Stuck on this proof that $ord(f) = ord(g)$

Question

Let $f, g: \mathbb R \to \mathbb R$ be smooth maps such that $f(a) = g(a') = 0$ and let $\tau, \sigma : \mathbb R \to \mathbb R$ be diffeomorphisms such that

$$ \tau \circ f = g \circ \sigma$$

Define $ord(f) = \min \{ n \in \mathbb N_{>0} \mid f^{(n)} (a) \neq 0\}$.

I am trying to show that $ord(f) = ord(g)$. I wanted to do induction but the induction step is broken since $ \tau \circ f = g \circ \sigma$ does not imply that there are diffeomorphisms $\phi, \pi$ such that

$ \phi \circ f' = g' \circ \pi$

Next I thought I could derive a closed form for the $n$-th derivative of $ \tau \circ f = g \circ \sigma$ but I can't see a pattern. The sums get rather long.

Now I am starting to wonder whether the thing I'm trying to show is even true. If it is I think the proof should be short and simple very unlike what I've tried so far.

Does anyone have any ideas on whether this is true and if it is on how to prove it?

Answer 1

Let's begin with properly defining the function ${\rm ord}$. Given a sufficiently differentiable function $f: \>{\mathbb R}\to{\mathbb R}$ one defines $${\rm ord\,}_a(f)=\inf\{n>0\>|\>f^{(n)}(a)\ne0\}\ .$$

Note that ${\rm ord}$ needs two inputs. In your question you introduced ${\rm ord\,}_a(f)$ but forgot to pin down the point $a_*$ at which ${\rm ord\,}_{a_*}(g)$ should be computed. This has to be the point $\sigma(a)$, because otherwise it is easy to give counterexamples, e.g., the following:

Assume $a'\ne a$ and put $$f(x):=(x-a)^2, \qquad g(x):=(x-a)^2-(a'-a)^2,\qquad \sigma:={\rm id},\qquad \tau (x):=x-(a'-a)^2\ .$$ Then one has $\tau\circ f=g\circ\sigma$, $f(a)=g(a')=0$, but $${\rm ord\,}_a(f)=2,\qquad {\rm ord\,}_{a'}(g)=1\ ,$$ since $g'(a')=2(a'-a)\ne0$.

(For this reason in my original answer I tacitly assumed $a'=\sigma(a)$ and then $a=a'=0$, which made $\sigma$ and $\tau$ diffeomorphisms with $\sigma(0)=\tau(0)=0$.)

Now the positive result: The function ${\rm ord}$ satisfies a sort of chain rule: $${\rm ord\,}_a(g\circ f)={\rm ord\,}_{f(a)}(g)\cdot{\rm ord\,}_a(f)\ .$$ Proof. We may assume $$a=f(a)=0,\qquad {\rm ord\,}_a(f)=p\geq1,\qquad {\rm ord\,}_{f(a)}(g)=q\geq1\ .$$ The following principle is an immediate consequence of Taylor's theorem: $${\rm ord\,}_0(f)=m<\infty\quad\Longleftrightarrow\quad f(x)=x^m\bigl(\alpha+r(x)\bigr)\quad {\rm with} \quad \alpha\ne0, \ \lim_{x\to0}r(x)=0\ .$$ So there are constants $\alpha\ne0$, $\beta\ne0$, and functions $x\mapsto r_1(x)$, $y\mapsto r_2(y)$ with $\lim_{x\to0}r_1(x)=\lim_{y\to0}r_2(y)=0$ such that $$f(x)=x^p\bigl(\alpha+r_1(x)\bigr),\qquad g(y)=y^q\bigl(\beta+r_2(y)\bigr)\ .$$ It follows that $$g\bigl( f(x)\bigr)=x^{pq}\bigl(\alpha+r_1(x)\bigr)^q\bigl(\beta+r_2\bigl(f(x)\bigr)\bigr)=x^{pq}\bigl(a^q\beta+r(x)\bigr)$$ for some function $x\mapsto r(x)$ with $\lim_{x\to0}r(x)=0.\qquad\square$

Since diffeomorphisms $\sigma:\>{\mathbb R}\to{\mathbb R}$ have ${\rm ord\,}_a(\sigma)=1$ at all points $a$ your original conjecture, interpreted correctly, follows.

