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For function $f:\mathbb{R}\rightarrow\mathbb{R}$ that satisfies $f\left(x+y\right)=f\left(x\right)f\left(y\right)$ and is not the zero-function I can prove that $f\left(1\right)>0$ and $f\left(x\right)=f\left(1\right)^{x}$ for each $x\in\mathbb{Q}$. Is there a way to prove that for $x\in\mathbb{R}$?

This question has been marked to be a duplicate of the question whether $f(xy)=f(x)f(y)$ leads to $f(x)=x^p$ for some $p$. I disagree on that. Both questions are answered by means of construction of a function $g$ that suffices $g(x+y)=g(x)+g(y)$. In this question: $g(x)=\log f(x)$ and in the other $g(x)=\log f(e^x)$. So the answers are alike, but both questions definitely have another startpoint.

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marked as duplicate by hardmath, Rick Decker, mau, Grigory M, Claude Leibovici Jan 21 '14 at 14:58

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

  • $\begingroup$ I think this has been answered here: math.stackexchange.com/questions/609603/… That question is imo incorrectly marked as a duplicate, because the one that it is claimed to be a duplicate of only considers continuous solutions. $\endgroup$ – Carsten S Jan 21 '14 at 13:03
  • $\begingroup$ @CarstenSchultz. If it is a duplicate, then sorry. Essential is indeed the question: 'is this also true if $f$ is not given to be continuous?' $\endgroup$ – drhab Jan 21 '14 at 13:08
  • $\begingroup$ I have to apologize, I misread your question (multiplicative seemed to indicate something else). Yours reduces more immediately to the additive case. $\endgroup$ – Carsten S Jan 21 '14 at 13:11
  • $\begingroup$ @CarstenSchultz. Accepted! I must have a better look on terms like 'multiplicative' before using them. $\endgroup$ – drhab Jan 21 '14 at 13:17
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No, because if $f$ is any of your functions, you may take any additive function $g : \mathbb{R} \to \mathbb{R}$ (that is, a function such that $g(x+y) = g(x) + g(y)$), and $f \circ g$ will still satisfy your assumption, as $f \circ g (x + y) = f(g(x+y)) = f(g(x) + g(y)) = f(g(x)) f(g(y)) = f \circ g(x) f \circ g(y)$.

And there are plenty such $g$, see under Hamel basis.

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    $\begingroup$ So you are reducing to additive functions and the underlying thought is that there are additive functions that cannot be written as $g(x)=c.x$. Is this interpretation correct? The logarithm of a multiplicative function is additive. $\endgroup$ – drhab Jan 21 '14 at 12:09
  • $\begingroup$ @drhab, that's absolutely correct, and very well formulated! $\endgroup$ – Andreas Caranti Jan 21 '14 at 12:16
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    $\begingroup$ I withdraw my last comment. Off course it is necessary. This follows directly from $f(x)=f(1)^x$. Sorry and thank you for your answer. $\endgroup$ – drhab Jan 21 '14 at 12:29
  • $\begingroup$ @drhab, thank you! $\endgroup$ – Andreas Caranti Jan 21 '14 at 12:55
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If $f(x_0)=0$ for some $x_0$ then $f(x)=f((x-x_0)+x_0)=f(x-x_0)f(x_0)=0$ for all $x$. As the zero function is excluded, $f(x)\ne 0$ for all $x$ and in fact $f(x)=f(\frac x2)^2>0$. Therefore we can define $g(x)=\ln f(x)$ and find the functional equation $$ g(x+y)=g(x)+g(y)$$ for $g$. As has been pointed out, this (i.e. Cauchy's) functional equation has precisely the affine linear functions as continuous solutions, but also many wild and non-continuous solutions. Each solution $f$ is of the form $f(x)=e^{g(x)}$ for such a $g$.

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  • $\begingroup$ I should have asked the analog for additive functions: does $g(x+y)=g(x)+g(y)$ garantee that $g(x)=c.x$? Thank you. $\endgroup$ – drhab Jan 21 '14 at 12:32
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Continuity (or continuity in some point or measurability) is required. See Cauchy's functional equation. Your problem is reducible to this.

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