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Let $\{e_n\}$ be an orthonormal basis for a Hilbert space $H$. Let $\{f_n\}$ be an orthonormal set in $H$ such that $\sum_{n=1}^{\infty}{\|f_n-e_n\|}<1$. How do I show that $\{f_n\}$ is also an orthonormal basis for $H$?

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It's already an orthonormal set, so it suffices to show it's a basis. Perhaps we can proceed by supposing $\langle x,f_n\rangle=0$ for every $f_n$ in the set for some $x\in H$... – anon Sep 15 '11 at 3:44
up vote 8 down vote accepted

Since $\mathcal S:=\{f_n,\;n\in\mathbb N\}$ is an orthonormal subset of $H$, it suffices to show that $\mathcal S^\perp=\{0\}.$

Having this in mind, pick $x\in H$ belonging to $\mathcal S^\perp$. Then for every $n\in\mathbb N$ one has $$0=\langle x,f_n\rangle=\langle x,f_n-e_n+e_n\rangle\Rightarrow \langle x,e_n\rangle=\langle x,e_n-f_n\rangle.$$ Now, since $\{e_n\}$ is an orthonormal basis, one has $$x=\sum_{n=1}^{+\infty}\langle x,e_n\rangle e_n=\sum_{n=1}^{+\infty}\langle x,e_n-f_n \rangle e_n.$$ If $x$ were not $0$, then one would obtain a contradiction as follows: $$\|x\|=\sum_{n=1}^{+\infty}|\langle x,e_n-f_n\rangle|\stackrel{C.S.}{\leq}\|x\|\sum_{n=1}^{+\infty}\|e_n-f_n\|.$$ From this last relation, one may divide out by $\|x\|\neq 0$ by our assumption and obtain $$1\leq\sum_{n=1}^{+\infty}\|e_n-f_n\|,$$ but this contradicts the initial hypothesis. Hence $x=0$ and $\mathcal S^{\perp}=\{0\}.$ This concludes the proof.

Edit Yes, i think i need some changes, thanks Matthew for pointing it out. Ok here is my fix: $$\|x\|^2=\sum_{n=1}^{+\infty}|\langle x,f_n-e_n \rangle|^2\leq \|x\|^2\sum_{n=1}^{+\infty}\|f_n-e_n\|^2,$$ again by Cauchy Schwarz, and if $x\neq 0$ we can divide out and obtain $$(\diamondsuit)\quad 1\leq \sum_{n=1}^{+\infty}\|f_n-e_n\|^2.$$ Now, since $$\sum_{n=1}^{+\infty}\|f_n-e_n\|<1,$$ readily implies that, for every $n\in\mathbb N$, $$\|f_n-e_n\|<1\Rightarrow \|f_n-e_n\|^2<\|f_n-e_n\|.$$ But this means $$(\diamondsuit)<\sum_{n=1}^{+\infty}\|f_n-e_n\|<1\Rightarrow 1<1. $$ Which is absurd. Again then $x=0$ and we conclude in the same way as before.

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Shouldn't you have $\|x\|^2 = \sum_n |\langle x,e_n-f_n\rangle|^2$?? – Matthew Daws Sep 16 '11 at 14:00
@Matthew: I've edited my message. Does it convince you now? P.S. thank you for your advice – uforoboa Sep 16 '11 at 15:40
Yep, looks good to me now! +1 – Matthew Daws Sep 19 '11 at 11:36

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