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

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.