Show a BMO function not in $W^{1,n}(Q, \mathbb{R})$

Question

As the title suggest, I am searching for a function that is BMO but not in $W^{1,n}(Q, \mathbb{R})$ with $Q \subset \mathbb{R}^n$ is a certain “cube”. I know the latter continuously embed in the former, but they should not be equal. It’s just that I’ve never been good at finding counterexamples, and I can’t find anything online

Answer 1

It is easy to consider counterexamples by remarking that $L^\infty\subset BMO$. Therefore, functions that are the characteristic functions of a set $A$, $f(x) = \mathbb{1}_{A}(x)$ work (the derivative is a measure, it is not even in $L^1_{\mathrm{loc}}$). The general reason is that to get functions in $W^{1,n}$ you need to have informations on the derivative, which is not the case for $BMO$.

However there are functions in $BMO$ that are not in $L^\infty$ and not in $W^{1,n}$. In dimension $1$, you can take $f(x) = \mathbb{1}_{[-1,1]}(x) \ln(|x|)$, that is not in $W^{1,n}$ since $|f'(x)| = \frac{1}{|x|}$ (out of $x=0$), not in $L^\infty$, but is in $BMO$ (see for example [Duoandikoetxea, Fourier Analysis, Chapter 6, Section 5.7]).