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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.

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  • $\begingroup$ 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. $\endgroup$ – randomgirl Apr 1 '15 at 19:11
  • $\begingroup$ If any of the two functions are $0$ at any point, they're both the constant function $0$. $\endgroup$ – Alice Ryhl Apr 1 '15 at 19:12
  • $\begingroup$ If $f$ is $1$ at any point, then $g(x)=f(x+\delta)$ for some $\delta\in\mathbb R$. $\endgroup$ – Alice Ryhl Apr 1 '15 at 19:13
  • $\begingroup$ 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. $\endgroup$ – Drew N Apr 1 '15 at 19:16
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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 $f(0) = 1$.

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    $\begingroup$ Ah, of course. Yes. A concrete example is $(2e^{x})(2e^{y}) = 4e^{x+y} = C*2e^{x+y}$, where C = 2. $\endgroup$ – Drew N Apr 1 '15 at 19:32
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    $\begingroup$ ...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 $f/C$. $\endgroup$ – Najib Idrissi Apr 1 '15 at 19:47
  • $\begingroup$ Nice addition @NajibIdrissi thanks! $\endgroup$ – BaronVT Apr 1 '15 at 20:08

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