*2017 Dec 8*

In previous posts, I've written about finding formulas for $f^{\circ n}$ in terms of $n$, but I've considered iterates of $f$ only for integer values of $n$. I will now discuss an extension of the definition of $f^{\circ n}$ to include rational non-integer values of $n$.

Find half-iterates of the functions and find a continuous half-iterate of the function

Recall that function iteration has the properties
and
The second property might lead us to say that
and so we shall define $f^{\circ 1/n}$ as any function satisfying
Notice that I say *any* function, as such functions are not always unique. For example, consider $\text{id}*{\mathbb R}^{\circ 1/2}$. The functions $\text{id}*{\mathbb R}$ and $-\text{id}*{\mathbb R}$ iterated twice both yield $\text{id}*{\mathbb R}$, providing an example of how half-iterates are not always unique. Thus we may think of the fractional iterate $f^{\circ 1/n}$ not as a function, but as as *set of functions*, each of which, when iterated $n$ times, yield the function $f$. Using our previous example, we might say that
and

First I will tackle the problem of finding all continuous functions in the set $\text{id}*{\mathbb R}^{\circ 1/2}$ - or the set of continuous involutions on the reals. Notice that I say continuous - in this post, I will not deal with any non-continuous function, since they can get pretty messy. We've already noted that $-\text{id}*{\mathbb R}$ is in this set. With a bit of further analysis, one may observe that all functions in the form
satisfy it as well, if $g:\mathbb R\mapsto\mathbb R$.

But is this set of solutions exhaustive? That is, if $f:\mathbb R\mapsto\mathbb R$ satisfies $f^{\circ 2}=\text{id}_{\mathbb R}$, does there necessarily exist a function $g:\mathbb R\mapsto\mathbb R$ such that $f(x)=g^{-1}(-g(x))$?

The answer is yes, but proving it will take a little bit of work. First we must prove a property that all half-iterates of the identity function share.

Notice first that if a function $f$ is non-injective, then the function $f^{\circ 2}$ is also non-injective. Thus, if $f$ is a half-iterate of the identity function, since the identity function is injective, $f$ must also be injective. Continuous and injective functions are always either increasing or decreasing, implying that $f$ is either increasing or decreasing. So for now, suppose that $f$ is increasing.

Suppose that $f(x)\gt x$ for some $x$. Then, since $f$ is increasing, $f^{\circ 2}(x)\gt f(x)$ and, since $f^{\circ 2}=\text{id}*{\mathbb R}$, we have $x\lt f(x)$. Contradiction! If we instead suppose that $f(x)\lt x$ for some $x$, we arrive at a similar contradiction. Thus we have that if $f$ is increasing, than $f$ cannot be greater than or less than its input for any input, meaning that $f=\text{id}*{\mathbb R}$. Thus, the only continuous increasing function in $\text{id}_{\mathbb R}^{\circ 1/2}$ is the identity function.

This tells us that the only continuous increasing involution on the reals is the identity function, meaning that all others must be *decreasing*. This is the property that I wanted to prove from the beginning. Since all such functions are decreasing, they must all have *exactly one fixed point*. If $f^{\circ 2}=\text{id}*{\mathbb R}$ and $f$ is continuous and decreasing, I will call its fixed point $\phi$. Let us now define
$$g(x)=$$
for any continuous, injective function $h$ with $h(\phi)=0$. From here, it is trivial to verify that both $g$ is continuous and that $g^{-1}(-g(x))=f(x)$, showing that for any function $f$ in $\text{id}*{\mathbb R}^{\circ 1/2}$, there exists a continuous function $g$ such that $g^{-1}(-g(x))=f(x)$, thus showing that the set of functions in the form $g^{-1}(-g(x))$ for continuous and injective $g$ provides an exhaustive set of continuous involutions on the real numbers.

Next I will consider continuous $\text{id}*{\mathbb R}^{\circ 1/3}$ - which, thankfully, is a much easier problem. Again, the key is to notice that any continuous function $f\in\text{id}*{\mathbb R}^{\circ 1/3}$ *must* be injective, and thus either increasing or decreasing. If $f$ is increasing, then we may use an argument similar to the one used previously to show that $f=\text{id}*{\mathbb R}$. If $f$ is decreasing, then for all $x,y\in\mathbb R$,
and so
and
making $f^{\circ 3}$ decreasing, which contradicts the fact that $f^{\circ 3}=\text{id}*{\mathbb R}$. Thus the *only* continuous function in $\text{id}_{\mathbb R}^{\circ 1/3}$ is the identity function.

