# Support of distribution $T$

Let $T$ a distribution in $\mathbb{R}^n$ and let $f \in C^{\infty}(\mathbb{R}^n)$ with values in $\mathbb{R}$. We put $Z(f)= \{x \in \mathbb{R}^n, f(x)=0\}$.

I have two questions please:

1. Proof that if $fT=0$ then $Supp \ T \subset Z(f)$.

2. Proof that if order of $T$ is 1 and if $Supp \ T \subset Z(f)$ then $fT=0$.

Can you help me please i'm lost, what's the relation of order of distribution with the support? And elements of Supp T are $x \in \mathbb{R}^n$?

Thank you in advance for the help

• $f$ is a vector in $\mathbb{R}^n$? Or a function? Nov 24, 2017 at 18:51
• Sorry i edit my question. $f$ is an function Nov 24, 2017 at 19:09
• Ok and I'm a little confused by the notation $fT$. $T$ is a distribution so I know what $T(f)$ means... is that you meant? Nov 24, 2017 at 19:20
• yes, $fT$ is defined by: $\forall \varphi \in C^\infty_c(\mathbb{R}^n): <fT,\varphi>= <T,f\varphi>$ Nov 24, 2017 at 19:28
• Ok thanks I understand now. Nov 24, 2017 at 19:31

1)First one is easy: take a test function $\phi$ with $\mathrm{supp}\,\phi \subset \mathbb{R}^n \setminus Z(f)$. Since $\mathrm{supp}\,\phi$ is a compact set and $f \ne 0$ on it there exists a smooth inverse $f^{-1}$ and the function $f^{-1}\phi$ is a well-defined test function. Now, $$<T, \phi> = <T, ff^{-1}\phi> = <fT, f^{-1}\phi> = 0.$$ Thus, by definition, $\mathrm{supp}\, T \subset Z(f)$.
2)I am not completely sure in my answer to the second question, unfortunately, but here are some thoughts. To prove the distribution is zero we take a test function $\phi$ and prove that $<fT, \phi> = <T, f\phi> = 0$. Now, we would have liked to say that $\mathrm{supp}\, f\phi \subset \mathbb{R}^n \setminus \mathrm{supp}\, T$ (in which case the desired equality follows) but it's false: $\mathrm{supp}\, f\phi$ lies only in the closure of $\mathbb{R}^n \setminus \mathrm{supp}\, T$. Thus, we need to write our function $f\phi$ as a sum $\psi_1+\psi_2$ with $\mathrm{supp}\, \psi_1 \subset \mathbb{R}^n \setminus \mathrm{supp}\, T$ (so $<T, \psi_1>=0$) and $<T, \psi_2>$ small. Since $T$ is of order one $<T, \psi_2>$ is small when $||\psi_2||_{C^1}$ is small. We can probably achieve that by choosing an appropriate partition of unity.