State and prove Bessel’s inequality.
Answers
Step-by-step explanation:
In mathematics, especially functional analysis, Bessel's inequality is a statement about the coefficients of an element {\displaystyle x} in a Hilbert space with respect to an orthonormal sequence. The inequality was derived by F.W. Bessel in 1828.[1]
Let {\displaystyle H} be a Hilbert space, and suppose that {\displaystyle e_{1},e_{2},...} is an orthonormal sequence in {\displaystyle H}. Then, for any {\displaystyle x} in {\displaystyle H} one has
{\displaystyle \sum _{k=1}^{\infty }\left\vert \left\langle x,e_{k}\right\rangle \right\vert ^{2}\leq \left\Vert x\right\Vert ^{2},}
where ⟨·,·⟩ denotes the inner product in the Hilbert space {\displaystyle H}.[2][3][4] If we define the infinite sum
{\displaystyle x'=\sum _{k=1}^{\infty }\left\langle x,e_{k}\right\rangle e_{k},}
consisting of "infinite sum" of vector resolute {\displaystyle x} in direction {\displaystyle e_{k}}, Bessel's inequality tells us that this series converges. One can think of it that there exists {\displaystyle x'\in H} that can be described in terms of potential basis {\displaystyle e_{1},e_{2},\dots }.
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 {\displaystyle x'} with {\displaystyle x}).
Bessel's inequality follows from the identity
{\displaystyle {\begin{aligned}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{aligned}}}