Is $\lim_{z\to0} \exp_{\sqrt{2}}^{\circ z}(\xi)$ continuous?

This question has arisen in a bunch of my research, to the side of my research actually, I keep on getting curious about how it should be answered. I'll frame it in an anachronistic sense, but the question is more general. I'm a firm believer that most general problems are solved by solving specific instances. Especially the most basic instances, they always seem to shed great insight at least.

We're going to look at $f(\xi) = \sqrt{2}^\xi$. This function has a bunch of nice properties and expresses how the general case behaves (at least the cases I'm interested in). We are dealing with the exponential, that makes a few of the questions straight forward, too.

Let $A$ be the immediate basin of attraction of $f(\xi)=\sqrt{2}^\xi$ about the fixed point $2$. That is, $f(2) = 2$ is a fixed point of $f$. If $\xi \in A$ then

$$f^{\circ n}(\xi) = (f \circ f \circ ...(n\,times)...\circ f)(\xi) =f(f(...(n\,times)...f(\xi))) \to 2\,\,\text{as}\,\, n \to \infty$$

Additionally, $A$ is connected and $2 \in A$, where further $A$ is the maximal of all sets to satisfy these properties. This uniquely defines the immediate basin of attraction.

Now it is no obvious fact, but it is possible to show there exists a unique holomorphic function in two variables $f^{\circ z}(\xi)$ such that

$$f^{\circ z}(\xi): \mathbb{C}_{\Re(z) > 0} \times A \to A$$

$$f^{\circ z_0}(f^{\circ z_1}(\xi)) = f^{\circ z_0 + z_1}(\xi)$$

$$f^{\circ 1}(\xi) = f(\xi)$$

$$f^{\circ z + \frac{2\pi i}{\log\log2}} = f^{\circ z}$$

This is what I like to call a complex iteration of the function $f$. Despite being a fractional iteration, I shy from the term 'fractional iteration' because I'm especially interested in complex behaviour. One should note there are many possible complex iterations of $f$, but this is the only one that has a purely imaginary period. As in $f^{\circ z}$ is multivalued but we are choosing the branch of $f^{\circ z}$ such that it has a purely imaginary period. This is very similar to the fact there is a bunch of branches of $e^z$ but only one that has a purely imaginary period, the principal branch.

We can write this function using a Newton series expansion, or a Mellin transform. It is a little cumbersome to show this unique solution can be expressed in this manner, but it is possible. Either way it gives a nice formula and makes the question more tangible. Here it is:

$$f^{\circ z}(\xi) = \sum_{n=0}^\infty \sum_{k=0}^n(-1)^{n-k} \dbinom{z-1}{n}\dbinom{n}{k}f^{\circ k+1}(\xi)$$

and for $0 < \Re(z) < 1$

$$f^{\circ z}(\xi)\Gamma(1-z) = \int_0^\infty \big{(}\sum_{n=0}^\infty f^{\circ n+1}(\xi)\frac{(-w)^n}{n!}\big{)}w^{-z}\,dw$$

if that perhaps helps.

Locally, for $|\xi -2|<\delta$, our complex iteration can be described rather simply. That is to say $$f^{\circ z}(\xi) = \Psi^{-1} \circ \log(2)^z \circ \Psi= \Psi^{-1}(\log(2)^z\Psi(\xi))$$

where $\Psi$ is the Schroder function about $2$. That is, $\Psi$ is the unique function in which $\Psi(f(\xi)) = \log(2)\Psi(\xi)$ and $\Psi'(2) = 1$. This is an important detail. This means the complex iteration is conjugate to $\log(2)^z$ locally around $2$.

What makes the following problem non-trivial, is that this complex iteration of $f(\xi)$ also has the periodicity that $f$ has in $\xi$. Namely just as $f(\xi + \frac{4 \pi i}{\log(2)}) = f(\xi)$ we get $f^{\circ z}(\xi + \frac{4 \pi i}{\log(2)}) = f^{\circ z}(\xi)$. This is apparent from its Newton series expansion. This is essentially what forms the question I'm asking.

