Bessel's inequality

Template:Short description In mathematics, especially functional analysis, Bessel's inequality is a statement about the coefficients of an element $x$ in a Hilbert space with respect to an orthonormal sequence. The inequality is named for F. W. Bessel, who derived a special case of it in 1828.

Conceptually, the inequality is a generalization of the Pythagorean theorem to infinite-dimensional spaces. It states that the "energy" of a vector $x$, given by $\Vert x{\Vert }^2$, is greater than or equal to the sum of the energies of its projections onto a set of perpendicular basis directions. The value $|⟨x,e_k⟩{|}^2$ represents the energy contribution along a specific direction $e_k$, and the inequality guarantees that the sum of these contributions cannot exceed the total energy of $x$.

When the orthonormal sequence forms a complete orthonormal basis, Bessel's inequality becomes an equality known as Parseval's identity. This signifies that the sum of the energies of the projections equals the total energy of the vector, meaning no energy is "lost." The inequality is a crucial tool for establishing the convergence of Fourier series and other series expansions in Hilbert spaces.

Statement of the inequality

Let $H$ be a Hilbert space and let $e_1,e_2,\dots$ be an orthonormal sequence in $H$. Then for any vector $x$ in $H$, Bessel's inequality states:

$\sum _{k=1}^{\mathrm{\infty }}{\left|⟨x,e_k⟩\right|}^2\le {\Vert x\Vert }^2$

where ⟨·,·⟩ denotes the inner product in the Hilbert space $H$, and $\Vert \cdot \Vert$ denotes the norm induced by the inner product.

The terms $⟨x,e_k⟩$ are the Fourier coefficients of $x$ with respect to the sequence $(e_k)$. The inequality implies that the series of the squared magnitudes of these coefficients converges. This allows for the definition of the vector $x^{\prime }$, which is the projection of $x$ onto the subspace spanned by the orthonormal sequence:

$x^{\prime }=\sum _{k=1}^{\mathrm{\infty }}⟨x,e_k⟩e_k$

Bessel's inequality guarantees that this series converges. If the sequence $(e_k)$ is a complete orthonormal basis, then $x^{\prime }=x$, and the inequality becomes an equality known as Parseval's identity.

Proof

The inequality follows from the non-negativity of the norm of a vector. For any natural number $n$, let

$x_n=\sum _{k=1}^n⟨x,e_k⟩e_k$

This vector $x_n$ is the projection of $x$ onto the subspace spanned by the first $n$ basis vectors. The vector $x-x_n$ is orthogonal to this subspace, and thus orthogonal to $x_n$ itself. By the Pythagorean theorem for inner product spaces, we have $\Vert x{\Vert }^2=\Vert x_n{\Vert }^2+\Vert x-x_n{\Vert }^2$. The proof proceeds by computing $\Vert x-x_n{\Vert }^2$:

$\begin{array}{rl}0\le {\Vert x-\sum _{k=1}^n⟨x,e_k⟩e_k\Vert }^2& =⟨x-\sum _{k=1}^n⟨x,e_k⟩e_k,x-\sum _{j=1}^n⟨x,e_j⟩e_j⟩\\ & =\Vert x{\Vert }^2-\sum _{k=1}^n\overline{⟨x,e_k⟩}⟨x,e_k⟩-\sum _{j=1}^n⟨x,e_j⟩⟨e_j,x⟩+\sum _{k=1}^n\sum _{j=1}^n\overline{⟨x,e_k⟩}⟨x,e_j⟩⟨e_k,e_j⟩\\ & =\Vert x{\Vert }^2-\sum _{k=1}^n|⟨x,e_k⟩{|}^2-\sum _{j=1}^n|⟨x,e_j⟩{|}^2+\sum _{k=1}^n|⟨x,e_k⟩{|}^2\\ & =\Vert x{\Vert }^2-\sum _{k=1}^n|⟨x,e_k⟩{|}^2\end{array}$

This holds for any $n\ge 1$. Since the partial sums are non-negative and bounded above by $\Vert x{\Vert }^2$, the series $\sum _{k=1}^{\mathrm{\infty }}|⟨x,e_k⟩{|}^2$ converges and its sum is less than or equal to $\Vert x{\Vert }^2$.

Fourier series

In the theory of Fourier series, in the particular case of the Fourier orthonormal system, we get if $f:\mathbb{ℝ}\to \mathbb{ℂ}$ has period $T$,

$\sum _{k\in \mathbb{\mathbb{Z}}}{|\underset{0}{\overset{T}{\int }}{e}^{-2\pi ikt/T}f(t)\mathrm{d}t|}^2\le T\underset{0}{\overset{T}{\int }}|f(t){|}^2\mathrm{d}t.$

In the particular case where $f:\mathbb{ℝ}\to \mathbb{ℝ}$, one has then

${|\underset{0}{\overset{T}{\int }}f(t)\mathrm{d}t|}^2+2\sum _{n=1}^{\mathrm{\infty }}{|\underset{0}{\overset{T}{\int }}\cos (2\pi kt/T)f(t)\mathrm{d}t|}^2+2\sum _{n=1}^{\mathrm{\infty }}{|\underset{0}{\overset{T}{\int }}\sin (2\pi kt/T)f(t)\mathrm{d}t|}^2\le T\underset{0}{\overset{T}{\int }}|f(t){|}^2\mathrm{d}t.$

Non countable case

More generally, if $H$ is a pre-Hilbert space and $(e_{\alpha }{)}_{\alpha \in A}$ is an orthonormal system, then for every $x\in H$

$\sum _{\alpha \in A}|⟨x,e_{\alpha }⟩{|}^2\le \Vert x{\Vert }^2$

This is proved by noting that if $F\subseteq A$ is finite, then

$\sum _{\alpha \in F}|⟨x,e_{\alpha }⟩{|}^2\le \Vert x{\Vert }^2$

and then by definition of infinite sum

$\sum _{\alpha \in A}|⟨x,e_{\alpha }⟩{|}^2=\sup \{\sum _{\alpha \in F}|⟨x,e_{\alpha }⟩{|}^2:F\subseteq A\text{ is finite}\}\le \Vert x{\Vert }^2.$

See also

• Cauchy–Schwarz inequality
• Parseval's theorem

External links

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

This article incorporates material from Bessel inequality on PlanetMath, which is licensed under the Creative Commons Attribution/Share-Alike License.