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$G := \{f : f:[0,1] \rightarrow [0,1]$ such that it is bijective function and strictly increasing }

Now the question is

  1. For any $ h \in G,$does there exist $g \in G$ such that $h=g \circ g $?

  2. Is such a $g$, if it exist , unique?

My observation :

  • $G$ is a group under function composition.(Is it helpful?)
  • Every function in $G$ is continuous.

Conjecture: if $h \in G$ has $n \in \mathbb{N}$ fixed points in (0,1) then it has $n+1$ 'square root' functions. Please help me to solve the question!

Akshay Hegde
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  • Maybe this could help in your analysis : https://en.wikipedia.org/wiki/Newton%27s_method –  Oct 08 '16 at 11:21
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    Functional square roots have been discussed before, here and at MO. http://math.stackexchange.com/questions/3633/square-root-of-a-function-in-the-sense-of-composition has some information, and also links to other discussions. See also the "Related" questions listed on the current page. – Gerry Myerson Oct 08 '16 at 11:56
  • Since its a homework problem for undergraduates I don't think it's that difficult....I saw the link... They are discussing much general problem... This is a very special case....Please note that its not a duplicate... – Akshay Hegde Oct 08 '16 at 14:09
  • All of your functions have $h(0)=0$ and $h(1)=1$, so there are always at least two fixed points. So, do you conjecture there are always at least 3 square root functions? Can you give an example of your conjecture, say, for $h(x) = x^{1/2}$? One square root of this is $g(x) = x^{\sqrt{1/2}}$, are there two more? – Michael Oct 12 '16 at 21:19
  • One minor observation: Suppose $h,g \in G$ are functions such that $h(x)=g(g(x))$ for all $x \in [0,1]$. Then if $y$ is a point in $[0,1]$ such that $h(y)=y$, then $g(y)=y$. In particular, the function $h(x)=x$ is in $G$ and has an infinite number of fixed points, but only one square root function $g(x)=x$. – Michael Oct 12 '16 at 21:43
  • @Michael Edited the question – Akshay Hegde Oct 14 '16 at 07:51
  • The edited conjecture is still false. Suppose the number of solutions $\mathcal{N}(h)$ is determined by the number $n$ of fixed points of $h$ by $\mathcal{N}(h)=f(n)$.In each interval with endpoints two adjacent fixed points, $h$ has no fixed points, and the number of solutions in this interval is $f(0)$. So we should get $f(n)=f(0)^n$ instead. – Cave Johnson Oct 14 '16 at 13:15

2 Answers2

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The square root always exists and, except for the special case where $f$ is the identity, there are uncountably many square roots.

Case I: $f$ has no fixed points in $(0,1)$. By the intermediate value theorem, we either have

  1. $f(x) > x$ for all $x\in(0,1)$ or,
  2. $f(x) < x$ for all $x\in(0,1)$.

I will consider (1) (the situation with $f(x) < x$ can be handled similarly). Choose any $x_0\in(0,1)$ and set $x_k=f^k(x_0)$, which is a strictly increasing sequence over $k\in\mathbb{Z}$. The limits of $x_k$ as $k\to\pm\infty$ are fixed points of $f$, so are equal to $1$ and $0$ respectively.

Now, choose any increasing homeomorphism $\theta\colon[0,1]\to[x_0,x_1]$. Extend $\theta$ to all of $\mathbb{R}$ by setting $$ \theta(k+x)=f^k(\theta(x)) $$ for $k\in\mathbb{Z}$ and $x\in[0,1)$. This defines a homeomorphism from $\mathbb{R}$ to $(0,1)$. Furthermore, $$ f(\theta(x))=\theta(x+1). $$ We can define a square root by $$ g(x) = \theta(\theta^{-1}(x)+1/2) $$ for $x\in(0,1)$, and $g(0)=0$, $g(1)=1$. Note that $g(x_0)=\theta(1/2)$, which can be chosen to be any value in $(x_0,x_1)$, so there are infinitely many square roots.

Case II: Now, for the general case.

Let $S\subseteq[0,1]$ be the set of fixed points of $f$. We define the square root to also be the identity on $S$. As $S$ is closed, its complement is a countable union of disjoint open intervals $(a,b)$ and, restricted to each such interval, $f$ gives a homeomorphism of $[a,b]$ with no fixed points in $(a,b)$. So, applying the construction above, there are uncountably many square roots on each such interval. So, $f$ has a square root and, except in the case where $S$ is all of $[0,1]$, there are uncountably many square roots.

George Lowther
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  • I doubt this answer is correct but have no time to deal with it, I think there is one root only. Not hard to prove $h'(0)=(g'(0))^2$, kinda smells of uniqueness – Djura Marinkov Oct 14 '16 at 17:50
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    @DjuraMarinkov : These square roots are not necessarily differentiable at the "switching points", and there are infinitely many switching points densely packed about $x=0$ and $x=1$. – Michael Oct 14 '16 at 19:57
  • The argument with the derivatives implies that there is at most one analytic square root, but there will be many non-analytic (and, possibly, non-differentiable) square roots. – George Lowther Oct 14 '16 at 22:46
  • Yes, you are all absolutely right. And the answer is great too – Djura Marinkov Oct 15 '16 at 07:25
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Basic details on my comment above:

Claim:

Suppose $g:[0,1]\rightarrow [0,1]$ is an increasing function. Let $y \in [0,1]$ be a point that satisfies $g(g(y))=y$. Then $g(y)=y$.

Proof:

Suppose $y<g(y)$. Since $g$ is increasing we conclude $$ g(y) < g(g(y))=y$$ a contradiction. A similar contradiction holds if $y>g(y)$. Hence $y=g(y)$. $\Box$

It follows that fixed points of $h$ correspond to fixed points of $g$.

In particular, if $h$ is the identity function $h(x)=x$ for all $x\in[0,1]$, then its only increasing square root is also the identity function $g(x)=x$ for all $x \in [0,1]$. In this example, $h$ has an infinite number of fixed points, but only one (increasing) square root function. This example seems to disprove the conjecture given in the question.

Michael
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