# discontinuous solutions of the functional equation $f\left(a^b\right) = f(a)^{f(b)}$ [duplicate]

This is a followup to this question.

It's well known that Cauchy's functional equation, $$f(x+y) = f(x) + f(y)\text,$$ has discontinuous solutions. In fact, any discontinuous solution is discontinuous everywhere (except perhaps 0).

Now let's consider the functional equation $$f\left(a^b\right) = f(a)^{f(b)}\text.$$ For simplicity we will restrict the domain to $$\mathbb{R}_{\ge 0}$$. We have seen in the original question that the only continuous solutions are trivial. However, we have not explored what discontinuous solutions there are.

Let $$U = \{a \in \mathbb{R}_{\ge 0} | f(a) \not\in \{0, 1\}\}\text.$$ One property of any discontinuous solution is that, because $$f\left(a^{1+\mathrm{d}x}\right) = f(a)^{f(1+\mathrm{d}x)}$$ for all $$a \in U$$, continuity at any such $$a$$ is equivalent to continuity at 1. Therefore any solution that is discontinuous anywhere on $$U$$ is discontinuous everywhere on $$U$$.

However, this property does not obviously prove or disprove the existence of such functions, not to mention provide us with an example of one if they exist. So the questions are: are there discontinuous solutions, how are they characterized, and what is a canonical example of one?

I was able to construct highly trivial example of a function that is discontinuous at one point and which satisfies your functional equation, it could be that you are not searching for the examples that are trivial as this one, but it is still an example.

(In this example I am assigning the value $1$ to the expression $0^0$.)

Now the example.

Define the function $f$ on $\mathbb R_{\geq0}$ as:

$f(x)=\begin{cases} 1 &\text{ if } x\neq0\\0 &\text{ if } x=0\end{cases}$

Now we have four cases to consider.

1) If $a=0$ and $b\neq0$ then we have $f(a^b)=f(0^b)=f(0)=0=0^1=f(a)^{f(b)}$.

2) If $a\neq0$ and $b=0$ then we have $f(a^b)=f(a^0)=f(1)=1=1^0=f(a)^{f(b)}$.

3) If $a=0$ and $b=0$ then we have $f(a^b)=f(0^0)=f(1)=1=0^0=f(a)^{f(b)}$.

4) If $a\neq0$ and $b\neq0$ then we have $f(a^b)=1=1^1=f(a)^{f(b)}$.

In the case four we used the fact that if $a\neq0$ and $b\neq0$ then $a^b\neq0$.