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How do we prove that the endomorphism of the multiplicative group of positive real numbers is unique (up to a complex variable)!? meaning: how do we prove that it has the following - and only the following - form: $$f(x)=x^{s}\;\;\;\;(x\in \mathbb{R}^{+} \;\;,s\in\mathbb{C})$$

Mohammad Al Jamal
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    There are other endomorphisms, but these are the continuous ones. The non-continuous ones are very non-continuous. :) And $s\in\mathbb R$ is enough, I think - any $s\in\mathbb C\setminus\mathbb R$ yields a map $\mathbb R^+\to\mathbb C$, but they aren't endomorphisms. – Thomas Andrews Aug 14 '13 at 01:08
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    This is trivially the Cauchy equational problem after passing to $\mathbb{R}$ via $\log$. – Alex Youcis Aug 14 '13 at 01:10
  • yeah, thanks, i should've mention this is the question : i want to prove that these are the only continuous endomorphism. – Mohammad Al Jamal Aug 14 '13 at 01:10
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    @ThomasAndrews I apologize. It's Cauchy's functional equation http://en.wikipedia.org/wiki/Cauchy's_functional_equation – Alex Youcis Aug 14 '13 at 01:12
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    @MohammadAlJamal Just an English issue. You are talking about an infinite number of endomorphisms, on for every $s$, so "this is the only continuous endomorphism" is confusing. "These are the only continuous endomorphisms" is a better phrasing. – Thomas Andrews Aug 14 '13 at 01:13
  • @AlexYoucis , but the Cauchy's functional equation doesn't have a unique solution. – Mohammad Al Jamal Aug 14 '13 at 01:13
  • @MohammadAlJamal What do you mean? – Alex Youcis Aug 14 '13 at 01:14
  • @MohammadAlJamal Neither does this problem. $s=1$ gives one endomorphism, $s=2$ gives another, $s=\pi$ gives a third, etc. – Thomas Andrews Aug 14 '13 at 01:14
  • i mean in terms of the "families" of solutions . – Mohammad Al Jamal Aug 14 '13 at 01:18
  • But Cauchy only has one family of solutions, too, when restricted to continuous functions. So your objection to Alex is confusing. @MohammadAlJamal – Thomas Andrews Aug 14 '13 at 01:22
  • And I'm not sure what it means to be a unique "family of solutions." That strikes me, again, as a poor use of the word "unique." What you mean is that this family is the complete set of continuous endomorphisms. – Thomas Andrews Aug 14 '13 at 01:24

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By passing to $(\mathbb{R},+)$ using $\log$ this becomes the Cauchy functional equation. This has been beat to death on math.se Look here for example.

For other similar questions, look here.

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Alex Youcis
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