Bessel's inequality

Template:Short description In mathematics, especially functional analysis, Bessel's inequality is a statement about the coefficients of an element <math>x</math> in a Hilbert space with respect to an orthonormal sequence. The inequality was derived by F.W. Bessel in 1828.<ref name="EoM">{{#invoke:citation/CS1|citation |CitationClass=web }}</ref>

Countable orthonormal sequence in a Hilbert space

Let <math>H</math> be a Hilbert space, and suppose that <math>e_1, e_2, ...</math> is an orthonormal sequence in <math>H</math>. Then, for any <math>x</math> in <math>H</math> one has

<math>\sum_{k=1}^{\infty}\left\vert\left\langle x,e_k\right\rangle \right\vert^2 \le \left\Vert x\right\Vert^2,</math>

where ⟨·,·⟩ denotes the inner product in the Hilbert space <math>H</math>.<ref>Template:Cite book</ref><ref>Template:Cite book</ref><ref>Template:Cite book</ref> If we define the infinite sum

<math>x' = \sum_{k=1}^{\infty}\left\langle x,e_k\right\rangle e_k, </math>

consisting of the "infinite sum" of the vector resolute <math>x</math> in the directions <math>e_k</math>, Bessel's inequality tells us that this series converges. One can think of it that there exists <math>x' \in H</math> that can be described in terms of potential basis <math>e_1, e_2, \dots</math>.

For a complete orthonormal sequence (that is, for an orthonormal sequence that is a basis), we have Parseval's identity, which replaces the inequality with an equality (and consequently <math>x'</math> with <math>x</math>).

Bessel's inequality follows from the identity

<math>\begin{align}

0 \leq \left\| x - \sum_{k=1}^n \langle x, e_k \rangle e_k\right\|^2 &= \|x\|^2 - 2 \sum_{k=1}^n \operatorname{Re} \langle x, \langle x, e_k \rangle e_k \rangle + \sum_{k=1}^n | \langle x, e_k \rangle |^2 \\ &= \|x\|^2 - 2 \sum_{k=1}^n |\langle x, e_k \rangle |^2 + \sum_{k=1}^n | \langle x, e_k \rangle |^2 \\ &= \|x\|^2 - \sum_{k=1}^n | \langle x, e_k \rangle |^2, \end{align}</math> which holds for any natural n.

Fourier series

In the theory of Fourier series, in the particular case of the Fourier orthonormal system, we get if <math>f \colon \mathbb{R} \to \mathbb{C} </math> has period <math>T</math>,

<math>

 \sum_{k \in \mathbb{Z}} \left\vert \int_0^T e^{-2 \pi i k t/T} f (t) \,\mathrm{d}t\right\vert^2 \le T \int_0^T \vert f (t)\vert^2 \,\mathrm{d}t.

</math> In the particular case where <math>f \colon \mathbb{R} \to \mathbb{R} </math>, one has then

<math>

\left\vert \int_0^T f (t) \,\mathrm{d}t\right\vert^2
+ 2 \sum_{n = 1}^\infty \left\vert \int_0^T \cos (2 \pi k t/T) f (t) \,\mathrm{d}t\right\vert^2
+ 2 \sum_{n = 1}^\infty \left\vert \int_0^T \sin (2 \pi k t/T) f (t) \,\mathrm{d}t\right\vert^2 \le T \int_0^T \vert f (t)\vert^2 \,\mathrm{d}t.

</math>

Non countable case

More generally, if <math>H</math> is a pre-Hilbert space and <math>(e_\alpha)_{\alpha \in A}</math> is an orthonormal system, then for every <math>x \in H</math><ref name="EoM" />

<math>

 \sum_{\alpha \in A}  | \langle x, e_\alpha \rangle |^2 \le \lVert x \rVert^2

</math> This is proved by noting that if <math>F \subseteq A</math> is finite, then

<math>

 \sum_{\alpha \in F}  | \langle x, e_\alpha \rangle |^2 \le \lVert x \rVert^2

</math> and then by definition of infinite sum

<math>

  \sum_{\alpha \in A}  | \langle x, e_\alpha \rangle |^2 = \Bigl\{\sum_{\alpha \in F}  | \langle x, e_\alpha \rangle |^2 : F \subseteq A \text{ is finite}\Bigr\}
  \le \lVert x \rVert^2.

</math>

See also

• Cauchy–Schwarz inequality
• Parseval's theorem

External links

• Template:Springer
• Bessel's Inequality the article on Bessel's Inequality on MathWorld.

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