Translation-invariant operators as convolutions

Show that for a linear continuous operator $A: \mathcal{D} \rightarrow \mathcal{E}$ the following properties are equivalent:

1. $A$ commutes with shifts: $(A \varphi )(x-a) = A(\varphi (x-a))$ for all $a \in \mathbb {R}$;

2. А commutes with differentiation: $(A \varphi )^{\prime } = A\left(\varphi^{\prime }\right)$;

3. $A \varphi = F * \varphi$ for all $\varphi$, where $F \in \mathcal{D}^{\prime }$ is some generalized function.

Express the differential operator with constant coefficients $L:f\to \sum_{k=0}^n c_k f^{(k)}$ as a convolution.

Let $f\in \mathcal{D}'(\mathbb {R}^{n})$. Prove that the operator $S(f):\mathcal{D}(\mathbb {R}^{n})\to \mathcal{E}(\mathbb {R}^{n})$ is continuous.

Here is another characterization of convolutions analogous to Theorem 6.33. Suppose $L$ is a continuous linear mapping of $\mathscr {D}$ into $C^{\infty }$ which commutes with every $D^{\alpha }$, that is, (a)

Then there is a $u \in \mathscr {D}^{\prime }$ such that

Suggestion: Fix $\phi \in \mathscr {D}$, put

let $D_{e}$ be the directional derivative used in the proof of Theorem 6.30, and show that

which is 0 if (a) holds. Thus $h(x) = h(0)$, which implies that $\tau_{x} L = L \tau_{x}$. Can the assumption that the range of $L$ is in $C^{\infty }$ be weakened?

Suppose $\Lambda : \mathscr {S}_{n} \rightarrow C\left(\mathbb {R}^{n}\right)$ is continuous, linear, and $\tau_{x} \Lambda = \Lambda \tau_{x}$ for every $x \in \mathbb {R}^{n}$. Does it follow that there exists $u \in \mathscr {S}_{n}^{\prime }$ such that

for every $\phi \in \mathscr {S}_{n}$?