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 $a,b \geq 0$.

  • $\begingroup$ Thank you! May I know the name of the inequality? It looks like Jensen's Inequality and Mean Value Theorem. $\endgroup$ – Evan William Chandra Mar 25 at 6:01
  • $\begingroup$ 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. $\endgroup$ – Kavi Rama Murthy Mar 25 at 6:05
  • $\begingroup$ For the 2nd inequality though, how can I apply it for the negative sign? It does not seem trivial $\endgroup$ – Evan William Chandra Mar 25 at 6:12
  • 1
    $\begingroup$ @EvanWilliamChandra Use: $|u_n|^{\alpha} \leq |u_n-u|^{\alpha}+|u|^{\alpha}$ and $|u|^{\alpha} \leq |u_n-u|^{\alpha}+|u_n|^{\alpha}$. $\endgroup$ – Kavi Rama Murthy Mar 25 at 6:17
  • $\begingroup$ Thank you! Now, everything is clear! $\endgroup$ – Evan William Chandra Mar 25 at 6:28

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