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?


marked as duplicate by user21467, Asaf Karagila axiom-of-choice May 29 '15 at 11:20

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  • $\begingroup$ @AsafKaragila Yes! Thanks. Evidently I should work on my searching skills. $\endgroup$ – user21467 May 29 '15 at 11:19
  • $\begingroup$ 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. $\endgroup$ – Asaf Karagila May 29 '15 at 11:22
  • $\begingroup$ (Also, math.stackexchange.com/questions/1032565/… was mentioned in a comment that was automatically deleted when the question was closed; and it is still worth mentioning here.) $\endgroup$ – Asaf Karagila May 29 '15 at 11:23

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?

  • $\begingroup$ 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. $\endgroup$ – user21467 May 29 '15 at 2:50
  • $\begingroup$ There is no constructive proof, because this is equivalent to the axiom of choice. I'm not sure what the issue is... $\endgroup$ – pre-kidney May 29 '15 at 2:50
  • $\begingroup$ Exhibit a proof that the existence of such a function implies the axiom of choice and we'll be done. See my third paragraph. $\endgroup$ – user21467 May 29 '15 at 2:51
  • $\begingroup$ 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. $\endgroup$ – pre-kidney May 29 '15 at 2:53
  • $\begingroup$ I accept that $\mathbb R$ is a vector space over $\mathbb Q$. How does this imply the axiom of choice? $\endgroup$ – user21467 May 29 '15 at 2:55