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I read a claim such that "being holomorphic at a single point does not imply to be $C^{\infty}$ at the point" which is the second answer in the following post Holomorphic functions and real functions: continuity of partial derivatives

The definition is from Wiki: "holomorphic at a point $z_0$" means not just differentiable at $z_0$, but differentiable everywhere within some neighbourhood of $z_0$ in the complex plane.

Can you give me an example for this claim?

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    $\begingroup$ There seems to be some confusion with definitions here. If $f$ is differentiable in a neighborhood of a point it is certainly $C^{\infty}$ in that neighborhood. $\endgroup$ – Kabo Murphy Jan 22 at 23:24
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This is just a matter of conflicting definitions. If a holomorphic function at a point is defined to be complex-differentiable in a neighborhood of the point (which is the standard definition), then a holomorphic function is always $C^\infty$. On the other hand, if a holomorphic function at a point is just required to be complex-differentiable at the point itself, then it need not be $C^\infty$. The claim which you linked is using the second definition, not the first.

For a simple example, let $g:\mathbb{R}\to\mathbb{R}$ be any function which is differentiable at $0$ with $g'(0)=0$, but not $C^\infty$ in any neighborhood of $0$ (for instance, $g(x)=x|x|$). Then $f:\mathbb{C}\to\mathbb{C}$ defined by $f(z)=g(\operatorname{Re}(z))$ is complex-differentiable at any $z$ with $\operatorname{Re}(z)=0$ (and $f'(z)=0$ at those points), but is not $C^\infty$ in any neighborhood of such a point.

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  • $\begingroup$ Can you give me an example such that $f^{\prime}(z_0)$ exist but $f^{\prime\prime}(z_0)$ does not exist? $\endgroup$ – user315531 Jan 23 at 7:16

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