Take the 2-minute tour ×
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.

This is copy of statement in this link.

If $f$ is an infinitely differentiable function defined on an open set $D \subset \mathbb{R}$, then the following conditions are equivalent.

  1. $f$ is real analytic.

  2. For every compact set $K \subset D$ there exists a constant $C$ such that for every $x \in K$ and every non-negative integer $k$ the following estimate holds: $$ \left| \frac{d^k f}{dx^k}(x) \right| \leq C^{k+1} k!$$

It is also said that a function is analytic if and only if its Taylor series converges to that function. please let me know how these two statements are equivalent. Also I don't understand the term 'estimate' in the second statement.

share|improve this question
If possible someone please make the text in the question uniform with equations visible. I am not able to figure out how to do this. –  Rajesh D Apr 2 '11 at 7:05
@Rajesh: Is this ok? –  t.b. Apr 2 '11 at 7:08
@Theo Buehler : thank you. Please let me know how you did this. –  Rajesh D Apr 2 '11 at 7:08
First of all, I removed the blanks at the beginning of the paragraphs (four blanks result in a code section). Second, I replaced 1) and 2) by 1. and 2. in order for it to be displayed as a ordered list. Third, I replaced the single dollar signs $...$ in the displayed formula by $$...$$ and dit some minor LaTeX-tweaking. The details can be seen by clicking on "x mins ago" above my name. –  t.b. Apr 2 '11 at 7:13
"Estimate" just means "inequality." Also, the definition of "analytic (at a point $x = a$)" is that the function's Taylor series (centered at $x = a$) converges to that function. –  Jesse Madnick Apr 2 '11 at 7:14
show 2 more comments

1 Answer

up vote 2 down vote accepted

We say that $f\colon D \to \mathbb{R}$ is analytic if for every $x_0 \in D$ we have

$$f(x)=\sum_{n=0}^\infty \frac{f^{(n)}(x_0)}{n!}(x-x_0)^n$$

for all $x$ in a neighborhood of $x_0$. If this is the case, then we can consider the following complex power series

$$F(z)=\sum_{n=0}^\infty \frac{f^{(n)}(x_0)}{n!}(z-x_0)^n$$

which is convergent in a complex neighborhood of $x_0$. This procedure yields an analytic complex function $F$ of one complex variable that extends $f$ and can be used to obtain claim $2$ from claim $1$.

To see this fix $x_0 \in D$ and $r >0$ so small that the complex disc $D(x_0, r)$ centered at $x_0$ and of radius $r$ is contained in the domain of $F$ with its boundary. By the Cauchy's integral formula we can express every derivative of $F$ in integral terms:

$$F^{(n)}(x_0)= \frac{n!}{2 \pi i}\int_{\{\lvert z- x_0\rvert = r\}} \frac{F(z)dz}{(z-x_0)^{n+1}}$$

yielding the estimate

$$\lvert F^{(n)}(x_0) \rvert \le \frac{n!}{2\pi r^{n+1}}2 \pi r \max_{z \in D(x_0, r)} \lvert F(z) \rvert =\frac{n!}{r^n} \lVert F \rVert_{\infty, D(x_0, r)}.$$

Now observe that $F$ and $f$ agree in $x_0$ with all their derivatives. So the last formula is essentially claim 2 if the chosen compact $K$ is the interval $[x_0-r, x_0+r]$.

To conclude the proof choose an arbitrary compact $K \subset D$. We can cover it with a finite number of intervals $[x_1-r_1, x_1 + r_1] \ldots [x_p - r_p, x_p +r_p]$ like the ones in the previous step. In particular every disc $D(x_j, r_j)$ is contained in the domain of $F$. Call $H$ the union of such discs: obviously $\lVert F \rVert_{\infty, D(x_j, r_j)} \le \lVert F \rVert_{\infty, H}$. So for all $x \in K$ we have

$$\lvert f(x) \rvert \le \frac{n!}{r^n}\lVert F \rVert_{\infty, H}$$

which is claim 2.

To see the converse we don't need any complex analysis. Fix $x_0 \in D$ and expand $f$ in a Taylor series up to order $n$ with Lagrange remainder form:

$$f(x)=\sum_{k=0}^n \frac{f^{(k)}(x_0)}{k!}(x-x_0)^k+ \frac{f^{(n+1)}(\xi)}{(n+1)!}(x-x_0)^{n+1}.$$

We need to prove that, for all $x$ in a neighborhood of $x_0$, $$\lim_{n \to \infty} \frac{f^{(n+1)}(\xi)}{(n+1)!}(x-x_0)^{n+1}=0.$$

This follows from claim 1 because

$$\left\lvert\frac{f^{(n+1)}(\xi)}{(n+1)!}(x-x_0)^{n+1}\right\rvert \le C^{n+2}\lvert x-x_0 \rvert^{n+1}=C(C \lvert x-x_0 \rvert)^{n+1}$$

and $(C \lvert x-x_0 \rvert)^{n+1}\to 0$ if $\lvert x-x_0 \rvert < 1/C$.

share|improve this answer
add comment

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.