A function with a property $f(x+y)=f(x)f(y)$ A function with the property $f(x+y)=f(x)f(y)$ is well known exponential function, $f(x)=a^x$. My question is, how do you prove if there is no other function with this kind of property?

Edit: I always find this in math contests. At first glance, it really is the exponential function. As I see in the comments section, there are many other functions. 
 A: The comments that suggest $f(x)=1$ and $f(x)=0$ as solutions are correct, but these are just special cases of $f(x)=a^x$ with $a=1$ or $a=0$. If $f$ is not continuous then $f$ need not be an exponential function on the irrationals, but on $\mathbb{Q}$ $f(x)=a^x$ if it satisfies this property.

To see this, let $f(1)=a$ and note that $f(n+1)=f(n)f(1)$ so that by induction $f(x)=a^n$ for all $a\in\mathbb{N}$. Assuming that $a\neq0$, from $f(x+(-x))=f(x)f(-x)$ where $f(x+(-x))=f(0)=1$ (the fact that $f(0)=1$ must be proven first from consideration of $f(1+0)$), we get that $f(x)=a^x$ for all $x\in\mathbb{Z}$. We also have that $f(\underbrace{\frac{1}{p}+\ldots+\frac{1}{p}}_\text{$p$ terms})=(f(\frac{1}{p}))^p=f(1)=a$ so that $f(\frac{1}{p})=a^\frac{1}{p}$ so it's easy to argue from this fact that $f(x)=a^x$ for all $x\in\mathbb{Q}$.

Now if we know that $f$ is continuous everywhere (I believe it is also sufficient for $f$ to be continuous merely at one point) it can be shown that $f(x)=a^x$ for all $x\in\mathbb{R}$.
A: HINT: Using the equation $f(x+y)=f(x)f(y)$,


*

*If $f(x)$ is polynomial, show that the degree of L.H.S. $\not =$ degree of R.H.S. (except if $f(x)$ is the zero polynomial)

*If $f(x)$ is trigonometric, show that it is not possible using the expansion formulae of $\sin,\cos,\tan$ of $(x+y)$.

A: The condition $f(x+y)=f(x)f(y)$ only implies $f(x)=a^x$ for all rational numbers $x\in\mathbb{Q}$ and for some $a\in\mathbb{R}$. You can get this equality for all real numbers if you have more conditions, for example, if $f$ is continuous in $\mathbb{R}$ or if $f$ is Lebesgue-measurable.
A: I proved 
at Proof of existence of $e^x$ and its properties
that,
if $f(x)$ is differentiable 
at $0$,
then
$f(x+y) 
=f(x)f(y)
$
implies that
$f'(x)
=f'(0) f(x)
$.
This leads immediately
to the exponential
or power function.
A: It would, assuming that $f$ is continuous and with this condition $(f(x+y)=f(x)f(y)) $ and $f(0)=1$ to prove that $\displaystyle\lim_{x\to-\infty} f(x)=0$ ?
My sketch
As $f$ is continuous, it suffices to consider a sequence $(a_n)_n$ with limit $-\infty$ and prove that
$\displaystyle\lim_{n\to+\infty} f(a_n)=0$. This is true ? Because then the only option I see is to prove that this is the exponential function and then conclude that the limit is $0$.
