We know that

$$ (\mathrm{e}^{b D} f)(x) = f(x + b) $$

$$ (a^{x D} f)(x) = f(x a) $$

Is there a similar expression for the following?

$$ \left(\frac{1}{1-D} f\right)(x) = \left(\sum_{n=0}^\infty D^n f\right) (x) $$

What are the properties of this operator?

Edit: Solving

\begin{align} (1 - D)^{-1} f &= g \\ f &= (1 - D) g \\ &= g - D g \end{align}

for $g$ yields

$$ g(x) = a \mathrm{e}^x - \mathrm{e}^x \int_b^x \mathrm{e}^{-t} f(t) \,\mathrm{d}t $$

as can be shown by taking the derivative of this expression.

  • $\begingroup$ $g = \frac{1}{1-D} f$ then $Dg = g-f$, so we get a first order ODE. Not much else is too interesting about this. $\endgroup$
    – user335907
    May 14, 2017 at 23:13
  • 1
    $\begingroup$ One might suspect that there isn't one on the basis that the operation $(1 - D)^{-1}$ can't be applied to most standard functions as the results don't converge. Take $f(x) = e^x$. The result is then an infinite series of $e^x$ which is just infinity. Take $f(x) = \sin(x)$ and you get an alternating series of sines and cosines which don't converge in the standards sense. I'm sure there are some interesting situations where this shows up but as I said it's probably not as neat. $\endgroup$
    – Squid
    May 14, 2017 at 23:21
  • $\begingroup$ @Squid What about $g(x) = c_1 \mathrm{e}^x - x \mathrm{e}^x$ and $g'(x) = c_1 \mathrm{e}^x - (x \mathrm{e}^x + \mathrm{e}^x) = c_1 \mathrm{e}^x - x \mathrm{e}^x - \mathrm{e}^x = g(x) - f(x)$? $\endgroup$
    – user76284
    May 15, 2017 at 2:51

1 Answer 1


Indeed, your formal (edit) expression $$ (1-D)^{-1} f (x)= e^x \int_x^b dt~ e^{-t} f(t) ~, $$ is standard in the seat-of-the-pants bag of tricks utilized in physics, associated with the name of Schwinger, and it is but the obverse of the very Lagrange translation operator, your first formula. I'm skipping your leading term, as it merely underscores the arbitrariness of the limit b, the kernel (homogeneous solution) of 1–D. That's why I'll be cavalier with the lower limit below--adjust as desired with apologies and a wink to mathematicians.

The standard Schwinger trick here amounts to the formal wisecrack, $$ \frac{1}{1-D} = \int_{-\infty}^{0} dt ~ e^{t(1-D)} $$ with the lower limit to be adjusted in comportance with circumstantial exigencies. Thus, from your Lagrange translation operator, $$ \frac{1}{1-D} ~f(x)= \int_{-\infty}^{0} dt ~ e^{t} ~f(x-t) =\int^{\infty}_{x} dt ~ e^{x-t} ~f(t), $$ the arbitrary upper limit adjusted to an arbitrary constant as discussed.

To test-drive it, apply to $f(x)=\exp(-x/2)$, so that $$ \frac{1}{1-D} ~ e^{-x/2}= e^x \left( a+\frac {2}{3} e^{-3x/2}\right ) = \frac{2}{3}e^{-x/2}, $$ where the constant a was chosen to vanish by the evident requirement of vanishing at infinite x.

  • $\begingroup$ Can this be expressed in terms of the convolution of $f$ and $\exp$? $\endgroup$
    – user76284
    Aug 12, 2017 at 21:24
  • $\begingroup$ I'm not sure what you have in mind with "this"? The action of the resolvent operator on f you wrote yourself is a convolution of f and exp, no? $\endgroup$ Aug 12, 2017 at 22:05

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