Real analyticity of a function 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.


*

*$f$ is real analytic.

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