# Limit of an integral

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)

-
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 $0$.