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This question already has an answer here:

I understand the other way around, where if a function is exponential then it will satisfy the equality $f(a+b)=f(a)f(b)$. But is every function that satisfies that equality always exponential?

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marked as duplicate by Chris Culter, user223391, Micah, GFauxPas, heropup Sep 27 '16 at 23:53

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.

  • $\begingroup$ Do we have continuity? If not, no. Take a discontinuous function $g:\mathbb R\to \mathbb R$ with $g(a+b)=g(a)+g(b)$. Then let $f(x)=e^{g(x)}$. But continuity is enough to get you where you want to be. $\endgroup$ – lulu Sep 27 '16 at 22:45
  • $\begingroup$ This wiki page on characterizations of the exponential function should be useful: en.wikipedia.org/wiki/… $\endgroup$ – Jonas Sep 27 '16 at 22:47
  • $\begingroup$ A trival counter example would be constant functions $ x \mapsto 1 $ and $ x \mapsto 0 $. $\endgroup$ – Q the Platypus Sep 27 '16 at 23:39
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    $\begingroup$ Why is this marked as a duplicate? One asks about $f(x + y) = f(x)f(y)$ vs $f(x+y) = f(x) + f(y)$ $\endgroup$ – Q the Platypus Sep 28 '16 at 2:23
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First see that $f(0)$ is either 0 or 1. If $f(0)=0$, then for all $x\in\mathbb R$, $f(x)=f(0)f(x)=0$. In this case $f(x)=0$ a constant function.

Let's assume $f(0)=1$. See that for positive integer $n$, we have $f(nx)=f(x)^n$ which means $f(n)=f(1)^n$. Also see that: $$ f(1)=f(n\frac 1n)=f(\frac 1n)^n\implies f(\frac 1n)=f(1)^{1/n}. $$ Therefore for all positive rational numbers: $$ f(\frac mn)=f(1)^{m/n}. $$ If the function is continuous, then $f(x)=f(1)^x$ for all positive $x$. For negative $x$ see that: $$ f(0)=f(x)f(-x)\implies f(x)=\frac{1}{f(-x)}. $$ So in general $f(x)=a^x$ for some $a>0$.


Without continuity, consider the relation: $xRy$ if $\frac xy\in \mathbb Q$ (quotient group $\mathbb R/\mathbb Q$). This relation forms an equivalence class and partitions $\mathbb R$ to sets with leaders $z$. In each partition the function is exponential with base $f(z)$.

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  • $\begingroup$ You can get by with just monotonicity rather than continuity. $\endgroup$ – Michael Hardy Sep 27 '16 at 23:22
  • $\begingroup$ That's also true; Thanks for mentioning it. $\endgroup$ – Arash Sep 27 '16 at 23:24
  • $\begingroup$ This made a lot of sense - thanks! $\endgroup$ – Vasting Sep 28 '16 at 2:50
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Suppose $f$

  • is a real-valued function of a real variable, and
  • is monotonic (i.e. either nowhere decreasing or nowhere increasing), and
  • satisfies $f(a+b)=f(a)f(b)$ for all $a,b\in\mathbb R$.

Then $f$ is an exponential function (and in partcicular, $f$ is continuous).

However, a complex-valued function of a complex variable can be continuous and satisfy $f(a+b)=f(a)f(b)$ without being an exponential function. Here is an example: $$ f(x+iy) = 2^x(\cos y^\circ + i\sin y^\circ). \qquad (\text{for }x,y\text{ real, where }x^\circ\text{ means } x\text{ degrees}.) $$ If that were exponential it would have base $2$, since its restriction to $y=0$ is $x\mapsto 2^x$. However, $$ 2^{x+iy} = 2^x(\cos(y\log_e 2) + i\sin(y\log_e 2)) \qquad (\text{with radians, not degrees}). $$ Arash's answer explains why in the real case it must be exponential (except that you should note that the hypothesis of continuity in Arash's answer can be weakened to monotonicity).

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  • $\begingroup$ Took me a while to figure out what $y^\circ$ meant (and I'm still not certain) :-) $\endgroup$ – arkeet Sep 27 '16 at 23:34

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