Determining the action of the operator $D\left(z, \frac d{dz}\right)$

Question

This question was motivated by a question by Tobias Kienzler and its wonderful answers.

I begin as in the linked question...

Using the Taylor expansion

$$f(z+a) = \sum_{k=0}^\infty \frac{a^k}{k!}\frac{d^k }{dz^k}f(z)$$

one can formally express the sum as the linear operator $e^{a\frac{d}{dz}}$ to obtain

$$f(z+a) = e^{a\frac{d}{dz}}f(z).$$

Other relationships were given in an answer by Tom Copeland:

$$ f(e^b z) = \exp\left(bz\frac d{dz}\right)f(z), $$

$$ f\left(\frac z{1-cz}\right) = \exp\left(c z^{2}\frac d{dz}\right)f(z). $$

My question is about the reverse. What if we start on the right-hand side with a function different from $\exp$ in the operator?

I asked about the specific case for $\sin$ in the comments of joriki's answer and Tobias found that

$$ \sin\!\left(a\frac{d}{dz}\right)f(z) = \frac{1}{2i}(f(z+ia) - f(z-ia)), $$

and similarly that

$$ \cosh\!\left(a\frac{d}{dz}\right)f(z) = \frac{1}{2}(f(z+a) - f(z-a)). $$

He also conjectured that the symmetrization of a function might be obtained from an operator like

$$\exp\left(i\frac\pi2\frac d{d\ln z}\right)\cosh\left(i\frac\pi2\frac d{d\ln z}\right)$$

I don't know much about Lie algebras so I apologize if this is too broad:

For which operators $\text{D}$ like these is $\text{D}f$ something 'nice' as in these examples?

Answer 1

I obtained those identities by using $\sin(x) = \frac{e^{ix}-e^{-ix}}{2i}$ etc. (without worrying about convergence of operators, I admit).

In general, you can use the Fourier-Transform of your operator as follows:

$$\begin{array}{rl} D\left(z,\frac d{dz}\right)f(z) &= D\left(z,\frac d{dz}\right)\frac1{2\pi}\int_{-\infty}^\infty dk \int_{-\infty}^\infty dw\ f(w)e^{ik(z-w)} \\ &= \int_{-\infty}^\infty dw \underbrace{\frac1{2\pi}\int_{-\infty}^\infty dk\ D\left(z,\frac d{dz}\right)e^{ik(z-w)}}_{=:W(z,w)}\ f(w) \end{array}$$

edit[ Note that you can replace $\frac d{dz}$ by $ik$ if it does not act on the $z$ in $D$ anymore, however for $\exp\left(bz\frac d{dz}\right)$ you can't! ]

So (by, once again, ignoring detailed discussions on convergence, whether swapping the integration is valid etc. (sorry, I'm a Physicist...)) you obtain a $W$ that for each $z$ gives you a weight distribution.

For demonstration, take $D=e^{a\frac d{dz}}$ to obtain $W(z,w)=\delta(w-(z+a))$.

On the other hand, if you want to know $\int_{-\infty}^\infty f(z)\,dz$ you can require $W=1$ and therefore $D = 2\pi\delta\left(\frac d{dz}\right)$. More generally, for a given weight $W(z,w)$ one obtains a differential operator $D\left(z,\frac d{dz}\right)$ via the inverse Fourier transform, so in summary:

$$ W(z,w) = \frac1{2\pi}\int_{-\infty}^\infty D\left(z,\frac d{dz}\right)e^{ik(z-w)}\,dk, \\D(z,ik) = \frac1{2\pi}\int_{-\infty}^\infty e^{-ikw}W(z,w+z)\,dw$$

The latter formula gives you $D$ such that $\frac d{dz}$ only acts to the right of it, so if you apply this to $\exp\left(bz\frac d{dz}\right)$ you will obtain a different expression (that can be reformulated).

