Let $\Omega \subset \mathbb{R}$ be a bounded domain and $\alpha > 0$ be fixed. Assume that $|| u_{n} - v||_{L^{\infty}(\Omega)}\to 0$ as $n\to\infty$. How can I show that $||\, |u_{n}|^{\alpha} - |v|^{\alpha}||_{L^{\infty}(\Omega)}\to 0$ as $n\to\infty$?
This is my attempt so far :
If $v\equiv 0$, then we immediately have $||\, |u_{n}|^{\alpha} - |v|^{\alpha}||_{L^{\infty}(\Omega)} = ||\, |u_{n}-v|^{\alpha}||_{L^{\infty}(\Omega)}\leq||u_{n}-v||_{L^{\infty}(\Omega)}^{\alpha} \to 0$ as $n\to\infty$.
Now, my problem is that I cannot show for the case $v\not \equiv 0$. Any hint will be much appreciated!
Hint: for $\alpha \geq 1$ use $|x^{\alpha} -y^{\alpha }| =|x-y|\alpha |\xi|^{\alpha -1}$ for some $\xi$ betwee $x$ and $y$. For $\alpha <1$ use the inequality $(a+b)^{\alpha} \leq a^{\alpha}+b^{\alpha}$ for all $a,b \geq 0$.