Let T be a distribution and define its derivative as
$$D^\alpha T:=\phi\ \rightarrow (-1)^{|\alpha|}T(D^\alpha \phi)$$
Show that the distributional derivative is a distribution as well.
So, I need to show that the map is linear and continuous. For linearity:
$$D^\alpha T(a_1x_1+a_2x_2)=(-1)^{|\alpha|}T(D^\alpha(a_1x_1+a_2x_2))$$ and since the differential operator $D^\alpha$ is linear and T is a distribution and hence linear, this is equal to $a_1(-1)^{|\alpha|}T(D^\alpha x_1)+a_2(-1)^{|\alpha|}T(D^\alpha x_2)$ which means that $D^aT$ is linear.
Is this proof correct? I struggle to show that the map is continuous. Any hints on this?
Take $\phi_k \in \mathcal{D}.$ Since $\phi_k \to \phi$ in $\mathcal{D}$ implies that $\partial^\alpha\phi_k \to \partial^\alpha\phi$ in $\mathcal{D}$ (show this claim!) it's easy: $$ \langle \partial^\alpha T, \phi_k \rangle = (-1)^{|\alpha|} \langle T, \partial^\alpha \phi_k \rangle \to (-1)^{|\alpha|} \langle T, \partial^\alpha \phi \rangle = \langle \partial^\alpha T, \phi \rangle. $$