# Function implicitly defined by composition is analytic

I have a function $\mu(x)$ which I know to be well defined on the positive real number line. Further it obeys $$\mu(h(x))=(1+px)\mu(x) \quad\text{and}\quad \mu(0)=1$$ where $$h(x)=\sqrt[m]{1-\Big(\frac{1-x}{1+px}\Big)^m}$$ where $p,m\ge 1$ are fixed to some choice of positive real numbers. And in case it does not follow from this definition, $\mu$ can also be expressed as $$\mu(x)=\hat\mu(x^m)\quad\text{for some } \hat\mu$$ Which I figure means we can write $\mu$ as $$\mu(x)=\sum_{k=0}^\infty a_kx^{km}$$

Is this enough to say the function is analytic in any domain?

I know the function is well defined for $x>0$, but showing that took tedious algebra and did not rely on this implicit definition. I suspect but have been unable to show that $\mu$ is continuous on the reals and possibly the open right half plane of the complex numbers.

I've also looked into the algebra of analytic functions. That is composing two analytic functions is analytic. Adding two analytic functions is analytic. Etc. But it seems that to use any such properties to show $\mu$ analytic begs the question.

• Unless you at least know that $\mu$ is continuous, this is false. – Paul Sinclair Nov 21 '17 at 1:19

Which I figure means we can write $\mu$ as $$\mu(x)=\sum_{k=0}^\infty a_kx^{km}$$

You figure incorrectly. Consider the case when $\hat \mu$ is the Dirichlet function: $$\hat \mu(t) = \begin{cases}0& t \in \Bbb Q\\1 & t \notin \Bbb Q\end{cases}$$

However, even if you demand that $\hat \mu$ is infinitely differentiable, this still doesn't follow, as there are infinitely differentiable functions that are still not analytic anywhere. If $\hat \mu$ is one of them, so will be $\mu$.

$$\mu(h(x))=(1+px)\mu(x) \quad\text{and}\quad \mu(0)=1$$
Suppose we know the value of $\mu(x_0)$ for some particular $x_0$. This tells us the value as well for $x_1 = h(x_0)$ and that tells us the value for $x_2 = h(x_1)$, etc. This gives us a nice sequence of known values. And you can also turn it around and find the value for any $x_{-1}$ with $h(x_{-1}) = x_0$. But even when you've traced out all the possible values derivable in this fashion, you only have a countable number of known points. Practically all of $\Bbb R$ is still undefined. You can choose a new $x_0'$, choose a value of $\mu(x_0')$ and do it all again. But there is nothing that demands the value for $\mu(x_0')$ behave nicely with the values derived from $x_0$. And you have to make uncountably many such choices before you've defined $\mu$ for all $\Bbb R$.
This is why functional equations generally specify that the function must be continuous. Then you can hope that the sequence $\{x_n\}$ is dense somewhere. Then you would have a shot at least of getting analycity.
The implicit function $\mu(\sqrt[m]{1-\Big(\frac{1-x}{1+px}\Big)^m})=(1+px)\mu(x)$ arises from defining $\mu$ as an iteration. $$\mu(x)=\overset{\text{lim}}{_{n\rightarrow\infty}} a_n, b_n$$ $$a_0:=1 \quad b_0:=\frac{1-x}{1+px}$$ $$a_{n+1}:=\frac{a_n+pb_n}{1+p}\quad\quad b_{n+1}=\sqrt[m]{a_{n+1}^m-(a_{n+1}-b_n)^m}$$ Note that both the definition by implicit formula and the definition by iteration necessarily determine a unique Taylor series expansion of $\mu$. Analyticity may be established in a way far easier than using the implicit formula above. We merely change the iteration of numbers to be an iteration of functions $$\mu(x)=\overset{\text{lim}}{_{n\rightarrow\infty}} a_n(x), b_n(x)$$ $$a_0(x):=1 \quad b_0(x):=\frac{1-x}{1+px}$$ $$a_{n+1}(x):=\frac{a_n(x)+pb_n(x)}{1+p}\quad\quad b_{n+1}(x)=\sqrt[m]{a_{n+1}(x)^m-(a_{n+1}(x)-b_n(x))^m}$$ $a_0$ and $b_0$ are clearly analytic. Inductively, each $a_n$ and $b_n$ can further be shown analytic. Thus $\mu$ is also analytic on any domain in which $a_n(x)$ and $b_n(x)$ converge to a common limit.