# Does Cauchy's estimate imply analyticity?

Komatsu says here (Proc. Japan Acad. Volume 36, Number 3 (1960), 90-93) that a smooth function which satisfies Cauchy's estimate is analytic.

How does one prove this?

Surely, if Cauchy's estimates hold for the derivatives of a function, then its Taylor series converges, but that is not enough for analyticity.

Is it necessary to consider the smooth function on a compact interval? In the second answer to this question, the above fact is used for a function defined on the whole real line.

If Cauchy's estimate $|f^{k}(x)| \le M \lambda^k k!$ with positive constants $M,\lambda$ is satisfied for all $k$ and all $x \in [a,b]$, and if $x_0\in [a,b]$ is arbitrary, let $T_k(x) = \sum_{k=0}^n \frac{f^{(k)}(x_0)}{k!} (x-x_0)^k$ be the $k$-th Taylor polynomial centered at $x_0$. By Taylor's theorem with the Lagrange form of the remainder we get $R_k(x) = f(x)-T_k(x) = \frac{f^{(k+1)}(\xi)}{(k+1)!}(x-x_0)^{k+1}$ with some $\xi$ between $x_0$ and $x$, so $$|R_k(x)| \le \frac{M \lambda^{k+1} (k+1)!}{(k+1)!} |x-x_0|^{k+1} = M \left( \lambda |x-x_0|\right)^{k+1}$$ This shows that $R_k(x) \to 0$ for $|x-x_0| < \lambda^{-1}$, as long as $x \in [a,b]$, and so the function is locally equal to its Taylor series and thus analytic.