Proof of differential operator identity in DLMF $16.3.5$

Question

DLMF $16.3.5$ gives the differential operator identity $$ (z\partial_z z)^nf(z)=z^n\partial_z^nz^nf(z),\quad n=1,2,\dots $$ where $\partial_z$ is differentiation w.r.t. $z$. I am having some issues proving this identity. It seems like this is a candidate for induction. Clearly, the identity holds for $n=1$ so assuming for $n$ we have $$ (z\partial_z z)^{n+1}=z\partial_z z^{n+1}\partial_z^nz^n. $$ At this point all I had to go on was trying to somehow use the product rule. I wrote $$ (z\partial_z z)^{n+1}=z\partial_z (z^{n+1}\partial_z^nz^n)=z((n+1)z^n\partial_z^nz^n+z^{n+1}\partial_z^{n+1}z^n), $$ which seems to lead to a dead end. How do I prove this identity? Perhaps a different proof technique altogether?

Answer 1

Define the linear operators $\, D(f) \!:=\! \partial_z f, \, Z(f) \!:=\! z f \,$ with $\, DZ(f) \!=\! f \!+\! Z D(f) .\,$ Since the operators $\, Z D\,$ and $\,DZ\,$ differ by the identity operator, they commute. Moreover, we can prove that $\,Z D\,$ and $\,D^n Z^n\,$ commute for all $\,n\,$ by induction.

Using simple algebra and a few steps we can prove that $$ Z DZ(Z^n D^n Z^n) \!=\! Z^{n+1}((n\!+\!1)D^n Z^n \!+\!Z D^{n+1}Z^n). \tag{1} $$ But now, again using simple algebra, $$ Z^{n+1}D^{n+1}Z^{n+1}\!=\! Z^{n+1}((n\!+\!1)D^n Z^n + D^n Z^{n+1}D). \tag{2}$$ Since $\,Z D\,$ and $\,D^n Z^n\,$ commute, the two quantities on the right side of equations $(1)$ and $(2)$ are equal. Thus, $\,(Z DZ)^n = Z^n D^n Z^n$ for all $\,n\,$ by induction.