Norm equivalence in Sobolev spaces I'm reading this old, and proof wise very minimalistic, paper, and I have a question about a change of norms. The setting is the following:
$\Omega \subset \mathbb{R}^d, \quad d =1,2,3, \Omega$ is open,bounded.
$f : \Omega \to \mathbb{R}^d$ and $f \in H^1(\Omega)$. 
Also, assume there exist an $x \in \Omega$ such that $f(x) = 0.$
The authors then use the following: 
$\|\nabla \cdot f \|_{L^2(\Omega)} = \|f\|_{H^1(\Omega)}$. 
I cannot fully convince myself about this equality, but I'm guessing it has something to do with $f$ not being constant(or if constant, then zero), and maybe related to the Poincare inequality. 
Help much appreciated. 
 A: The Poincare inequality says
$$\|u\|_{L^p(\Omega)}\leq C\|\nabla u\|_{L^p(\Omega)}$$
provided $u$ vanishes in some strong sense in $\Omega$ to rule out constant functions. The inequality holds when $u=0$ on $\partial \Omega$ (or on a positive measure subset of $\partial \Omega$), when $\int_\Omega u \, dx = 0$ (the average of $u$ is zero), or when $u$ is zero on a positive measure subset of $N\subset \Omega$ (and the constant depends on the measure of $N$).
When we consider a subspace of $H^1(\Omega)$ for which the Poincare inequality holds, the usual norm on $H^1$ is equivalent to the norm $\|\nabla u\|_{L^p(\Omega)}$.
With that being said, vanishing at a single point $x\in \Omega$ is not enough for the Poincare inequality to hold (unless $p>d$, but here $p=2$ and $d=1,2,3$). Remember Sobolev space functions are only defined up to sets of measure zero, so it is not even well-defined to say that $u(x)=0$ for a particular $x \in \Omega$ when $p\leq d$. It is even non-trivial to say $u=0$ on $\partial \Omega$, since $\partial \Omega$ has measure zero (the Trace Theorem is required for this). 
So in your paper something is missing. If they really mean $u(x)=0$ at just one $x\in \Omega$, then this is not well-defined, and the norm $\|\nabla u\|_{L^p}$ is merely a semi-norm and is not equivalent to the full norm on $H^1$.
