36, 37
fresh_42 said:
I follow you until the epsilontic starts with "Suppose". My questions are:
What is the idea behind the construction, esp. why distinguish the two cases and where does the asymmetry in ##s## come from?
Where do we get the existence of ##f_{\frac{1}{2}}## from, since you start with the necessary condition?
How does uniqueness follow from a recursive construction? My guess is ##f_{\frac{1}{2}}(x) = \lim_{\varepsilon \to x} \displaystyle{\int_0^\varepsilon} \ldots dt ## but we don't know anything about ##f_{-\frac{1}{2}}## on ##[0,\varepsilon]##.
Edit: I tried to follow your argumentation on ##f(x)=4x^3-6x^2+3x \in G_+##.
Apologies for the more-than-somewhat incomplete post. The idea is to construct ##f_{\frac{1}{2}}(x)## from a given ##f(x)## by solving the 'differential equation' ## f_\frac{1}{2}'(f_\frac{1}{2}(x))f_\frac{1}{2}'(x)=f'(x)## on ##[0,1]##, or whatever the existence interval turns out to be, starting at ##x=0##. The interesting part is that the left hand side depends on derivatives of ##f_\frac{1}{2}## evaluated at two points that are typically different (##f_\frac{1}{2}(x)## and ##x##), and one of these, namely ##f_\frac{1}{2}(x)##, could be outside the interval ##[0,\epsilon]## where ##f_\frac{1}{2}## has been defined. To get around this, one evaluates the derivative of ##f_\frac{1}{2}## at the rightmost edge in two different ways, depending on whether ##f_\frac{1}{2}(\epsilon)## is greater or less than ##\epsilon##. If ##f_\frac{1}{2}(\epsilon)\leq\epsilon##, set ##f_\frac{1}{2}'(\epsilon)=\frac{f'(\epsilon)}{f_\frac{1}{2}'(f_\frac{1}{2}(\epsilon))}##. Otherwise, by continuity (and positivity of ##f_\frac{1}{2}'(x)## on ##[0,\epsilon]##) there must exist a point ##s\in[0,\epsilon]## where ##f_\frac{1}{2}(s)=\epsilon##. ##f_\frac{1}{2}'(\epsilon)## can then be expressed in terms of predefined quantities by evaluating the differential equation at ##s## instead of ##\epsilon##. Incidentally, ##s(\epsilon)## is just the inverse of ##f_\frac{1}{2}## evaluated at ##\epsilon##: if ##f_\frac{1}{2}(x)## is defined on ##[0,\epsilon]##, then ##f_{-\frac{1}{2}}(x)## is defined on ##[0,f_\frac{1}{2}(\epsilon)]## (which includes ##[0,\epsilon]## in the case where ##f_{-\frac{1}{2}}(x)## is needed.)
For your example (the function ##4x^3-6x^2+3x##), one starts with ##f_\frac{1}{2}(0)=0##, ##f_\frac{1}{2}'(0)=\sqrt{f'(0)}=\sqrt{3}>1##, so in a neighborhood of ##0## we expect ##f_\frac{1}{2}(x)>x##, and the relevant equation (at least up to the first point where ##f_\frac{1}{2}(x)=x##) is the one involving ##f_{-\frac{1}{2}}(x)##, which can be used to compute ##f_\frac{1}{2}## using e.g. Euler's method. An especially tricky case is ##f(x)=1+\frac{1}{2}\arcsin\Big(\pi(x-\frac{1}{2})\Big)## (or ##(f(x))^n##), whose derivative diverges at ##x=0##.
Here's another approach to establishing conditions for uniqueness of the 'square root', which has the added benefit of elucidating how ambiguities can arise depending on what amount of regularity is assumed. Suppose ##h## and ##p## are both in ##G_+##, and that ##h(h(x))=p(p(x))## for all ##x\in[0,1]##. Note that if ##h(h(x))=x##, then ##h(x)=x## (and ##p(x)=x##), and if ##{x_j}_{j\in\Lambda}## is the (ordered) set of all fixed points of ##h^2=p^2##, then ##h(x)## and ##p(x)## can be decomposed into self-maps of the intervals ##[x_j,x_{j+1}]##, which themselves can be identified with elements of ##G_+##. So without loss of generality, we can assume that ##h^2(x)>x## and ##p^2(x)>x##.
Prop: under the conditions mentioned above, ##h(x)>x##, ##p(x)>x##, and there exists at least one point ##x_0\in(0,1)## where ##h(x_0)=p(x_0)##.
Proof:
(By contradiction.) First, suppose that ##h(x)<x## for some ##x##. Then ##h(h(x))>x>h(x)##, and so ##h(x)## is decreasing somewhere on ##[h(x),x]##. We know, however, that ##h\in G_+## (##\Rightarrow \Leftarrow##.)
