0
$\begingroup$

I have an equation of the form:

$f(s)=g(s)\exp(ih(s))f(b-s), \qquad (b \in \mathbb{R}, s \in \mathbb{C})$

where $f:\mathbb{C} \to \mathbb{C}$, $g(s)>0$, $h(s) \in (-\pi,\pi]$.

We can prove that if $h(s) \neq 0$ then $f(s)=0$, $f(b-s)=0$ (I know that if $f(s)=0$ then $f(b-s)=0$). However the case $h(s)=0$ does not imply this case directely. But we can see also that we can have $f(s)=0$, $f(b-s)=0$ if $h(s)=0$.

(a) My question is how to deal with the case when $h(s)=0$.

(b) How I can solve this functional equation: $g(1-s)g(s)=1$ with respect to $g$ for all $s$.

  • 0
    Yes, Done. The equation hlods for all $s$ and some $b$ including $b=1$. I am searching for the roots of $f(s)=0$ when $b=1$.2012-12-30

1 Answers 1

1

Fix $b \in \mathbb{R}$ and suppose that $f(s) = g(s)\exp(ih(s))f(b-s)$ for all $s \in \mathbb{C}$. As $g(s) > 0$ and $\exp(ih(s)) \neq 0$, $f(s) = 0$ if and only if $f(b - s) = 0$. Depending on $f$, there may not be any zeroes, but if there is a zero $s_0$, then $b - s_0$ is another zero.

Now suppose $b = 1$ as in your comment, then we have

$f(s) = g(s)\exp(ih(s))f(1-s)$

and

$f(1-s) = g(1-s)\exp(ih(1-s))f(s)$

so

\begin{align*} f(s) &= g(s)\exp(ih(s))f(1-s)\\ &= g(s)\exp(ih(s))[g(1-s)\exp(ih(1-s))f(s)]\\ &= g(s)g(1-s)\exp(ih(s) + ih(1-s))f(s). \end{align*}

Provided $f$ has isolated zeroes (as in the case of a non-zero holomorphic function), we see that

$1 = g(s)g(1-s)\exp(ih(s) + ih(1-s))$

and therefore

$g(s)g(1 - s) = \frac{1}{\exp(ih(s) + ih(1-s))} = \exp(-ih(s)-ih(1-s)).$

For an arbitrary $b$, the same manipulations give

$g(s)g(b - s) = \exp(-ih(s)-ih(b-s)).$