Prove if $f(x+y)=f(x)f(y)$ then $f(x)=a^x$

Question

Let $f:\mathbb{R} \rightarrow \mathbb{R}$ be a continuous function such that $f(x+y)=f(x)f(y), \ \forall x,y\in \mathbb{R}$. Prove: if $f \not \equiv 0$, then there exists constant $a$ such that $f(x)=a^x.$

I tried to deduce the result from this question and this question, but had hard time with it.

Any help appreciated.

score 9 · Accepted Answer · answered Mar 08 '17 at 17:19

9

First note that $f(0+0) = f(0)^2$, thus $f(0)$ is either $1$ or $0$. If it was $0$ then $f(x+0) = f(x)f(0) = 0$ and then $f\equiv 0$ which contradicts our hypothesis. It must be that $f(0) = 1$.

Let $a = f(1)$. Then $f(2) = a^2$. $f(3) = f(1)f(2) = a^3$ and inductively, $f(n) = a^n$ for all positive integer $n$.

Conversely, $f(1-1) = f(1)f(-1) = 1$, so $f(-1) = a^{-1}$ and now one can reason as before to conclude that $f(n) = a^n$ for any integer $n$.

Now to compute $f(p/q)$ where $p$ and $q$ are integers and $q$ is positive, we have that $a^p = f(p) = f(\underbrace{p/q + \ldots + p/q}_{q\text{ times}}) = f(p/q)^q$, thus $f(p/q) = \sqrt[q]{f(p)} = a^{p/q}$.

Now, we know that $f(x) = a^x$ for any rational number $x$. Since the set of rationals is dense in the set of reals, then by the continuity of $f$, it must be that $f(x) = a^x$ for any real number $x$.

answered Mar 08 '17 at 17:19

Darth Geek

12,296

Let's pretend that we have not yet defined the exponential function for real values of the exponent. Is it possible to reinvent it from the functional equation? – Massimo Mar 08 '17 at 18:49
@Massimo Yes. Let $x$ be a real number and ${x_n}$ a Cauchy sequence of rational numbers with limit $x$ (such a sequence exists since $\mathbb{Q}$ is dense in $\mathbb{R}$). Since $f$ is continuous, then it preserves limits, i.e.
$$\lim_{n\to\infty} f(x_n) = f\left(\lim_{n\to\infty} x_n\right) = f(x).$$
– Darth Geek Mar 08 '17 at 18:54
Thank you!. And what about the series expansion of the exponential function? In other words, is it possible to find the well known series directly from the functional equation? – Massimo Mar 08 '17 at 18:58
@Massimo I think that series derives from it's Taylor expansion, and I'm not sure how to even prove that the function in the equation is derivable. – Darth Geek Mar 08 '17 at 21:30
Thanks, I posted a question: http://math.stackexchange.com/questions/2178948/power-series-solution-of-fxy-fxfy-functional-equation – Massimo Mar 09 '17 at 10:41

score 1 · Answer 2 · answered Mar 08 '17 at 17:10

1

Hints:

First prove that $f(x)>0$ for all $x\in\mathbb{R}$.
Consider $g(x)=\ln f(x)$. What can you say about $g(x)$?

answered Mar 08 '17 at 17:10

Guy

2,414

Should hints be answers or comments? – toliveira Mar 08 '17 at 17:18
Depends. The op asked for any help thus this is an answer to his question – Mar 08 '17 at 17:23

Prove if $f(x+y)=f(x)f(y)$ then $f(x)=a^x$

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