Prove that $supp\, T=\{(x,-x); \, x \in \mathbb R\}$?

Question

Let denote by $T$ the distribution of order $0$ on $\mathbb R$ given by $$ T(\phi) = \int_{\mathbb R} \phi(x,-x) dx.$$ How can prove that the support of $T$ is $supp\, T=D:=\{(x,-x); \, x \in \mathbb R\}$ ?

Thank you in advance

Answer 1

I presume you mean that $T$ is a distribution on $\Bbb R^2$.

Any $\phi$ with support not meeting $D$ is annihilated by $T$, so supp$\,T$ is a closed subset of $D$. Let $(x,-x)\in D$ and $U$ an open neighbourhood of $x$. Taking $\phi$ to be a bump function centred at $(x,-x)$ with arbitrarily support in $U$ shows that $(x,-x)$ is not in the support of $T$.