# Functions defined by convergent power series

Let $r \in (0,\infty)^n$ be a polyradius. For $F \in \Pc{R}{X}_r$, we denote by $F_r:\bar B(r) \into \RR$ the function defined by $F_r(x):= F(x)$.

We denote by $C^\infty(r)$ the ring of all $C^\infty$ functions on $B(r)$. For $f \in C^\infty(r)$, we let $T_r(f) \in \Ps{R}{X}$ be the Taylor expansion of $f$ at $0$; note that the map $T_r:C^\infty(r) \into \Ps{R}{X}$ is a ring homomorphism. Moreover, if $s \lt r$, then $C^\infty(r) \subseteq C^\infty(s)$ and $T_s$ extends $T_r$.

Lemma 1
For $F \in \Pc{R}{X}_r$, the function $F_r$ is continuous on $\bar B(r)$ and $C^\infty$ on $B(r)$. The map $F \mapsto F_r\rest{B(r)}:\Pc{R}{X}_r \into C^\infty(r)$ is an injective ring homomorphism and, for $F \in \Pc{R}{X}_r$, we have $F = T(F_r)$ and $$\sup_{x \in B(r)} \left|F_r(x)\right| \le \|F\|_r.$$

(Show proof)
Proof. We let $D(r)$ be the polydisk $D(0,r_1) \times \cdots \times D(0,r_n)$ in $\CC^n$. Note that each $F \in \Pc{R}{X}_r$ defines a function $\bar{F}_r:D(r) \into \CC$. Moreover, each partial sum of the series $F$ defines a holomorphic function on $D(r)$, and the absolute convergence of $F$ implies that these holomorphic functions converge absolutely and uniformly on compact subsets of $D(r)$ to $\bar F_r$. Hence $\bar F_r$ is holomorphic, and since $F_r \rest{B(r)}$ is the restriction of $\bar F_r$ to $B(r)$, the lemma follows. $\qed$

In view of Lemma 1, we define $$\C_r:= \set{F_r:\ F \in \Pc{R}{X}_r}$$ and $$\C_{\gt r}:= \set{F_r:\ F \in \Pc{R}{X}_{\gt r}}.$$ To define a ring corresponding to $\Pc{R}{X}$, though, we need to talk about germs of functions: two functions $f:U \into \RR$ and $g:V \into \RR$, with $U$ and $V$ neighbourhoods of $0$, are called equivalent at 0 if there exists a neighbourhood $W \subseteq U \cap V$ of $0$ such that $f\rest{W} = g\rest{W}$. The corresponding equivalence classes of functions defined on neighbourhoods of $0$ are called germs at 0 of such functions.

Exercise

1. Show that being equivalent at 0 is indeed an equivalence relation on the set of all functions defined on neighbourhoods of 0.
2. Show that the set of germs at 0 is a ring with unit, where addition and multiplication is induced from the usual addition and multiplication of functions.

We now let $C^\infty_n$ be the set of all germs at 0 of functions in $\bigcup_{r\gt 0} C^\infty(r)$. Note that, if $g \in C^\infty(r)$ and $h \in C^\infty(s)$ are equivalent at 0, then $T_rg = T_sh$. Therefore, there exists a unique ring homomorphism $T:C^\infty_n \into \Ps{R}{X}$ such that $$T[f] = T_r f$$ for any $f \in C^\infty(r)$, where $[f]$ denotes the germ at 0 of $f$.

Corresponding to our definition of $C^\infty_n$, we now let $\C_n$ be the set of all germs at 0 of functions in $\bigcup_{r\gt 0} \C_r$, a subring of $C^\infty_n$.

Corollary 1 (quasianalyticity)
The restriction of $T$ to $\C_n$ is a ring isomorphism onto $\Pc{R}{X}$. $\qed$

We now establish a few more properties of $\C_n$.

