Analytic functions

Recall Def 154.1. We have seen that every analytic function is smooth but not every smooth function is analytic, that is, $C^{\omega }⊊C^{\infty }$.

When is a smooth function analytic?

Let $f:(a,b)\to \mathbb{R}$ be smooth. The Taylor series for $f$ at $x\in (a,b)$ is

$$\sum _{k=0}^{\infty }\frac{f^{(k)}(x)}{k!}{h}^k.$$

Recall that if $f$ is analytic then the $\sum a_r{h}^r$ of Def 154.1 must be the Taylor series for $f$ at $x$.

Therefore, for $f$ to be analytic on $(a,b)$, it is sufficient and necessary that for all $x\in (a,b)$, the Taylor series for $f$ at $x$

1. have positive radius of convergence, and
2. actually converge to $f$ on some interval $(x-\delta ,x+\delta )$.

Let $I=[x-\sigma ,x+\sigma ]$ be a subinterval of $(a,b)$, $\sigma >0$. Denote by $M_r$ the maximum of $|f^{(r)}(t)|$ for $t\in I$. The derivative growth rate of $f$ on $I$ is

$$\alpha =\underset{r\to \infty }{\mathrm{limsup}}\sqrt[r]{\frac{M_r}{r!}}.$$

Clearly, $\sqrt[r]{|f^{(r)}(x)|/r!}⩽\sqrt[r]{M_r/r!}$, so the radius of convergence

$$\begin{array}{r}R=\frac{1}{{\mathrm{limsup}}_{r\to \infty }\sqrt[r]{\frac{|f^{(r)}(x)|}{r!}}}\end{array}$$

of the Taylor series at $x$ satisfies

$$\frac{1}{\alpha }⩽R.$$

Lemma 364.1(Lemma).

If $\alpha \sigma ⩽1$, then the Taylor series converges uniformly to $f$ on the interval $I$.

Lemma 364.2(Lemma).

If $f$ is expressed as a convergent power series $f(x+h)=\sum c_k{h}^k$ with radius of convergence $R>\sigma$, then $f$ as bounded derivative growth rate on $I$.

Theorem 364.3(Theorem).

A smooth function is analytic iff it has locally bounded derivative growth rate.

Corollary 364.4(Corollary).

A smooth function is analytic if its derivatives are uniformly bounded.

Proof.

If $|f^{(r)}(\theta )|⩽M$, for all $r$ and $\theta$, then the derivative growth rate of $f$ is bounded. In fact, $\alpha =0$ and $R=\infty$.□