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Show that the only continuous function on $(-1,+1)$, which is not identically zero and satisfies the equation $f(x + y) = f(x)f(y)$ for all $x,y \in \mathbb{R}$, is the exponential function $f(x) = a^x$ with $a = f(1) > 0$.

user10024395
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2 Answers2

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Assume $f(x_0)=0$ for some $x_0\in(-1,1)$. Then from $0=f(x_0)=f(x_0/2+x_0/2)=f(x_0/2)^2$ we conclude $f(x_0/2)=0$. By induction, $f(2^{-n}x_0)=0$ and by continuity $f(0)=0$. Then $f(x)=f(x+0)=f(x)f(0)=0$ for all $x\in(-1,1)$. As $f$ is not supposed to be the zero function, we conclude $f(x)\ne 0$ for all $x\in (-1,1)$. Together with $f(x)=f(x/2)^2\ge0$ we conclude $f(x)>0$ for all $x\in(-1,1)$. Specifically $f(0)=f(0+0)=f(0)^2$ implies $f(0)=1$. Pick $x_1\in (0,1)$ and let $a=\exp(\frac1{x_1}\ln f(x_1)))>0$. Let $$ A=\{\,x\in(-1,1)\mid f(x)=a^x\,\}.$$ We have $x_1\in A$ and $0\in A$. If $x,y,z\in(-1,1)$ with $x+y=z$ and two of $x,y,z$ are in $A$ then all are in $A$: For example if $x,y\in A$, then $f(z)=f(x)f(y)=a^xa^y=a^{x+y}=a^z$; and if $x,z\in A$, then $f(y)=\frac{f(z)}{f(x)}=\frac {a^z}{a^x}=a^{z-x}=a^y$.

Especially, $x\in A$ implies $-x\in A$ and then $nx\in A$ for all $n\in\mathbb Z$ with $nx\in(-1,1)$. Also $x\in A$ implies $\frac12x\in A$ and by induction $2^{-n}x\in A$ for all $n\in\mathbb N$. We conclude that $\frac m{2^n}x_1\in A$ for all $n\in \mathbb N$, $m\in\mathbb Z$ with $-1<\frac{m}{2^n}x_1<1$. Then $A$ is dense in $(-1,1)$ and by continuity of $f$, it is all of $(-1,1)$.

Note that this implies $\lim_{x\to 1}f(x)=a$.

Hagen von Eitzen
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  • Hi, I don't really understand why $-1< \frac{m}{2^n} x_1 < 1$ shows that A is dense? Can you explain more how the result is arrived since I am fairly new to analysis. On top of that, your solution shows that the exponential function satisfy the condition but how does it show that it is the only one that satisfy it? How is uniqueness proven here? Thanks – user10024395 May 03 '14 at 02:20
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This answer is basically in the same spirit as Hagen von Eitzen's answer but probably in a more "algorithmic" way. From his answer we know that $f(0)=1$.

Step 1: For all $n\in\mathbb N$, we have: $$ f(2)=f(1)^2\\ \vdots\\ f(n)=f(1)^n $$ So for all positive integers $n$ , we have $f(n)=a^n$ with $a=f(1)$.

Step 2: We know that $f(nx)=f(x)^n$ for all positive integers $n$. Therefore $f(\frac 1n)^n=a$. Also: $$ f(\frac mn)=f(\frac 1n)^m=a^{\frac mn} $$ So for all positive rationals $\frac mn$ , we have $f(\frac mn)=a^{\frac mn}$ with $a=f(1)$.

Step 3: For an arbitrary positive real number $r$, there is a sequence of $\{a_n\}$ of rational numbers such that $\lim_{n\to\infty}a_n=r$. Now $f$ is supposed to be continuous and we have: $$ f(r)=f(\lim_{n\to\infty}a_n)=\lim_{n\to\infty}f(a_n)=\lim_{n\to\infty}a^{a_n}=a^r. $$ So for all positive real numbers $r$ , we have $f(r)=a^{r}$.

Step 4: Use $f(-x)=\frac 1{f(x)}$ to prove the result for all $r\in\mathbb R$.

Usually, when we deal with a functional equation, we can use these steps to find the function. Frist we find the value of function for some obvious choice of inputs (like $x=0$ here). Then we try to find the function for all integers and then all rationals. Finally, if the function is assumed to be continuous, we can use a sequence of rationals to find the function at all irrational number. That's why I called this approach "algorithmic". No need to say that there is no universal solution applicable to all functional equations.

Arash
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