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I am trying to prove a theorem, and I have been able to reduce it to the following question. I feel that this should be easy, but I can't see the solution.

If $(g_n)_{n\geq 1}$ is a sequence of functions in $L^1$ such that $\lim_{n\to\infty} \|g_n\|_1 = 0$, then there exists an $N$ such that $g_N$ is in $L^2$. Here, $\|\cdot\|_1$ is the norm in $L^1$.

I am not even sure this is true, but any help would be appreciated.

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In what spaces are you working? This is important since, for example, your assertion is trivially true in $\mathbb{N}$ with counting measure (in this case $L^1 \subset L^2$), however it is false in $\mathbb{R}^n$ with Lebesgue measure. –  Jose27 Oct 22 '12 at 1:40
    
Thanks! I work in $[0,1]$ with Lebesgue measure. –  Ferenc Oct 22 '12 at 1:41
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A counterexample in $[0,1]$ with Lebesgue measure would be something like $f_n = \frac{1}{n\sqrt{x}}$ –  Deven Ware Oct 22 '12 at 1:43
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@DevenWare You could post your comment as an answer. –  Davide Giraudo Oct 24 '12 at 21:04
    
@DavideGiraudo Okay, done. –  Deven Ware Oct 25 '12 at 0:47
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1 Answer 1

up vote 3 down vote accepted

The claim is actually false in $[0,1]$ with Lebesgue measure.

A counterexample can be seen by consider the functions $f_n = \frac{1}{n\sqrt{x}}$, then the integral is $\frac{2}{n}$ so each are in $L^1$ and $\left|\left|f_n\right|\right|_1 \rightarrow 0$.

However, for all $n$ we have $f_n^2 = \frac{1}{n^2x}$, so $f_n \notin L^2$ for any $n$.

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