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Let $(X,m)$ be a measure space, $(f_n)_n,f \in L^1(m)$ all non-negative and $f_n \rightarrow f$ in $L^1$. Is it true that then $\sqrt{f_n} \rightarrow \sqrt f$ in $L^2$?

My try: $$ \int (\sqrt{f_n} - \sqrt f)^2 = \int f_n + f - 2 \sqrt{f f_n} = \left(\int f_n - f \right) + 2\int f - \sqrt{ff_n}, $$ so I need just to prove that $\int f - \sqrt{ff_n} \rightarrow 0$. But why does this holds?

Aspera
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1 Answers1

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Suppose by contradiction $\sqrt{f_n} \not\rightarrow \sqrt f $ in $L^2$. Then, up to subsequences, we can suppose $$ \int (\sqrt{f_n} - \sqrt{f})^2 \rightarrow C > 0, $$ and $f_n \rightarrow f$ a.e.. But in this situation by reverse Fatou we would have $$ C= \lim \int (\sqrt{f_n} - \sqrt f)^2 \leq \int \lim (\sqrt{f_n} - \sqrt f)^2 = 0,$$ which is absurd.

Aspera
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  • What exactly do you mean by "reverse Fatou"? Fatou's Lemma yields exactly the reverse estimate of your estimate. – PhoemueX Jun 06 '15 at 00:24
  • @PhoemueX Yes of course, it's just a variation of Fatou's lemma: wiki link – Aspera Jun 06 '15 at 09:32
  • For the reverse version, you need to know $f_n \leq g$ for all $n$ with $\int g d\mu <\infty$! – PhoemueX Jun 06 '15 at 11:31
  • If $|g_n - g| \not \to 0$, then you do not necessarily have $|g_n - g| \to C$. Sometimes $|g_n - g|$ does not converge at all. – Alex M. Jun 06 '15 at 12:25
  • @PhoemueX Indeed consider the following inequality: $$ (\sqrt{f_n} - \sqrt f)^2 \leq 4(f_n + f), $$ which is a special case of the first inequality of the first answer here – Aspera Jun 06 '15 at 12:35
  • @AlexM. Yes, but I am passing up to a subsequence – Aspera Jun 06 '15 at 12:36
  • Passing to subsequences only helps if your sequences (or nets) are contained in some compact subset, which is not the case here. – Alex M. Jun 06 '15 at 12:37
  • @AlexM. But $[0,\infty]$ is compact, no? – Aspera Jun 06 '15 at 12:39
  • Not to mention that your functions don't live in $[0,\infty]$ but in $L^1$, so you'd have to find a compact set therein. – Alex M. Jun 06 '15 at 12:41
  • @AlexM. If a non-negative sequence is bounded then it has a convergent subsequence. Otherwise it has a subsequence going to $\infty$. I don't understand what is your problem – Aspera Jun 06 '15 at 12:42
  • What you say is true for sequences of numbers, not of functions (because $L^1$ is not locally compact). And you deal with sequences of functions, this is my problem (and yours). – Alex M. Jun 06 '15 at 12:45
  • @AlexM. Please read the text carefully. I am taking a subsequence of $h_n$ s.t. $\int h_n$, which is in $[0, \infty]$, converges to something. – Aspera Jun 06 '15 at 12:46