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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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