# 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! :)

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What does Taylor's theorem in several variables say? – froggie May 10 '12 at 16:07
To prove that $f$ is analytic, you need to show that the Taylor series of $f$ converges to $f$ (this is the definition of analytic). Explicitly, if $P_n(x)$ is the partial Taylor series of $f$ consisting of only terms of degree $\leq n$, you must show that $P_n\to f$. But Taylor's theorem tells you how to estimate $|f - P_n|$. – froggie May 10 '12 at 16:10
@froggie sorry but I don't get it: do you mean the remainder term???? – user31035 May 10 '12 at 16:42
I do mean the remainder! You know that $f = P_n + R_n$, where $R_n$ is the $n$th remainder term. If you can show that $R_n\to 0$, this is the same as saying that $P_n\to f$. The estimates given in the problem allow you to control the remainder term $R_n$. – froggie May 10 '12 at 16:45
Added an answer to clarify my hints. – froggie May 10 '12 at 19:05

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!