# Prove that a function is zero

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?

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 $$x\in [-1,1]$$.
shows that we actually have $$f(x)=\sum \frac {f^{(k)}(0)} {k!} x^{k}$$ for $$|x| <1$$ so $$f \equiv 0$$.
• 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 $R_k(x) \to 0$. – Kavi Rama Murthy Jul 4 at 12:25