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I'm messing around with exponentials & logarithms to deepen my familiarity with the basics.

It occurred to me to wonder if $f(x+y) = f(x)\cdot f(y)$ is sufficient to define an exponential, for $f:\mathbb{R}\rightarrow\mathbb{R}^+$ (and the obvious flip for defining logarithms).

The one spot I'm stuck on is with showing that $(f(x))^\lambda = f(\lambda x)$.

It's straightforward to show that this holds for $\lambda \in \mathbb{Q}$. And I have a clear intuition that if $f$ is continuous then it must hold. (It's been a while since I worked with formal proofs of continuity, but the gist looks something like $f(\lambda x) = \lim f(\lambda_n x) = \lim f(x)^{\lambda_n} = f(x)^\lambda$.)

But it seems like I should be able to prove $f(x)^\lambda = f(\lambda x)$ directly, prove it via proving the continuity of $f$, or disprove it by finding a (discontinuous) counterexample.

And I haven't been able to make headway on any of that. The most I've done is noted that neither the characteristic function for the rationals nor a function that's one exponential (say $2^x$) on rationals and another (say $3^x$) on irrationals satisfy the definition of $f$.

So what am I missing?

Morphenius
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    $f(x + y) = f(x)f(y)$ has solutions that aren't exponentials. However, they are extremely pathological, and even mild conditions like "bounded on at least one finite interval" rule them out. – eyeballfrog Mar 31 '23 at 19:55
  • @eyeballfrog Please detail such a solution. – K.defaoite Mar 31 '23 at 20:03
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    @K.defaoite I can't, as they require the Axiom of Choice to prove they even exist. See here for more details. – eyeballfrog Mar 31 '23 at 20:10
  • Looks like true....$f(x)^\lambda = f(x).f(x)...f(x)$. Since $f(x+y)=f(x)+f(y)$ then $f(x+x+....+x)= f(x)+f(x)...+f(x)$ Which is $f(x\lambda)=f(x)^{\lambda}$. – NoChance Mar 31 '23 at 22:53
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    @NoChance This is part of the proof that it works for $\lambda \in \mathbb{Q}$. You've just shown that it works for $\lambda \in \mathbb{N}$. The part I'm getting stuck on is when $\lambda\in\mathbb{R}-\mathbb{Q}$. Like for $\sqrt2$ or $\pi$. – Morphenius Apr 01 '23 at 00:10
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    @eyeballfrog Thank you! I don't feel like I have an answer yet, but it's helpful to know I'm looking for something called "Cauchy's exponential functional equation". Googling that finds me this post, which looks promising. – Morphenius Apr 01 '23 at 00:13
  • @eyeballfrog The linked article talks about the functional equation $f(x+y)=f(x)+f(y)$ which is not what is being asked here. – K.defaoite Apr 01 '23 at 12:51
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    @K.defaoite It looks like the relevant Wikipedia article doesn't exist anymore, but it's linked to & named as "Cauchy's exponential functional equation". Searching for that is how I found the post on this Stack Exchange that I linked to. – Morphenius Apr 02 '23 at 01:46

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