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I'd love your help with the following question:

let $f\in C^{\infty}$ and there's $M$ such that $|f^{(j)}(x)|\leq M$ for every $j\in \mathbb{Z}_{+}$ and for all $ x\in [-1,1]$.

I need to prove tht if $f(1/k)=0 $ for every $k\in \mathbb{N}$, so $f=0$.

O.k, so to be honest I tried Lagrange theorem, Taylor series, I tried to get some things together, but Didn't get any good conclusion.

From the continuous of $f$ we get that $f(0)=0$.

Thank you so much guys.

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    Consider this (I have not thought it out fully): Try Taylor series about 0 (making adjustments for $0^1$ if necessary): 0=f(0)+M($x+x^2/2+.....+x^n/n!......)$ What does the series remind you of?2011-06-18
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    Correction: the $o^1$ should be $0^-$, the left-limit at 0.2011-06-18
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    @gary: that is not a correct statement of Taylor's theorem. Where are the higher-order derivatives?2011-06-19

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Taylor's theorem states that $$ f(x) = f(a) + f'(a)(x-a) + \frac{f''(a)}{2!}(x-a)^2 + \cdots + \frac{f^{(k)}(a)}{k!}(x-a)^k + \frac{f^{(k+1)}(\xi_L)}{(k+1)!} (x-a)^{k+1}, $$ where $\xi_L$ is between $a$ and $x$.

This will help us finding the function if we take $a=0$. First note that by Lagrange's theorem for $x >0$ we have $f(x)-f(0)=f'(c)x, \ c \in [0,x]$. Therefore, there exists $c_k \in (0,1/k)$ with $f'(c_k)=0$. We can pick a decreasing subsequence of this sequence, and denote it for simplicity with (the same) $(c_k)$. Since $f'(c_k)=0$ and $c_k \to 0$ it follows that $f'(0)=0$.

Doing the same think with $(c_k)$ instead of $1/k$ we get that $f''(0)=0$ and inductively $f^{(n)}=0$, for all positive integers $n$.

Pick $x \in [-1,1]$. By Taylor's formula around zero we get that $f(x)=\frac{f^{(k+1)}(\xi_L)}{(k+1)!} x^{k+1}$ and using the bound on the derivatives we get $$ |f(x)| \leq \frac{M}{(k+1)!}|x|^{k+1}$$ for all positive integers $k$.

Taking $ k \to \infty$ we get that $f(x)=0$. Since $x$ was arbitrary, the problem is solved.