Weak Con. of Orthonormal Sequences to 0 for Inner Product Spaces
Weak Convergence of Orthonormal Sequences to 0 for Inner Product Spaces
Recall from the Bessel's Inequality for Inner Product Spaces page that Bessel's inequality states that if $H$ is an inner product space and if $(e_k)$ is an orthonormal sequence in $H$ then for all $h \in H$ we have that:
(1)\begin{align} \quad \sum_{k=1}^{\infty} \langle e_k, h \rangle^2 \leq \| h \|^2 \end{align}
We will now use Bessel's inequality to show that if $H$ is an inner product space and $(e_n)$ is an orthonormal sequence then $(e_n)$ weakly converges to $0 \in H$.
| Proposition 1: Let $H$ be an inner product space. If $(e_n)$ is an orthonormal sequence in $H$ then $(e_n)$ weakly converges to $0$. |
- Proof: Since $(e_n)$ is an orthonormal sequence in $H$ we have by Bessel's inequality that for all $h \in H$:
\begin{align} \quad \sum_{n=1}^{\infty} \langle e_n, h \rangle^2 \leq \| h \|^2 \end{align}
- Since $\| h \|^2 < \infty$, the above series converges and so:
\begin{align} \quad \lim_{n \to \infty} \langle e_n, h \rangle ^2 = 0 \end{align}
- Therefore, for every $h \in H$:
\begin{align} \quad \lim_{n \to \infty} \langle e_n, h \rangle = 0 = \langle 0, h \rangle \end{align}
- So $(e_n)$ weakly converges to $0$ by the characterization of Weak Convergence in Hilbert Spaces. $\blacksquare$
