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Suppose $f \in C^{\infty}[-1, 1]$, $f^{(k)}(0) = 0, k = 0, 1, 2, \dots$ and there is some $C \in \mathbb{R}$ such that $\sup_{x \in [-1, 1]}|f^{(k)}(x)| \leq k!C$ for any $k \in \mathbb{N}$. Then $f$ is constantly 0 on $[-1, 1]$.

Ok, I've been trying to solve this for many hours now. The mean value theorem gives me $f(x_0)=f'(x_1)x_0=f''(x_2)x_1x_0=\dots$ and thus $|f(x)|\leq |f^{(k)}(\hat{x})||x|^k$, so if there is an $x \in (-1, 1)$ such that $f(x) = A \neq 0$ then for any $k \in \mathbb{N}$ there is another point $|\hat{x}| < |x|$ such that $|f^{(k)}(\hat{x})| > \frac{A}{|x|^k}$.

The zero Taylor series at 0 means that $f(x) = o(x^k)$ $\forall k \in \mathbb{N}$ as $x \to 0$ and the converging Taylor sum of an expansion at any point gives us, if I get this right, that for all $x, x_0 \in [-1, 1]$ $|f(x)-f(x_0)| \leq \frac{C|x-x_0|}{1-|x-x_0|} + o((x-x_0)^k)$ again for any $k \in \mathbb{N}$ as $x \to x_0$.

Not much but that's all I could think of and I still don't see a contradiction with $A \neq 0$. So any clue?

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Let $\varepsilon\in (0,1)$ and $x\in [-1+\varepsilon,1-\varepsilon]$. The Taylor expansion (about 0) with Lagrange remainder tells you that for any positive integer $n$, there exists $\xi_n\in[-1+\varepsilon,1-\varepsilon]$ such that $$ f(x)=\frac{f^{(n)}(\xi_n)}{n!} x^n$$ and thus $$ |f(x)|\le \frac{|f^{(n)}(\xi_n)|}{n!} |x|^n\le C|x|^n\le C(1-\varepsilon)^n\overset{n\to \infty}{\longrightarrow} 0.$$

This shows that $f(x)=0$ for all $x\in [-1+\varepsilon,1-\varepsilon]$. Since $\varepsilon$ was arbitrary, you conclude that $f(x)=0$ for all $x\in (-1,1)$ and by continuity for all $x\in [-1,1]$.

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The integral form of reminder: https://en.wikipedia.org/wiki/Taylor%27s_theorem

shows that we actually have $f(x)=\sum \frac {f^{(k)}(0)} {k!} x^{k} $ for $|x| <1$ so $f \equiv 0$.

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  • $\begingroup$ Did you look at the integral form of remainder in that article? You can easily see that $|R_k(x)| \leq \frac C {k+1}$ for all $x$ with $|x| <1$. Hence $R_k(x) \to 0$. $\endgroup$ – Kavi Rama Murthy Jul 4 at 12:25

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