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I asked a similar question here but only received a response within the context of real analysis. Since I am mainly interested in the context of complex analysis, I am posting a modified version here.

Suppose we have a function $f$ that is analytic in some region $R$, and the point $z=0$ may be on the boundary $\partial R$. It seems intuitive to me that if $\lim_{z \to 0} f(z)$ exists, then $\lim_{z \to 0} z f'(z) = 0$.

Here is an argument for this that would convince a typical physicist like myself. $f(z)$ cannot have an essential singularity at $0$, because $\lim_{z \to 0} f(z)$ would not exist. So as $z \to 0$, it scales like $f(z) = c + O( z^\alpha )$ for some constants $c$ and $\alpha$. Moreover, $\alpha > 0$ since the limit exists. Therefore, $z f'(z) = O( z^\alpha ) \to 0$.

Is this true? If so, is there a simpler or more elegant proof? If not, what additional assumptions would be needed?

Edit: The case I am most interested in is where $R$ satisfies the necessary conditions for the singularity classification theorem (see, for example, the Wikipedia page for an essential singularity). That is, assume that for every open neighborhood $N$ of $z=0$, $R \cap N$ is non-empty.

Edit: Sangchul Lee's response shows that this is not sufficient, but as I noted in my comment to his answer, intuitively, the counterexample suffers from defining the function on a very artificial domain that requires $z=0$ to be approached basically along the $x$-axis. If the function is analytically continued to a natural domain of some kind (so that essential singularities can be seen for what they are), is the result true?

sasquires
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  • I think expanding $f$ as a series, then derivating it and multiplying by $z$ and finally taking the limit would be a proof. – Gonzalo A. Benavides Jul 17 '18 at 22:21
  • @GonzaloBenavides: not in the boundary. – Martin Argerami Jul 17 '18 at 22:22
  • @MartinArgerami which boundary? – Gonzalo A. Benavides Jul 17 '18 at 22:22
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    You may want to read the question. – Martin Argerami Jul 17 '18 at 22:22
  • You're right, I missed a part. – Gonzalo A. Benavides Jul 17 '18 at 22:23
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    You cannot talk about the type of singularity of $f$ at $0$. For example if $R=B(1,0)$ then $0$ is a boundary point but not an isolated singularity. You are probably thinking of $R$ as a set obtained by taking an open set containing $0$ and then removing $0$. That would make a big difference, so please clarify. – Kavi Rama Murthy Jul 17 '18 at 23:31
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    If $f$ can be expanded as a power series on some ball $B=B(c,|c|) $ then $$ zf'(z) = \frac{z}{z-c} \sum_{k=1}^\infty \sum_{n=k}^\infty a_n (z-c)^n$$ so if $\sup_{z\in B \cap B(0,\epsilon)} | \sum_{n=k}^\infty a_n (z-c)^n | $ is summable, say $O(k^{-2})$, the result holds. Feels too strong though – Calvin Khor Jul 18 '18 at 00:07
  • @KaviRamaMurthy Thanks for clarifying. The case I am most interested in is where $R=B(1,1)$ (which I think is probably what you meant when you wrote $R=B(1,0)$). – sasquires Jul 18 '18 at 16:23
  • I'm not sure what you mean in your first edit. Every open set eith $0$ in its boundary has that property. – zhw. Jul 19 '18 at 15:05
  • @zhw Sure, I think that is an equivalent statement. My edit just reflects what is said on the Wikipedia page that is linked (right underneath "Alternate descriptions"). It is a sufficient condition for being able to identify any point where the function is undefined as a removable singularity, a pole, or an essential singularity. – sasquires Jul 19 '18 at 16:12
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    Those are isolated singularities - very special case of the setup you described. Are you saying that you are most interested in the case where $0$ is an isolated singularity? (It's easy in that case.) – zhw. Jul 19 '18 at 16:25
  • @zhw I see! The Wikipedia paragraph is a little bit unclear, and I had forgotten that the classification scheme only applies to isolated singularities. (It has been a long time since I've studied complex analysis seriously.) Anyway, this is the implicit assumption in my "proof" based on scaling. So the point is: (1) If the singularity is isolated, my statement is true. (2) Sangchul Lee showed that you can make an isolated singularity into a non-isolated singularity by changing the domain. (3) The question of a non-isolated singularity in a natural domain is still undecided. – sasquires Jul 20 '18 at 16:39

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We can adapt one of the counter-examples in real case to this question.

Let $R = \{ x+iy : x > 0 \text{ and } |y| < x^2 \} $ and $f(z) = z \sin(1/z)$. If $z = x+iy \in R$, then

$$ \left| \operatorname{Im}\left(\frac{1}{z}\right) \right| = \frac{\lvert y \rvert}{x^2+y^2} < \frac{x^2}{x^2+y^2} < 1. $$

This proves that we have $|\sin(1/z)| \leq C$ on $R$ for some constant $C > 0$. So $f(z) \to 0$. But

$$ z f'(z) = f(z) - \cos\left(\frac{1}{z}\right) $$

and therefore $zf'(z)$ does not converge as $z\to 0$ in $R$.

Sangchul Lee
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    This is a great adaptation of the counterexample! But the physicist in me wants to say, "We all know that in this case $f(z)$ can be defined on a bigger region (the whole plane minus $z=0$), making $z=0$ an essential singularity." That is, it seems like a trick to make $z=0$ into a removable singularity by changing the domain of the function. So the answer is definitely correct and valuable, it makes me wonder whether a sufficient condition is that the function is analytically continued to some sort of maximal domain. But I do not know enough to formalize this statement. – sasquires Jul 18 '18 at 16:29
  • It seems to me like the reason why this counterexample works is that the domain is defined in such a way that you have to approach $z=0$ basically along the real axis, since each of the boundary curves $y=\pm x^2$ has derivative zero at $x=0$. Therefore, it is not surprising that this is equivalent to the real case. It seems to me like this technique can be used quite broadly to hide essential singularities. Do you have any thoughts on a way to incorporate the "natural domain" of the function and get around this counterexample? – sasquires Jul 19 '18 at 16:19
  • @sasquires, Frankly I have no good idea for general cases. At least I think I can create examples which are maximal in the sense of analytic continuation and still serve as counter-example. (Composing $f$ with a suitable function will do the job.) – Sangchul Lee Jul 20 '18 at 05:07
  • Thanks for thinking about it at least. – sasquires Jul 20 '18 at 16:37
  • @SangchulLee - If $f(z)$ and $zf'(z)$ both converge, must the latter converge to $0$? I'm interested in general absolute-valued fields, not just $\mathbb R$ or $\mathbb C$. This might be applicable to my recent Question. – mr_e_man May 09 '23 at 01:56