# distribution theory $\ \ \text{order}(T) = \max(\text{order}(T')-1,0)\$?

I would like to know in what case is it possible that $\text{order}(T) = \text{order}(T') \ge 1$. So I came to write this :

• $\|\varphi\| = \sup_{x \in \mathbb{R}} |\varphi(x)|$

• Let $T \in D'(\mathbb{R})$ be a distribution compactly supported on $\Omega$ and $\text{order}(T') = k \ge 1$, that is $$\forall \varphi \in C^\infty_c(\mathbb{R}),\qquad |\langle T', \varphi \rangle| \le C\sum_{m \le k} \|\partial_m \varphi\|$$

Then pick $\phi \in C^\infty_c([a,b])$ with $[a,b]$ slightly larger than $\Omega$ and $\phi(x) = 1$ for $x \in \Omega$, and let $\displaystyle I[\varphi](x) = \phi(x) \int_a^x \varphi(t)dt$.

Since $\partial_m I[\varphi] = \sum_{l=0}^m {m \choose l} (\partial_{m-l}\phi)(\partial_l \int \varphi) = (\partial_m \phi)(\int \varphi)+\sum_{l=1}^m {m \choose l} (\partial_{m-l}\phi)(\partial_{l-1} \varphi)$ we have

$\begin{eqnarray} \|\partial_m I[\varphi]\| &\le& \|(\partial_m \phi)({\textstyle\int} \varphi)\|+\sum_{l=1}^m {m \choose l} \|(\partial_{m-l}\phi)(\partial_{l-1} \varphi)\| \\ &\le& \|\partial_m \phi\|\, |b-a|\,\| \varphi\|+\sum_{l=1}^m {m \choose l} \|\partial_{m-l}\phi\|\|\partial_{l-1} \varphi\| \\ &\le & C_2\sum_{l\, \le\, \max(0,m-1)} \|\partial_{l-1} \varphi\|\end{eqnarray}$

And hence $$|\langle T, \varphi \rangle| = |\langle T', I[\varphi] \rangle| \le C\sum_{m \le k} \|\partial_m I[\varphi]\| \le C_3\sum_{l\, \le\, \color{red}{k-1}} \|\partial_m \varphi \|$$

whence $\text{order}(T) \le k-1$

Is it correct ?

• Your test function $I[\varphi]$ is not compactly supported in $\Omega$, so the inequality $|\langle T', I[\varphi] \rangle| \le C\sum_{m \le k} \|\partial_m I[ \varphi]\|_{\infty, \Omega}$ looks cheesy. – TZakrevskiy Oct 11 '16 at 21:39
• My remark was about that subscript $\Omega$ (in $|\langle T', I[\varphi] \rangle| \le C\sum_{m \le k} \|\partial_m I[ \varphi]\|_{\infty, \Omega}$). As you have just shown, it should have been $[a,b]$, with $[a,b]$ containing the support of $\phi$. – TZakrevskiy Oct 12 '16 at 8:02
• @TZakrevskiy So you agree that at least when $T$ is compactly supported, then $order(T) = order(T')-1$ ? And when $T$ isn't, we can reduce it to a compactly supported distribution for some given set of test functions, so it should stay true ? – reuns Oct 12 '16 at 8:09
• I don't really see where you use the fact that $T$ is compactly supported. – TZakrevskiy Oct 12 '16 at 8:10
• @TZakrevskiy at $-\langle T',\phi \int \varphi \rangle = \langle T,\phi' \int \varphi + \phi \varphi \rangle = \langle T, \varphi \rangle$ if $\phi = 1$ on $support(T)$ – reuns Oct 12 '16 at 8:13