Weak convergence in different $L^p$ spaces

Question

Consider $p \ge \alpha \ge 1.$

If a sequence converges weakly in $L^p,$ say $u_n \rightharpoonup u$, is it true that: $$u_n^{\alpha} \rightharpoonup u^{\alpha} \text{ in $L^{p/ \alpha}$}$$

This question came to me, while trying to solve the following:

Composition of a weakly convergent sequence with a nonlinear function

We can always argue that $u_n^{\alpha}$ has a converging subsequence, but I couldn't go anywhere with this.

Answer 1

Assume that $p=\alpha=2$ and that we work on the unit interval. If the sequence $\left(u_n^2\right)_{n\geqslant 1}$ converges weakly to $u^2$ in $\mathbb L^1$, then using the fact that the constant function equal to $1$ belongs to $\left(\mathbb L^1\right)'$, we derive that $$\tag{*}\lVert u_n\rVert_2\to \lVert u\rVert_2.$$ If $\left(u_n\right)_{n\geqslant 1}$ converges to $u$ weakly in $\mathbb L^2$, then we can derive with the help of (*) that $\lVert u_n-u\rVert_2$ strongly in $\mathbb L^2$.

Therefore, any sequence in $\mathbb L^2(0,1)$ with converges in $\mathbb L^2$ weakly but not strongly qualifies for a counter-example.

Answer 2

Since $u_n\in L^p$ it holds $u_n^\alpha\in L^\frac{p}{\alpha}$. That should be correct so far. For $p\not=\alpha$ your Claim holds since $L^q$ ist reflexiv for $1<q<\infty$. If $p=\alpha$ I have no Idea sorry. ;)