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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!

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

1 Answer 1

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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. $$|x| < \frac1{\exp\exp M - 1}.$$

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  • $\begingroup$ Thanks for the answer but I wanted to use the definition of essential boundness...So, in this case , which is the subset of the disk with strictly positive measure in which my function isn't bounded from above? $\endgroup$
    – kaithkolesidou
    Apr 4, 2018 at 16:39
  • $\begingroup$ @kaithkolesidou, see the edit. $\endgroup$
    – Martín-Blas Pérez Pinilla
    Apr 4, 2018 at 16:45

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