Let $f:\mathbb{R}^n\to\mathbb{R}^n$ be an affine transformation, with $f(y)=Ay+\xi, \forall y\in\mathbb{R}^n$ and $\det A \ne 0, \xi\in\mathbb{R}^n$. Let $T\in \mathcal{D}'(\mathbb{R}^n)$ be a distribution, we then define $T^{\ast}=T\circ f$ as $$ \langle T^{\ast},\phi\rangle:=\frac1{|\det A|} \langle T_x,\phi(f^{-1}(x))\rangle, \forall \phi\in \mathcal{D}(\mathbb{R}^n).$$
I want to prove that $T^{\ast}$ is continuous. Suppose that $\phi_j\to \phi$ in $\mathcal{D}$ (i.e. $\exists K\subset\subset \mathbb{R}^n: [\phi]\subset K, [\phi_j]\subset K,\forall j$ and $\forall \alpha\in\mathbb{N}^n: D^{\alpha}\phi_j \to D^{\alpha}\phi$ uniformly on $K$). It suffices to show that $\phi_j\circ f^{-1}\to \phi\circ f^{-1}$ in $\mathcal{D}$.
We have that $f^{-1}(x)=A^{-1}\cdot (x-\xi)$. Since $f$ is essentially defined on the whole space $\mathbb{R}^n$, we can claim that $[\phi_j\circ f^{-1}]\subset K,\forall j$ and $[\phi\circ f^{-1}]\subset K$ (with the same compactum $K$ from the definition of convergence in $\mathcal{D}$). Is this correct?
Also, $D^{\alpha} (\phi\circ f^{-1})(x) = D^{\alpha}\phi(f^{-1}(x))\cdot A^{-1}$. So uniform converges follows as well (clearly for $f^{-1}(x)\not\in K$, since values in this point would be zero; and by definition of $\phi_j\to_{\mathcal{D}}\phi$ for points $f^{-1}(x)\in K$).
Is this general idea correct? My main question regarding this question is whether $K$ has to be transformed as well under the affine transformation. Would such a transformation result in a compact set?