This is a follow-up on [another question](https://mathoverflow.net/questions/110741/weak-convergence-of-the-image-of-an-l1-converging-sequence-under-a-convex-func).


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 $\int f(u\_k)\to\int f(u)$?



Weak convergence of the image of a weakly $L^1$ converging sequence
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.


