find the non constant polynomial so that P(x) δ_1' = δ_1'

Question

This is the last one for today:

I am trying to find a non-constant polynomial $P(x)$ so that the following equation is true:

$$P(x) δ_1' = δ_1'$$

where $δ_1$ is the Dirac distribution supported in the point $x = 1$, and $δ_1′$ is the derivative of $δ_1$.

I started like this:

for a test function $φ$ we can write:

$$P(x)δ_1'(φ)=δ_1'(Pφ) $$

since $P(x)$ is a polynomial and then a function of class C^infinite, Right? if so then:

$$δ_1'(Pφ)=-δ_1(P'φ+Pφ')=-P'(1)φ(1)-P(1)φ(1)'$$

And now I don't know what to do.

Answer 1

As you write, for each $\phi \in C^\infty_c(\mathbb R)$, we must have \begin{align*} (P\delta_1')(\phi) &= \delta_1'(P\phi)\\ &= -\delta_1\bigl((P\phi)'\bigr)\\ &= -(P\phi)'(1)\\ &= -P'(1)\phi(1) - P(1)\phi'(1)\\ &= -P'(1)\delta_1(\phi) + P(1)\delta_1'(\phi) \end{align*} This equals $\delta_1'(\phi)$ for all $\phi$ iff $-P'(1) = 0$ and $P(1) = 1$. Can you find a $P$ that satisfies this two conditions?