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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
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"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

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

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