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It is possible to find a finite upper bound for the derivatives of a "Flat function"?

From the comments looks like I have a conceptual mistake. Please explain where it is and why works different from my current believes. Thanks beforehand.

Intro________

In this another question I explored why the derivatives of Flat functions looks to grow so fast, and in this answer it is explained that it is indeed the case.

Following Wikipedia example: $$f(x) =\begin{cases} 0\,\,\,\quad,\, x=0 \\ \displaystyle{e^{-\frac{1}{x^2}}},\,x\neq 0 \end{cases}$$ this flat function is an example of a non-analytic smooth function like smooth bump functions are, so, because of being smooth all their derivatives must exist so they must be bounded on any compact finite-size interval on the real line, let call it $K$ (as in Wikipedia).

So, there should exist some constant $0\leq C<\infty$ such as $$C=\sup_{\begin{array}{c} x\,\in\, K\\ m\,\in\,\mathbb{Z_0^+}\end{array}}\left\|f^{(m)}(x)\right\|_{\infty}$$ with $\|f\|_\infty$ the uniform norm, and I think in this case I can also consider that $K\equiv (-\infty,\,+\infty)$, but let consider initially that $K:= [-10,\,10]$.

The problem is that this constant $C$ should exist for $f(x)$ since is smooth class $C^\infty$, but if I take their first derivatives, as done in the mentioned question, I can see their maximum values increases rapidly, even faster than a exponential function, as can be seen in Desmos: $$\begin{array}{c | c | l} m & \sup & \text{in the order of} \\ \hline 0 & f & \qquad\sim 1 \\ 1 & f' & \qquad\sim 1 \\ 2 & f'' & \qquad\sim 2^2 \\ 3 & f^{(3)} & \qquad\sim 2^4 \\ 4 & f^{(4)} & \qquad\sim 2^8 \\ 5 & f^{(5)} & \qquad\sim 2^{12} \\ 6 & f^{(6)} & \qquad\sim 2^{17} \\ 7 & f^{(7)} & \qquad\sim 2^{22} \\ \hline \end{array}$$ when already for $m=7$ the supreme is in the order of $2^{22} = 4,194,304$ so it look like is growing unbounded $C\overset{m\to\infty}{\to}\infty$, against the intuition given by it's smoothness.

Question________

It is possible to find $C=\sup \left\|f^{(m)}(x)\right\|_{\infty}$ for this example $f(x)$? Or at least to prove that $C<\infty$?

Joako
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  • Why should smoothness remotely suggest that $C\ne\infty$ when $K$ isn’t compact? – Ted Shifrin Nov 22 '24 at 01:30
  • @TedShifrin in the case of this specific $f(x)$ I think it holds since it becomes flat far away from zero, so their derivatives maximums are contained on the interval $[-2,,2]$ (near the point where it is flat $x=0$)... but if you feel more comfortable, assume $K=[-10,,10]$ – Joako Nov 22 '24 at 01:39
  • But $K \times \mathbb Z_0^+$ is still not compact. Or consider a simpler function like $f(x) = e^{2x}$; then $f^{(m)}(x) = 2^m e^{2x}$, and even if we limit $K$ to some very tiny interval, this is unbounded as $m \to \infty$. – Misha Lavrov Nov 22 '24 at 01:50
  • @MishaLavrov but $f(x)\in C^\infty$ so the $\sup |f^{(m)}|_\infty$ should exist for all $\forall m \in \mathbb{Z}_0^+$ included $m\to\infty$ (at least if Wikipedia is right) – Joako Nov 22 '24 at 02:27
  • For each $m$, it does indeed exist and is finite, but "$m \to \infty$" is not a value of $m$. – Misha Lavrov Nov 22 '24 at 02:47
  • @MishaLavrov mmmm... maybe I have a misunderstanding, but $f(x)$ it is presented as class $C^\infty$ and not just as class $C^m$, so I don't understand why $m$ should not work for $m\equiv \infty$ if $f^{(m)}$ should be bounded for any possitive integer value of $m$? – Joako Nov 22 '24 at 03:09

1 Answers1

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$f \in C^\infty$ means by definition that $f\in C^m$ for every $m\in\mathbb N$. So for each $m$, $\| f^{(m)}\|_\infty<\infty$. Hopefully you agree this says nothing about the limit $m\to\infty?$

In fact, if it was true that $\sup_m\|f^{(m)}\|_\infty < C <\infty$ then from Taylor's theorem with e.g. the Lagrange form of the remainder $f^{(m)}(\xi_m)(x-x_0)^m/m!$ we see that on any compact subset $K\Subset\mathbb C$ the Taylor series converges uniformly to $f$. That is, such a function is analytic, even entire. :)

It is in fact a requirement that the function derivatives on compact sets grow faster than $n!$ (otherwise the function is analytic, similarly to the above.) So flat functions are of Gevrey type.

Calvin Khor
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