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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.

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    $\begingroup$ How about the zero function? $\endgroup$ – Element118 Nov 27 '15 at 5:54
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    $\begingroup$ Or the constant function $f(x) = 1$. $\endgroup$ – Ethan Alwaise Nov 27 '15 at 5:56
  • $\begingroup$ There are also wildly discontinuous functions, obtained by using a discontinuous solution of the Cauchy functional equation, followed by exponentiation. $\endgroup$ – André Nicolas Nov 27 '15 at 5:56
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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 $x\in\mathbb{R}$.

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HINT: Using the equation $f(x+y)=f(x)f(y)$,

  1. 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)
  2. If $f(x)$ is trigonometric, show that it is not possible using the expansion formulae of $\sin,\cos,\tan$ of $(x+y)$.
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    $\begingroup$ But there are more types of functions than just polynomial and trig? $\endgroup$ – Ovi Nov 27 '15 at 6:25
  • $\begingroup$ @Ovi Yes sure. I just helped him prove the case for polynomials and trig functions. $\endgroup$ – SchrodingersCat Nov 27 '15 at 6:41
  • $\begingroup$ Even showing that $f$ is not a polynomial or a trig function is not enough to prove that $f$ is an exponential function. $\endgroup$ – Ángel Valencia Nov 27 '15 at 13:46
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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.

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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.

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  • $\begingroup$ It holds if $f(0)=1$. $\endgroup$ – Empty Nov 27 '15 at 6:49

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