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It is well-known that the only continuous functions $f\colon\mathbb R\to\mathbb R^+$ satisfying $f(x+y)=f(x)f(y)$ for all $x,y\in\mathbb R$ are the familiar exponential functions. (Prove $f(x)=f(1)^x$ successively for integers $x$, rationals $x$, and then use continuity to get all reals.)

The usual example to show that the identity $f(x+y)=f(x)f(y)$ alone doesn't characterize the exponentials requires the axiom of choice. (Define $f$ arbitrarily on the elements of a Hamel basis for $\mathbb R$ over $\mathbb Q$, then extend to satisfy the identity.)

Is there an explicit construction of a discontinuous function satisfying the identity? On the other hand, does the existence of such a function imply the axiom of choice or some relative?

  • @AsafKaragila Yes! Thanks. Evidently I should work on my searching skills. –  May 29 '15 at 11:19
  • Another way to see that it is consistent to have only continuous functions would be to not that $\Bbb R^+$ is a Polish group, homeomorphic using $e^x$, to $\Bbb R$ with addition. Under some circumstances (every set of reals has the Baire property) every homomorphism between Polish groups is continuous. This saves the part where we use $\log$ to reduce the question back to a linear function. – Asaf Karagila May 29 '15 at 11:22
  • (Also, http://math.stackexchange.com/questions/1032565/functions-that-satisfy-fxy-fxfy-and-f1-e was mentioned in a comment that was automatically deleted when the question was closed; and it is still worth mentioning here.) – Asaf Karagila May 29 '15 at 11:23

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Let $g(x)=\ln f(x)$, so that $g(x+y)=g(x)+g(y)$. The solutions to this equation are precisely the ring homomorphisms from $\mathbb R\to\mathbb R$ and they are in bijection with solutions to your original equation. Since $\mathbb R$ is an infinite dimensional $\mathbb Q$-vector space, there are infinitely many such ring homomorphisms. They are determined by a Hamel basis for $\mathbb R$.

Does this answer your question?

pre-kidney
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  • No, it doesn't. The existence of a Hamel basis for $\mathbb R$ over $\mathbb Q$ is usually proved by Zorn's lemma, so it's not the constructive proof I'm after. Compare my second paragraph. –  May 29 '15 at 02:50
  • There is no constructive proof, because this is equivalent to the axiom of choice. I'm not sure what the issue is... – pre-kidney May 29 '15 at 02:50
  • Exhibit a proof that the existence of such a function implies the axiom of choice and we'll be done. See my third paragraph. –  May 29 '15 at 02:51
  • Suppose you have a non-trivial homomorphism. This induces a non-zero $\mathbb Q$-module structure on $\mathbb R$. Thus since $\mathbb Q$ is a ring, we have shown that $\mathbb R$ is a $\mathbb Q$-vector space, which implies the axiom of choice. – pre-kidney May 29 '15 at 02:53
  • I accept that $\mathbb R$ is a vector space over $\mathbb Q$. How does this imply the axiom of choice? –  May 29 '15 at 02:55
  • Because every vector space has a basis; this allows you to construct a Hamel basis from the $\mathbb Q$-module structure on $\mathbb R$ induced by $f$. – pre-kidney May 29 '15 at 03:02
  • How do you prove that every vector space has a basis? –  May 29 '15 at 03:02
  • (The usual proof uses Zorn's lemma.) –  May 29 '15 at 03:03
  • I see your point. It looks like a more refined argument is needed. I'm not used to working without choice. – pre-kidney May 29 '15 at 03:05
  • Just the fact that $\mathbb{R}$ has a basis over $\mathbb{Q}$ is very likely to be strictly weaker then AC, just due to the fact that you only need "small" AC i.e. you only need to be able to have choice functions for collections of size $\mathfrak{c}$ – DRF May 29 '15 at 08:31
  • @DRF: It is much much *much* weaker than the axiom of choice. Yes. Whether or not it implies some sort of other known choice principle, we don't know (other than relatively trivial implications through the existence of discontinuous functionals which in turn imply the existence of somewhat pathological sets of reals). – Asaf Karagila May 29 '15 at 11:29