why is a measure distribution of order 0? Question is from distribution theory and fourier transform. Let $\Omega$ be an open domain in $\mathbb{R}^{n}$. A distribution $u$ in $\Omega$ is a linear functional on $C_{0}^{\infty}$ such that for every compact set $K \subset \Omega$ there are constants C and k such that $u(\phi) \leq C \sum_{|\alpha| < k} ||\partial^{\alpha} \phi ||_{\infty}$ for all $ \phi \in C_{0}^{\infty}$ with supp $\phi \subset K$. if same k can be used for all $K$ we say u has order less than equal to k.
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1Given a measure $\mu$ and test function $\phi$ with support contained in $K$, there is a very obvious bound on $\langle\mu,\phi\rangle:=\int_{\Omega}\phi,d\mu=\int_K\phi,d\mu$. – peek-a-boo Feb 18 '24 at 11:39