Suppose $f$ is a function satisfying $$ f(x)f(y)=f(x+y) \label{*}\tag{*} \\ $$ for all $x$ and $y$. If we require that $f$ be continuous, then it can be shown that $f(x)=a^x$ for some constant $a$, or $f=0$. In Michael Spivak's Calculus, it is mentioned that there in fact infinitely many discontinuous functions satisfying $\eqref{*}$. What are the most notable examples?
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1See here. – Arturo Magidin May 25 '21 at 20:43
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2See also Overview of basic facts about Cauchy functional equation. – dxiv May 25 '21 at 20:51
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Take the logarithm of both sides and it will be converted into the Cauchy functional equation. – Vivaan Daga May 26 '21 at 13:16
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@VivaanDaga: I don't understand what allows us to take logs of both sides. That only seems to work if we are given that $f(x)>0$ for all $x$. Even if we allow complex numbers to enter the mix, we still have to ensure $f(x) \neq 0$, don't we? – Joe May 26 '21 at 18:17
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@Joe You can prove that if such a function is $0$ somewhere then it is $0$ everywhere so you can discard that case , also note that $f(x+x)=(f(x))^2$ which is always positive – Vivaan Daga May 27 '21 at 02:23
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@Joe does my answer answer your question ? Or is there still other questions ? – Vivaan Daga May 29 '21 at 14:43
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@VivaanDaga: Sorry, I haven't gone through your answer yet, but thanks for posting it. – Joe May 29 '21 at 21:49
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Since there exists a discontinuous function satisfying the Cauchy functional equation $g(x)$ ,$e^{g(x)}$ will satisfy $f(x+y)=f(x)f(y)$ and $k(x)=e^{g(x)}$ must be discontinuous since composition of continuous functions are always continuous but $\ln(k(x))=g(x)$ which is discontinuous. Since there are infinitely many discontinuous functions satisfying the Cauchy functional equation there are infinitely many discontinuous functions satisfying your equation .
Vivaan Daga
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