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I am working on a proof and I would like to make the following assertion:

If a real function $f: [z,\infty) \to (z,\infty)$ is analytic on $[z,\infty)$, and I know that $f(z) \neq 0$, then $f$ is analytic at $z$.

(I know that f is completely monotone on its domain.) Any help with thinking about whether or not this assertion may be true would be greatly appreciated. Thank you!

So I think the answer to my question is "no." For $f$ to be analytic at $z$, $z$ would have to be an interior point of $[z,\infty)$, which is not. It would be great if I were wrong. Thanks!

Anna
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  • Since $f$ is defined on $[z,\infty),$ then of course $f(z)$ exists; you don't need to assume it again. – zhw. Mar 30 '18 at 17:41
  • Thanks. I made a mistake, I wanted to say that $f(z) \neq 0$. I corrected this now. – Anna Mar 30 '18 at 18:02
  • you need to add that $f$ is continuous also. And you are saying that $f$ is already analytic at $z$, so there is nothing to show. – Masacroso Mar 30 '18 at 18:34
  • If $f$ is only defined on $[z,\infty),$ then how can $f$ be analytic at $z?$ – zhw. Mar 30 '18 at 19:38

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An analytic function $f$ on an open interval can be continuously extended to the boundary if $f$ is bounded. However this doesnt imply that $f$ would be analytic in the boundary.

A classical example is the function

$$f:(0,\infty)\to\Bbb R,\quad x\mapsto e^{-1/x}$$

continuously extended to the point zero with $f(0):=0$. Then the extended $f$ is smooth but it is not analytic at zero because $f^{(k)}(0)=0$ for all $k\in\Bbb N$.

Im not sure if this answer your question.

Masacroso
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  • I see, thank you! I realized that I had made a mistake in the question. I do impose that the function is non-zero at z. I corrected this now. – Anna Mar 30 '18 at 18:03
  • @Anna if $f(z)\neq 0$ it doesnt change something. From my function set $g(x):=f(x)+1$. Then $g(0)\neq 0$ but $g$ is still not analytic. Monotony doesnt help either, my function $g$ is monotone – Masacroso Mar 30 '18 at 18:36