# (Weak convergence $\implies$ strong convergence) $\implies \mathcal{H}$ finite-dimensional

Let $$\mathcal{H}$$ be a Hilbert space. Show:

$$(\forall \psi \in \mathcal{H}: \lim \langle \psi, \phi_{n}\rangle = \langle \psi, \phi_{n}\rangle \implies \lim \phi_{n}=\phi )\implies \mathcal{H}$$ finite-dimensional. $$(*)$$

The idea:

Assume that $$\mathcal{H}$$ is infinite dimensional, then in particular there exists a countable orthonormal system $$(e_{n})_{n\in \mathbb N}$$ such that by the Bessel inequality, we have:

$$\sum\limits_{n \in \mathbb N}\lvert \langle e_{n}, \phi_{m}\rangle\rvert^{2}\leq \lvert \lvert \phi_{m}\rvert \rvert^{2}<\infty$$

Thus for any $$m \in \mathbb N$$, we obtain $$\lim\limits_{n \to \infty}\langle e_{n}, \phi_{m}\rangle=0$$.

I do not see how this shows that the left-hand side of $$(*)$$ is false

$$e_n \to 0$$ weakly because $$\sum (\langle e_n, x \rangle)^{2} <\infty$$ which implies $$\langle e_n, x \rangle \to 0$$ for all $$x$$. But $$\|e_n\|=1$$ so $$e_n$$ does not tend to $$0$$ in the norm..
Assume that $$(x_n)_n$$ is a sequence in the closed unit ball of $$\operatorname{Ball}(\mathcal{H})$$. Then $$(x_n)_n$$ has a subsequence $$(x_{p(n)})$$ which converges weakly to some $$x \in \operatorname{Ball}(\mathcal{H})$$. By the assumption we then get $$x_{p(n)} \to x$$ strongly so $$\operatorname{Ball}(\mathcal{H})$$ is strongly (sequentially) compact. Therefore $$\mathcal{H}$$ is finite-dimensional.
• I like that this argument makes use of the statement: $\operatorname{dim}X < \infty \iff \overline{B}_{1}^{X}(0)$ compact Commented Jul 12, 2020 at 8:43