Proposition 6.2.21EF937410C2F

Proof. Let $(x_ j)$ be a sequence with $\| x_ j\| \le C$, and let $V$ be the closed linear span of its terms. The finite linear combinations of the $x_ j$ with coefficients in $\mathbb Q+\mathrm i\mathbb Q$ form a countable dense subset of $V$. Applying the Gram–Schmidt process to it, and discarding the vectors that depend linearly on earlier ones, we obtain a finite or countable orthonormal family $(e_ k)$ whose linear span is dense in $V$.

For each $k$ the sequence $(\langle x_ j,e_ k\rangle )_ j$ is bounded by $C$, so by a diagonal argument we may pass to a subsequence, still denoted $(x_ j)$, such that $\langle x_ j,e_ k\rangle $ converges to a number $a_ k$ for every $k$. For every $N$, Bessel's inequality gives $\sum _{k\le N}|\langle x_ j,e_ k\rangle |^2\le C^2$; letting $j\to \infty $ and then $N\to \infty $, we get $\sum _ k|a_ k|^2\le C^2$. Since $\| \sum _{N<k\le N'}a_ ke_ k\| ^2=\sum _{N<k\le N'}|a_ k|^2$ by Pythagoras' theorem, the partial sums of the series $x=\sum _ ka_ ke_ k$ form a Cauchy sequence; the series therefore converges in the closed, hence complete, subspace $V$, and $\| x\| ^2=\sum _ k|a_ k|^2\le C^2$.

By the continuity of the inner product and the orthonormality of the $e_ k$, we have $\langle x,e_ k\rangle =a_ k=\lim _ j\langle x_ j,e_ k\rangle $ for every $k$; hence $\langle x_ j,v\rangle \to \langle x,v\rangle $ for every $v$ in the linear span of the $e_ k$. If $v\in V$ and $v'$ lies in that span with $\| v-v'\| <\delta $, then $|\langle x_ j-x,v\rangle |\le |\langle x_ j-x,v'\rangle |+2C\delta $, so the convergence extends to all $v\in V$. For an arbitrary $v\in H$, write $v=v_1+v_2$ with $v_1\in V$ and $v_2\in V^\perp $ by proposition 6.2.1; since $x_ j$ and $x$ lie in $V$, we get $\langle x_ j-x,v\rangle =\langle x_ j-x,v_1\rangle \to 0$.

For the last assertion we may assume that $x\ne 0$. By the Cauchy–Schwarz inequality, $\| x\| ^2=\lim _ j\langle x_ j,x\rangle \le \varliminf _ j\| x_ j\| \, \| x\| $, and we divide by $\| x\| $. $\square$