- Let $B_1$ the unit ball closed in $\mathbb{R}^n$
- $u: B_1 \to \mathbb{R}$ is a function in $H^1(B_1)$
If I have the following inequality:
(*) $\|u\|_{L^{\gamma_i}(B_{\frac{1}{2}})} \leq C\|u\|_{L^2(B_1)}$ where $\gamma_i = 2(\frac{n}{n-2})^i$ for $n>2$ and for all $i=1,2,3,\dots$ and $C$ is a positive constant.
How do I show that
$$\sup_{B_{\frac{1}{2}}} \leq C\|u\|_{L^2(B_1)} \text{ ?}$$
If I take $i \to \infty$ , I know $\gamma_i \to \infty$, and as the domain is $B_{\frac{1}{2}}$ which has finite Lebesgue measure, then $\|u\|_{L^{\gamma_i}(B_{\frac{1}{2}})} \to \|u\|_{L^\infty(B_{\frac{1}{2}})}$. Hence, I do not know how to get out of the impasse that has just appeared, because the only thing I know is that the $L^\infty$ norm of $u$ in that ball, dominates or is equal to $|u|$ ae. How then does this supremum appear?
Thanks
If $\|u\|_{L^{\gamma_i}(B_{1/2})} \leq C\|u\|_{L^2(B_1)}$ for all $i$, and if $\|u\|_{L^{\gamma_i}(B_{1/2})} \to \|u\|_{L^\infty(B_{1/2})}$ as $i\to\infty$, then it would seem that $\|u\|_{L^\infty(B_{1/2})}\le C\|u\|_{L^2(B_1)}$. You say that $|u|$ is dominated a.e. on $B_{1/2}$ by $\|u\|_{L^\infty(B_{1/2})}$ (which I agree with). Doesn't it then follow that $\mathrm{ess}\sup_{B_{1/2}} |u|\le C\|u\|_{L^2(B_1)}$?
Is the issue that is worrying you the difference between supremum and essential supremum? It seems obvious that we can't bound the supremum of $u$ if we only know things about $u$ which don't change if $u$ changes on a set of measure zero.
One caveat: for my argument to make sense, in $\|u\|_{L^{\gamma_i}(B_{1/2})} \leq C\|u\|_{L^2(B_1)}$ we must have one $C$ for all the $i$.