Question 12.5: Find the distributional products a) x^p · δ^(q) (p, q ∈ N); ......

Find the distributional products

a) x^p \cdot \delta^{(q)} \quad(p, q \in \mathbf{N} )       b) e^{a x} \cdot \delta^{(q)} \quad(a \in \mathbf{R} , q \in \mathbf{N} )

where \delta^{(q)} is the q-th distributional derivative of δ for q ∈ N.

Step-by-Step
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a) For \varphi \in \mathcal {D} ( \mathbf{R} ) it holds

\left\langle x^p \cdot \delta^{(q)}, \varphi\right\rangle=\left\langle\delta^{(q)}, x^p \varphi(x)\right\rangle=(-1)^q\left\langle\delta,\left(x^p \varphi(x)\right)^{(q)}\right\rangle .

Assume first p > q. Then it holds

\begin{aligned}\left(x^p \varphi(x)\right)^{(q)} & =\sum_{j=0}^q\left(\begin{array}{l}q \\j\end{array}\right)\left(x^p\right)^{(j)} \varphi^{(q-j)}(x) \\& =\sum_{j=0}^q\left(\begin{array}{l}q \\j\end{array}\right) p(p-1) \cdots(p-j+1) x^{p-j} \varphi^{(q-j)}(x) .\end{aligned}          (12.17)

So we have

\begin{aligned}\left\langle x^p \cdot \delta^{(q)}, \varphi\right\rangle & =(-1)^q \sum_{j=0}^q\left\langle\delta,\left(\begin{array}{l}q \\j\end{array}\right) p(p-1) \cdots(p-j+1) x^{p-j} \varphi^{(q-j)}(x)\right\rangle \\& =\left.(-1)^q \sum_{j=0}^q\left(\begin{array}{l}q \\j\end{array}\right) p(p-1) \cdots(p-j+1)\left(x^{p-j} \varphi^{(q-j)}(x)\right)\right|_{x=0} \\& =0=\langle 0, \varphi\rangle .\end{aligned}

Assume next p ≤ q. Then for all x ∈ R it holds for

\left(x^p\right)^{(j)}= \begin{cases}0 & \text { for } j>p \\ p ! & \text { for } j=p\end{cases}

and therefore we have from (12.17)

\begin{aligned}\left\langle x^p \cdot \delta^{(q)}, \varphi\right\rangle= & (-1)^q\left\langle\delta,\left(x^p \varphi(x)\right)^{(q)}\right\rangle \\= & (-1)^q\left\langle\delta, \sum_{j=0}^{p-1}\left(\begin{array}{l}q \\j\end{array}\right) p(p-1) \cdots(p-j+1) x^{p-j} \varphi^{(q-j)}(x)\right\rangle \\& +(-1)^q\left\langle\delta,\left(\begin{array}{l}q \\p\end{array}\right) p ! \varphi^{(q-p)}(x)\right\rangle \\& +\left(-1^q\right)\left\langle\delta, \sum_{j=p+1}^q\left(\begin{array}{l}q \\j\end{array}\right)\left(x^p\right)^{(j)} \varphi^{(q-j)}(x)\right\rangle\end{aligned}
\begin{aligned}& =0+(-1)^q q(q-1) \cdots(q-p+1)\left\langle\delta, \varphi^{(q-p)}\right\rangle+0 \\& =q(q-1) \cdots(q-p+1)(-1)^p\left\langle\delta^{(q-p)}, \varphi\right\rangle .\end{aligned}

b) Similarly as in a), we get

e^{a x} \cdot \delta^{(q)}=\overset{q}{\underset{j=0}{\sum}}(-a)^j \delta^{(q-j)}

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