Using $W(z,w) = \chi_{[-\infty,z]}(w)$ one can therefore also express the antiderivative via an operator $\frac1{2\pi}\int_{-\infty}^ze^{-w\frac d{dz}}\,dw$.

Answer 2

Another angle on characterizing the action of D(z,d/dz):

With f(z) expressible as a Taylor series, you could also look at the umbral operator

$U(c.y;d/dz) f(z) = exp(c.yd/dz) f(z) = f(z+c.y)$, formally, with the special cases,

A) $exp(c.d/dz)|_{z=0} f(z) = f(c.)$ ,

B) $exp[-(1-c.) d/dz]|_{z=1} f(z) = f(c.)$,

C) $exp(c.:zd/dz:) f(z) = f[(1+c.)z]$, and

D) $exp[-(1-c.) :zd/dz:] f(z) = f(c. z)$

where $c.^n=c_n$, $(:zd/dz:)^n=z^n(d/dz)^n$, and, e.g., $(z+c.y)^n=\sum_{j=0}^n \binom{n}{j} c_j y^j z^{n-j}$.

For the sine operator above, in U let $y=a$ and $c_n=sin(\pi n/2)$.

For the cosh operator above, in U let $y=a$ and $c_n=|cos(\pi n/2)|$.

For the scaling operator, in D let $c.= e^b$ and see my notes in MSQ 116633 on $S_0$.

Answer 3

Along a different vein, complementing Tobias' formulation:

For $z>0$ and appropriate $\sigma$, formally

$$K \left(z\frac{d}{dz} \right)f(z)=\displaystyle\frac{1}{2\pi i} \int_{\sigma-i\infty}^{\sigma+ i\infty} \frac{\pi}{\sin(\pi s)} K(-s)g(-s) \frac{z^{-s}}{(-s)!} ds$$

with

$$\displaystyle\int^{\infty}_{0}{f(z) \frac{z^{s-1}}{(s-1)!} dz} = g(-s)$$

using a modified Mellin transform and its inverse.

For action on $f(z)=e^{-z}$ with $K(\omega)=\binom{\omega+\alpha+\beta}{\beta}$, see my notes "The Inverse Mellin Transform, Bell Polynomials, a Generalized Dobinski Relation, and the Confluent Hypergeometric Functions".

Also formally,

$$K \left(z\frac{d}{dz} \right)f_{LPT}(z)=\displaystyle\frac{1}{2\pi i} \int_{\sigma-i\infty}^{\sigma+ i\infty} \frac{\pi}{\sin(\pi s)} K(-s)f_{MT}(1-s) \frac{z^{-s}}{(-s)!} ds$$

for $f_{LPT}(z)$ the Laplace transform of $f(x)$ and $f_{MT}(s)$ the unmodified Mellin transform of $f(x)$; i.e.,

$$\displaystyle\int^{\infty}_{0}{f(x) e^{-xz} dx} = f_{LPT}(z)$$ and $$\displaystyle\int^{\infty}_{0}{f(x) x^{s-1} dx} = f_{MT}(s)$$

This is derivable formally from

$$e^{-xz}=\displaystyle\frac{1}{2\pi i} \int_{\sigma-i\infty}^{\sigma+ i\infty} \frac{\pi}{\sin(\pi s)} \frac{(xz)^{-s}}{(-s)!} ds \text{ for }\sigma>0$$

Answer 4

Consider a compositional inverse pair of functions, $h$ and $h^{-1}$, analytic at the origin with $h(0)=0=h^{-1}(0)$.

