Bessel's Inequality

Theorem
Let $H$ be a Hilbert space with inner product $\innerprod \cdot \cdot$ and inner product norm $\norm \cdot$.

Let $E = \set {e_n: n \in \N}$ be a countably infinite orthonormal subset of $H$.

Then, for all $h \in H$:


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

Proof
Note that for any natural number $n$ we have, applying sesquilinearity of the inner product:

We have:

Therefore:

Since:


 * $\ds \norm {h - \sum_{k \mathop = 1}^n \innerprod h {e_k} e_k}^2 \ge 0$

we have:


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

Since:


 * $\size {\innerprod h {e_k} }^2 \ge 0$ for each $k$

we have that:


 * the sequence $\ds \sequence {\sum_{k \mathop = 1}^n \size {\innerprod h {e_k} }^2}_{n \in \N}$ is increasing.

So:


 * the sequence $\ds \sequence {\sum_{k \mathop = 1}^n \size {\innerprod h {e_k} }^2}_{n \in \N}$ is bounded and increasing.

So from Monotone Convergence Theorem (Real Analysis): Increasing Sequence, we have that:


 * the sequence $\ds \sequence {\sum_{k \mathop = 1}^n \size {\innerprod h {e_k} }^2}_{n \in \N}$ converges.

Since:


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

we then have:


 * $\ds {\norm h}^2 \ge \lim_{n \mathop \to \infty} \sum_{k \mathop = 1}^n \size {\innerprod h {e_k} }^2 = \sum_{k \mathop = 1}^\infty \size {\innerprod h {e_k} }^2$