# Showing $\;\ln(\ln(1+\frac{1}{\vert x \vert}))\notin L^{\infty}(\{\vert x \vert \lt 1\})\;$

I'm trying to prove that $\;\ln(\ln(1+\frac{1}{\vert x \vert}))\;$ is a counterexample for the non embedding of $\;W^{1,2}(\{\vert x \vert\ \lt 1\})\;$ space into $\;L^{\infty}(\{\vert x \vert \lt 1\})\;$.

I've got trouble showing $\;\ln(\ln(1+\frac{1}{\vert x \vert}))\notin L^{\infty}(\{\vert x \vert \lt 1\})\;$. In class, we've mainly studied $\;L^p-$norms for $\;1 \le p \lt \infty\;$, thus I 'm a little bit confused on how I should proceed.

I know that $\;f \notin L^{\infty}(\{\vert x \vert \lt 1\})\;$ means that for all $\;M \gt 0\;$ there exists $\;A\subset \{\vert x \vert \lt 1\}\;$ with $\;μ(A)\gt 0\;$ and $\;\vert f \vert \gt M\;$ on $A\;$. But how do I find such $\;A\;$ for my chosen function?

Furthermore, in class when it comes to $\;f \in L^{\infty}\;$, we mostly handle it as it suffices to find an upper bound for $\;\vert f \vert\;$, but this doesn't imply necessary essentially boundness, does it?

I would really appreciate if somebody could help me because I feel lost. Thanks in advance!

• If $|x|<\frac{1}{n}$ then $\log\log\left(1+\frac{1}{|x|}\right)>\log\log n$. If we pick $n$ as $\exp\exp M$... Apr 4, 2018 at 16:31
• @JackD'Aurizio $\;μ(\{|x|<\frac{1}{n}\})\gt 0\;$ since $\;\{ |x|<\frac{1}{n} \}\;$ is a non-empty open set? Apr 4, 2018 at 16:34
• Exactly.$\phantom{}$ Apr 4, 2018 at 16:42
• @JackD'Aurizio Thanks a lot! Apr 4, 2018 at 16:44

Use that $\log\circ\log$ is increasing and $$\lim_{r\to+\infty}\log(\log r) = +\infty.$$ EDIT: Let be $M > 0$: $$\log(\log r) > M\iff r > \exp\exp M,$$ or $$1 + 1/|x| > \exp\exp M,$$ i.e. $$|x| < \frac1{\exp\exp M - 1}.$$