0
$\begingroup$

Does anyone know if there is a way to determine the possible solutions to the equation

$f(z-g(z))=g(z)$

For example, one for which the solutions behave as

$f(z), g(z) \sim \frac{1}{z}, \qquad z\rightarrow \infty $

plus subleading terms in $z$.

$\endgroup$
1
$\begingroup$

Here is how you can produce a lot of examples.

Let $h(z)$ be any onto function. Define $$g(z)=z-h(z)$$ and hence $z-g(z)=h(z)$.

Therefore, the functional equation becomes $f \circ h = g$.

Since $h$ is onto, we can find some $h'$ such that $h \circ h'=Id$.

This gives the solution $$g(z)=z-h(z) \\ f=g \circ h'$$

There are very likely many other solutions.

Note In this construction, if we ask for $g$ to be entire, then $g$ is linear. Then $g(z)=az+b$ with $a \neq 1$, $h(z)=(1-a)z-b$ and $h'(z)=\frac{1}{1-a}(z-b)$.

Then $$f(z)=\frac{a}{1-a}(z-b)+b=\frac{a}{1-a}z+\frac{b}{1-a}$$

$\endgroup$
  • $\begingroup$ Since one of your tags is Complex Analysis you may be looking for entire functions. If you just want some functions satisfying your equation you can take $f(z)=1/z$ or $g(z)=1/z$ and find the other function from the given equation. The question may become interesting if you are very clear about the domain of the functions and the properties they are expected to satisfy. There are no entire functions satisfying the given equation. $\endgroup$ – Kabo Murphy Feb 15 '18 at 6:36
  • $\begingroup$ @KaviRamaMurthy There are entire functions satisfying the equation. $\endgroup$ – N. S. Feb 15 '18 at 15:18
  • $\begingroup$ @NS I meant entire functions satisfying the condition that they behave like 1/z at $\infty$. $\endgroup$ – Kabo Murphy Feb 16 '18 at 22:36
  • $\begingroup$ @KaviRamaMurthy Then, the answer is easy: by Liouville Theorem any entire function behaving like $1/z$ at infinity is constant, and the fuctional equation is irrelevant. $\endgroup$ – N. S. Feb 17 '18 at 0:16

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.