Bessel's Inequality/Corollary 2

Corollary to Bessel's Inequality
Let $\struct {V, \innerprod \cdot \cdot}$ be an inner product space. Let $E$ be a orthonormal subset of $V$.

Then, for all $h \in V$:
 * $\ds \sum_{e \mathop \in E} \size {\innerprod h e}^2 \le \norm h^2$

Proof
From, we have that:
 * $\innerprod h e \ne 0$ for only countably many $e \in E$.

So the set
 * $X = \set {e \in E : \innerprod h e \ne 0}$

is countable.

Hence, there exists a sequence $\sequence {e_k}_{k \in \N}$ such that:
 * $X = \set {e_k : k \in \N}$

For brevity, write:
 * $X_n = \set {e_k : k \le n}$

for each $n \in \N$.

We have, by Bessel's Inequality, that:
 * $\ds \sum_{k \mathop = 1}^\infty \size {\innerprod h {e_k} }^2$ converges

with:
 * $\ds \sum_{k \mathop = 1}^\infty \size {\innerprod h {e_k} }^2 \le {\norm h}^2$

We now argue that:
 * $\ds \sum_{e \mathop \in E} \size {\innerprod h e}^2 = \sum_{k \mathop = 1}^\infty \size {\innerprod h {e_k} }^2$

Let $\epsilon > 0$.

From the definition of net convergence, we aim to show that for each $\epsilon > 0$, there exists a finite set $S \subseteq X$ such that:
 * $\ds \cmod {\sum_{e \mathop \in F} \size {\innerprod h e}^2 - \sum_{k \mathop = 1}^\infty \size {\innerprod h {e_k} }^2} < \epsilon$

for all finite sets $F \supseteq S$.

Let $F$ be a finite set containing $X_n$.

Since:
 * $\innerprod h e = 0$ for $h \not \in X$

we have:
 * $\ds \sum_{e \mathop \in F} \size {\innerprod h e}^2 = \sum_{k \mathop = 1}^n \size {\innerprod h {e_k} }^2 + \sum_{k \in F_\ast} \size {\innerprod h {e_k} }^2$

where $F_\ast$ is some subset of $\set {k \in \N : k \ge n + 1}$.

Since $\size {\innerprod h {e_k} }^2 \ge 0$ for each $k \in \N$, we have:
 * $\ds \sum_{k \mathop = 1}^n \size {\innerprod h {e_k} }^2 + \sum_{k \in F_\ast} \size {\innerprod h {e_k} }^2 \le \sum_{k \mathop = 1}^\infty \size {\innerprod h {e_k} }^2$

so that:
 * $\ds \sum_{e \mathop \in F} \size {\innerprod h e}^2 \le \sum_{k \mathop = 1}^\infty \size {\innerprod h {e_k} }^2$

For the lower bound, note that:
 * $\ds \sum_{k \mathop = 1}^n \size {\innerprod h {e_k} }^2 + \sum_{k \in F_\ast} \size {\innerprod h {e_k} }^2 \ge \sum_{k \mathop = 1}^n \size {\innerprod h {e_k} }^2$

Pick $N_\epsilon \in \N$ such that:
 * $\ds \sum_{k \mathop = 1}^{N_\epsilon} \size {\innerprod h {e_k} }^2 \ge \sum_{k \mathop = 1}^\infty \size {\innerprod h {e_k} }^2 - \epsilon$

So, for all finite sets $F \supseteq X_{N_\epsilon}$, we have:
 * $\ds \cmod {\sum_{e \mathop \in F} \size {\innerprod h e}^2 - \sum_{k \mathop = 1}^\infty \size {\innerprod h {e_k} }^2} < \epsilon$

So:
 * $\ds \sum_{e \mathop \in E} \size {\innerprod h e}^2$ converges as a generalized sum to $\ds \sum_{k \mathop = 1}^\infty \size {\innerprod h {e_k} }^2$

Since we have that:
 * $\ds \sum_{k \mathop = 1}^\infty \size {\innerprod h {e_k} }^2 \le \norm h^2$

we are done.