This is a follow-on of sorts to this question, but is self-contained.

Let $F_1 := \{f \in C^\infty(\mathbb{R}) \mid \|\frac{df}{dx}\|_\infty \le c_1\}$ ($c_i > 0$ throughout).

Given $f, g \in F_1$, we have by the chain rule that $g \circ f \in F_1$ if

$\|\frac{dg}{df}\|_\infty \|\frac{df}{dx}\|_\infty \le c_1$

and this is automatically true if $c_1 \le 1$.

Now let $F_2 := \{f \in F_1 \mid \|\frac{d^2f}{dx^2}\|_\infty \le c_2\}$.

Given $f, g \in F_2$, we have by the generalized chain rule (Faà di Bruno's formula) that $g \circ f \in F_2$ if it is in $F_1$ and

$\|\frac{d^2g}{df^2}\|_\infty \|\frac{df}{dx}\|_\infty^2 + \|\frac{dg}{df}\|_\infty \|\frac{d^2f}{dx^2}\|_\infty \le c_2$

and this is automatically true if $c_1, c_2$ satisfy the corresponding inequality $c_2 c_1^2 + c_1 c_2 \le c_2$, i.e. if $c_1^2 + c_1 \le 1$ (in which case $c_1 \le 1$ also).

In general, let $F_n := \{f \in F_{n-1} \mid \|f^{(n)}\|_\infty \le c_{n} \}$.

Given $f, g \in F_n$, we have by the generalized chain rule that $g \circ f \in F_n$ if it is in $F_{n-1}$ and

$\sum_{\pi\in\Pi(n)} \|g^{(|\pi|)}\|_\infty \prod_{B\in\pi} \|f^{(|B|)}\|_\infty \le c_n$ (notation from the Wikipedia article)

and this is automatically true if the $c_i$ satisfy the corresponding inequalities

$\sum_{\pi \in \Pi(i)} c_{(|\pi|)} \prod_{B \in \pi} c_{(|B|)} \le c_i \forall i \in \{1, ..., n\}$.

My question then is: can $F_\infty := \bigcap_{n \in \mathbb{N}} F_n$ be closed under composition without being trivial? In other words:

Does there exist a sequence of values $c_i > 0$ which satisfy the inequalities

$\sum_{\pi \in \Pi(i)} c_{(|\pi|)} \prod_{B \in \pi} c_{(|B|)} \le c_i \forall i \in \mathbb{N}$?


After thinking about this a bit more, I realized there's a simple proof that the answer is yes:

For $n > 1$, the only terms containing $c_n$ in the $n$th inequality are $c_n c_1^n$ and $c_1 c_n$ on the left, and $c_n$ on the right. So we can collect these terms on the right as $c_n (1 - c_1 - c_1^n)$.

For $n = 2$ this leaves nothing on the left, so it tells us that $1 - c_1 - c_1^2 \ge 0$, as above. For $n > 2$, the coefficient $1 - c_1 - c_1^n$ is positive ($c_1^n < c_1^2$ since $c_1 < 1$, so $1 - c_1 - c_1^n > 1 - c_1 - c_1^2 \ge 0$) so we can divide through by it to obtain a tight lower bound for $c_n$ purely in terms of the previous $c_i$.

In summary, we're free to choose any (positive) values for $c_1$ (subject to $c_1^2 + c_1 \le 1$) and for $c_2$; then each successive $c_n$ can be chosen to satisfy the lower bound given by the $n$th inequality.

There's a natural follow-up question: assuming each $c_n$ takes the minimum possible value for $n > 2$, what's the asymptotic behaviour of $c_n$ in terms of $c_1, c_2$? Can $c_n$ be uniformly bounded? I suspect the answer is no, but I might post that as a separate question.


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