= Solution
The <Taylor series> definition says that $f\in C^\infty(\mathbb R)$ and, for every $a$, there is $r>0$ such that
$$
f(x)=\sum_{n=0}^\infty\frac{f^{(n)}(a)}{n!}(x-a)^n\qquad(|x-a|<r).
$$
The equivalent <factorial derivative criterion for real analyticity> says that, for every $a$, there are a neighborhood $I$ of $a$ and constants $C,A>0$ such that
$$
\boxed{\sup_{x\in I}|f^{(n)}(x)|\leq CA^n n!\quad(n\geq0).}
$$
The uniformity over $I$ matters: bounds only at $a$ do not exclude a <flat function>.
Assume the <factorial derivative criterion for real analyticity>. The <Taylor theorem with Lagrange remainder> gives
$$
\left|f(a+h)-\sum_{n=0}^{N-1}\frac{f^{(n)}(a)}{n!}h^n\right|\leq C(A|h|)^N
$$
when the segment from $a$ to $a+h$ is contained in $I$. For sufficiently small $|h|<A^{-1}$ the <Taylor remainder> tends to zero, proving the <Taylor series> definition.
Conversely, write the convergent <power series> at $a$ as $\sum b_k h^k$. Choose $\rho$ strictly inside its radius of convergence; then $|b_k|\leq M\rho^{-k}$ for some $M$. Termwise <differentiation> on $|h|\leq\rho/2$ gives
$$
|f^{(n)}(a+h)|\leq M\rho^{-n}n!\sum_{j=0}^\infty\binom{j+n}{n}2^{-j}
=2M(2/\rho)^n n!.
$$
Here the sum is $(1-1/2)^{-n-1}$, obtained by differentiating the <geometric series>. This is the required locally uniform bound. \b[The two definitions of a <real analytic function> are equivalent.]
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