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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.

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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
2  
"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
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1 Answer 1

up vote 3 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

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

Observe that $F$ and $f$ agree in $x_0$ with all their derivatives, so the last formula is very similar to claim 2. 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 (edited shaded area)

$$ \begin{split} \left\lvert \frac{d^n f}{dx^n}(x) \right\rvert &\le n!\left(\dfrac{1}{\displaystyle \operatorname{min}_{j=1\ldots p}r_j}\right)^n\lVert F \rVert_{\infty, H}\\ &=n! (C_{K, f})^n \|F\|_{\infty, H}, \end{split} $$ where $C_{K, f}$ stands for "a positive number that depends only on $K$ and $f$". Now observe that $\|F\|_{\infty, H}\le \|F\|_{\infty, K}$ and that $$(C_{K, f})^n \|F\|_{\infty, K}\le\tilde{C}_{K, f}^{n+1}, $$ with $\tilde{C}_{K, f}=\max(C_K, 1)\cdot\max(\|F\|_{\infty,K}, 1)$. (We put an exponent $n+1$ to encompass the case $n=0$ as well). We conclude that $$\left\lvert \frac{d^n f}{dx^n}(x) \right\rvert\le \tilde{C}^{n+1}_{K, f} n!, $$ which is claim 2. (Note that the "constant" is not really so since it depends on $K$ and $f$. The crucial fact is that it does not depend on $n$).

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 $\lvert x-x_0 \rvert < 1/C$.

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I bumped into this and I just want to ask: How did you conclude "So the last formula is essentially claim 2 if the chosen compact K is the interval $[x_0−r,x_0+r]$?" The inequality holds for $x_0$. But choosing a different $x$ from that interval will have effect on the Cauchy estimate; namely, the $r$ will have to be made smaller. –  Alex Strife Jul 6 at 2:39
    
Hoping to get a reply from you. Thanks. –  Alex Strife Jul 9 at 6:47
    
@AlexStrife: Hi, sorry for the delay. I edited the answer, trying to improve clarity. The compactness argument that follows that remark you noticed is there precisely to eliminate that dependence on $r$. –  Giuseppe Negro Jul 9 at 15:51
    
Let's see if I am getting this right. I also really want to understand this. Let's consider $[x_1 - r_1, x_1 + r_1]$. It is clear to me that $|\frac{d^n f}{dx^n} (x_1)| \leq \frac{n!}{r_1^n} \cdot \lVert F \rVert_{\infty, D(x_1, r_1)}$. Now, what if we have $x_1' \neq x_1$ but which is also in $(x_1 - r_1, x_1 + r_1)$? I believe the inequality should now be $$\left| \frac{d^n f}{dx^n} (x_1') \right| \leq \frac{n!}{(r_1-x_1')^n} \cdot \lVert F \rVert_{\infty, D(x_1, r_1)},$$ which might not be consistent with the uniform bound on $K$ that you gave. –  Alex Strife Jul 10 at 2:17
    
@AlexStrife: I edited again, trying to be as careful as I can with constants and bounds. Please let me know if you find it satisfactory now, thank you –  Giuseppe Negro Jul 14 at 9:59
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