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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?

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nice question! Suggestion for a title: "How to determine the action of an operator $D\left(z, \frac d{dz}\right)$?" – Tobias Kienzler Apr 1 '12 at 19:38
Thanks @TobiasKienzler, you knew what I was looking for better than I did! – Antonio Vargas Apr 1 '12 at 19:44
@Tobias This is so interesting! – Pedro Tamaroff Apr 4 '12 at 13:32
up vote 4 down vote accepted

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 $\frac1{2\pi}\int_{-\infty}^ze^{-w\frac d{dz}}\,dw$.

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This is really cool. – Antonio Vargas Apr 1 '12 at 19:39
Thanks! I hope you're happy about making me use my brains on a Sunday :-) – Tobias Kienzler Apr 1 '12 at 19:43
I added the inversion, hope it's correct despite it being midnight... – Tobias Kienzler Apr 1 '12 at 21:30

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 $S_0$.

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Along a different vein, complementing Tobias' formulation:

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

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


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

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

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

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@Peter, thanks for improving the LaTex code. I'll try to make good use of it in the future. – Tom Copeland Apr 10 '12 at 22:27
You're welcome. Great triad of answers by the way. I hope someday I'll understand them fully! – Pedro Tamaroff Apr 10 '12 at 22:32

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


with the limiting case for $m=0$ being

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


Note for the Witt algebra that the actions are given by


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I think it would be best to condense your three answers into one. – Antonio Vargas Apr 6 '12 at 23:54
@Antonio, why? Three different approaches, three different paragraphs (four including Tobias') makes it easier to comment on the different approaches and express interest or disinterest in each, and also suggests how broad the question and answers are. – Tom Copeland Apr 8 '12 at 1:04
@Peter, thanks for the better formatting. (However, you also introduced a substantial math error that I had to undo.) – Tom Copeland Apr 10 '12 at 22:20

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