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It is as follows:

If $f \in C^{(\infty)}\left([-1,1]\right) $and $f^{(n)}(0)=0$ for $n=0,1,2,...,$ and there exists a number $C$ such that $\sup_{-1 \leq x\leq 1}|f^{(n)}(x)|\leq n!C$ for $n \in \Bbb N$, then $f \equiv 0 $ on$[-1,1]$.

If we take the Taylor polynomial of $f$ at $0$ then each term is $0$ except the remainder which means the function has the form (Lagrange):$$f(x)=\frac{1}{(n+1)!}f^{(n+1)}(\xi)x^{n+1}$$ for some $\xi \in [-1,1]$. Surely there isn't a constant $A_n$ such that $f(x) =A_nx^n$, except $A_n = 0$ for all $n$.

What am I doing wrong here? Isn't the other hypothesis redundant?

user110503
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1 Answers1

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The problem is that $\xi$ depends on $x$, while you seem to be treating it as a constant, dependent on $n$ only.

Here's my attempt. Let $x\in(-1, 1)$. By assumption, $$ \left|f(x)\right| = \left|\frac{f^{(n+1)}(\xi_n)}{(n+1)!}x^{n+1}\right|\leq \frac{\left|f^{(n+1)}(\xi_n)\right|}{(n+1)!}\left|x\right|^{n+1}\leq \frac{C(n+1)!}{(n+1)!}\left|x\right|^{n+1}= C \left|x\right|^{n+1}\to 0 $$ since $\left|x\right|<1$, and so $f(x)=0$. By continuity, $f(-1)=f(1)=0$ as well.

As an example of $C^\infty$ function with zero derivatives which is not identically $0$, consider $e^{-1/x}$ for $x>0$ and $0$ for $x\leq 0$. It's an easy exercise to show that all the derivatives vanish at $0$, but the function clearly is nonzero for $x>0$. It's one of standard examples of a smooth function that is not analytic.

Marcin Łoś
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