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

Let $\{f_{n}(x)\}$ be a sequence of continuous positive real valued functions on $\mathbb R$. If $a_{n}=\sup_{x\in \mathbb R}|f_{n}(x)|$, such that $a_{n}\to 0$ as $n\to\infty$, and $a_{n}$ is a decreasing sequence, with $a_{n}\in (0,1), \forall n$, and $$\int_{\mathbb R}|f_{n}(x)|^{2}dx\leq A$$ for some $A$, for all $n\geq 1$.

Is it true that $\lim_{n\to\infty}\int_{\mathbb R}|f_{n}(x)|^{2}dx=0$?

My guess: Since $a_{n}\to 0$, this means that the sequence $|f_{n}(x)|$ converges to 0 uniformly on $\mathbb R$, hence $|f_{n}(x)|^{2}$ also converges to 0 uniformly on $\mathbb R$, this will imply the result somehow!

I asked this before but I got one answer which is not applied for the edit (uniform-convergence-and-integration)

share|cite|improve this question
No. Take $f_n(x)=1/\sqrt n$ on $[-n,n]$, $0$ off $[-n,n]$, and "smooth it out at $n$ and $-n$" appropriately. – David Mitra Jun 11 '12 at 20:04

Try $f_n(x) = \frac{1}{n} \sqrt{|x-n|} 1_{[-n,n]} (x)$. Then $\int |f_n(x)|^2 \, dx = 1$, $\forall n$. It is easy to check that $a_n = \frac{1}{\sqrt{n}}$, which converges to $0$.

So the answer is no, the integral does not converge to $0$.

share|cite|improve this answer

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.