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$$\forall x,y \in \mathbb{R} :f(x+y)=f(x)f(y) ,f(0)\neq0$$ Exponential function $f(x)=a^x $ works fine here .
My question :Is there any $\bf\color{red} {\text{other function}}$ exept of exponential functions work for this property ?

$\bf\text{I know like this question asked before}$ ,but My question is about uniqueness type of functions which work with this property.
On the other hand is there a proof that only exponential function work for that property ?

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  • $\begingroup$ If $f$ is differentiable (and maybe even if it is continuous), the answer is yes. $\endgroup$ – Bernard Sep 11 '17 at 19:00
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    $\begingroup$ Without continuity, there are others. Take any additive non-linear function $g(x)$. Then let $f(x)=\exp(g(x))$. $\endgroup$ – lulu Sep 11 '17 at 19:02
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    $\begingroup$ If $f$ is continuous at one point, then $f(x)=a^x$. $\endgroup$ – Kenny Lau Sep 11 '17 at 19:10
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Let $g$ be any function satisfying $g(x + y) = g(x) + g(y)$. Then if we set $f = \exp g$,

$$f(x + y) = \exp\big(g(x) + g(y)\big) = f(x) f(y)$$

Functions $g$ are solutions to the Cauchy functional equation, of which there are many (typically constructed using choice), including anything of the form $ax$.

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    $\begingroup$ :Did you read my question? I know exp. func. work over there ... $\endgroup$ – Khosrotash Sep 11 '17 at 19:06
  • $\begingroup$ Drats, you beat me to it... $\endgroup$ – Simply Beautiful Art Sep 11 '17 at 19:06
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    $\begingroup$ You also assume that $f>0$ here. $\endgroup$ – Simply Beautiful Art Sep 11 '17 at 19:07
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    $\begingroup$ @Khosrotash this is only an exponential function if $g(x) = ax$ for some $a$. $\endgroup$ – Omnomnomnom Sep 11 '17 at 19:08
  • $\begingroup$ @Khosrotash See Wikipedia. $\endgroup$ – Simply Beautiful Art Sep 11 '17 at 19:24
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From this one proves that $f(rm)=\left[f(m)\right]^r$ with $r$ rational.

It only has to be $a^x$ for each equivalence class of $m$ under the relation $x \sim y \iff xy \ne 0 \land \dfrac xy \in \Bbb Q$.

Partition the reals under this equivalence relation, and set the value of the function at each representative to your favourite real number, and you'll have a solution not in the form $a^x$ for all $x$.

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  • $\begingroup$ I've never been clear on that last point. You are requiring that $\mathbb R$ have a basis as a vector space over $\mathbb Q$. How do you prove that without the axiom of choice? $\endgroup$ – lulu Sep 11 '17 at 19:25
  • $\begingroup$ @lulu That specific function itself does not require the axiom of choice, but to build general functional solutions you need choice. $\endgroup$ – Kenny Lau Sep 11 '17 at 19:31
  • $\begingroup$ Well, your function does not work. What is $f(\pi+1)$? You'd say it is $0$ yes? but that is incompatible with "linearity". $\endgroup$ – lulu Sep 11 '17 at 19:35
  • $\begingroup$ @lulu $f(\pi+1) = f(\pi)f(1) = e \cdot 0 = 0$ $\endgroup$ – Kenny Lau Sep 11 '17 at 19:36
  • $\begingroup$ Ah, bad example on my part. Try $f(\pi)=f(\pi+1-1)=f(\pi+1)f(-1)=0$. Exponential-like functions can never be $0$ unless they are identically $0$. $\endgroup$ – lulu Sep 11 '17 at 19:37
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It's very easy to show that $f(0) = 1$ and if $f(1)= b$ then by induction $f(n)= b^n$ for any natural $n$ and that $f(\frac 1n) = \sqrt[n]b$, and so for all $q \in \mathbb Q$ that $f(q) = b^q$ ($b^{\frac nm} = (\sqrt[m]{b})^n$).

And so as $\mathbb Q $ is dense in $\mathbb R$ it follows that if $f$ if continuous, that $f(x) = b^x$.

But $f$ is not continuous it does not. For any irrational $k$ we could set $f(k)= c$ by any value and we could have and $f(qk) = f(k)^q$ for all rational $q$ but as $qk$ is not rational no contradiction need arise. ($f(r + qk) = b^rc^q \not \in \mathbb Q$ (unless $q=0$) for all $r,q \in \mathbb Q$.)

But we will have for any $w \in \mathbb R$ that for and $q_i\in \mathbb R$ that $f(\sum q_i*w^i) = \prod f(w^i)^{q_i}$.

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