# What are all pairs of functions f and g so that $f(x)f(y) = g(x+y)$?

It can be shown, and is a problem in Rudin's Principles of Mathematical Analysis (Chapter 8), that when $f$ is continuous, and $f(x)f(y) = f(x+y)$, $f$ is a function of the form $e^{cx}$.

Must this necessarily be true when the right hand side is a different function $g$, and then $g = f$? If so, is there a good reason why, in this particular functional equation, it doesn't matter whether it's $f$ or $g$ on each side--but it matters in others?

EDIT: $f$ should also be assumed to be never zero.

• Well I would say $e^{x} \cdot e^{y}=e^{x+y}$ still satisfies $f(x) \cdot f(y)=g(x+y)$ (because g could equal f) but there could be other functions to satisfy it also. For some reason I was thinking about some trig identities but I haven't found any that work yet. – randomgirl Apr 1 '15 at 19:11
• If any of the two functions are $0$ at any point, they're both the constant function $0$. – Alice Ryhl Apr 1 '15 at 19:12
• If $f$ is $1$ at any point, then $g(x)=f(x+\delta)$ for some $\delta\in\mathbb R$. – Alice Ryhl Apr 1 '15 at 19:13
• Then let's set them nonzero. In fact Rudin does, in that problem, but I neglected to write it. The question's edited now. Also, randomgirl, what I'm wondering is not what the solution is, but if the solution can be DIFFERENT from $e^{cx}$. That's edited too. – Drew N Apr 1 '15 at 19:16

For every $x$,

$$f(x)f(0) = g(x + 0) = g(x)$$

so $g(x) = Cf(x)$, where $C = f(0)$.

Then, you have

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

and further, we can obviously write $f(x) = \dfrac{1}{C}f(x)f(0)$.

Now, if we use the limit definition of the derivative:

$$f'(x) = \lim_{h \to 0} \frac{f(x+h) - f(x)}{h} = \frac{\dfrac{1}{C}f(x)f(h) - \dfrac{1}{C}f(x)f(0)}{h} = \dfrac{1}{C}f(x)\dfrac{f(h) - f(0)}{h} = \dfrac{1}{C}f'(0)f(x)$$

So solutions are just $f(x) = Ce^{\alpha x}$; i.e. you get basically the same result, except that you are not forced to take $f(0) = 1$.

• Ah, of course. Yes. A concrete example is $(2e^{x})(2e^{y}) = 4e^{x+y} = C*2e^{x+y}$, where C = 2. – Drew N Apr 1 '15 at 19:32
• ...And what if $f$ is not differentiable? Just replace the equation $f(x) f(y) = Cf(x+y)$ by the equivalent equation $\frac{f(x)}{C} \frac{f(y)}{C} = \frac{f(x+y)}{C}$ and apply the already-known result to $f/C$. – Najib Idrissi Apr 1 '15 at 19:47
• Nice addition @NajibIdrissi thanks! – BaronVT Apr 1 '15 at 20:08