X, X^*)$-continuous semi-norm is actually a norm.
What I have done so far: Let $U$ be any neighborhood of $0$ in the weak topology. Since the weak topology is not locally bounded, it is known that $U$ contains an infinite dimensional subspace. Thus by the Kolmogorov normability criterion, the space $X$ is not normable.
I have two questions:
- Is the use of the Kolmogorov normability criterion justified here? It requires us to show that $0$ has no bounded convex neighborhood of the origin. However, in my approach I am using only a subset of all possible seminorms (only those which are weakly continuous).
- I have shown that $X$ is not normable. But I think this is incomplete, as it does not rule out one particular seminorm being a norm which is what I want to show. Is this correct?
Let $\rho$ be a weakly-continuous seminorm. This implies there exists some $\delta>0$ and $\Lambda_1,\dots\Lambda_n\in X^*$ such that for all $x\in X$ $$\left(\forall i\in\{1,\dots,n\}:|\Lambda_i(x)| < \delta\right) \implies \rho(x)< 1\quad (1)$$ Let $A:X\to\Bbb{F}^n$ be the map $$A(x)=\left(\Lambda_1(x),\dots,\Lambda_n(x)\right)^\top$$ Since $A$ is a linear map from an infinite-dimensional space to a finite-dimensional space, it cannot be injective. Therefore, there exists some $v\in X$, $v\ne 0$ such that $A(v)=0$. This means $\Lambda_1(v)=\dots=\Lambda_n(v)=0$.
So for any $a\in \Bbb{F}, a>0$, $\Lambda_1(av)=\dots=\Lambda_n(av)=0$ and therefore from $(1)$ it follows that $a\rho(v)=|a|\rho(v)=\rho(av) < 1$. Since $a$ is an arbitrary positive number, this can only be if $\rho(v)=0$. Hence $\rho$ is not a norm.