Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

Suppose $\{u_n\}_{n=1}^{\infty}$ is an orhtonormal basis in $L^2[0,1]$, prove that $\sum_{n=1}^{\infty}|u_n(x)|^2=\infty$ for almost every $x\in [0,1]$.

Any hint on this problem?

I tried to prove the set $Y$ has measure zero, where $Y=\{x\in [0,1]: \sum_{n=1}^{\infty}|u_n(x)|^2<\infty\}$. Then by decomposition, WLOG, we only need to show that $Y_k=\{x\in [0,1]: k\leq \sum_{n=1}^{\infty}|u_n(x)|^2<k+1\}$ has measure zero for any $k\in \mathbb{Z}^+$, then I tried to prove by contradiction. But I can only show that $m(Y_k)\geq \frac{1}{k+1}$, then I have no idea how to proceed the proof.

share|cite|improve this question
Yes, fixed, thanks. – ougao Dec 2 '12 at 20:55
If you've shown that $m(Y_k)\geq 1/(k+1)$, then $m(Y)=\infty$ since $Y$ is a the disjoint union of $Y_k$, so something seems off... – Alex R. Dec 2 '12 at 21:06
@Alex, I showed this under the assumption that $m(Y_k)$ is nonzero(like the type $m(Y_k)\leq m(Y_k)^2(k+1)$), so it is possible that we only have finite $Y_k$ whose measure is nonzero. – ougao Dec 2 '12 at 21:13
I suspect your argument for proving $m(Y_k)\le m(Y_k)^2(k+1)$ also works when $Y_k$ is replaced with any of its measurable subsets(just like my answer). Then you can complete your proof. – 23rd Dec 3 '12 at 9:21
up vote 1 down vote accepted

Given $k\in\mathbb{N}$, let $$X_k=\{x\in[0,1]:\sum_{n=1}^\infty|u_n(x)|^2\le k\}.$$ To prove your conclusion, it suffices to show that for every $k\in\mathbb{N}$, $m(X_k)=0$. To prove $m(E_k)=0$, it suffices to show that if $E$ is a measurable subset of $X_k$ with $km(E)<1$, then $m(E)=0$.

Fixing such a set $E$, since $\{u_n\}_{n=1}^\infty$ is an orthonormal basis of $L^2[0,1]$, $\sum_{n=1}^\infty\int_{E}\bar{u}_ndm\cdot u_n$ converges to $\chi_E$ in $L^2[0,1]$. Therefore, $\sum_{n=1}^\infty\int_{E}\bar{u}_ndm\cdot u_n\chi_E$ also converges to $\chi_E$ in $L^2[0,1]$. Since $E\subset X_k$, by Cauchy-Schwarz inequality, for every $x,y\in E$, $\sum_{n=1}^\infty|u_n(x)\bar{u}_n(y)|\le k$. It follows that
$$\chi_E=|\sum_{n=1}^\infty\int_{E}\bar{u}_ndm\cdot u_n\chi_E|\le km(E)\chi_E<\chi_E \quad\mbox{a.e. on } E,$$ which implies that $m(E)=0$.

share|cite|improve this answer
as in your notation, it is clear that for any measurable subset $E\subset X_k$, I can prove that $m(E)\leq m(E)^2k$, which implies that if $km(E)<1$, then $m(E)=0$, but why does this corollary imply $m(X_k)=0$? – ougao Dec 3 '12 at 12:49
@ougao: Because $E$ is arbitray, you can choose, for example, $E_i=X_k\cap[\frac{i-1}{k+1}\frac{i}{k+1}]$, $i=1,\dots,k+1$. – 23rd Dec 3 '12 at 12:56
oh, got it, the point is to do further decomposition as you said. Thanks! – ougao Dec 3 '12 at 16:29
@ougao: You are welcome! – 23rd Dec 3 '12 at 16:33

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.