This question entirely revolves around the following limit, what I call the null-limit of $f^{\circ z}$. The limit is

$$\lim_{z \to 0} f^{\circ z}(\xi) = G(\xi)$$

and the question addresses why it seems so mysterious to describe $G$. $G$ is the null-limit of the iteration, it is not easy, but I can prove the limit exists pointwise in $\xi$.

Here's why it is interesting and what there is to care about. Despite initial assumptions $G(\xi) \neq \xi$. Thinking that since $f^{\circ 0}(f^{\circ z}) = f^{\circ 0+z} = f^{\circ z}$ implies that $G(\xi)$ must be the identity function is a false conclusion. Here's why.

$f^{\circ z}(\xi + \frac{4\pi i}{\log(2)}) = f^{\circ z}(\xi)$

This screws everything up. This makes the whole function act all wonky. It is not hard to show for $|\xi - 2| < \delta$ that $G(\xi) = \xi$. This is because

$$G(\xi) = \lim_{z\to0}\Psi^{-1}(\log(2)^{z}\Psi(\xi)) = \Psi^{-1}(\Psi(\xi))= \xi$$

for $|\xi - 2| < \delta$, where $\Psi$ is the Schroder function of $f$ and in a small enough neighborhood about $2$ the limit can be pulled through. However, by the same argument $G(\xi + \frac{4k\pi i}{\log(2)}) = \xi$ for all $k \in \mathbb{Z}$. This is because of the periodicity of $f^{\circ z}(\cdot)$. So now we have a whole bunch of disks centered about $2 + 4k \pi i/\log(2)$ where $G$ sends all of these disks to a disk about $2$. It obviously can't be the identity, it can't even be holomorphic on $A$.

What makes $G$ even more absurd is that

$$f(G(\xi)) = f(\lim_{z \to 0} f^{\circ z}(\xi)) = \lim_{z\to 0} f(f^{\circ z}(\xi)) = \lim_{z \to 0}f^{\circ z+1}(\xi) = f(\xi)$$

by the holomorphy of $f^{\circ z+1}$ about $z=0$ and the continuity of $f$. This seems to suggest that $G$ actually does behave like a right identity function... on $f$ least of all. This, again, implies $G$ cannot be holomorphic on $A$. It is going to look a little wonky. Therefore understanding $G(\xi)$ on a larger domain must be more subtle. It obviously can't be analytic, but is it at least continuous?

Is $G(\xi)$ continuous on $A$?

The second question is based on related evidence and not on clear fact. I want to further understand how $G$ behaves. I'm assuming for this next question that $G$ is in fact continuous. Is it true that at most points $G(\xi)$ is a linear function, and that at some points we have a discontinuity in the first derivative? To explain this statement it helps to look at the fractional iteration of $\sin$.

It is well known $\sin^{\circ t}(x) \to T(x)$ as $t \to 0$, where $x \in \mathbb{R}$ and $T(x)$ is a triangle wave with period $2\pi$. We are taking the analytic solution to $\sin^{\circ t}$ that is real valued. On $- \pi \le x \le \pi$ the function $T(x)$ looks like the following

\begin{eqnarray*} T(x) &=& -\pi - x\,\,\,&\text{for}&\,\,\,-\pi \le x \le -\pi/2\\ &=& x\,\,\,&\text{for}&\,\,\,-\pi/2 \le x \le \pi/2\\ &=& \pi - x \,\,\, &\text{for}&\,\,\,\pi/2 \le x \le \pi\\ \end{eqnarray*}

$T(x)$ is linear at most points, with discontinuities in the first derivative at $\pi/2 + k \pi$.

