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For $f\in C^{\infty}[0,1]$ such that $\max_{[0,1]}|f^{(n)}|\le M^n\cdot n!$ prove that $f$ can be extended to an analytic function on domain $[0,1]\subset G$.

What I am quite not sure about is a claim I came by several times, according to which, if a function has a radius of convergence such that its Taylor series converges around a real point, then it converges for the disk of the same radius around that point, but I find that pretty new, thinking that I can't even tell whether the function can work with complex inputs. Can you generally help me understand the above strategy relevance, and the logic according to which the above is true?

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    By $f^n$ do you mean $n$th power of $f$ or $n$th derivative of $f$? – 2017-01-19
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    What does $G$ mean in this context ? – 2017-01-19
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    $G$ stands for a connected open subset of $\Bbb{C}$ containing $ [0,1]$. As for the $f^n$, it was a typo... – 2017-01-19
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    What are you given about $M$? – 2017-01-19
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    The convergence of a Taylor series for a $C^\infty$ function $f$ does not imply that the Taylor series converges to $f$. As an example, for smooth functions with compact but nonempty support the Taylor series is usually identically $0$ at boundary points of the support. What you are actually supposed to show (and it's not so easy if I recall correctly) is that the inequality you have implies that the Taylor series of $f$ not only converges but converges to $f$ (locally near every point). – 2017-01-19
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    Actually, a $C^\infty-$function in a closed interval extends analytically iff $$\limsup_{n\to\infty}\frac{\|f^{(n)}\|^{1/n}}{n}<\infty.$$ – 2017-01-19

2 Answers 2

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Here is a slightly different line of reasoning:

  • Set $r = 1/M$ and define $B(a, r) = \{ z \in \Bbb{C} : |z - a| < r \}$ as the open ball of radius $r$ at $a$. Now for each fixed $a \in [0, 1]$, define $\tilde{f_a} : B(a, r) \to \Bbb{C}$ by

    $$ \tilde{f_a} (z) := \sum_{n=0}^{\infty} \frac{f^{(n)}(a)}{n!} (z-a)^n. $$

    Since $|f^{(n)}(a)/n!| \leq r^{-n}$, it follows that this series converges absolutely on $B(a,r)$ and hence defines a holomorphic function on $B(a,r)$.

  • Now we claim that $f \equiv \tilde{f_a}$ on the interval $(a-r,a+r)\cap [0,1]$ for each $a \in [0, 1]$. Once this claim is proved, then a simple application of identity theorem tells that $\tilde{f_a} \equiv \tilde{f_b}$ whenever $B(a,r)$ and $B(b,r)$ intersects, and hence we can patch them together without ambiguity to create a holomorphic function on $G := \cup_{a \in [0,1]} B(a,r)$ which extends $f$.

  • Proving the claim amounts to establishing that the Taylor series of $f$ at $a$ converges to $f$ on the prescribed interval. By the Taylor's theorem, we can write

    $$ f(x) = \sum_{n=0}^{N} \frac{f^{(n)}(a)}{n!} (x-a)^n + \frac{f^{(N+1)}(\xi)}{(N+1)!}(x-a)^{N+1} $$

    for some $\xi = \xi(a,x,N)$ between $a$ and $x$. Consequently, for $x \in (a-r,a+r)\cap[0,1]$, the remainder term satisfies the bound

    $$ \left| \frac{f^{(N+1)}(\xi)}{(N+1)!}(x-a)^{N+1} \right| \leq \left( \frac{|x-a|}{r} \right)^{N+1} \xrightarrow[]{N\to\infty} 0. $$

    Therefore the Taylor series converges to $f$ on $(a-r,a+r)\cap[0,1]$ and the claim follows.

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    I had just started to write essentially this answer. My only complaint here is that you should have posted two minutes earlier, so I wouldn't have wasted time starting to type. – 2017-01-20
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    @DanielFischer, I feel sorry for that, especially thinking that I wasted a couple of minutes to check if this is a duplicate answer or not. :s – 2017-01-20
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Set $$ g(x+iy)=\sum_{n=0}^\infty \frac{f^{(n)}(x)(iy)^n}{n!}. $$ Then, the series converges for all $z=x+iy$, with $$ |y|<\frac{1}{M}\quad\text{and}\quad x\in [0,1],\tag{1} $$ and defines $C^\infty$ function for those $x,y$. In particular, $g$ coincides with $f$ in $[0,1]$ and $g$ satisfies the Cauchy-Riemann equations: $$ g_y=ig_x. $$ Thus $g$ is holomorphic in $(1)$.

A detailed proof, and generalisation, appears in:

G. Akrivis, D. T. Papageorgiou and Y.-S. Smyrlis, On the analyticity of certain dissipative–dispersive systems, Bull. London Math. Soc. 45 (2013) 52–60 doi:10.1112/blms/bds061

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    You can't really set $g$ equal to that series until you know it converges. – 2017-01-20
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    It is convergent. $$ |f^{(n)}(x)|\le n! M^n\,\,\,\text{and} \,\,\, |y| – 2017-01-20