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Conversely, supposeand defineFor every $N$, Bessel inequality and passage to the limit giveHence ${\sum }_i|a_i{|}^2\le C^2$. The partial sums ${\sum }_{i=1}^N{a}_i{e}_i$ are therefore Cauchy; completeness of the Hilbert space gives an element

Let $P_{N}h={\sum }_{i=1}^N⟨h,e_i⟩e_i$. For fixed $N$, coordinate convergence givesThe Hilbertian basis property gives $P_{N}h\to h$, andFirst choose $N$ large and then $n$ large. This provesfor every $h\in H$, so $x_n\rightharpoonup x_{\infty }$. This establishes the coordinate criterion for weak convergence in a separable Hilbert space.

Let $\Vert x_n\Vert \le C$. For each $i$, Bessel inequality givesso the $i$th coordinate sequence is bounded. By the Bolzano-Weierstrass theorem, choose a subsequence on which the first coordinate converges. From it choose a further subsequence on which the second coordinate converges, and continue.

Take the diagonal subsequence $(x_{\varphi (n)})$. For every fixed $i$, its $i$th coordinate converges, and it remains bounded by $C$. The coordinate criterion for weak convergence in a separable Hilbert space supplies an $x_{\infty }\in H$ such thatThis proves the theorem on a weak subsequence of a bounded Hilbert-space sequence.

Every bounded sequence in a separable Hilbert space has a weakly convergent subsequence. Successive subsequences make each basis coordinate converge, and a diagonal subsequence converges in every coordinate; the coordinate criterion for weak convergence in a separable Hilbert space finishes the proof.