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 :

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 ?

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