Let $T\in \mathcal D'(X)$ be a distribution of order $m$. ($X$ is an open set in $\mathbb R^n$.) Let $\alpha$ be a multi-index. My question is: what is the order of $\partial^\alpha T$?
It is really obvious that the order of $\partial^\alpha T$ $\leq m+|\alpha|$, by direct calculation from the definition $$ |\langle T,\phi\rangle|\leq C_K\sum_{|\alpha|\leq m} \sup_{x\in K}|\partial^\alpha \phi (x) |, $$ but can I prove that the equality holds? After some attempt, I start to think that the inequality "the order of $\partial^\alpha T$ $\leq m+|\alpha|$" might be strict in some cases.
The inequality can be strict. The easiest example is a distribution that "doesn't depend" on one variable and has positive order on the other.
Take for example $$ \langle T,\phi \rangle = \int\limits_{-\infty}^{\infty} \frac{\partial{\phi}}{\partial x}(0,y)dy. $$ Then, obviously, $$ \langle \frac{\partial T}{\partial y} ,\phi \rangle = -\int\limits_{-\infty}^{\infty} \frac{\partial^2 {\phi}}{\partial x \partial y}(0,y)dy = 0. $$ It's easy to see that order of $T$ is $1$ and the derivative of $T$ is $0$. You can add function $f(x,y) = y$ and then $\frac{\partial (T + f)}{\partial y}$ has exactly order $0$ (it is equal to regular function $1$) and $T + f$ still has order $1$.
Edit: $T = \alpha \otimes \beta$ is tensor product of $\alpha(x) = -\delta'(x)$ and constant function $\beta(y) = 1$.