# If $f_n$ converges to $f$ in $L^p$ then does $f_n^2$ converge to $f^2$?

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?

• Do you mean, convergence in $L^{p/2}$? Feb 1 at 19:03
• 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 $f_n(x)=x^{-1/2} \chi_{[1/n,1]}$.
– Ian
Feb 1 at 19:04
• We have $f_n^2\to f^2$ in $L^{p/2}.$ Feb 1 at 19:07
• @FShrike I didn't know that would be a relevant one. I was wondering if it implies convergence in any other sense. Edited, thanks. Feb 1 at 19:24
• @Ian Thanks, glad to know it's not trivial. What are some of the relevant theorems or references? Feb 1 at 19:25

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 $$L^{p/2}$$.
• Not sure if this affects anything, but $p/2$ can be less than 1 Feb 1 at 20:42
• 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. Feb 1 at 20:53
• Thanks! Do you know of such results for more general continuous functions of $f_n$ and $f$? Feb 1 at 20:54
• 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 $L^{p/m}$. Feb 1 at 21:15