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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

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2 Answers 2

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$e_n \to 0$ 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..

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Here is another argument.

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.

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  • $\begingroup$ I like that this argument makes use of the statement: $\operatorname{dim}X < \infty \iff \overline{B}_{1}^{X}(0)$ compact $\endgroup$
    – SABOY
    Commented Jul 12, 2020 at 8:43

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