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Let $f \ \colon \ [0,\infty) \to \mathbb{R}$ be a function satisfying:

  • $f$ is differentiable infinitely many times in $(0,\infty)$, and has a right-derivative of any order at $0$.
  • $f$ satifsfies the condition (condition 3 here) for analyticity: for every compact $K \subset [0,\infty)$ there exists a constant $C_K$ such that $$\forall x \in K:\forall n \geq 0:|f^{(n)}(x)|\leq C_K^{n+1}n!$$ where in the last formula, if $K$ contains $0$ and $x=0$, then the $n$-th derivative in the formula is the $n$-th right derivative in $0$.

Is it true in this case that $f$ is analytic in $[0,\infty)$ and that for some $\epsilon > 0$,

$$\forall x \in [0,\epsilon) \ \colon \ f(x) = \sum_{n=0}^{\infty}\frac{f^{(n)}(0)}{n!}x^n$$

?

If in addition we have a constant $C$ such that $$\forall x \in [0,\infty):\forall n \geq 0:|f^{(n)}(x)|\leq C^{n+1}n!$$

does the following hold: $$\forall x \in [0,\infty) \ \colon \ f(x) = \sum_{n=0}^{\infty}\frac{f^{(n)}(0)}{n!}x^n$$

?

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    $\begingroup$ You could try re-using the known result for $g(x) = f(x^2)$: one should be able to prove that $g$ satisfies the condition that you mentioned by using Faà di Bruno's formula. But it is perhaps simpler to just use Taylor's theorem and estimate the remainder as in the two-sided case. $\endgroup$
    – Mateusz Kwaśnicki
    Commented Jan 18, 2019 at 9:10

2 Answers 2

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The answer to the first question is yes, as explained by Liviu Nicolaescu. The answer to the second question is no. Consider a bounded analytic function in the strip $\{z :|\Im z|<2/C\}$ for which the boundary of the strip is the natural boundary. Then the radius of convergence of the Taylor series at $0$ is $2/C$, and it cannot represent the function on the positive ray. On the other hand, $$|f^{n}(x)|=\left|\frac{n!}{2\pi}\int_{|z-x|=1/C}\frac{f(z)dz}{(z-x)^{n+1}}\right|\leq C^nn!.$$

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  • $\begingroup$ How did you use in this case the existence of the global constant $C$ in $|f^{(n)}(x)|\leq C^{n+1}n!$ ? Does a simple explicit example exist? $\endgroup$
    – co.sine
    Commented Jan 18, 2019 at 14:41
  • $\begingroup$ By Cauchy's inequality. $\endgroup$
    – Alexandre Eremenko
    Commented Jan 18, 2019 at 14:42
  • $\begingroup$ Alexandre Eremenko please help me understand, from the Cauchy inequality we have $|f^{n}(x)|=\left|\frac{n!}{2\pi}\int_{|z-x|=1/C}\frac{f(z)dz}{(z-x)^{n+1}}\right|= \int_{|z-x|=1/C}\frac{|f(z)|dz}{|(z-x)|^{n+1}}\leq n!\cdot \max_{z\colon|z-x|\leq1/C}|f(z)|\cdot \frac{1}{2\pi} \cdot \int_{|z-x|=1/C}\frac{dz}{\frac{1}{C}^{n+1}} \leq n! \cdot \max_{z\colon|z-x|\leq1/C}|f(z)| \cdot \frac{1}{2\pi} \cdot 2\pi \frac{1}{C}\ \cdot C^{n+1}= n! \cdot \max_{z\colon|z-x|\leq1/C}|f(z)| \cdot C^{n}$ $\endgroup$
    – co.sine
    Commented Jan 18, 2019 at 15:15
  • $\begingroup$ Please show me where I'm wrong, how do I get the $\max$ of $f$ in $|z-x|\leq \frac{1}{C}$ $\endgroup$
    – co.sine
    Commented Jan 18, 2019 at 15:17
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    $\begingroup$ For an explicit example, one may take $f(z)=1/(1+z^2)$: this is bounded in the strip $|\operatorname{Im}z|\le1/2$ (or in any region bounded away from $\pm i$, for that matter), but the Taylor series at 0 has radius of convergence 1 due to the poles at $\pm i$. $\endgroup$
    – Emil Jeřábek
    Commented Jan 18, 2019 at 18:44
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The Taylor formula states that, under your assumptions, for any $x>0$ and any positive integer $n$ we have $$ f(x)= f(0)+f'(0)x+\frac{1}{2!}x^2+\cdots +\frac{1}{n!} f^{(n)}(0)x^n+R_n(x), $$ where according to Lagrange $$ R_n(x)=\frac{1}{(n+1)!}f^{(n+1)}(\xi)x^{n+1}, $$ for some $\xi\in (0,x)$. A reference of this fact under your specific assumptions is Theorem 2, Section 5.3.3 of

V.A. Zorich: Mathematical Analysis I, Springer 2004.

If in your estimate for the derivatives you take $K=[0,1]$ and set $C=C_K$ then one deduces that the series $$ \sum_{n\geq 0}\frac{f^{(n)}(0)}{n!}x^n $$ converges uniformly to $f(x)$ for $0\leq x\leq \frac{1}{2C}$.

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  • $\begingroup$ Liviu Nicolaescu how did you get the uniform convergence in $0\leq x \leq \frac{1}{2C}$? Can the convergence be extended to a larger domain? $\endgroup$
    – co.sine
    Commented Jan 18, 2019 at 15:26
  • $\begingroup$ $R_n(x)$ converges uniformly to $0$ on any compact $K\subset[0,1/C)$. $\endgroup$
    – Liviu Nicolaescu
    Commented Jan 18, 2019 at 15:54
  • $\begingroup$ $$\left| \;f(x)-\sum_{k=0}^n\frac{ f^{(k)}(0)}{k!}x^k\;\right|=|R_n(x)|\leq |(Cx)^{n+1}| $$ $\endgroup$
    – Liviu Nicolaescu
    Commented Jan 18, 2019 at 19:20

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