Answer 2

First, you do not define $ord(g)$; I shall presume it to be like $ord(f)$ but with $a$ replaced by $a'$.

Second, your statement is true if and only if $\sigma(a) = a'$ (read the proof for $org(g) = 1$ to understand why).

Third, we shall make essential use of the following formula which generalizes Leibniz's formula:

$$(f_1 f_2 \dots f_n) ^{(p)} = \sum \limits _{i_1 + i_2 + \dots + i_n = p} \frac {p!} {i_1 ! i_2 ! \dots i_n !} f_1 ^{(i_1)} f_2 ^{(i_2)} \dots f_n ^{(i_n)}$$

(if you want, it is easy to prove it by induction).

Note that since $\tau$ is a diffeomorphism, we may invert it, getting $f = \tau ^{-1} \circ g \circ \sigma$.

Let us see what happens when $ord(g) = 1$ (i.e. $g'(a') \ne 0$). We have

$$f'(a) = (\tau ^{-1})' \circ g \circ \sigma (a) \cdot g' \circ \sigma (a) \cdot \sigma ' (a) = (\tau ^{-1})' (g(a')) \cdot g' (a') \cdot \sigma ' (a) = \\ (\tau ^{-1})' (0) \cdot g' (a') \cdot \sigma ' (a) .$$

Note that since $\tau ^{-1}$ and $\sigma$ are diffeomorphisms, their derivatives cannot cancel anywhere; since $g'(a') \ne 0$ by assumption, this gives $f'(a) \ne 0$, i.e. $ord(f) = 1$.

Now let us see the induction step: assume $ord(g) = n \ge 2$. We shall prove that $f^{(k)} (a) = 0 \; \forall k<n$ and $f^{(n)} (a) \ne 0$.

Note that

$$f^{(k)} (a) = (f')^{(k-1)} (a) = ((\tau ^{-1})' \circ g \circ \sigma \cdot g' \circ \sigma \cdot \sigma ') ^{(k-1)} (a) = \sum \limits _{i_1 + i_2 + i_3 = k-1} \frac {(k-1)!} {i_1 ! i_2 ! i_3 !} \big( (\tau ^{-1})' \circ g \circ \sigma \big) ^{(i_1)} \cdot \big( g' \circ \sigma \big) ^{(i_2)} \cdot \big( \sigma ' \big) ^{(i_3)} .$$

Now stop calculating and think: in the above sum, the only derivatives of $g$ showing up will be of order at most $k$, because $i_1$ and $i_2$ can go up to at most $k-1$. Since $ord(g) = n > k$, all these derivatives are $0$, from which follows $f^{(k)} (a) = 0 \; \forall k<n$.

In order to see that $f^{(n)} (a)\ne 0$ we shall use the same formula above, but with $k$ replaced by $n$:

$$f^{(n)} (a) = (f')^{(n-1)} (a) = ((\tau ^{-1})' \circ g \circ \sigma \cdot g' \circ \sigma \cdot \sigma ') ^{(n-1)} (a) = \sum \limits _{i_1 + i_2 + i_3 = n-1} \frac {(n-1)!} {i_1 ! i_2 ! i_3 !} \big( (\tau ^{-1})' \circ g \circ \sigma \big) ^{(i_1)} \cdot \big( g' \circ \sigma \big) ^{(i_2)} \cdot \big( \sigma ' \big) ^{(i_3)} .$$

The only non-zero term will be the one with $i_2 = n-1$ (which implies $i_1 = i_3 = 0$), which is $(g' \circ \sigma) ^{(n-1)} (a)$. Now, you either understand at a glance what happens here, or you can use induction once more to show that, in fact, this is equal to $( g^{(n)} \circ \sigma ) (a) = g^{(n)} (a') \ne 0 $ by assumption. Therefore, $f^{(n)} (a) = g^{(n)} (a') \ne 0$, so $ord(f) = n$.