Does convergence to zero follow from $x\cdot a_n \to 0,\ x\in H$?

Question

Let $H$ be an arbitrary Hilbert space. Suppose that for some sequence $\{a_n\} \subseteq H$ we have $$\forall x\in H,\quad x\cdot a_n \to 0\in\mathbb K.$$

Does it follow that $a_n\to 0$?

I have tried various manipulations of the expression $a_n\cdot a_n$ to no avail. We could also take an orthonormed basis $E\subseteq H$, then $H \cong \ell _2(E)$. But the assumption would only justify coordinate-wise convergence to zero, which I don't think is equivalent to convergence, in general.

Answer 1

No. If $(e_n)$ is an orthonormal basis for $H$ then $x.e_n \to 0$ for every $x$ but $\|e_n\|=1$ so $e_n$ does not tend to $0$.

Answer 2

As the others pointed out, the answer is no. What you have is that $a_n$ converges to the zero vector weakly. The definition of weak convergence is that $a_n$ converges weakly to $a$ if $a_n\cdot x \to a \cdot x$ for all $x$ in $H$.

But we have the following theorem:

Let $H$ be a Hilbert-space. A sequence $a_n$ is convergent to a vector $a$ if and only if $a_n$ converges to $a$ weakly and $\lVert a_n \rVert \to \lVert a \rVert$.

One direction is trivial, and the other one comes from the following equality: $$\lVert a_n -a \rVert^2=\lVert a \rVert^2 + \lVert a_n \rVert^2-a_n \cdot a -a\cdot a_n$$