Bessel's Inequality in L2 Space
Theorem
Let $\left\{ \phi_{n} \right\}_{n=1}^{\infty}$ be an orthonormal set in $L^{2}(a,b)$. And let $f\in L^{2}(a,b)$. Then the following inequality holds.
$$ \sum \limits_{n=1}^\infty \left| \left\langle f, \phi_{n} \right\rangle \right|^{2} \le \| f \|^{2} $$
Explanation
This is called Bessel’s inequality.
A function satisfying the following equation is said to be square-integrable.
$$ \int_{a}^b |f(x)|^2 dx < \infty $$
The set of square-integrable functions on the interval $[a,b]$ is denoted $L^2(a,b)$.
$$ L^2(a,b) := \left\{ f\ \Bigg|\ \int_{a}^b |f(x)|^2 dx < \infty \right\} $$
Consider the finite-dimensional vector space $\mathbb{R}^n$. If we let $\left\{\mathbf{e}_{1},\ \cdots,\ \mathbf{e}_{n} \right\}$ be a basis, then for any $\mathbf{a} \in \mathbb{R}^n$ the following equation holds trivially.
$$ \sum_{1}^n | \left\langle \mathbf{a}, \mathbf{e}_{i} \right\rangle |^2=|a_{1}|^2+\cdots+|a_{n}|^2=| \mathbf{a} |^2 $$
The significance of Bessel’s inequality is that a similar statement can be made for infinite-dimensional spaces as well. In fact, even equality holds. Furthermore, we want to give $\left\{ \phi_{n} \right\}_{1}^\infty$ a role similar to the basis defined in the finite-dimensional case. That is, we want to know whether the following expression can be used, and if so, under what conditions.
$$ f=\sum \limits_{1}^{\infty}\left\langle f,\phi_{n}\right\rangle\phi_{n} $$
Meanwhile, Bessel’s inequality admits a generalization to Hilbert spaces.
Proof
Note that $\left\langle f,\phi_{n} \right\rangle$ is a scalar. Since by the properties of the inner product $\left\langle f,ag \right\rangle = \overline{a}\left\langle f,g\right\rangle$,
$$ \left\langle f, \left\langle f,\phi_{n} \right\rangle \phi_{n} \right\rangle=\overline {\left\langle f,\phi_{n} \right\rangle}\left\langle f,\phi_{n}\right\rangle = \left| \left\langle f,\phi_{n} \right\rangle \right|^{2} $$
Also, by the Pythagorean theorem, the following equation holds.
$$ \left\| \sum \limits_{1}^{N} \left\langle f,\phi_{n}\right\rangle \phi_{n} \right\|^2=\sum \limits_{1}^{N} \left\| \left\langle f,\phi_{n}\right\rangle \phi_{n} \right\|^2 = \sum \limits_{1}^{N} \left| \left\langle f,\phi_{n}\right\rangle \right|^2 $$
Lemma
Let $\mathbf{a},\ \mathbf{b}\in \mathbb{C}^n$. Then
$$ \left\| \mathbf{a} + \mathbf{b} \right\|^{2} = \left\| \mathbf{a} \right\|^{2} + 2\text{Re}\left\langle \mathbf{a}, \mathbf{b} \right\rangle + \left\| \mathbf{b} \right\|^{2} $$
Therefore
$$ \begin{align*} 0 \le& \left\| f- \sum \limits_{1}^{N} \left\langle f,\phi_{n}\right\rangle \phi_{n} \right\|^2 \\ =&\ \left\| f \right\|^{2} - 2\text{Re} \left\langle f,\ \sum \limits_{1}^{N} \left\langle f,\phi_{n}\right\rangle \phi_{n} \right\rangle + \left\| \sum \limits_{1}^{N} \left\langle f,\phi_{n}\right\rangle \phi_{n} \right\|^2 \\ =&\ \left\| f \right\|^{2} -2\sum \limits_{1}^{N} \left| \left\langle f, \phi_{n}\right\rangle \right|^2 + \sum \limits_{1}^{N} \left| \left\langle f,\phi_{n}\right\rangle \right|^2 \\ =&\ \left\| f \right\|^{2} -\sum \limits_{1}^{N} \left| \left\langle f,\phi_{n}\right\rangle \right|^2 \end{align*} $$
In the second equality the lemma was used, and in the third equality the two equations derived earlier were used. Therefore
$$ \sum \limits_{1}^{N} \left| \left\langle f,\phi_{n}\right\rangle \right|^2 \le \left\| f \right\|^{2} $$
Taking the limit $N \rightarrow \infty$ on both sides,
$$ \sum \limits_{1}^\infty | \left\langle f, \phi_{n} \right\rangle|^2 \le \left\| f \right\|^{2} $$
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