Square-Summable Complex Families Form a Hilbert Space

For each $i\in I$, the singleton finite-sum bound gives $\mathrm{\mid }{x}_i^{(m)}-x_i^{(n)}\mathrm{\mid }\le \mathrm{\parallel }{x}^{(m)}-x^{(n)}{\mathrm{\parallel }}_2$, so $(x_i^{(m)}{)}_m$ is Cauchy in $\mathbb{C}$. By [F2] it has a unique limit $x_i$. These uniquely determined coordinates form a function $x:I\to \mathbb{C}$ by [A1]; no choices are made.

Let $x\in {\mathrm{\ell }}^2(I,\mathbb{C})$ and $\epsilon >0$. By [F1], $s=\mathrm{\parallel }x{\mathrm{\parallel }}_2^2$ is the supremum of its finite coordinate sums, so there is a finite $F\subseteq I$ with ${\sum }_{i\in F}\mathrm{\mid }{x}_i{\mathrm{\mid }}^2>s-{\epsilon }^2\mathrm{/}4$. For every finite $G\subseteq I\setminus F$, disjointness gives ${\sum }_{i\in G}\mathrm{\mid }{x}_i{\mathrm{\mid }}^2\le s-{\sum }_{i\in F}\mathrm{\mid }{x}_i{\mathrm{\mid }}^2<{\epsilon }^2\mathrm{/}4$. Taking the supremum over such $G$ yields $\mathrm{\parallel }x-x{\mathrm{\mid }}_F{\mathrm{\parallel }}_2\le \epsilon \mathrm{/}2<\epsilon$, proving finite-support families are dense. The coordinate vectors have inner products $⟨e_i,e_j⟩={\delta }_{ij}$ by the canonical inner product in [F1], and their finite span is exactly the finitely supported families, so they form an orthonormal basis.

Since $(x^{(m)})$ is Cauchy, choose $N$ such that $\mathrm{\parallel }{x}^{(m)}-x^{(n)}{\mathrm{\parallel }}_2<1$ whenever $m,n\ge N$. Then $\mathrm{\parallel }{x}^{(m)}{\mathrm{\parallel }}_2\le \mathrm{\parallel }{x}^{(N)}{\mathrm{\parallel }}_2+1$ for $m\ge N$; the finitely many earlier norms have a finite maximum, so there is $M<\mathrm{\infty }$ with $\mathrm{\parallel }{x}^{(m)}{\mathrm{\parallel }}_2\le M$ for every $m$. For any finite $F\subseteq I$, coordinate convergence and continuity of the finite sum give ${\sum }_{i\in F}\mathrm{\mid }{x}_i{\mathrm{\mid }}^2={\lim }_m{\sum }_{i\in F}\mathrm{\mid }{x}_i^{(m)}{\mathrm{\mid }}^2\le M^2$. Taking the supremum over finite $F$ in [F1] shows $x\in {\mathrm{\ell }}^2(I,\mathbb{C})$.

Given $\epsilon >0$, choose $N$ so that $\mathrm{\parallel }{x}^{(m)}-x^{(n)}{\mathrm{\parallel }}_2<\epsilon \mathrm{/}2$ for all $m,n\ge N$. Fix $m\ge N$. For every finite $F\subseteq I$, coordinate convergence yields ${\sum }_{i\in F}\mathrm{\mid }{x}_i^{(m)}-x_i{\mathrm{\mid }}^2={\lim }_n{\sum }_{i\in F}\mathrm{\mid }{x}_i^{(m)}-x_i^{(n)}{\mathrm{\mid }}^2\le {\epsilon }^2\mathrm{/}4$. Taking the supremum over finite $F$ gives $\mathrm{\parallel }{x}^{(m)}-x{\mathrm{\parallel }}_2\le \epsilon \mathrm{/}2<\epsilon$. Hence $x^{(m)}\to x$ in ${\mathrm{\ell }}^2(I,\mathbb{C})$ and every Cauchy sequence converges.

By [F1], ${\mathrm{\ell }}^2(I,\mathbb{C})$ is an inner-product space, and step 3.1 proves it is complete. It is therefore a Hilbert space by [F3]. The proof used no family of choices.