Convergence of sequence of $L^{p}$ function

Question

Given that $\Omega \subset \mathbb{R}^{n}$ is bounded. If you are given that $u_{k} \rightarrow u$ in $L^{p- \epsilon}(\Omega)$ and a functions $f: \mathbb{R} \rightarrow \mathbb{R}$ where $\{f(u_{k})\}_{k \in \mathbb{N}}$ is bounded in $L^{p'+\epsilon}(\Omega)$, where $\epsilon > 0$ and $p' = \frac{p}{p-1}$. How does it follow that $$\int_{\Omega}f(u_{k})(u_{k}-u)dx \rightarrow 0$$ Given that $p-\epsilon > 1$ annd $p'+ \epsilon > 1$. Can anyone see how this follows?

Answer 1

$p'$ is the conjugate exponent of $p$. Let $(p-\epsilon)'$ be the conjugate exponent of $p-\epsilon$. Then $$ (p-\epsilon)'=p'+\frac{\epsilon}{(p-1-\epsilon)(p-1)}=p'+\delta,\quad\delta>0. $$ Then, by Hölder's inequality: $$ \Bigl|\int_{\Omega}f(u_{k})\,(u_{k}-u)\,dx\Bigr|\le\Bigl(\int_{\Omega}|f(u_{k})|^{p'+\delta}\,dx\Bigr)^{1/(p'+\delta)}\Bigl(\int_{\Omega}|u_k-u|^{p-\epsilon}\,dx\Bigr)^{1/(p-\epsilon)}. $$ This gives the result if $\Omega$ is bounded and $\delta\le\epsilon$, or for any $\Omega$ if the conditions are in the sense that they hold for all sufficiently small $\epsilon$.