Question 12.7: Let ρ be an infinitely differentiable function on R with sim......

Clearly, it is enough to analyze the case when ρ has only one simple zero at some point a. In that case, we have to prove that equation (12.18) is equivalent with equation

(x-a) \cdot T=0,          (12.20)

and its solution is the distribution T=A \delta_a, for some constant A.
Let us introduce the function \rho_1 by

\rho(x)=(x-a) \rho_1(x) \text {. }        (12.21 )

Then it holds \rho_1(a) \neq 0 \text { and the mapping } \varphi \mapsto \psi=\rho_1 \varphi \text { is a bijection from } \mathcal {D} (\mathbf{R} ) onto itself. Putting (12.21) into (12.18) we obtain

which implies the equivalence of the equations (12.18) and (12.20).
Let us find now the solution of (12.20). To that end, note that the mapping \psi \mapsto(x-a) \psi from the set

\mathcal {A} =\{\psi \in \mathcal {D} ( \mathbf{R} ) \mid a \notin \operatorname{supp} \psi\}

into itself is, in fact, a bijection. Thus for every test function \varphi with the property a \notin \operatorname{supp} \varphi it holds

\langle T, \varphi\rangle=0

which is equivalent with the statement supp T = {a}. Any distribution T whose support is a single point a is necessarily of the form

T=A \delta_a+\overset{p}{\underset{k=1}{\sum}} A_k \delta_a^{(k)}          (12.22)

for some constants A and A_k, k=1,2, \ldots, p \text {. Choose now } k \in\{1,2, \ldots, p\}. Then for \varphi \in \mathcal {D} (\mathbf{R} ) it holds:

Since there exists a \varphi \in \mathcal {D} (\mathbf{R} ) such that \varphi^{(k-1)}(a) \neq 0, it follows that. the distribution \delta_a^{(k)} is not a solution of (12.20), hence also not of (12.18). Thus T from (12.22) is a solution of (12.18) \text { (for } m=1 \text { and } a_1=a \text { ) if and only if } A_1=A_2=\ldots=A_p=0 \text {, } which finally gives us the solution

T=A \delta_a \text { for some constant } A \text {. }