Bessel's Inequality/Corollary 2

Theorem
Let $H$ be a Hilbert space. Let $E$ be a orthonormal subset of $H$.

Then, for all $h \in H$:


 * $\ds \sum_{e \mathop \in E} \size {\innerprod h e}^2 \le \norm h^2$

Proof
From Bessel's Inequality: Corollary 1, we have that:


 * $\innerprod h e \ne 0$ for only countably many $e \in E$.

So the expression:


 * $\ds \sum_{e \mathop \in E} \size {\innerprod h e}^2$

is indeed well-defined.

Since the set:


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

is countable, there exists a sequence $\sequence {e_k}_{k \in \N}$ such that:


 * $X = \set {e_k : k \in \N}$

We then have, by Bessel's Inequality, that:


 * $\ds \sum_{k \mathop = 1}^\infty \size {\innerprod h {e_k} }^2$ converges

with:


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

Since only terms with $e \in X$ contribute to the sum, (those with $e \in E \setminus X$ are zero) we have:


 * $\ds \sum_{e \mathop \in E} \size {\innerprod h e}^2 = \sum_{e \mathop \in X} \size {\innerprod h e}^2 = \sum_{k \mathop = 1}^\infty \size {\innerprod h {e_k} }^2$

and so:


 * $\ds \sum_{e \mathop \in E} \size {\innerprod h e}^2 \le {\norm h}^2$

as required.