properties of a real analytic function If there are a radius $r>0$ and constants $M,C\in\mathbb R$ for all $y\in U$ with
$$|\partial^if(x)|\leq M\cdot i!\cdot C^{|i|}\space\space\space\space \forall x\in\mathbb B_r(y),i\in\mathbb N_0^n$$
then $f\in C^\infty(U)$ is real analaytic.
But I don't have any idea how to prove this.
I just know a function is called analytic if there are power series (convergent) in each point of U.
thanks for helping! :)
 A: Fix $y\in U$, and let $M$, $C$, $r$ be as given in the problem. For each $m\geq 1$, let $P_m(x)$ be the $m$th Taylor polynomial $$P_m(x) = \sum_{|i| = 0}^m\frac{(\partial^if)(y)}{i!}(x - y)^i.$$ We want to show that $P_m(x)\to f(x)$ whenever $x$ is sufficiently close to $y$, because this is what analytic means. In order to do this, we will apply Taylor's theorem in several variables, which says that for $x\in \mathbb{B}_r(y)$ $$\tag{$*$}f(x) - P_m(x) = \sum_{|j| = m + 1}R_j(x)(x - y)^j.$$ The right hand side of this expression is the "remainder term" in Taylor's theorem. We want to show that it goes to $0$ as $m\to \infty$. Using the explicit formula for $R_j(x)$ in the Wikipedia link given, $$R_j(x) = \frac{m+1}{j!}\int_0^1(1-t)^m(\partial^jf)(y + t(x - y))\,dt.$$ Here's where we use the estimates given in the problem: $$|R_j(x)|\leq \frac{m+1}{j!}\int_0^1(1-t)^mM\cdot j!\cdot C^{m+1}\,dt = MC^{m+1}.$$ Plugging this into equation ($*$) gives $$|f(x) - P_m(x)|\leq \sum_{|j| = m+1}MC^{m+1}|x-y|^{m+1} = \binom{m+n}{n-1}MC^{m+1}|x-y|^{m+1}.$$ Thus we only have to show the right hand side of this expression goes to $0$ when $|x - y|$ is small enough. One has the very crude estimate $$\binom{m+n}{n-1}\leq (m+n)^{n-1},$$ which is $\leq 2m^{n-1}$ when $m$ is large. Thus for large $m$, $$|f(x) - P_m(x)|\leq 2Mm^{n-1}C^{m+1}|x-y|^{m+1}.$$ If $|x - y|<C^{-1}$, the right hand side of this expression $\to 0$ locally uniformly as $m\to \infty$. Hope this helps!
