Let \((u_n)_{n=1}^\infty\) be an orthonormal sequence in a Hilbert space \(H\).
For \(x\in H\), prove that for every \(N\), \[ \left\|x-\sum_{n=1}^{N}\langle x,u_n\rangle u_n\right\|_H^2 = \|x\|_H^2- \sum_{n=1}^{N}|\langle x,u_n\rangle|^2, \] and deduce Bessel’s inequality \[ \sum_{n=1}^{\infty}|\langle x,u_n\rangle|^2\le \|x\|_H^2. \]
If \((a_n)\in\ell^2(\mathbb N)\), prove that there is \(x\in H\) such that \[ \langle x,u_n\rangle=a_n \quad\text{for every }n, \] and \(x\) may be chosen so that \[ \|x\|_H=\left(\sum_{n=1}^{\infty}|a_n|^2\right)^{1/2}. \]
Prove that if \((u_n)\) is complete, then Bessel’s inequality is an equality.
Solution
1. \(\left\|x-\sum_{n=1}^{N}\langle x,u_n\rangle u_n\right\|^2 = \|x\|^2-\sum_{n=1}^{N}|\langle x,u_n\rangle|^2\) for every \(N\), and \(\sum_{n=1}^{\infty}|\langle x,u_n\rangle|^2\le\|x\|^2\).
Set \[ s_N:=\sum_{n=1}^N\langle x,u_n\rangle u_n. \] For each \(1\le k\le N\), \[ \langle x-s_N,u_k\rangle = \langle x,u_k\rangle- \sum_{n=1}^N\langle x,u_n\rangle\langle u_n,u_k\rangle =0. \] Thus \(x-s_N\) is orthogonal to \(s_N\). By Pythagoras, \[ \|x\|^2 = \|x-s_N\|^2+\|s_N\|^2. \] Since the \(u_n\) are orthonormal, \[ \|s_N\|^2 = \sum_{n=1}^N|\langle x,u_n\rangle|^2. \] Therefore \[ \left\|x-\sum_{n=1}^{N}\langle x,u_n\rangle u_n\right\|^2 = \|x\|^2- \sum_{n=1}^{N}|\langle x,u_n\rangle|^2. \] The left side is nonnegative, so \[ \sum_{n=1}^{N}|\langle x,u_n\rangle|^2\le\|x\|^2 \] for every \(N\). Letting \(N\to\infty\) gives \[ \boxed{ \sum_{n=1}^{\infty}|\langle x,u_n\rangle|^2\le\|x\|^2.} \]
2. For \((a_n)\in\ell^2\), the series \(\sum_n a_nu_n\) converges in \(H\) to some \(x\) with \(\langle x,u_k\rangle = a_k\) for every \(k\) and \(\|x\|^2 = \sum_n|a_n|^2\).
Let \((a_n)\in\ell^2\) and define \[ S_N:=\sum_{n=1}^N a_nu_n. \] For \(M>N\), orthonormality gives \[ \|S_M-S_N\|^2 = \sum_{n=N+1}^M|a_n|^2. \] Since \((a_n)\in\ell^2\), the right side tends to \(0\) as \(M,N\to\infty\). Hence \((S_N)\) is Cauchy. Completeness of \(H\) gives some \(x\in H\) with \[ S_N\to x. \] For each fixed \(k\), continuity of the inner product yields \[ \langle x,u_k\rangle = \lim_{N\to\infty}\langle S_N,u_k\rangle =a_k. \] Also \[ \|x\|^2 = \lim_{N\to\infty}\|S_N\|^2 = \sum_{n=1}^{\infty}|a_n|^2. \] Thus \(x\) has the required coefficients and norm.
3. If \((u_n)\) is complete, then \(\|x\|^2=\sum_{n=1}^{\infty}|\langle x,u_n\rangle|^2\) for every \(x\in H\).
Fix \(x\in H\). By Bessel’s inequality, the coefficient sequence \[ a_n:=\langle x,u_n\rangle \] belongs to \(\ell^2\). By step 2, the series \[ s:=\sum_{n=1}^{\infty}a_nu_n \] converges in \(H\). For every \(k\), \[ \langle x-s,u_k\rangle = a_k-a_k=0. \] Thus \(x-s\) is orthogonal to every \(u_k\). Completeness of the orthonormal sequence means that the only vector orthogonal to every \(u_k\) is \(0\), so \(x=s\). Therefore step 2 gives \[ \boxed{ \|x\|^2= \sum_{n=1}^{\infty}|\langle x,u_n\rangle|^2.} \]
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P-XKFPDBessel's inequality and reconstruction from an orthonormal sequencerelated-to