Bessel's Inequality for Inner Product Spaces

Bessel's Inequality for Inner Product Spaces

Recall from the The Pythagorean Identity for Inner Product Spaces page that if $H$ is an inner product space and $\{ x_1, x_2, ..., x_n \}$ is an orthonormal subset of $H$ then for all $c_1, c_2, ..., c_n \in \mathbb{C}$ we have that:

(1)
\begin{align} \quad \biggr \| \sum_{k=1}^{n} c_kx_k \biggr \|^2 = \sum_{k=1}^{n} |c_k|^2 \end{align}

We now use the Pythagorean identity to prove the very important Bessel's inequality.

Theorem 1 (Bessel's Inequality for Inner Product Spaces): Let $H$ be an inner product space and let $(x_n)_{n=1}^{\infty}$ be an orthonormal sequence in $H$. Then for every $y \in H$ we have that $\displaystyle{\sum_{n=1}^{\infty} |\langle y, x_n \rangle|^2 \leq \| y \|^2}$.
  • Proof: For each $N \in \mathbb{N}$ let:
(2)
\begin{align} \quad y_N = \sum_{n=1}^{N} \langle y, x_n \rangle x_n \end{align}
  • Since $\{ x_1, x_2, ..., x_N \}$ is an orthonormal subset of $H$, by the Pythagorean identity for inner product spaces we have that:
(3)
\begin{align} \quad \| y_N \|^2 &= \biggr \| \sum_{n=1}^{N} \langle y, x_n \rangle x_n \biggr \|^2 \\ &= \sum_{n=1}^{N} |\langle y, x_n \rangle|^2 \end{align}
  • Therefore:
(4)
\begin{align} \quad 0 &\leq \| y - y_N \|^2 \\ & \leq \langle y - y_N, y - y_N \rangle \\ & \leq \| y \|^2 - 2 \mathrm{Re} \langle y, y_N \rangle + \| y_N \|^2 \\ & \leq \| y \|^2 - 2 \mathrm{Re} \langle y, y_N \rangle + \sum_{n=1}^{N} |\langle y, x_n \rangle|^2 \quad (*) \end{align}
  • Observe that:
(5)
\begin{align} \quad \mathrm{Re} \langle y, y_N \rangle &= \mathrm{Re} \sum_{n=1}^{N} \overline{\langle y, x_n \rangle} \langle y, x_n \rangle \\ &= \mathrm{Re} \sum_{n=1}^{N} |\langle y, x_n \rangle|^2 \\ &= \sum_{n=1}^{N} |\langle y, x_n \rangle|^2 \end{align}
  • So from $(*)$ we see that:
(6)
\begin{align} \quad 0 \leq \| y \|^2 - \sum_{n=1}^{N} |\langle y, x_n \rangle|^2 \end{align}
  • So for each $N \in \mathbb{N}$
(7)
\begin{align} \quad \sum_{n=1}^{N} |\langle y, x_n \rangle|^2 \leq \| y \|^2 \end{align}
  • Taking the limit as $N \to \infty$ gives us that:
(8)
\begin{align} \quad \sum_{n=1}^{\infty} |\langle y, x_n \rangle|^2 \leq \| y \|^2 \end{align}
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