Let $(H, \langle \cdot, \cdot \rangle)$ be an infinite-dimensional, real, separable Hilbert space and $\Vert \cdot \Vert_X$ any norm on $H$. Is it true that the weak topology on $H$ is always weaker than the norm topology induced by $\Vert \cdot \Vert_X$?
If the weak topology and the $\Vert \cdot \Vert_X$-topology are comparable then this is true: Since $H$ is reflexive, by Banach-Alaoglu the $\Vert \cdot \Vert_X$-unit ball is compact in the weak topology, but (by the Riesz-Lemma) not in the $\Vert \cdot \Vert_X$-topology. However, it is not clear that the $\Vert \cdot \Vert_X$-topology and the weak topology are comparable i.e. that one is contained in the other.
Equivalently: Is any functional of the form $\langle \cdot, f \rangle: H \rightarrow \mathbb{R}, f \in H$ continuous w.r.t. the $\Vert \cdot \Vert_X$-topology i.e. is it true that for any sequence $(h_n)_{n = 1}^{\infty} \subseteq H$ s.t. $\Vert h_n \Vert_X \rightarrow 0$ we have $\langle h_n, f \rangle \rightarrow 0$?
If this is not the case (which I would guess) I would appreciate an explicit counter example.
Best Answer
No, certainly not. If you're putting a new norm on $H$, that means all you know about $H$ is its vector space structure, so you can do practically anything with a new norm. For instance, let $(x_n)$ be any sequence of linearly independent vectors in $H$ that does not converge weakly and let $(y_n)$ be any sequence of linearly independent vectors that does converge with respect to the Hilbert space norm. Then there is a vector space isomorphism $T:H\to H$ such that $T(x_n)=y_n$. Pulling back the Hilbert space norm along $T$ gives a new norm on $H$ (even a norm that makes $T$ a Hilbert space!) with respect to which $(x_n)$ converges. Since $(x_n)$ does not converge weakly, this means the norm topology is not stronger than the weak topology.