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Let $I=[0,1]$, and let $(u_n)$ be a sequence in $L^\infty(I)$ converging to $u\in L^\infty(I)$ in the weak$\vphantom{}{^*}$ topology of $L^\infty(I)$. Let $f:\mathbb R\to\mathbb R$ be a $C^2$ function with $f''(t)>0$ for any real $t$. Assume that $$\lim_{n\to\infty}\int_I f(u_n(x))\mathrm dx=\int_If(u(x))\mathrm dx.$$ Prove then that $u_n$ converges strongly in $L^2(I)$ to $u$.

EDIT since the question has been raised in the comments I want to remark that the $f$ in the problem is not arbitrary but it is a specific strictly convex function..

EDIT 2. I've thought quite a long time about this problem but yet I didn't came up with anything interesting. But a reliable way to follow it seems to me Taylor Expansion in the sense that one can write $$f(u(x))=f(u_n(x))+f'(u(x))(u_n(x)-u(x))+\frac 12 f''(u(x))(u_n(x)-u(x))^2+h_2(u(x))(u_n(x)-u(x))^2$$ and then try to integrate and pass to the limit and see what happens. Is this useful?

Please let me know because i feel lost in front of this problem.

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Maybe just a stupid idea... If you write your expansion as $$f(u(x))=f(u_n(x))+f'(u(x))(u_n(x)-u(x))+\frac{1}{2}f''(\xi_n(x)){(u_n(x)-u(x))‌​}^{2}$$ for some $\xi_n(x)$ lying between $u(x)$ and $u_n(x)$, and if you know that $\xi_n(x)$ is confined to a bounded interval, then you can exploit the fact that $f'' \geq C>0$ on that interval, and pass to the limit as $n \to +\infty$. Do you think this may be useful? –  Siminore Sep 2 '12 at 12:23
    
I was thinking the same as you did Siminore. But how would you conclude that $\xi_n(x)$ remains bounded not only as $n\to\infty$ but also as $x$ varies in $I$? –  uforoboa Sep 2 '12 at 12:45

1 Answer 1

up vote 4 down vote accepted
+200

Since $u_n$ converges to $u$ weak* in $L^\infty(I)$, we know that $\|u_n\|_{L^\infty}$ is uniformly bounded. Indeed, the weak* convergence implies $$ |\langle u_n,\phi\rangle| \leq |\langle u,\phi\rangle| + 1 \leq \|u\|_{L^\infty} + 1, $$ for $\phi\in L^1$ such that $\|\phi\|_{L^1}=1$, and for $n$ large. This gives a uniform bound on $|\langle u_n,\phi\rangle|$ for any fixed $\phi\in L^1$, and an application of the uniform boundedness principle, followed by the duality between $L^\infty$ and $L^1$, shows that $\|u_n\|_{L^\infty}$ is uniformly bounded.

Then as Siminore observed in the comments, $$ f(u_n(x))-f(u(x))-f'(u(x))(u_n(x)−u(x))=\frac12f''(\xi_n(x))(u_n(x)−u(x))‌^2\geq c|u_n(x)-u(x)|^2, $$ for almost every $x\in I$, with some constant $c>0$. Integrating over $I$, we get $$ \int \left(f(u_n)-f(u)\right) - \int f'(u) (u_n-u) \geq c\|u_n-u\|_{L^2}^2. $$ In the limit $n\to\infty$, the first integral on left hand side goes to $0$ by assumption, and so does the integral term by weak* convergence. Hence $u_n\to u$ in $L^2$.

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How does your argument show that $\|u_n\|_{L^\infty}$ is bounded? –  Byron Schmuland Sep 2 '12 at 17:39
    
@Byron: Please have a look at the update. –  timur Sep 2 '12 at 22:04
    
I don't see it. Your bound is only valid for $n\geq N(\phi)$, where $N(\phi)$ depends on $\phi$. –  Byron Schmuland Sep 2 '12 at 22:17
    
@Byron: Thanks for the catch! I guess it is ok now. –  timur Sep 2 '12 at 23:03
    
That's what I was looking for! –  Byron Schmuland Sep 2 '12 at 23:05

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