Lemma 2
Let $F \in \Pc{R}{X}_r$, $s \lt r$ and $i \in \{1, \dots, n\}$. Then $\partial F/\partial X_i \in \Pc{R}{X}_s$ and $$\left(\frac{\partial F}{\partial X_i}\right)_s = \frac{\partial F_r}{\partial x_i}\rest{B(s)}.$$

(Show proof)
Proof. Say $F = \sum a_\alpha X^\alpha$. Then, with $t:= \max\set{s_1/r_1, \dots, s_n/r_n} \lt 1$, we have $$\left\|\frac{\partial F}{\partial X_i}\right\|_s = \sum \left|a_\alpha\right| \left|s^\alpha\right| \le \left|a_0\right| + t \cdot \sum_{\alpha \gt 0} \left|a_\alpha\right| \left|r^\alpha\right| \lt \infty,$$ as required. $\qed$

Corollary 2 (closure under differentiation)
For $f \in \C_n$ and $i \in \{1, \dots, n\}$, we have $\partial f/\partial x_i \in \C_n$ and $$T\left(\frac{\partial f}{\partial x_i}\right) = \frac{\partial (Tf)}{\partial X_i}.\ \qed$$

Exercise 13

1. Let $F^i \in \Pc{R}{X}_r$, for $i \in \NN$, and let $F \in \Pc{R}{X}_r$, and assume that $F^i \to F$ in the $\|\cdot\|_r$-norm. Show that $F^i_r(x) \to F_r(x)$ uniformly for $x \in \bar B(r)$.
2. Let $F \in \Pc{R}{Y}_s$, with $Y = (Y_1, \dots, Y_m)$ and $s = (s_1, \dots, s_m)$, and let $G_1, \dots, G_m \in \Pc{R}{X}_r$ be such that $\left\|G_i\right\|_r \le s_i$ for each $i$. Show that $F\left(G_1, \dots, G_m\right) \in \Pc{R}{X}_r$ with $\left\|F\left(G_1, \dots, G_m\right)\right\|_r \le \|F\|_s$.
3. Let $F \in \Pc{R}{X}_r$ be such that $F = X_i \cdot G$, for some $i \in \{1, \dots, n\}$ and $G \in \Ps{R}{X}$. Show that $G \in \Pc{R}{X}_s$, for $s \lt r$.

Corollary 3 (closure under composition)
Let $f \in \C_m$ and $g_1, \dots, g_m \in \C_n$. Then $f(g_1, \dots, g_m) \in \C_n$ and $$T \big(f(g_1, \dots, g_m)\big) = (Tf)(Tg_1, \dots, Tg_m).$$

Proof. Since $f \in C^\infty_m$ and $g_1, \dots, g_m \in C^\infty_m$, we have $f(g_1, \dots, g_m) \in C^\infty_n$ with $(Tf)(Tg_1, \dots, Tg_m)$. Since $Tf$ and $Tg_1, \dots, Tg_m$ are convergent, it follows from Exercise 13.2 that $(Tf)(Tg_1, \dots, Tg_m)$ is convergent. Since the finite partial sums of the series $F$ converge to $F$ in the $\|\cdot\|_r$-norm, for sufficiently small $r$, we get from Exercise 13.1 and the morphism properties of the map $H \mapsto H_r$ that $f(g_1, \dots, g_m) = \big((Tf)(Tg_1, \dots, Tg_m)\big)_r \in \C_n$. $\qed$

Corollary 4 (closure under division by monomials)
Let $f \in \C_n$, and assume that $Tf = X_i \cdot G$ for some $i \in \{1, \dots, n\}$ and $G \in \Ps{R}{X}$. Then there exists $g \in \C_n$ such that $f = x_i \cdot g$ and $Tg = G$.

Proof. By Exercise 13.3 and the morphism properties of the map $H \mapsto H_r$. $\qed$

Proposition (closure under implicit functions)
Let $f \in \C_n$ be such that $f(0)=0$ and $(\partial f/\partial x_n)(0) \ne 0$. Then there exists a unique $g \in \C_{n-1}$ such that $g(0)=0$ and $f\left(x_1, \dots, x_{n-1}, g\left(x_1, \dots, x_{n-1}\right)\right) = 0$.

For the proof of this proposition, see for instance Section 1.8 in Krantz and Parks’s book.

## 3 thoughts on “Functions defined by convergent power series”

1. Peter says:

Looking at Part 3 of Exercise 13, don’t we have that $G\in \R\{X\}_r$? If $G = \sum a_\alpha X^\alpha$ then
$\|G\|_r = \sum |a_\alpha||r^\alpha| = \frac1{|r_i|} \|F\|_r < \infty$
since $\|F\|_r 0$.

1. Peter says:

Sorry, I got caught by the less than issue. The last line should be $\|F\|_r < \infty$ and $0<r$. (Also, I meant $\mathbb R\{X\}_r$).

2. Patrick says:

Hmmm, if you put it this way… Yes, this appears to be correct.

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