# Bessel's Inequality/Corollary 1

## Corollary

Let $\struct {V, \innerprod \cdot \cdot}$ be an inner product space.

Let $E$ be a orthonormal subset of $V$.

Then, for each $h \in V$, the set:

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

is countable.

## Proof

Let:

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

For each natural number $n$, define:

$\ds X_n = \set {e \in E : \size {\innerprod h e} > \frac 1 n}$

We have that:

$\ds X = \bigcup_{n \mathop = 1}^\infty X_n$

We can show that for each $n \in \N$, the set $X_n$ is finite.

Aiming for a contradiction, suppose, suppose that for some $m \in \N$, the set $X_m$ is infinite.

Then, there exists an countable orthonormal subset of $X_m$, say $\set {e_k : k \in \N}$, such that:

$\ds \size {\innerprod h {e_k} } > \frac 1 m$

for each $k$.

By Bessel's Inequality, we have that:

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

and:

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

Then, for each $n \in \N$, we have:

$\ds \sum_{k \mathop = 1}^n \size {\innerprod h {e_k} }^2 > \frac n {m^2}$

So, for any natural number $n$ with $n > m^2 {\norm h}^2$, we have:

$\ds \sum_{k \mathop = 1}^n \size {\innerprod h {e_k} }^2 > {\norm h}^2$

This contradicts Bessel's Inequality, so we must have that each $X_m$ is finite.

We therefore see that:

$\ds X = \bigcup_{n \mathop = 1}^\infty X_n$
$X$ is countable

as required.

$\blacksquare$