Suppose that $u\_k$ is a sequence of $L^1$ functions defined on a compact $K\subset R^n$ and a function $f:[0, \infty)\to[0, \infty)$ with the following properties


* $u\_k\ge 0$
* $\|u\_k\|\_{L^1}=\int u\_k=1$
* $u\_k\to u$ strongly in $L^1$
* $f$ is convex, $f(0)=0$ and has superlineair growth at $+\infty$ (that is: $\lim\_{z\to+\infty} \frac{f(z)}{z}=+\infty$)
* $\int f(u\_k)< C$ for all $k$


Note that the conditions on $f$ imply that the functional $v\mapsto\int f(v)$ is lower semicontinuous with respect to weak $L^1$ convergence.


Does this imply that $f(u\_k)$ converges to $f(u)$ weakly in $L^1$, possibly under the extra assumption that $\int f(u\_k)\to \int f(u)$?


**EDIT**: Thanks for the answers, both were helpful and received an upvote.



Weak convergence of the image of an $L^1$ converging sequence under a convex function
With the extra assumption it is true, and only continuity on $f:[0,\infty)\rightarrow [0,\infty)$ is needed. Of course, it is sufficient to show that some subsequence of $f(u\_k)$ converges. So we can also assume w.l.o.g. that $u\_k$ converges a.e. to $u$. 


Consider the sequence of non-negative measurable functions on $K$, $v\_k:=f\circ u\_k$. Because $f$ is continuous, it converges a.e. to $v:=f\circ u$, and by the extra hypothesis, $\int\_ K v\_k \to\int\_K v < \infty $. And, as an immediate consequence of the Fatou's lemma, this also implies $\int\_ S v\_k \to\int\_S v$ for every measurable $S\subset K$ (this is the key point, after all, coming from the equality $\int\_K v\_k =\int\_S v\_k +\int\_{K\setminus S} v\_k $: if one does not lose mass globally, one does not lose mas locally). 


Now, as a general fact, on a finite measure space $K$ this situation 
implies the $L^1$ convergence. 


Indeed, let be given a number $\epsilon > 0$. There is a number $\delta > 0 $ such that $\int\_ S v < \epsilon$ whenever $S\subset K$ has measure $\big| S\big| < \delta$.
By the Severini-Egorov theorem $v\_k$ converges almost uniformly, so there is some $S\subset K$ of measure less than $\delta$ such that $v\_k$ converges uniformly to $v$ on $K\setminus S$. So we have:


$$ \|v \_ k - v\| \_ {1,K}=\|v \_k-v\| \_{1,S}+\|v \_k-v\| \_{1,K\setminus S}$$ $$\le \int \_S v \_ k + \int \_S v + \big|K\big| \|v \_ k-v\| \_ {\infty,K\setminus S} \le 2\epsilon + o(1)\, , \quad \mathrm{as}\quad k\to\infty \, .$$


Since this is true for any $\epsilon > 0$ we have $\limsup\_{k\to\infty} \|v \_ k - v\| \_ {1,K}=0$ proving the convergence . Actually, the finiteness assumption on $K$ may also be dropped.


Here's another proof: consider the sequence $ w\_k : = \sqrt v\_k \in L^2(K)$. It is norm-bounded, converges to $w : = \sqrt{v}$ a.e., hence weakly in $L^2$; moreover, 
$\|w\_ k \| \_ 2 \to \|w \| \_ 2 $ . In a Hilbert space, this implies strong convergence (this is immediately seen just expanding $\|w \_ k - w\| \_ 2 ^2$). The map $L^2\ni w\mapsto w^2\in L^1$ is continuous, and we conclude $v\_k \to v$ in $L^1$.