Now, suppose that no solution to ##h(x)=p(x)## exists, apart from ##x=0## and ##x=1##. Then without loss of generality, assume ##h(x)>p(x)## on ##(0,1)##. However, if ##h(s)>p(s)## for all ##s\in(0,1)##, and ##h(h(x))=p(p(x))##, then ##h(p(x))>p(p(x))=h(h(x))##, which would imply that ##h## is decreasing on ##[p(x),h(x)]## and thus not in ##G_+## (##\Rightarrow\Leftarrow##.)
Hence, there must be at least one nontrivial solution to ##h(x)=p(x)##, and this generates a countable set of additional solutions from the orbit under powers of ##h(x)## (or ##p(x)##) and its inverse.
Prop: Let ##s_{j}##, ##j\in\mathbb Z##, be the orbit described above. Then the functions ##h(x)## and ##p(x)## can each be associated with a countable set of maps in ##G_+##, ##h\rightarrow \{f_j\}_{j\in\mathbb Z}##, ##p\rightarrow\{g_j\}_{j\in\mathbb Z}##, where ##f_j(x) = \frac{1}{h(s_{j+1})-h(s_j)}\Big(h((s_{j+1}-s_j)x+s_j)-h(s_j)\Big)## and ##g_j(x)## are defined analogously. The functions ##f_j(x)## and ##g_j(x)## are related by ##f_{j+1}f_j=g_{j+1}g_j## for all ##j\in\mathbb Z##.
Proof:
This is just another way of expressing the fact that ##h(h(x))=p(p(x))## and that both ##h## and ##p## map ##[s_j,s_{j+1})## bijectively to ##[s_{j+1},s_{j+2})##.
The above proposition gives insight into the extent to which ##p(x)## can differ from ##h(x)##.
Prop: Given ##f_j## and ##g_j##, ##j\in\mathbb Z## as above, let ##C_j=f_{j-1} f_{j-2} \cdots f_1 f_0## for ##j>0##. Then for ##i>0##,
$$
g_i=\begin{cases}
f_i(C_i(g_0^{-1}f_0)C_i^{-1}),\quad i\text{ odd}\\
f_i(C_i(f_0^{-1}g_0) C_i^{-1}),\quad i\text{ even}
\end{cases}
$$
and for ##i<0##, letting ##D_j=f_0f_{-1}\cdots f_{j+1}##,
$$
g_i=\begin{cases}
(D_i^{-1}(f_0g_0^{-1})D_i )f_i\quad i\text{ odd}\\
(D_i^{-1}(g_0f_0^{-1})D_i )f_i\quad i\text{ even}
\end{cases}
$$
Proof:
(By induction, applying ##g_{j+1}=f_{j+1}f_jg_j^{-1}##.)
Note that ##g_i=f_i## (and ##p=h##) if and only if ##g_0f_0^{-1}=\mathbb I(x)##. This is probably at a point where I think I can begin referring to published literature, but due to my ignorance I shall stumble onward.
First, an example might be helpful. Consider the function
$$
h(x)=\begin{cases}
\frac{3}{2}x,\quad x<\frac{1}{2},\\
\frac{1}{2}x+\frac{1}{2},\quad x\geq \frac{1}{2}.
\end{cases}
$$
and orbits of the point ##x_0=\frac{1}{2}##. In this case, the functions ##f_j## are all equal to ##\mathbb{I}(x)##, and the orbits partition ##[0,1]## into two collections of exponentially shrinking subintervals.
Let's suppose (adhering to the principle of least action) that ##g_0=h##, so that ##g_i=h## if ##i## is even and ##g_i=h^{-1}## for ##i## odd. The associated graph of ##p(x)## has a somewhat irregular fractal shape, but clearly ##p^2=h^2## and ##p\neq h##. Moreover, ##p(x)## is locally as well-behaved as ##h(x)## on the interior ##(0,1)##, apart from having an infinite number of cusps that accumulate near the end points, and has a bounded first derivative. So boundary conditions are essential to uniqueness.
If we require that functions have continuous first derivatives everywhere, then uniqueness follows from the observation that the intervals ##[s_i,s_{i+1})## shrink to zero width near end-points. Continuity of first derivatives in ##h## then implies ##f_j## converge to ##\mathbb I## as ##j\rightarrow \pm\infty##, so any bumps in ##f_mg_m^{-1}##, with ##[s_m,s_{m+1})## near an end-point, are conserved in ##g_j## for arbitrarily large ##|j|##, causing (lower-bounded) oscillations in the derivative over vanishingly small intervals, in contradiction to the ##C^1## condition. (The extent to which regularity can be weakened at all from ##C^1## is TBD.)
) that it be adopted as retroactive; viz, the constraint that all terms or words used be absent of names of persons, and instead be named after by-reason-intellegible characteristics, so that. for example, the term 'non-Abelian' does not occur -- I don't have an agenda against crediting Prof. Abel, who contributed so much of great worth; however, one can understand the term 'non-commutative' without having to learn much about who discovered or invented what.
