It is a homework problem.
Function $f:\mathbb{R}^2\rightarrow\mathbb{R}$,$\Omega$ is a disk at origin with radius as $\dfrac{1}{2}$.
$f(x) = \log(\log(\dfrac{1}{|x|}))$, where $x\in\Omega$(i,e. $|x|\in (0,1/2)$),
it is easy to show that $f\in H^1(\Omega)$, the problem says, find a optimal $s\ge 1$, such that
$f(x)\in H^s(\Omega)$.
Just some hints will be ok.
The proof for it be to in $H^1(\Omega)$ is here
Thanks!
I think there is a problem at $0$ that you should look at. Also, first check your assertion that this function belongs in $H^1$. A general suggestion is: use polar coordinates, substitute $r = e^{-e^t}$, so that $f(r) = t$, check Sobolev space inclusions for integer points and use the Sobolev interpolation results to conclude for numbers in between integers.
I think $s=1$ is the optimal solution.
If $s>1$, since we can also manage to show that $f$ does not belong to $H^2$, then $1<s<2$.
$$\|f\|_{H^s} = \|f\|_{H^1}+(\int_D\int_D \dfrac{|Df(x)-Df(y)|^2}{|x-y|^{2+2(s-1)}}dxdy)^{1/2}=\|f\|_{H^1}+(\int_D\int_D \dfrac{|Df(x)-Df(y)|^2}{|x-y|^{2s}}dxdy)^{1/2}$$
where $x\in D,y\in D$.
$f(x) = \log\log(\dfrac{1}{r})$, $Df = -\dfrac{\mathbf{r}}{r^2\log r}$, we can see that even we choose $(s-1)$ very small, the integral will give you something as
$$\int_{0}^{1/2} \dfrac{1}{r^4(\log r)^2}dr$$
which is not bounded.
Thus $f\in H^1(\Omega)$, which is the optimal sobolev space.