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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?

Martin Sleziak
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Vasting
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  • 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. – lulu Sep 27 '16 at 22:45
  • This wiki page on characterizations of the exponential function should be useful: https://en.wikipedia.org/wiki/Characterizations_of_the_exponential_function – Jonas Sep 27 '16 at 22:47
  • A trival counter example would be constant functions $ x \mapsto 1 $ and $ x \mapsto 0 $. – Q the Platypus Sep 27 '16 at 23:39
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    Why is this marked as a duplicate? One asks about $f(x + y) = f(x)f(y)$ vs $f(x+y) = f(x) + f(y)$ – Q the Platypus Sep 28 '16 at 02:23

2 Answers2

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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 $f(z)$.

Arash
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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).

Michael Hardy
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