If $u_n^p \rightharpoonup v$ in $L^1$, then does it follow that $u_n \rightharpoonup v^{\frac 1p}$ in $L^p$?

Question

Let $\Omega$ be a bounded domain. Suppose that $u_n^p \rightharpoonup v$ in $L^1(\Omega)$. Does it follow that $u_n \rightharpoonup v^{\frac 1p}$ in $L^p(\Omega)$?

Answer 1

No. Nonlinear transformations and weak convergence go together like drinking and driving. For example, let $r_k$ be the $k$th Rademacher function on $[0,1]$, that is $r_k = \operatorname{sign}\sin ( 2^k \pi x) $. Then $2^p r_k \rightharpoonup 2^{p-1}\mathbf {1}$ in $L^1$, where $\mathbf{1}$ is the constant function equal to $1$. On the other hand, $2r_k\rightharpoonup \mathbf{1}$ in $L^p$.