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This is a follow-up on another question.

Can something be said about the image of a weakly converging sequence in $L^1$? More precisely

  • $u_k\ge 0$
  • $\|u_k\|_{L^1}=\int u_k=1$
  • $u_k$ converges to $u$ weakly in $L^1$
  • $f$ is convex, $f(0)=0$ and has superlinear growth at $+\infty$ (that is: $\lim_{z\to+\infty}\frac{f(z)}{z}=+\infty$)
  • $\int f(u_k)$ is bounded

I don't think that it can be shown that $f(u_k)\to f(u)$ weakly in $L^1$. But can this be shown if it is also assumed that $\int f(u_k)\to\int f(u)$?

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Not an answer, but I vaguely recall that the only functions $f$ for which $u_k$ weakly to $u$ implies $f(u_k)$ weakly to $f(u)$ are affine linear $f$'s. –  Dirk Nov 13 '12 at 10:49
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2 Answers 2

up vote 4 down vote accepted

I think with extra assumption $\int f(u_{k}) \to \int f(u)$ one can prove weak convergence. Crucial observation, already made by Pietro Majer, is that $f(u \chi_{A}) = f(u) \chi_{A}$, since $f(0)=0$. If I recall correctly, convex functions are lower semicontinuous with respect to weak topology (it should be easily verifiable), so our assumptions give $$ \int f(u) \chi_{A} \leqslant \liminf_{n\to \infty} \int f(u_{n}) \chi_{A}.$$ But we can apply exactly the same reasoning to $\chi_{A'}$ to arrive at

$$ \int f(u) \leqslant \liminf_{n \to \infty} \int f(u_n) \chi_{A} + \liminf_{n \to \infty} \int f(u_n) \chi_{A'} \leqslant \liminf \int f(u_n) = \int f(u).$$ Hence,

$$ \liminf \int f(u_n) \chi_A = \int f(u) \chi_A$$ $$ \liminf \int f(u_n) \chi_{A'} = \int f(u) \chi_{A'}$$

What is more, $\liminf \int f(u_n) \chi_{A'} =\lim \int f(u_n) - \limsup \int f(u_n) \chi_{A}$, so $ \lim \int f(u_n) \chi_{A} = \int f(u) \chi_{A}$. Characteristic functions are linearly dense in $L^{\infty}$ and sequence $f(u_n)$ is bounded in $L^1$, so this implies weak convergence.

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Yesterday evening, I already thought that the solution might go along this lines. Thanks for posting this idea! –  Martijn Nov 14 '12 at 9:03
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First point: indeed, without the extra assumption, $f(u_k)$ does not converge weakly to $f(u)$ in general, even if $u_k$ converges strongly. Take $u_k:=k \chi_ {\big[0,\frac{1}{f(k)}\big]}$. So $\|u_k\|_1= k/f(k)\to0$. Since $f(0)=0$ one has $f(u _ k) = f(k) \chi_ {\big[0,\frac{1}{f(k)}\big]}$, so $\int f(u _ k) = 1$, but $f(u _k)$ does not converge weakly to $0$; actually it has no weakly-$L^1$ convergent subsequences at all.

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(actually, the question has $\int u_k = 1$, so this is a counterexample to something else) –  Pietro Majer Nov 13 '12 at 17:17
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