Verify that if $n > 1$, the unbounded function $u = \log \log \left(1+\frac 1{|x|}\right)$ belongs to $W^{1,n}(U)$, for $U=B^0(0,1)$.
This is PDE Evans, 2nd edition: Chapter 5, Exercise 14.
In order to show that $u \in W^{1,n}$, where $W^{1,n}$ is a Sobolev space containing the function $u$ itself and its first-order derivative.
Differentiating $u=\log \log\left(1+\frac 1{|x|}\right)$, I obtain (and please verify if you would like) $$u_{x_i}=\frac 1{\log \left(1+\frac 1{|x|}\right)}\frac 1{1+\frac 1{|x|}} \frac{x_i}{|x|^2}.$$
Given $u'=v$ in the weak sense, should I show, according to definition, that $$\int_{B(0,1)} u \phi' \, dx = -\int_{B(0,1)} v\phi dx$$ for all test functions $\phi \in C_c^\infty(U)$? The integration by parts formula comes to my mind here.
Or, should I be employing material from Section 5.6 (Sobolev inequalities) of the textbook? In particular, they discussed on page 280 the borderline case $p=n$, which mentions:
The borderline case p=n. We assume next that $$p=n.$$ Owing to Theorem 2 (page 279) and the fact that $p*=\frac{np}{n-p}\to+\infty$ as $p \to n$, we might expect $u \in L^\infty(U)$, provided $u \in W^{1,n}(U)$. This is however false if $n>1$: for example, if $U=B^0(0,1)$, the function $u=\log \log\left(1+\frac 1{|x|}\right)$ belongs to $W^{1,n}(I)$ but not to $L^\infty(U)$.