I am currently studying for an analysis qualifying exam, and this problem has been bothering me: Suppose we have a sequence $\{f_n\}$ in $L^2(\mathbb{R})$ such that $\sum_{n=1}^{\infty}||f_n||_2^2<\infty$ and $\sum_{n=1}^{\infty} f_n(x)=0$ for almost every $x\in\mathbb{R}$. Then for every $g\in L^2(\mathbb{R})$, $\sum_{n=1}^{\infty}\int_{\mathbb{R}}f_n(x)g(x)d\mu(x)$ exists and is equal to zero.

My first thought was to use Cauchy-Schwarz, but the problem is that the norm of $f_n$ is squared in the sum above. My other thought was to try and use something like the Dominated Convergence Theorem on a function $h_m(x)=\sum_{n=1}^{m}f_n(x)$, but I think I am forgetting something easy about the relationship between $|f_n(x)|$ and $||f_n||_2$. Any pointers would be appreciated. Thank you in advance!

  • $\begingroup$ I take it that $\mu$ refers to Lebesgue measure? $\endgroup$
    – Andrew
    Aug 18, 2015 at 16:31
  • $\begingroup$ @Andrew Yes indeed. $\endgroup$ Aug 18, 2015 at 16:31
  • $\begingroup$ I don't suppose you're also given that the $f_n$ are orthogonal? $\endgroup$ Aug 18, 2015 at 16:33
  • $\begingroup$ @DavidC.Ullrich I wish, but unfortunately no. $\endgroup$ Aug 18, 2015 at 16:34

1 Answer 1


Pretty sure it's false - maybe you should check with the guys who wrote the exam.

It's going to be a counterexample in $L^2([0,1])$, with $g=1$.

Say $(I_n)_{n=1}^\infty$ is a sequence of disjoint intervals with $|I_n|=2^{-n}$. Define $$F_1=\chi_{I_1}$$and $$F_n=2^{n-1}\chi_{I_n}-2^{n-2}\chi_{I_{n-1}}\quad(n>1).$$

Then $\sum F_n=0$ almost everywhere. But $\int F_1=1/2$ and $\int F_n=0$ for $n>1$, so $\sum\int F_n\ne 0$.

(If that last sentence is false it's an off-by-one error, getting tired; this is just one of the standard examples where $\int\sum\ne\sum\int$.)

So that's a counterexample, except that $\sum||F_n||_2^2=\infty$. We fix that:

Say $F\in L^2$. Let $f_j=F/n$. Then $$\sum_{j=1}^n||f_j||_2^2=\frac1n||F||_2^2.$$So: Let $(f_j)$ be the sequence consisting of $n_1$ copies of $F_1/n_1$ followed by $n_2$ copies of $F_2/n_2$, etc. The other things we want get inherited from the sequence $(F_n)$ (details on request), while if $n_j\to\infty$ fast enough we also have $\sum||f_j||_2^2<\infty$.

  • $\begingroup$ I think this holds if it is sum for the L^2 norm converges, rather than their squares? Just apply c-s followed by minkowski? $\endgroup$
    – Lost1
    Aug 19, 2015 at 10:17
  • 1
    $\begingroup$ Oh certainly if the sum of the norms is finite there's no problem. If the $f_n$ are orthogonal there's no problem. So it's not clear what the "intended" correct version was... $\endgroup$ Aug 19, 2015 at 13:17

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .