# Proving $supp (D^\alpha \phi) \subseteq supp (\phi)$ for $\phi \in C^\infty$

I am currently learning about distributions and test functions. Test functions are $C^\infty (\Omega)$ ($\Omega \subseteq \mathbb{R}^n$ open) functions which have compact support, say the set of all test functions is $C^\infty_c$. Now I am interested in the derivatives of such test functions which shall be test functions as well.

So let $\phi \in C^\infty_c$ and $\alpha$ a multiindex. Now $D^\alpha \phi$ clearly is a $C^\infty$ function. Furthermore as $supp (D^\alpha \phi) \subseteq supp (\phi)$ and the fact that supports are by definition closed we get that $D^\alpha \phi$ has compact support because in Euclidean spaces we have that closed and bounded implies compact.

I hope that so far my thoughts are correct. What is left to show is \begin{align*} supp (D^\alpha \phi) \subseteq supp (\phi). \end{align*} Unfortunately I do not know how to show this and I could not find any proofs. I guess that it is not very difficult to show that but I can't get my head around it. Could you give me some hints on how to proof that or maybe even provide me with a short sketch of the proof?

Kind regards!

• A function is identically $0$ outside of its support, and the derivatives of the $0$ function are $0$, thus the support of the derivatives must be contained in the support of the original function. – SquirtleSquad Jun 3 '17 at 8:49

## 1 Answer

Note that $D:= R^n \setminus supp(\phi)$ is open and $\phi(D)=0$ Therefore

$$D^{\alpha}\phi(D) =0$$