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Let $H$ be a Hilbert space. Here is the definition of weak Cauchy sequence: a sequence $\{x_n\} \subset H$ is a weak Cauchy sequence if for every $y\in H$, the sequence $\{\langle x_n,y\rangle\}$ is a Cauchy sequence.

Can anyone help me to show every Hilbert space is (sequentially) complete with respect to weak topology?

Dean Miller
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    I don't understand your question. Completeness is a property of metric spaces, but no infinite-dimensional Banach space is metrizable in the weak topology. – Ben W Feb 07 '20 at 17:13
  • @Surb Can I use the fact that if a sequence converges strongly, then it converges weakly as well and then apply your definition of Hilbert space to complete the proof? –  Feb 07 '20 at 17:44
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    Infinite dimensional Hilbert spaces are weakly sequentially complete but not weakly complete! – Jochen Feb 10 '20 at 14:34
  • @BenW: No, completeness can be stated very well in uniform spaces (that generalize metric spaces). Even more, completeness is in fact a property of Cauchy spaces, which in turn generalize uniform spaces. Since the weak topology on $H$ is given by a family of seminorms (the seminorms $v \mapsto | \langle v, v_0 \rangle |$ for all $v_0 \in H$), it follows that $H$ is a uniform space, therefore it does make sense to talk about is completeness. – Alex M. Jul 07 '21 at 07:14

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Here is a proof using only uniform boundness: If $x_n$ is a weak Cauchy sequence, then for each $n \in \mathbb{N}$ we can define a bounded functional $\phi_n \in H'$ by $\phi_n(y) = \langle y,x_n\rangle$. Then $\phi_n$ are pointwise bounded, and hence by uniform boundness $||\phi_n|| = ||x_n||$ are bounded. But then $\phi_n \to \phi$ pointwise for some $\phi \in H'$ and thus there exists some $x \in H$ with $\phi(y) = \langle y,x \rangle$. Obviously then $x_n \to x$ weakly.

Andrei Kh
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Let $(x_n)$ be a weakly Cauchy sequence. By Uniform Boundedness Theorem $(x_n)$ is bounded. Since $H$ is reflexive, its unit ball $B_H$ is weakly compact. By Eberlein Smulian Theorem, $B_H$ is weakly sequentially compact, so $(x_n)$ has a weakly convergent subsequence $(x_{n_k})$. Say $x_{n_k}$ converges to $x$ weakly. Then $x_n$ also converges to $x$ weakly because: Given $y\in H$, $$\lim_{n\to\infty}\langle x_n,y\rangle =\lim_{k\to\infty}\langle x_{n_k},y\rangle=\langle x,y\rangle$$

Dean Miller
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Nick
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If $(x_n)$ converges weakly towards $x$, the uniform boundedness principal implies that $(x_n)$ is bounded. The Alaoglu theorem says that the unit ball of the dual $H^*=H$ of $H$ is compact. This implies that $(x_n)$ converges.

Tsemo Aristide
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