Example when equality holds in Bessel Inequality

Question

Example of $x\in l^2$ such that $\sum_{k=1}^{\infty}|\langle x,e_k\rangle|^2\leq \|x\|^2$ has strict inequality where $(e_k)$ is an orthonormal sequence in $l^2$.

My thinking: I think it's not possible As $\|x\|\ _{2}=\left(\sum_{k=1}^{\infty}|x_k|^2\right)^{1/2}$ and so by Bessel inequality we have $$ \sum_{k=1}^{\infty}|\langle x,e_k\rangle|^2\leq \left(\left(\sum_{k=1}^{\infty}|x_k|^2\right)^{1/2}\right)^2 $$ $$ \sum_{k=1}^{\infty}|\langle x,e_k\rangle|^2\leq \left(\sum_{k=1}^{\infty}|x_k|^2\right)$$ But aren't both the things same, I mean there should be an equality

Kreyzig: Introduction to Functional Analysis, Ch-3, 3.4 Ques 4

Have you seen Parseval's theorem? It simply says that equality holds for all $x$ if and only if $(e_k)$ is an orthonormal basis. — Prahlad Vaidyanathan, Apr 12 '17 at 07:16
I got it, sir, if I take any orthonormal sequence which is not a basis, I will get a strict inequality. — Abhinav Jha, Apr 12 '17 at 07:52

score 9 · Accepted Answer · answered Apr 12 '17 at 07:16

I think the confusion here is that the orthonormal sequence does not necessarily need to be a complete orthonormal sequence. (In other words, the closure of the span of $\{e_k \}$ need not be the whole of $l^2$.)

For example, consider the sequence $$e_1 = (0,1,0,0,0\dots), \ \ \ e_2 = (0,0,1,0,0\dots), \ \ \ e_3 = (0,0,0,1,0\dots), \ \ \ \dots$$ which is orthonormal, but NOT complete.

Then consider $$x = (1,0,0,0,\dots).$$ You have $\langle x, e_i \rangle = 0$ for all $i$, so $\sum_i |\langle x, e_i \rangle |^2 = 0$. And yet, $|| x ||^2 = 1$.

Example when equality holds in Bessel Inequality

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