A real valued function $f:\Re\rightarrow\Re$ is said to be analytic at a point $x_0 \in \Re$ if $f$ can be represented by a convergent power series near $x_0$, that is, for some $\epsilon >0$ and coefficients $a_n \in \Re$, one has $$f(x) = \sum_{n=0}^{\infty} a_n(x-x_0)^n$$ $$|x-x_0|<\epsilon$$
Prove that if such an $f$ is given and is analytic at $x_0$, then there is a holomorphic function $F$ defined in the open disk $D(x_0,\epsilon)$ such that $F(x)=f(x)$ for all $x$ in the interval $(x_0-\epsilon , x_0+\epsilon)$
My Proof:
Since $f:\Re \rightarrow \Re$ be an analytic function at $x=x_0$ therefore, there exists $\epsilon >0$, such that $$f(x) = \sum_{n=0}^{\infty} a_n(x-x_0)^n$$ $$|x-x_0| < \epsilon $$
where, $(a_n)_{n=0}^\infty \subseteq \Re$
Now, $D(x_0,\epsilon)=\{z \in \mathbb{C} \mid |z-x_0|<\epsilon \}$
set $z \in D(x_0,\epsilon)$
then $|z-z_0|<\epsilon$
By density property of $\Re$, there exists a real number $p$ such that $|z-x_0|<|p|<\epsilon$
therefore, $|z-x_0|<|p|<\epsilon$
then, $|z-x_0|<|x_o -p-x_0|<\epsilon$ where $|-p|=|p|$
set $a=x_0-p$, then $a \in \Re$,
therefore, $|z-x_0|<|a-x_0|<\epsilon$ $(1)$
since, $|a-x_0|<\epsilon$ therefore, $f(a)=\sum _{n=0}^\infty a_n(a-x_0)^n$
then, the series $\sum _{n=0}^\infty a_n(a-x_0)^n$ is convergent.
Hence, $\lim_{n\rightarrow\infty}a_n(a-x_0)^n=0$
therefore, $[a_n(a-x_0)^n]_{n=0}^\infty$ is a bounded sequence
therefore, there exists $M>0$ such that, $|a_n(a-x_0)^n|\leq M$, $\forall n\geq 0$ $(2)$
Now, $$|a_n(z-x_0)^n|=\left|a_n\left(\frac{z-x_0}{a-x_0}\right)^n(a-x_0)^n\right|$$ $$=|a_n(a-x_0)^n|\left|\left(\frac{z-x_0}{a-x_0}\right)^n\right|$$ $$\leq M\left|\left(\frac{z-x_0}{a-x_0}\right)^n\right|$$ $$=M\left|\frac{z-x_0}{a-x_0}\right|^n$$
therefore, $$|a_n(z-x_0)^n|\leq M\left|\frac{z-x_0}{a-x_0}\right|^n$$
set $r=\left|\frac{z-x_0}{a-x_0}\right|$
then, by $(1)$ we get, $0<r<1$
therefore, $\left|a_n(z-x_0)^n\right|\le Mr^n$ where $0<r<1$
since $\sum_{n=0}^\infty r^n$ is convergent (geometric series convergent if $0<r<1$),
by comparison test, $\sum_{n=0}^\infty |a_n(z-x_0)^n|$ is convergent
then, $\sum_{n=0}^\infty a_n(z-x_0)^n$ is convergent
therefore, $\sum_{n=0}^\infty a_n(z-x_0)^n \in \mathbb {C}$ $\forall z\in D(x_0,\epsilon)$
Define: $F:D(x_0,\epsilon)\rightarrow \mathbb{C}$ by, $$F(z)=\sum_{n=0}^\infty a_n(z-x_0)^n$$
By previous discussion $F$ is a well defined function on $D(x_0,\epsilon)$, and since $F$ be a power series on $D(x_0,\epsilon)$, therefore $F$ is holomorphic on $D(x_0,\epsilon)$.
Also, $F(x)=\sum_{n=0}^\infty a_n(x-x_0)^n$ $\forall x \in (x_0-\epsilon , x_0+\epsilon )$
Therefore, $F(x)=f(x)$ $\forall x \in (x_0-\epsilon , x_0+\epsilon )$
Does it make sense?