Lemma 12.3.7 (Facts about $C^{\om}$ functions).label

1. (1)

   $f$ is analytic near $x_{0}$ then there exists a cube $\OM=\curl{x:\abs{x_j-x_j^0}<r,1\leq j\leq n}$ such that $\sums{}{\al}\frac{1}{\al!}\parens{\pd^\al f}(x_{0})(x-x_{0})^{\al}$ converges to $f(x)$ for all $x\in\OM$ and convergence is uniform and absolute on compact subsets of $\OM$.

2. (2)

   Let $f(x)=\sums{}{}a_{\al}(x-x_{0})^{\al}$ be convergent enar $x_{0}$ and $x$ be an analyhtic function of $\xi$

   \begin{align*}x=\sums{}{\be}b_{\be}(\xi-\xi_{0})^{\be}\end{align*}

   
   

   with $x(\xi_{0})=x_{0}$. Then

   \begin{align*}F(\xi)=f(x(\xi))\end{align*}

   
   

   is analytic at $\xi_{0}$ and power series representation of $F$ is obtained by substitution

   \begin{align*}F(\xi)=\sums{}{\gam}c_{\gam}(\xi-\xi_{0})^{\gam}\end{align*}

   
   

   where

   \begin{align*}c_{\gam}=P_{\gam}(a_{\al},b_{\be})\end{align*}

   
   

   are universal polynomials with non-negative coefficients. (This is an algebraic fact about rings of power series)

3. (3)

   Given $M,r>0$ the function

   \begin{align*}f(x)=\frac{Mr}{r-\parens{\sums{n}{i=1}x_i}}\end{align*}

   
   

   is analytic in the rectangle $\curl{x:\MAX{i}\abs{x_j}<r/n}$ and

   \begin{align*}f(x)=M \sums{\infty}{k=0}\frac{\parens{\sums{n}{i=1}x_i}^{k}}{r^{k}}=M \sums{}{\abs{\al}\geq0}\frac{\abs{\al}!}{\al!r^{\abs{\al}}}x^{\al}\end{align*}

   
   
4. (4)

   We say that $\sums{}{\al}a_{\al}(x-x_{0})^{\al},a_{\al}\geq 0$ majorizes $\sums{}{\al }b_{\al}(x-x_{0}^{\al})$ if $a_{\al}\geq\abs{b_\al}$. In which case $\sums{}{\al }b_{\al}(x-x_{0}^{\al})$ converges absolutely whenever $\sums{}{\al }a_{\al}(x-x_{0}^{\al})$ does.

5. (5)

   Suppose that $\sums{}{\al }a_{\al}(x-x_{0}^{\al})$ in a rectangle $\curl{x:\MAX{i}\abs{x_i}<R }$ then $\exists$ geoemetric series as in (3) majorizing $\sums{}{\al }a_{\al}(x-x_{0}^{\al})$. Indeed let $x=(r,...,r),r<R$, $x\in$ rectangle. So $\exists M>0$ such that $\abs{a_\al r^{\abs{\al}}}\leq M$, $\fall \al$. So $\abs{a_\al}\leq\frac{M}{r^{\abs{\al}}}\leq \frac{M\abs{\al}!}{r\al!}$