Prove that for $n>1$,the non-bounded function $u=\ln\ln\left(1+\frac{1}{|x|}\right)\in W^{1,n}(\Omega)$,Here $\Omega=B(0,1)\subset \mathbb{R}^{n}$
I think we have to prove that $$ \int_{B(0,1)}\left|\ln\ln\left(1+\frac{1}{|x|}\right)\right|^{n} dx<\infty,$$ and $$ \int_{B(0,1)}\left|D\ln\ln\left(1+\frac{1}{|x|}\right)\right|^{n} dx<\infty,$$ but it hard for me to compute these integrates.
First, we need to show that $u \in L^n(\Omega)$. You can show by a change of variable that
$$\int_\Omega |\ln \ln (1 + \frac{1}{|x|})|^n dx \leq C \int_0^1 |\ln \ln (1 + \frac{1}{r})|^n r^{n-1} dr$$
The function $|\ln \ln (1 + \frac{1}{r})|^n r^{n-1}$ is continuous on $]0,1]$, so the only problem may be on 0. But we have that for r smaller than 1/e
$$1 < \ln (1 + \frac{1}{r}) < \frac{1}{r}$$
Hence
$$0<|\ln \ln (1 + \frac{1}{r})|^n < \ln(\frac{1}{r})^n$$
So
$$0 < |\ln \ln (1 + \frac{1}{r})|^n r^{n-1} < \ln(\frac{1}{r})^nr^{n-1}$$ and this converge to 0. So the function is bounded near 0, and integrable near 0
Then for $\nabla f$, you can see that it's differentiable in the classical sense, so computing $\nabla f$ is not too problematic. If I'm not mistaken,
$$\partial_i f(x) = \frac{x_i}{|x|^2(1+|x|)\ln(1+\frac{1}{|x|})}$$
You get that
$$\int_{\Omega} \left( \partial_i f(x)\right)^n dx \leq \int_0^1 \left| \frac{r}{r^2(1+r)\ln(1+\frac{1}{r})} \right|^n r^{n-1} dr$$
$$= \int_0^1 \frac{r^{n-1}}{(r^2(1+r))^n } \frac{1}{\ln(1+\frac{1}{r})^n} dr $$
Same thing, the only problem is on 0, but the function is positive, so we can take an equivalent
$$\frac{r^{2n-1}}{(r(1+r))^n \ln(1+\frac{1}{r})^n} \sim_0 \frac{1}{r\ln(1+\frac{1}{r})^n} $$
This can be show to be equivalent to
$$\sim_0 \frac{1}{r|\ln(r)|^n}$$
And this function is classically know to be integrable near 0 (looks for Bertrand integral if you don't know it)