Properties of convergence in $L^{\infty}$

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 $$a,b \geq 0$$.
• The first one is just MVT. The second is also fairly standard bit it doesn' t have a name. To prove it consider RHS $-$ LHS as a function of $a$. Check that it is an increasing function on $[0,\infty)$ and that it has the value $0$ when $a=0$. hence it is non-negative. – Kavi Rama Murthy Mar 25 at 6:05
• @EvanWilliamChandra Use: $|u_n|^{\alpha} \leq |u_n-u|^{\alpha}+|u|^{\alpha}$ and $|u|^{\alpha} \leq |u_n-u|^{\alpha}+|u_n|^{\alpha}$. – Kavi Rama Murthy Mar 25 at 6:17