Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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...

share|improve this question
add comment

3 Answers 3

up vote 1 down vote accepted

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 $L^{2/a}$.

Inequality

share|improve this answer
    
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
show 1 more comment

A partial answer .

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 $f$.

Perhaps other participants of this forum can show us how to remove this assumption or give us a better argument.

share|improve this answer
    
Of course, if $a>3$ the boundeness condition can be easily removed. –  Leandro Mar 29 '11 at 3:48
add comment

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 $p > 1$.

share|improve this answer
    
Kudos for the upper bound on $p$! –  user1736 Mar 29 '11 at 15:53
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.