Some notes about the Bessel inequality and Parseval equation

1. $Bessel$ Inequality : set $X$ of inner product space, $M=\left \{e_{n}:n\geqslant 1 \right \}$ as a standard positive intersection, $Bessel$ the following inequation:

$\sum_{i=1}^{\infty} \left |<x,e_{i}> \right |^{2} \leqslant \left \| x \right \|^{2}$

Proved relatively simple, first order:

$y=\sum_{i=1}^{n} <x,e_{i}> e_{i}$

$\forall x\in X$ , There are (proof omitted):

$<y,x-y>= ... =0\Rightarrow \left \| x \right \|^{2}=\left \| y \right \|^{2}+\left \| x-y \right \|^{2}$

So there: $\left \| y \right \|^{2} \leq \left \| x \right \|^{2}$ that

$\sum_{i=1}^{n} \left |<x,e_{i}> \right |^{2} \leqslant \left \| x \right \|^{2}$

Use bounded monotonic sequence has limit, get:

$\sum_{i=1}^{\infty} \left |<x,e_{i}> \right |^{2} \leqslant \left \| x \right \|^{2}$

Note: $\sum_{i=1}^{\infty} \left |<x,e_{i}> \right |^{2}$ Convergence than $\sum_{i=1}^{\infty} <x,e_{i}> e_{i}$ a little stronger. It can be shown $Hilbert$ both the same lower space convergence, more generally, are:

$\ Sum_ {i \ geq 1} <a_ {i}, e_ {i}>$ The $\sum_{i\geq 1} \left | a_{i} \right |^{2}$ same convergence.

2. Consider a special inner product space, i.e. $Hilbert$ space:

The following infinite series must be convergent:

$\sum_{i=1}^{\infty} <x,e_{i}> e_{i}$

If $M$ is an orthonormal basis (completely orthogonal sets standard), then the formula converges to $x$
If $M$ is an orthonormal basis, the inequality is further expressed as Bessel Parseval equation:

$\sum_{i=1}^{\infty} \left |<x,e_{i}> \right |^{2} = \left \| x \right \|^{2}$

3. Comprehensive: Parseval equation is Bessel inequality complete inner product space + standard completely orthogonal sets under a special case.

4. Application of Fourier analysis.

Continuous update. . .

Fanfan TT West

Published 14 original articles · won praise 19 · views 30000 +

Private letter concerns

Some notes about the Bessel inequality and Parseval equation

Guess you like