$\sin^{\circ t}(x) \to T(x)$ is the standard example of a fractional iteration that does not tend to the identity as the iteration tends to zero. But still, it kind of looks like the identity. $\sin(T(x)) = \sin(x)$. Plus, least of all, it is linear almost everywhere. This leads me to ask, in a more sophisticated language

If $G(\xi)$ is continuous on $A$, is it complex-differentiable and linear almost everywhere for $\xi \in A$?

I mean to ask if $G(\xi(x,y))$ satisfies the Cauchy-Riemann equations for $(x,y) \in A \subset \mathbb{R}^2$ almost everywhere under the $\mathbb{R}^2$ Lebesgue measure, and if $\frac{\partial G}{\partial x} = \text{Constant}$ and $\frac{\partial G}{\partial y} = \text{Constant}$ almost everywhere.

This would imply that $G(\xi)$ is a weird triangle wave in its own right. Sadly though, comparing the situation to $\sin$ again, the discontinuities of the derivative of $T(x)$ are at the critical points of $\sin(x)$. This makes things a bit easier to classify. $T'(x)$ is discontinuous when $\sin'(x) = 0$. Unfortunately $f(x)$ has no critical points, $f'(x) \neq 0$, so we don't get something as simple. Forcing us to ask, if $G$ is continuous and differentiable outside a set of measure zero with no limit points,

Where are the discontinuities in the first derivative of $G(\xi)$? How do we classify them?

I feel like I have all the pieces to put this together, but I'm probably missing something. The first problem depends on $f^{\circ z}(\xi(x,y)) = u(z,x,y) + i v(z,x,y)$. Namely how can we show that $u(z,x,y_0)$ converges uniformly in $x$ as $z \to 0$ when $y_0$ is fixed? And how can we show $u(z,x_0,y)$ converges uniformly in $y$ as $z \to 0$? How do we do the same thing with $v$? Since I usually deal with holomorphic functions as a whole and not their individual real parts, this has got me a little stooped.

Hopefully, I'm not missing something obvious here!

• Do we know: if $\xi \in A$, then also $\xi+4\pi i/\log(2) \in A$ – Gerald Edgar Jul 10 '17 at 21:00
• Sadly, I've never seen a nice picture of the Fatou set of $f$ (is there some app to produce your own?) so it's hard to even get intuition. I've never found a proof detailing this fact. But you may be on to something here. Maybe if $\xi \in A$ then necessarily $\xi + 4k\pi i/\log(2) \not\in A$ for $k \neq 0$ and in fact $G(\xi) = \xi$ for $\xi \in A$. Any anomalies get wiped out because $f$ would be univalent on $A$. I kind of just thought this too good to be true, if I'm being frank. – user78249 Jul 10 '17 at 22:19
• I did some numerical computations. As far as I could tell from the pictures, $f^{\circ k}(\xi) \to 2$ for all complex numbers $\xi$. Well, I know that is wrong since $f(4) = 4$. But I guess the pictures suggest the set with $f^{\circ k}(\xi) \to 2$ is dense in $\mathbb C$. And that doesn't tell us anything about $A$. – Gerald Edgar Jul 11 '17 at 13:42
• @GeraldEdgar, that sounds right. I think one can show using a theorem in Milnor's Complex Dynamics that it's dense--I'll pick up the book and double check. The trouble is $A$ is somewhat more mysterious. I've been mostly dealing with dynamics on the real line, but for complex things get trickier to visualize. I have been drawing mental comparisons between $\sin^{\circ t}$ on the real line and the such (both periodic, might have something in common). Also, IFF $G\Big{|}_{A} = \text{Id}\Big{|}_{A}$, then $A$ has only one set of periods in it, which describes it a bit more – user78249 Jul 12 '17 at 0:42
• Thinking about it, maybe if this result is open, one should actually prove $G(\xi) = \xi$ to prove if $\xi \in A$ then $\xi + 4k\pi i/\log(2) \not \in A$. Because I have a quasiproof which may actually give this result; I just assumed it wrong because I never considered this consequence a possibility (too good to be true). – user78249 Jul 12 '17 at 7:27