# Being holomorphic at a single point does not imply to be $C^{\infty}$ at the point

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?

• 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. – Kabo Murphy Jan 22 at 23:24

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