When we have a nonnegative, bounded sequence $f_n\in L^p[0,1]$, we can get a nonnegative, bounded sequence $f_n^{p-1}$ in $L^q[0,1]$ (where $q$ is the Hölder conjugate of $p$). Assume furthermore that both sequences are weakly convergent if we consider $L^p$ and $L^q$ as duals of one another (we can do this by Banach-Alaoglu theorem).
The question is, if we take for $f$ the weak limit of $f_n$, is it true that $f^{p-1}$ is the (weak) limit of $f_n^{p-1}$? How do I prove such thing? This is trivial for $p=2$, but I need a little more generality...
The motivation comes from a differential equation problem. $f_n$ I consider are actually (absolute values of) weak derivatives, but I don't think it really matters for this question.
This is not true unless $p = 2$. Let $$f_n(x) = \begin{cases}2 & \text{if } 2^n\, x \in [2\,m,2\,m+1] \text{ for some } m \in \mathbb N \\ 0 & \text{else}.\end{cases}$$ Then, $f_n \rightharpoonup 1$ in every $L^p(0,1)$, $p < \infty$. That is, $f \equiv 1$. However, $f_n^\lambda \rightharpoonup 2^{\lambda-1}$ in every $L^p(0,1)$ and all $p < \infty$ for every $\lambda \in \mathbb R$.
For $p \ne 2$, you have $p - 2 \ne 0$, hence $f_n^{p-1} \to 2^{p-2} \ne 1 = 1^{p-1}$.
The crucial part is that $x \mapsto x^{p-1}$ is non-linear and does not preserve weak convergence.