Let $\Omega \subset \mathbb{R}$ be a bounded domain, $v \in H_{0}^{1}(\Omega)$ such that $||u_{n}-v||_{H_{0}^{1}(\Omega)}\to 0$ as $n\to\infty$ for a bounded sequence $\{u_{n}\}_{n\in\mathbb{N}} \subset H_{0}^{1}(\Omega)$, and $2<p<\infty$. I want to show that $\forall \phi \in C_{0}^{\infty}(\Omega),\, \int_{\Omega}u_{n}|u_{n}|^{p-2}\phi dx \to \int_{\Omega}v|v|^{p-2}\phi dx$

This is my attempt so far $$|\int_{\Omega}(u_{n}|u_{n}|^{p-2} - v|v|^{p-2})\phi dx| \leq\sup\limits_{\Omega}|\phi|\int_{\Omega}|u_{n}|u_{n}|^{p-2} - v|v|^{p-2} |dx $$ Then, I am not sure how to proceed from that line to make use of the bounded sequence. I was only given hints to use mean value theorem and Holder inequality (which I have used by taking the supremum for $\phi$).

Any help is much appreciated! Thank you very much!

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    $\begingroup$ Since your domain is one-dimensional, the convergence in $H_0^1(\Omega)$ implies convergence in $L^\infty(\Omega)$. $\endgroup$ – gerw Mar 20 at 6:33
  • $\begingroup$ I want to ensure. Do you mean I can take $\sup_{\Omega}|u_{n}|u_{n}|^{p-2} - v|v|^{p-2}|\to 0$ as $n\to\infty$? How do I show this claim? Thank you for your hint. $\endgroup$ – Evan William Chandra Mar 25 at 4:18
  • 1
    $\begingroup$ This should follow from $\sup_\Omega | u_n - v| \to 0$ (together with the uniform boundedness of all involved functions). $\endgroup$ – gerw Mar 25 at 7:02

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