Convergence Criterion for Series in Hilbert Spaces

Recall from the Bessel's Inequality for Inner Product Spaces page that if $H$ is an inner product space and $(x_n)_{n=1}^{\infty}$ is an orthonormal sequence of points in $H$ then for every $y \in H$ we have that:

(1)

\begin{align} \quad \sum_{n=1}^{\infty} |\langle y, x_n \rangle|^2 \leq \| y \|^2 \end{align}

The following theorem will give us a nice convergence criterion for a particular type of series of points in a Hilbert space.

Theorem 1: Let $H$ be a Hilbert space and let $(x_n)_{n=1}^{\infty}$ be an orthonormal sequence of points in $H$. Then for every $y \in H$ the series $\displaystyle{\sum_{n=1}^{\infty} \langle y, x_n \rangle x_n}$ converges in $H$ to some $z \in H$ such that $z - y \perp \{ x_1, x_2, ... \}$.

• Proof: For each $N \in \mathbb{N}$ define $s_N$ to be:

(2)

\begin{align} \quad s_N = \sum_{n=1}^{N} \langle y, x_n \rangle x_n \end{align}

• Then $(s_N)_{N=1}^{\infty}$ is the sequence of partial sums for the series $\displaystyle{\sum_{n=1}^{\infty} \langle y, x_n \rangle x_n}$. We will show that the sequence of partial sums converges. Now for $M \geq N+1$ we have that:

(3)

\begin{align} \quad \| s_M - s_N \|^2 = \biggr \| \sum_{n=1}^{M} \langle y, x_n \rangle x_n - \sum_{n=1}^{N} \langle y, x_n \rangle x_n \biggr \| = \biggr \| \sum_{n=N+1}^{M} \langle y, x_n \rangle x_n \end{align}

• By The Pythagorean Identity for Inner Product Spaces, since $(x_n)_{n=1}^{\infty}$ is a sequence of orthonormal points in $H$ we have that:

(4)

\begin{align} \quad \| s_M - s_N \| = \sum_{n=N+1}^{M} |\langle y, x_n \rangle|^2 \end{align}

• By Bessel's inequality (referenced at the top of this page), we have that $\displaystyle{\sum_{n=1}^{\infty} |\langle y, x_n \rangle|^2 \leq \| y \|^2}$ and hence converges. Therefore the difference $\| s_M - s_N \|$ can be made as small as possible, and so $(s_N)_{N=1}^{\infty}$ is a Cauchy sequence. Since $H$ is a Hilbert space, $H$ is complete. So the series $\displaystyle{\sum_{n=1}^{\infty} \langle y, x_n \rangle x_n}$ converges to some $z \in H$.

• To show that $z - y \perp \{ x_1, x_2, ... \}$ we will show that $z - y \perp x_m$ for every $m \in \mathbb{N}$. For each $m \in \mathbb{N}$ let $N \in \mathbb{N}$ be such that $N \geq m$. Then:

(5)

\begin{align} \quad \langle s_N, x_m \rangle = \langle \sum_{n=1}^{N} \langle y, x_n \rangle, x_m \rangle = \sum_{n=1}^{N} \langle y, x_n \rangle \langle x_n, x_m \rangle = \langle y, x_m \rangle \underbrace{\langle x_m, x_m \rangle}_{= 1} = \langle y, x_m \rangle \end{align}

• Therefore:

(6)

\begin{align} \quad \langle s_N - y, x_m \rangle = 0 \end{align}

• Since the inner product is a continuous function, taking the limit as $N \to \infty$ gives us that:

(7)

\begin{align} \quad 0 = \lim_{N \to \infty} \langle s_N - y, x_m \rangle = \biggr \langle \lim_{N \to \infty} s_N - y, x_m \biggr \rangle = \langle z - y, x_m \rangle \end{align}

• Therefore:

(8)

\begin{align} \quad z - y \perp \{ x_1, x_2, ... \} \quad \blacksquare \end{align}