Obviously $a^x$ is a solution of $f(x+y)=f(x)f(y)$. But how do I prove that it is the only solution of that functional equation (apart from $f(x)=0$ or $f(x)=1$)
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1$f(x)=0$, $f(x)=1$. – Descartes Before the Horse Jan 05 '20 at 16:35
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4Is $f$ a function on real numbers? Complex numbers? Integers? Rational numbers? Natural numbers? Is it continuous? – Milo Brandt Jan 05 '20 at 16:35
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1There are lots of solutions that are not continuous: See Cauchy functional equation. Any solution $g$ of Cauchy's functional equation gives a solution $f(x)=e^{g(x)}$ of your equation. – MoonLightSyzygy Jan 05 '20 at 16:36
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$x$ can be an arbitrary complex number. – UraUra Jan 05 '20 at 16:36
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2Is this question a duplicate of your question? – Axion004 Jan 05 '20 at 16:38
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@Axion004 It's a duplicate. Thanks for the link. – UraUra Jan 05 '20 at 16:40
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With $x=y$, we have $f(2x)=f^2(x)$ which shows that $f$ is non-negative. $f$ is also nonzero, otherwise if $f(y)=0$ for some $y$, then $f(x+y)=0$ and $f$ is the constant $0$. Then we can take the logarithm and $\log f(x+y)=\log f(x)+\log f(y)$ is the linearity condition. – Jan 05 '20 at 16:44
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@MoonLightSyzygy I'm not completely familiar with all the mathematics used within the article. Could you give me an example of such $g(x)$? – UraUra Jan 05 '20 at 16:50