Non-exponential functions $f(x)$ satisfying $f(x+c)=\gamma(c)f(x)$

Question

Question:

what can be said about the existence of functions
\begin{align} f:x\mapsto f(x)&\implies x+c\mapsto \gamma(c)f(x)\\ f(x)\ne f(y)&\iff\frac{f(x)}{\gamma(x)}\ne\frac{f(y)}{\gamma(y)} \end{align}

These functions would generalize the functional equation $e^{x+y}=e^xe^y$

I am especially interested in the case $x,f(x)\,\in\,\mathbb{R}$, but also appreciate answers related to different types of functions and arguments.

Answer 1

Assume that $f(x)$ and $g(x)$ are real functions satisfying the relationship $f(a+b)=f(a)g(b)$. Then one must have $f(x)=Kg(x)$ for some constant $K$.

Proof: Observe that $f(0)=f(0+0)=f(0)g(0)$. So we have that either $f(0)=0$ or $g(0)=1$.

Consider the case where $g(0)=1$. Then we have that $f(a+0)=f(a)g(0)$ but $f(a+0)=f(0+a) = f(0)g(a)$. So $f(a)g(0)=f(0)g(a)$ and since $g(0)=1$ we have $f(a)=f(0)g(a)$ for all $a$.

Now, consider the case $f(0)=0.$ Then we have $f(x-x)=f(x)g(-x)=0$ for all $x$. But if $f(x)$ is zero for any $x$, then we must have $f(x)$ is always zero. So we must have $g(x)=0$ for all $x$ which also implies that $f(x)=0$.

So our only possible non-trivial cases are $f(x)=Kg(x)$ for some non-zero $K$.

Answer 2

Let us consider only the equation $f(x+c)=\gamma(c)f(x)$ for all $x,c$. Compute

$$ \gamma(a+b)f(x) = f(x+a+b) = \gamma(a)f(x+b) = \gamma(a)\gamma(b)f(x). $$ Assume there is at least one $x$ with $f(x) \ne 0$. Then we have, for all $a,b$, $$ \gamma(a+b) = \gamma(a)\gamma(b) . \tag{2}$$ Now assume we are in the case $\gamma : \mathbb R \to \mathbb R$.

If $\gamma(a_0) = 0$ for some $a_0$, then $\gamma(a_0 + b) = 0$ for all $b$, and then $\gamma(x) = 0$ for all $x$. And then $f(x) = 0$ for all $x$.

Assume that $\gamma(x) \ne 0$ for some $x$ so that $\gamma(x) \ne 0$ for all $x$. From $(2)$ we get $\gamma(x) = \gamma(x/2)^2 > 0$, so $\gamma$ has positive values. And $\gamma(0) = 1$.

According to the Axiom of Choice, there are badly-behaved solutions of $(2)$. Let us rule them out by assuming $\gamma$ is differentiable (or, equivalently, continuous, or Borel measurable, or bounded on some interval, or ...).

Compute $$ \frac{\gamma(x+h)-\gamma(x)}{h} = \gamma(x)\frac{\gamma(h)-\gamma(0)}{h} $$ so $\gamma'(x) = k\gamma(x)$ for some constant $k$. Thus $$ \gamma(x) = e^{kx}\quad\text{for some constant } k \in \mathbb R . $$

The OP wants non-exponential functions, so this is not one.

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For the case $\gamma : \mathbb C \to \mathbb C$, we get solutions: $$ \gamma(x) = \exp\left(k x + l \,\overline{x}\right) \quad\text{for some complex constants }k, l . $$ in addition to many wild, discontinuous solutions