# Convergence of functions in $L^p$

Let $\{f_k\} \subset L^2(\Omega)$, where $\Omega \subset \mathbb{R}^n$ is a bounded domain and suppose that $f_k \to f$ in $L^2(\Omega)$. Now if $a \geq 1$ is some constant, is it possible to say that $|f_k|^a \to |f|^a$ in $L^p$ for some $p$ (depending on $a$ and also possibly depending on $n$)?

Showing the statement is true would probably require a smart way of bounding $\left| |f_k|^a - |f|^a \right|$ by a term including the factor $|f_k - f|^2$. However, I don't really know what to do with the fact that $a$ doesn't have to be an integer...

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First, one cannot expect better results than $p=\frac 2 a$ because we only know $f\in L^2(\Omega)$. And I do think it is true for $p=\frac 2 a$. Proof is as follows.

Note that $x^a$ is a convex increasing function for $x\ge 0$, hence (draw a picture and you can see this) $$0\le \frac{u^a-v^a}{u-v}\le a\max\{u^{a-1},v^{a-1}\}, \forall u, v\ge 0, u\neq v.$$ Plugging in $u=|f_k|, v=|f|$ and noticing that $||f_k|-|f||\le |f_k-f|$, we have $$||f_k|^a-|f|^a|\le a|f_k-f|\max\{|f_k|^{a-1},|f|^{a-1}\}.$$ Then, raising the last inequality to the power $p=\frac 2 a$, by Hölder's inequality, $$\||f_k|^a-|f|^a\|_{L^p}^p\le a^p \|f_k-f\|_{L^2}^{2/p}\|\max\{|f_k|^{a-1},|f|^{a-1}\}\|_{L^{\frac{2}{a-1}}}^{\frac a {a-1}}.$$ (When we apply Hölder, $|f_k-f|^p\in L^a,$ and $\max\{|f_k|^{a-1},|f|^{a-1}\}^p\in L^r, r=a^*=\frac a {a-1}.$ You have to check the exponents to see if I made any mistake.)

Since $f_k$ have bounded $L^2$-norm (they converge),$$\|\max\{|f_k|^{a-1},|f|^{a-1}\}\|_{L^{\frac{2}{a-1}}}$$ is bounded by some constant. Sending $k\to\infty$, we have $|f_k|^a\to|f|^a$ in $L^{2/a}$.

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Hm, I think you actually get $||\max\{|f_k|^{a-1},|f|^{a-1}\}||_{L^\frac{2}{a-1}}^\frac{a-1}{a}$ after the application of Holder's inequality. I haven't convinced myself of the convex inequality yet, but other than that, your result looks really nice! – user1736 Mar 29 '11 at 16:40
Well, I'm not sure about the exponent. But to see the inequality, draw a picture and compare the slope of the secant line and that of the tangent lines at u and v. – GWu Mar 29 '11 at 20:13
Um, isn't the expression a 3 dimensional image? How do you draw that? – user1736 Mar 31 '11 at 16:29
I'm taking about graph of $y=x^a$. I uploaded a picture. Hope it helps. – GWu Apr 1 '11 at 4:31
You can also just use the mean value theorem: ${u^a - v^a \over u - v} = a w^{a-1}$ for some $w$ between $u$ and $v$. Since $ax^{a-1}$ is increasing, this is at most $a\max{(u^{a-1},v^{a-1})}$ – Zarrax Apr 1 '11 at 14:31

Some answers for $a = 1, 2$:

If $a = 1$, then $\big||f_k| - |f|\big| \leq |f_k - f|$, so you have convergence in $L^2$. And you'd also have convergence in $L^p$ for $p < 2$ since $L^p$ norms increase with $p$ on bounded domains. But never for $p > 2$ since $|f|^{p}$ needs to be integrable.

If $a = 2$, then you can use $|f_k^2 - f^2| = |f_k - f| |f_k + f|$. So by Cauchy Schwarz $$\int_{\Omega}|f_k^2 - f^2| \leq (\int_{\Omega}|f_k - f|^2)^{1 \over 2} (\int_{\Omega}|f_k + f|^2)^{1 \over 2}$$ The left factor goes to zero by assumption, and the right factor is bounded by a constant since $|f_k + f|^2 \leq 2|f_k|^2 + 2|f|^2$. Thus the product on the right goes to zero, and the same therefore must be true for the product on the left. So $f_k^2$ goes to $f^2$ in $L^1$. It can't coverge in any higher $L^p$ since $f$ is not assumed to be in $L^{2p}$ for any $p > 1$.

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Kudos for the upper bound on $p$! – user1736 Mar 29 '11 at 15:53

Fix $a\geq 1$ and define $\varphi(x)=x^a$. By the Mean value Theorem we have $$|\varphi(|f_k|)-\varphi(|f|)|\leq a \sup_{t\in[0,1]} \big| t|f_k|+(1-t)|f|\big |^{a-1}\cdot \big| |f_k|- |f|\big|$$ Assume that $f_k$ and $f$ are bounded. From the previous inequality we get that $$|\varphi(|f_k|)-\varphi(|f|)|\leq M \big| |f_k|-|f|\big|\leq M|f_k-f|$$ the last inequality is obtained by the second triangular inequality. Taking square in both sides we end up with $$||f_k|^a-|f|^a|^2\leq \tilde{M} |f_k-f|^2$$ therefore you can chose $p=2$ and you have the convergence you are looking for.
Right now I don't know how to remove the strong assumption I made about the boundedness of $f_k$ and $f$.
Of course, if $a>3$ the boundeness condition can be easily removed. – Leandro Mar 29 '11 at 3:48