One can use decreasingness to prove the nonexistence of continuous half-iterates of many other functions, even if they aren't strictly decreasing. For example, suppose a differentiable function $f$ has a fixed point at $x=\phi$ such that $f'(\phi)$ is less than zero. Then, for some neighborhood $N$ of $\phi$, $f$ must be decreasing, since its derivative at $x=\phi$ is negative. If we now assume that a continuous $h\in f^{\circ 1/2}$ exists, we may state immediately that it is injective in $N$, since $f$ is injective in $N$. Thus it is either increasing or decreasing in $N$. But h being increasing or decreasing in $N$ *both* imply that $h^{\circ 2}=f$ is increasing in $N$, leading to a contradiction. This method can be used, for example, to show that the cosine function has no continuous half-iterate, since it is decreasing at its fixed point.

So far, all I've done is *disprove* the existence of half-iterates of functions. Now I will demonstrate how one might *construct* a specific half-iterate of a function.

Some functions have nice and neat closed-form half-iterates. For example, consider the function Because we can rewrite this as we can easily find the following half-iterate $h\in a^{\circ 1/2}$:

Some functions have half-iterates that are a bit harder to find. Consider for example the function Some half-iterates of this function can be defined as piecewise functions. An example is $h_1,h_2\in b^{\circ 1/2}$:

Sometimes, piecewise construction of half-iterates can get a little bit wild. Before I demonstrate a construction of a half-iterate of the exponential function, I must explain the graphical construction technique that I used for it. Imagine that $f:\mathbb R\mapsto\mathbb R$ is a function with half-iterate $h:\mathbb R\mapsto\mathbb R$, and suppose we know the value of $h(x_0)$ for some $x_0$. Let $h_0$ be this value. Then we have that which implies or and we know another value of $h$! But we can take this even further and say that and so on. Let's think of this information in terms of the graph of $y=h(x)$. We were given initially that the point $(x_0,h_0)$ was on the graph, and from this, we determined that the points $(h_0,f(x_0))$ and $(f(x_0),f(h_0))$ were also on the graph. In general, if the point $(a,b)$ is on the graph of $y=h(x)$, then the point $(b,f(a))$ is also on the graph. Visually, this means that if we assume some point $P$ is on the graph of $y=h(x)$, we may use the following steps to find another point $P'$ on the graph:

- Plot the point $P$ and graph the curves $y=f(x)$ and $y=x$.
- Draw the horizontal line segment $PQ$, where $Q$ is on the graph of $y=x$.
- Draw the vertical line seqment $PR$, where $R$ is on the graph of $y=f(x)$.
- Construct rectangle $PQP'R$. Then the point $P'$ is also on the graph of $y=h(x)$.

Here is a picture illustrating this process:

Though it may sound complicated, this process is as easy as drawing a rectangle, and it can lead to some interesting results. For example, consider the function $f$ in my example graph (it does not matter exactly what function it is). By assuming that the point $P$ was on the graph of $y=h(x)$, we found the point $P'$ also on $y=h(x)$. Notice that $P$ lies *below* the curve $y=f(x)$ and $P'$ lies above it. Thus, if we assume $h$ to be continuous, then by the intermediate value theorem, the curves $y=f(x)$ and $y=h(x)$ must intersect at some $x$-value between the x-values of $P$ and $P'$. Thus, for some $x'$ in that interval,
or
implying that $h$ has a fixed point at $x=h(x')$, and thus that $f$ also has this fixed point. But since $f$ only has one fixed point between the y-values of $P$ and $P'$ (call it $\phi$), then
This could go on - for example, we could keep drawing rectangles to find $P''$, and $P'''$, et cetera.

This method can be used to prove the following interesting result:

If a function $f:\mathbb R\mapsto\mathbb R$ has no fixed points and has a continuous half-iterate $h:\mathbb R\mapsto\mathbb R$, then $h(x)$ is between $x$ and $f(x)$ for all $x\in\mathbb R$.

This is a rather nice result, but the proof is unpleasant and uses a lot of casework, so I will omit it.

Now I will lay out my construction of a half-exponential function, as written in one of my answers on MSE:

Define a sequence $a_n$ as follows:
and define a sequence of functions $H_n$ as follows:
Then one may verify rather simply that the function
or
is a half-iterate of $e^x$. Amazingly, it can be proven by induction that this function is *continuous* and *differentiable everywhere*. Unfortunately, it is not twice-differentiable. Here is a graph of this function (in orange) alongside a graph of $y=e^x$ (in purple):

One might wonder if there exists a half-iterate of $e^x$ that is *infinitely differentiable* as $e^x$ is... but that topic will have to be left for a later blog post.