let $m\in \mathbb N$ and define $T : C_c^\infty (\mathbb R)\rightarrow \mathbb R$ by $T(\phi)=\phi^{(m)}(x_0)$.Then show that $T$ is a distribution with order "m".
i had proved that $T$ is a distribution as $|T(\phi)|\le Sup|\phi^{(m)}|$ where $\phi \in C_C^\infty (K)$ and $K$ is compact in $ \mathbb R $. From here it is clear that order of the distribution is $\le$ m but i am not able to get why it is equal to m.
Any type of help will be appreciated. Thanks in advance.
If you meant $\phi^{(m)}$ then the order is $m$ almost by definition :
Find $\phi \in C^\infty_c$ such that $T(\phi) = \phi^{(m)}(x_0)= 1$ and let $\phi_\epsilon = \epsilon^m \phi(x_0+(x-x_0)/\epsilon)$.
Then $\sum_{n =0}^{m-1} \|\phi_\epsilon^{(n)}\|_\infty = \mathcal{O}(\epsilon)$ while $T(\phi_\epsilon) = 1$ thus $T$ can't be a distribution of order $< m$.