5

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.

ougao
  • 3,671
  • Yes, fixed, thanks. – ougao Dec 02 '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 02 '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 02 '12 at 21:13
  • 1
    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 03 '12 at 09:21

1 Answers1

1

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 $m(E)=0$.

23rd
  • 16,234
  • 44
  • 70
  • 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 03 '12 at 12:49
  • 1
    @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 03 '12 at 12:56
  • oh, got it, the point is to do further decomposition as you said. Thanks! – ougao Dec 03 '12 at 16:29
  • @ougao: You are welcome! – 23rd Dec 03 '12 at 16:33