Then with $\omega=h(z)$ and $g(z)=1/[dh(z)/dz]$,

$$\exp \left[ {t \cdot g(z)\frac{d}{{dz}}} \right]f(z) = \exp \left[ {t\frac{d}{{d\omega }}} \right]f[{h^{ - 1}}(\omega )] = f[{h^{ - 1}}[t + h(z)]] = f[L(t,z)]$$

(see OEIS A145271 and A139605),

so with $D_{FT}(\alpha)$ the Fourier transform of $D(z)$, formally

$$D\left( {t \cdot g(z)\frac{d}{{dz}}} \right)f(z) = \int\limits_{ - \infty }^\infty {{D_{FT}}} (\alpha )\exp \left[ {2\pi i\alpha t\cdot g(z)\frac{d}{{dz}}} \right]d\alpha f(z)$$

$$ = \int\limits_{ - \infty }^\infty {{D_{FT}}} (\alpha )f\left\{ {{h^{ - 1}}\left[ {2\pi i\alpha t + h(z)} \right]} \right\}d\alpha = \int\limits_{ - \infty }^\infty {{D_{FT}}} (\alpha )f\left[ {L\left( {2\pi i\alpha t,z} \right)} \right]d\alpha $$

For the special case $D(z)=\sin(2\pi a z)$, $D_{FT}=\dfrac{\delta(\alpha-a)- \delta(\alpha+a)}{2i}$,

and so

$$sin\left( {2\pi a \cdot g(z)\frac{d}{{dz}}} \right)f(z) = \frac{{f\left\{ {{h^{ - 1}}\left[ {h(z) + 2\pi ia} \right]} \right\} - f\left\{ {{h^{ - 1}}\left[ {h(z) - 2\pi ia} \right]} \right\}}}{{2i}}$$

(For a consistency check, try $h(z)=z$.)

Similarly, switch to the inverse Laplace transform to obtain formally

$$D\left( {t \cdot g(z)\frac{d}{{dz}}} \right)f(z) = \frac{1}{{2\pi i}}\int\limits_{\sigma - i\infty }^{\sigma + i\infty } {{D_{LPT}}} (p)\exp \left[ {pt \cdot g(z)\frac{d}{{dz}}} \right]dpf(z)$$

$$ = \frac{1}{{2\pi i}}\int\limits_{\sigma - i\infty }^{\sigma + i\infty } {{D_{LPT}}} \left( p \right)f\left\{ {{h^{ - 1}}\left[ {pt + h(z)} \right]} \right\}dp = \frac{1}{{2\pi i}}\int\limits_{\sigma - i\infty }^{\sigma + i\infty } {{D_{LPT}}} \left( p \right)f\left[ {L\left( {pt,z} \right)} \right]dp$$

For the special case $D(z)=\cosh(az)$, ${{\text{D}}_{LPT}}{\text{ = }}\frac{1}{2}\left[ {\frac{1}{{p - a}}{\text{ + }}\frac{1}{{p + a}}} \right]$,

and purely formally

$${\text{cosh}}\left[ {ag(z)\frac{d}{{dz}}} \right]f(z) = \frac{1}{{2\pi i}}\int\limits_{\sigma - i\infty }^{\sigma + i\infty } {\frac{1}{2}} \left[ {\frac{1}{{p - a}} + \frac{1}{{p + a}}} \right]f\left\{ {{h^{ - 1}}\left[ {p + h(z)} \right]} \right\}dp$$

$=\frac{1}{2}[f[h^{-1}[a+h(z)]]+ f[h^{-1}[-a+h(z)]]$.

Examples can be constructed from

$g(z)=(1+z)^{m+1}$, $h^{-1}(z)=(1-mz)^{-1/m}-1$, $h(z) = - \dfrac{{{{(1 + z)}^{ - m}} - 1}}{m}$, and

$L(t,z)=h^{-1}[t+h(z)]=[(1+z)^{-m}-mt]^{-1/m}-1$

with the limiting case for $m=0$ being

$g(z)=(1+z)$, $h^{-1}(z)= \exp(z)-1$, $h(z)= \log(1+z) $, and

$L(t,z)=h^{-1}[t+h(z)]=(1+z)e^{t}-1$.

Note for the Witt algebra that the actions are given by

$exp[tz^{m+1}d/dz]f(z)=f[z(1-mtz^{m})^{-1/m}]$.