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Let $a_{n}=\sup_{x\in \mathbb R}|f_{n}(x)|$, such that $a_{n}\to 0$ as $n\to\infty$, If $a_{n}$ is a decreasing sequence and $a_{n}\in (0,1), \forall n$, 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!

Edit: The functions $f_{n}(x)$ are continuous on $\mathbb R$.

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No. Uniform convergence does not imply that the integral converges uniformly.

Consider the function $g_n(x) = \frac{1}{n} \chi_{[0,n]}(x)$. That is, the function $g_n(x)$ equals $1/n$ if $x\in [0,n]$ and $0$ otherwise. We have that $g_n \to 0$ uniformly: for $\epsilon > 0$, if $n > \epsilon^{-1}$ we have that $\sup |g_n(x)| < \epsilon$. But $\int g_n(x) \mathrm{d}x = n \cdot \frac1n = 1$. This gives us a sequence of function which converges uniformly to 0 but not in integral.

To answer your question above, you just need to take $f_n(x) = \sqrt{g_n(x)} = \frac{1}{\sqrt{n}} \chi_{[0,n]}(x)$. The limit of the integrals is 1, which is not zero.

Now, if you are working over a measure space with finite total measure (such as a bounded interval), then uniform convergence does in fact imply convergence in $L^1$ (or in $L^p$ for any $p < \infty$). In the case of a measure space with infinite total measure, you need something like the Dominated Convergence Theorem.

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I forgot to mention that the functions are continuous on $\mathbb R$, this makes your example not working! – Jude Jun 11 '12 at 16:05
So, does it change anything?! – Jude Jun 11 '12 at 16:59
It doesn't change at all. Integration properties are often agnostic of continuity and differentiability. Instead of $g_n(x)$ as above, let $\tilde{g}_n(x)$ be just like $g_n$, except that it is linear between $(-1,0)$ and $(n,n+1)$. That is, $\tilde{g}_n(x) = \frac{1}{n}(1+x)$ if $-1<x < 0$ and $\frac{1}{n}(n+1 - x)$ if $n<x < n+1$. Then $\tilde{g}_n$ is continuous, and their integrals are strictly larger than $g_n$, and hence we have that $\lim_{n\to\infty}\int \tilde{g}_n(x)\mathrm{d}x \geq 1 \neq 0$. – Willie Wong Jun 12 '12 at 7:36
In fact, if you start with any bounded function $h(x)$ (with any additional property you want: continuous, smooth, etc.) such that $\int |h(x)|^2 \mathrm{d}x = 1$, and let $a_n$ and $b_n$ be two strictly decreasing sequence of positive real numbers, both of which converging to 0, you can consider the sequence of functions $f_n(x) = a_n h(b_n x)$. It is an elementary exercise to show that $\int |f_n(x)|^2 \mathrm{d}x = \frac{a_n^2}{b_n}$. So by suitably arranging your choice of $a_n$ and $b_n$, you can make $\lim \int|f_n|^2\mathrm{d}x$ converge to any non-negative real value you want. – Willie Wong Jun 12 '12 at 7:42

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