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**J.R** hello,

f is a function $\displaystyle C^{\infty}$ from [a,b] to $\displaystyle \mathbb{R}$.

We suppose :

$\displaystyle \exists x_0 \in ]a,b[ \ and \ k \ real \ >0 | \forall n\in \mathbb{N},f^{(n)}(x_0)=0 \ \ and \ \ sup_{[a,b]}\{|f^{(n)}(x)|\}\le k^nn!$

1) Show that f is the function null from $\displaystyle ]x_0-\frac{1}{k},x_0+\frac{1}{k}[$.

I think that we must use Taylor's inegalities which say that (I don't know if it's the same name ...) :

for $\displaystyle u \in [a,b]$

$\displaystyle |f(b)-\sum_{k=0}^n\frac{(x-u)^k}{k!}f^{(k)}(u)|\le M\frac{(b-u)^{n+1}}{(n+1)!}$

where M majors $\displaystyle f^{n+1}$ from [a,b].

so, here, we have:

for $\displaystyle x\in ]x_0-\frac{1}{k},x_0+\frac{1}{k}[$

$\displaystyle |f(x)|\le \frac{|x-x_0|^{n+1}}{(n+1)!}M$

howerver we can take $\displaystyle M=k^{n+1}(n+1)!$

and $\displaystyle |x-x_0|\le \frac{1}{k}$

so we have:

$\displaystyle |f(x)|\le 1$ ...

and I have no idea to keep on the demonstration.