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I'm wondering what kind of function preserves exponentiation, i.e., what is an $f$ such that $f(a^b)=f(a)^{f(b)}$?

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  • This probably won't help, but the identity function is a trivial example of a function that preserves exponentiation. – nukeguy Feb 05 '15 at 22:16
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    Note that for all $a$, $f(a) = f(a)^{f(1)}$ so either $f(x) \in {0, 1}$ or $f(1) = 1$. In the latter case, for all $a \ne 0$, $1 = f(1) = f(a)^{f(0)}$ so $f = 1$ or $f(0) = 0$. – Reinstate Monica Feb 05 '15 at 22:33
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    Also, if $f$ is continuous then $f$ commutes with infinite power towers so $f$ maps $[e^{-e}, e^{\frac{1}{e}}]$ to itself. By this property and the aforementioned ones, I conjecture that the only continuous, non-constant solution to the equation is the identity. – Reinstate Monica Feb 05 '15 at 23:55
  • @Solomonoff'sSecret That's interesting. Maybe function is too restrictive for this question, what about things like modular arithmetic? – nullgraph Feb 06 '15 at 04:15
  • One more observation. The set of such functions is closed under composition, for if $f$ and $g$ are such functions, $f(g(a^b))=f(g(a)^{g(b)})=f(g(a))^{f(g(b))}$. @nullgraph I'm not sure what you mean by your comment. Perhaps post another question or clarify this question. – Reinstate Monica Feb 06 '15 at 13:38
  • @Solomonoff'sSecret This is still an interesting question, I'll post another one elsewhere. – nullgraph Feb 06 '15 at 21:41
  • @Solomonoff'sSecret $f(a^{xy})=f(a^x)^{f(y)}=f(a)^{f(x)f(y)}=f(a)^{f(xy)}\Rightarrow f(xy)=f(x)f(y)$. I think this proves your conjecture. – Martin Nicholson Dec 02 '15 at 21:51
  • @MartinNicholson Filling in the details: By your comment and this answer: https://math.stackexchange.com/a/44006/147200 , if $f$ is not degenerate then we know that $f(x) = x^c$ for some constant $c$. Plugging into the original equation gives $(a^b)^c = (a^c)^{b^c}$ for all $a$ and $b$ so $c \in {0, 1}$. Finally, the only 3 functions that work are $0$, $1$, and the identity. Do you want to write this up as an answer or should I? – Reinstate Monica Dec 03 '15 at 03:43
  • @MartinNicholson Obviously I meant to say the only 3 continuous functions that work. – Reinstate Monica Dec 03 '15 at 03:53
  • @MartinNicholson Comments are transient so it's appropriate to put the answer to the question in an answer. – Reinstate Monica Dec 03 '15 at 16:33

1 Answers1

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First off, the constant functions $-1$, $0$ (where $0^0 = 0$), and $1$ satisfy the functional equation. Let us now find the remaining continuous functions that satisfy the equation.

According to Martin Nicholson's comment, for all $a$, $x$, $y$: $$f(a)^{f(xy)} = f(a^{xy}) = f(a^x)^{f(y)} = f(a)^{f(x)f(y)}.$$ Therefore if there is some $a$ for which $1 \ne f(a) > 0$ then $f(xy) = f(x)f(y)$. By this answer, $f(x)$ must then be of the form $x^c$ for some constant $c$. So: $$(a^b)^c = f(a^b) = f(a)^{f(b)} = (a^c)^{b^c}$$ which implies that $c \in \{0, 1\}$. If $c$ is $0$ then $f = 1$; otherwise $f$ is the identity function.

Now we show that such an $a$ mentioned above exists. Note that $f$ preserves power towers: $$f\left(x^{x^\ldots}\right) = f(x)^{f\left(x^{x^\ldots}\right)}$$ so $$f\left(x^{x^\ldots}\right) = f(x)^{f(x)^\ldots}.$$ As power towers are defined for exactly those $x \in I := \left[e^{-e}, e^\frac{1}{e}\right]$ and as $f\left(x^{x^\ldots}\right)$ is defined for any $x \in I$, $f$ maps this interval to itself. If $f$ were $1$ on this interval then the functional equation would require $f$ to be $1$ everywhere. Thus if $f \ne 1$ then there must be some $a \in I$ for which $f(a) \in I$, $f(a) \ne 1$, which gives the required $a$.

In summary, the continuous functions that satisfy the functional equation are $-1$, $0$ (if $0^0 = 0$), $1$, and the identity.

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