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Suppose $\Omega$ be an open subset of $\mathbb R^n$ and $a_h$ be a sequence of $L^\infty(\Omega)$ functions such that there exist two constants $0<c_1\leq c_2$ such that $$c_1\leq a_h(x)\leq c_2$$ for almost any $x\in\Omega$ and any $h\in\mathbb N$. Suppose moreover that there exists $a,b\in L^\infty(\Omega)$ such that $$\begin{aligned}a_h\rightharpoonup a,\\ \frac{1}{a_h}\rightharpoonup\frac 1b. \end{aligned}$$

in $w^*-L^\infty(\Omega)$.

Prove that in general $b\leq a$ and $b=a$ implies that $a_h\to a$ strongly in $L^2(\Omega)$.

I'm pretty sure this should be aneasy consequence of Banach Steinhaus, however in my computations I always end up with the equality reversed. Probably I'm messing something with limit superiors and inferiors. Can anybody help me?


Ok so thanks to the comment below the map $x\mapsto \frac 1x$ is CONVEX then by Jensen and the Weak star convergence in $L^\infty(\Omega)$ we now that for almost any $x\in\Omega$ everything is differentiable according to the LEbesgue differentiation theorem and then $$|B_r|\left(\int_{B_r}a_h\mathrm d\mu\right)^{-1}\leq \frac{1}{|B_r|}\int_{B_r}\frac{1}{a_h}\mathrm d\mu.$$ Passing to the limit as $r\to 0$ around a Lebesgue point we eventually get $b\leq a$. Am I right?

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I would average over a small interval and use Jensen. – mike Nov 5 '12 at 16:40
@guido the answer is right since $x\mapsto 1/x$ is CONVEX.. – uforoboa Nov 5 '12 at 18:11
lol you are right... – guido giuliani Nov 5 '12 at 18:48

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