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Is there a name for such kind of function?

The question is: is there a nice characterization of all nonnegative functions $f:\mathbb{R}\rightarrow \mathbb{R}$ such that $f(x+y)=f(x)f(y)$.

If $f$ is continuously differentiable, then I can prove that $f$ is exponential, but I don't know this in general. For what it's worth, in my particular case, I can assume that $f$ is right continuous, i.e. $\lim _{x\to c^+}f(x)=f(c)$, but that is all. From this, what can I deduce about the form of $f$?

Thanks much!

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marked as duplicate by Qiaochu Yuan Jun 7 '11 at 22:13

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

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Look up this math.stackexchange.com/questions/22069/… Possible duplicate –  user17762 May 11 '11 at 2:13
    
Right continuity should be sufficient to conclude f is exponential –  user17762 May 11 '11 at 2:14
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3 Answers 3

up vote 12 down vote accepted

Given any function $g\colon \mathbb{R}\to\mathbb{R}$ such that $g(x+y)=g(x)+g(y)$, you obtain a function of the type you want by composing with an exponential function, $z\mapsto e^{az}$, since $f(x) = e^{ag(x)}$ satisfies $$f(x+y) = e^{ag(x+y)} = e^{ag(x)+ag(y)} = e^{ag(x)}e^{ag(y)} = f(x)f(y).$$

Conversely, any function $f\colon\mathbb{R}\to\mathbb{R}$ such that $f(x+y) = f(x)f(y)$ must be nonnegative, since $f(x) = f(\frac{1}{2}x+\frac{1}{2}x) = \left(f(\frac{1}{2}x)\right)^2$. If $f(x)=0$ for some $x$, then $f(y) = 0$ for all $y$, since $f(y) = f(x+(y-x)) = f(x)f(y-x) = 0$. So one solution is $f(x)=0$ for all $x$. If $f(x)\gt 0$ for all $x$, then composing $f(x)$ with a logarithm function gives a function $g\colon\mathbb{R}\to\mathbb{R}$ such that $g(x+y) = g(x)+g(y)$.

So your question reduces to determining the functions $g\colon\mathbb{R}\to\mathbb{R}$ that are additive. Such functions satisfy $g(q) = g(1)q$ for all $q\in\mathbb{Q}$. Under very mild conditions one can conclude that the function is of the form $g(x) = ax$ with $a=g(1)$ for all $x\in\mathbb{R}$, but if you assume the axiom of choice, there are functions that are not of this form: pick any Hamel basis $\beta$ for $\mathbb{R}$ as a vector space over $\mathbb{Q}$, and fix $\alpha\in\beta$, $\alpha\notin\mathbb{Q}$. Define $g\colon\mathbb{R}\to\mathbb{R}$ by mapping $\alpha$ to $1$ and all other basis elements to $0$, and extend $\mathbb{Q}$-linearly. This map is additive, but not of the form $g(x)=g(1)x$.

(Any additive map from $\mathbb{R}$ to $\mathbb{R}$ must be $\mathbb{Q}$-linear, of course).

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would similar reasoning be used to find all solutions to f(f(f(f(x)))) = f(f(f(x))) + 2x ? I'm involved in a discussion here: mymathforum.com/viewtopic.php?f=13&t=20520 and we are at a stand-still. Thanks –  The Chaz 2.0 May 11 '11 at 2:38
    
@The Chaz: I'm not sure what you mean by "similar reasoning"; I don't see how the reasoning that changed this problem from one that translates sums to products into a purely additive problem applies to your problem. –  Arturo Magidin May 11 '11 at 2:42
    
It probably would have helped if I left less undeveloped! Take, for instance, your observation that x = (1/2)x + (1/2)x. Maybe with enough clever/intuitive observations about compositions, we could arrive at a solution. Thanks for your time, as always! –  The Chaz 2.0 May 11 '11 at 2:51
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If $f(c) = 0$ for some $c$, then $f$ is identically zero.

So assume $f(x) \gt 0 \ \forall x \in \mathbb{R}$.

Let $g(x) = \log f(x)$ and consider the Cauchy Functional Equation.

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Consider $\log f$ and see Cauchy's functional equation.

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