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I’m having some trouble solving the following equation for $f: A \rightarrow B$ where $A \subseteq \mathbb{R}$ and $B \subseteq

\mathbb{C}$ such as:

$$f(x)f(y) = f(x+y) \quad \forall x,y \in A$$

The solution is supposed to be the form $f(x) = e^{kx}$ where k can be complex but I don’t see how to show it.

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Note: I’m trying to solve this equation to find irreductible representation of U(1) but it doesn’t really matter I think

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The general solution of the given functional equation depends on conditions imposed to $f$.

We have:

The general continuous non vanishing complex solution of $f(x+y)=f(x)f(y)$ is $\exp(ax+b\bar x)$, where $\bar x$ is the complex conjugate of $x$. Note that this function is not holomorphic.

If we request that $f$ be differentiable then the general solution is $f(x)=\exp (ax)$, as proved in the @doetoe answer.

Added:

here I give a similar but a bit simpler proof:

Note that $f(x+y)=f(x)f(y) \Rightarrow f(x)=f(x+0)=f(x)f(0) \Rightarrow f(0)=1$ ( if $f$ is not the null function). We have also: $f'(x+y)=f'(x)f(y)$ (where $f’$ is the derivative with respect to $x$).

Now, setting $x=0$ and dividing by $f(y)$ yields:

$$

f'(y)=f'(0)f(y) \Rightarrow f(y)=k\exp\left(f'(0)y\right)

$$

and, given $f(0)=1$ we must have $k=1$.

The continuity request can be weakened but, if $f$ is not measurable then, with the the aid of an Hamel basis, we can find more general ”wild” solutions.

For a proof see: *J.Aczél : lectures on functional equations and their applications, pag. 216*.

If we assume differentiability on the outset, there is one approach that I like that is quite general to turn it into a differential equation: consider the map $\Bbb R^2\to\Bbb R^2$ defined by

$$(x,y)\mapsto (f(x)f(y), f(x+y)).$$

The functional equation implies that the image is one dimensional, so the Jacobian is everywhere degenerate. Working it out we see:

$$f'(x)f(y)f'(x+y) = f'(x+y)f(x)f'(y)$$

from which you can easily argue that

$$\frac d{dx}\log f(x) = \frac{f'(x)}{f(x)} = \frac{f'(0)}{f(0)} = constant$$

so that the result follows (in the differentiable case).

Let $y=1$ and $f(1)=e^k$ for some $k\in\mathbb{C}$, then

$$

f(x+1)=e^kf(x)\\

f(x)=e^kf(x-1)\\

f(x)=e^{2k}f(x-2)\\

f(x)=e^{3k}f(x-3)\\

\vdots\\

f(x)=e^{(x-1)k}f(1)\\

f(x)=e^{kx}\\

$$

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