Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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! :)

share|cite|improve this question
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
up vote 2 down vote accepted

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!

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.