I want to say no, but what is a counterexample? More generally, does $L^p$ convergence of a sequence of functions imply any sort of convergence (i.e. in some different $L^{p'}$ space) of a function of that sequence of functions?
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$\begingroup$ Do you mean, convergence in $L^{p/2}$? $\endgroup$– FShrikeFeb 1 at 19:03
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1$\begingroup$ There's a whole well-developed theory about which of the usual modes of convergence imply which other modes of convergence, and when $f \circ g_n$ converges if $g_n$ does. It's a lot to just write down let alone prove it all. But just to give one example to help out, look at $f_n(x)=x^{-1/2} \chi_{[1/n,1]}$. $\endgroup$– IanFeb 1 at 19:04
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1$\begingroup$ We have $f_n^2\to f^2$ in $L^{p/2}.$ $\endgroup$– Ryszard SzwarcFeb 1 at 19:07
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$\begingroup$ @FShrike I didn't know that would be a relevant one. I was wondering if it implies convergence in any other sense. Edited, thanks. $\endgroup$– 900edgesFeb 1 at 19:24
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$\begingroup$ @Ian Thanks, glad to know it's not trivial. What are some of the relevant theorems or references? $\endgroup$– 900edgesFeb 1 at 19:25
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Yes, it does, just write $$ \|f_n^2-f^2\|_{L^{p/2}} = \|(f_n-f)(f_n+f)\|_{L^{p/2}} $$ and so by Hölder's inequality and the triangle inequality $$ \|f_n^2-f^2\|_{L^{p/2}} ≤ \|f_n-f\|_{L^p}\left(\|f_n\|_{L^p}+\|f\|_{L^p}\right). $$ Since $f_n\to f$ in $L^p$, the sequence $(\|f_n\|_{L^p})_{n\in\Bbb N}$ is bounded and so one deduces that $f_n^2\to f^2$ in $L^{p/2}$.
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$\begingroup$ Not sure if this affects anything, but $p/2$ can be less than 1 $\endgroup$ Feb 1 at 20:42
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$\begingroup$ Yes, indeed. If $p<2$, then all the above manipulations above still make sense noting just that $\|g\|_{L^{p/2}} = (\int |g|^{p/2})^{2/p}$ is not a norm anymore. In this case, $\|g\|_{L^{p/2}}^{p/2} = \int |g|^{p/2}$ is however a distance (it verifies the triangle inequality) so there is still a convergence. $\endgroup$– LL 3.14Feb 1 at 20:53
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$\begingroup$ Thanks! Do you know of such results for more general continuous functions of $f_n$ and $f$? $\endgroup$– 900edgesFeb 1 at 20:54
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$\begingroup$ Not particularly, but notice that the same reasoning is easy to generalize to higher powers. Using $$ |a^m-b^m| \leq |a-b| \sum_{k=1}^m |a|^{k-1} |b|^{m-k} \leq |a-b| \sum_{k=1}^m \frac{k-1}{m-1} |a|^{m-1} + \frac{m-k}{m-1} |b|^{m-1} $$ gives $$ |a^m-b^m| \leq \frac{m}{2} \left(|a|^{m-1} + |b|^{m-1}\right) |a-b| $$ and you deduce as in my post that $f_n^m \to f^m$ in $L^{p/m}$. $\endgroup$– LL 3.14Feb 1 at 21:15