Show that $u(x)=\ln\left(\ln\left(1+\frac{1}{|x|}\right)\right)$ is in $W^{1,n}(U)$, where $U=B(0,1)\subset\mathbb{R}^n$.

Question

The entire problem statement is:

Let $n>1$ and let $U=B(0,1)\subset\mathbb{R}^n$. Show that $u:U\to\mathbb{R}$ given by $$u(x)=\ln\left(\ln\left(1+\frac{1}{|x|}\right)\right)$$ is in $W^{1,n}(U).$

My attempt at the proof is as follows:

To show that $u\in W^{1,n}(U)$ it suffices to show that $|Du|\in L^{n}(U)$. Consider $$u_{x_i}=\frac{1}{\ln\left(1+\frac{1}{|x|}\right)}\frac{1}{1+\frac{1}{|x|}}|x|^{-2}\frac{x_i}{|x|},$$ which simplifies to, $$u_{x_i}=\frac{1}{\ln\left(1+\frac{1}{|x|}\right)}\frac{1}{1+\frac{1}{|x|}}\frac{x_i}{|x|^3}.$$ Thus, $$|Du|=\frac{1}{\ln\left(1+\frac{1}{|x|}\right)}\frac{1}{1+\frac{1}{|x|}}\frac{1}{|x|^2}$$ which can be manipulated to, $$|Du|=\frac{1}{\ln\left(1+\frac{1}{|x|}\right)}\frac{1}{1+|x|}\frac{1}{|x|}.$$

Moreover, since $U=B(0,1)$ we can bound the second term from above which gives, $$|Du|\leq\frac{1}{\ln\left(1+\frac{1}{|x|}\right)}\frac{1}{|x|}$$

Thus, we can instead show that $$\int_U\left(\frac{1}{\ln\left(1+\frac{1}{|x|}\right)}\frac{1}{|x|}\right)^n<\infty.$$

Converting this into polar coordinates we have with $r=|x|$, $$\int_0^R\int_{\partial B(0,r)}\left(\frac{1}{\ln\left(1+\frac{1}{r}\right)}\frac{1}{r}\right)^n\,dSdr=\int_0^R\left(\frac{1}{\ln\left(1+\frac{1}{r}\right)}\frac{1}{r}\right)^n\int_{\partial B(0,r)}dSdr$$ Note that $\int_{\partial B(0,r)}\,dS=n\alpha(n)r^{n-1}$, which is the surface area of $B(0,r)$. Thus, we have then $$n\alpha(n)\int_0^R\left(\frac{1}{\ln\left(1+\frac{1}{r}\right)}\frac{1}{r}\right)^nr^{n-1}\,dr=n\alpha(n)\int_0^R\left(\frac{1}{\ln\left(1+\frac{1}{r}\right)}\right)^n\frac{1}{r}\,dr$$

And so this is where I get stuck in the problem, since I don't how I can evaluate that last integral to show it's finite.

Thank you for any help or feedback!

Answer 1

First of all, the problem should be $u(x):=\ln\left(\ln\left(1+\frac{1}{|x|}\right)\right) $ but not $u(x)=\ln\left(\ln\left(\frac{1}{1+|x|}\right)\right)$ as you stated.

Next, we have

$$ \partial_i u(x) = - \frac{1}{\ln(1+\frac{1}{|x|})}\frac{x_i}{|x|^3}\frac{1}{1+\frac{1}{|x|}}$$ Notice that $$ \frac{x_i}{|x|^3}\frac{1}{1+\frac{1}{|x|}}=\frac{x_i}{|x|^3}\frac{|x|}{|x|+1}=\frac{x_i}{|x|^2}\frac{1}{|x|+1} $$ and the term $$ \frac{1}{|x|+1}$$ does not matter and the term $$\left\lvert\frac{x_i}{|x|^2}\right\rvert=\frac{1}{|x|}$$ Hence we have $$ |\nabla u| \approx -\frac{1}{\ln|x|}\frac{1}{|x|}$$ and the integrand, after change to polar, is $$ -\frac{1}{\ln^n r}\frac{1}{r} $$ and it is integrable since $n>1$.