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May
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Sep
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awarded | Autobiographer |

May
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A question about $L^p$ integral of an entire function on $\mathbb{C}$
@Pooper: How do you prove that $|f|^p$ is subharmonic when $p\in (0,1)$ even in the case of $f$ entire? |

Jan
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On analytic function differentiable on the circle of convergence of its Taylor series
@David Speyer:Your two definitions about differentiablity are not equivalent. (1) implies (1'). $f(z)$ is differntiable on the whole $\overline \Delta $ but it is possible that $f(z)$ is not $C^1$ on the boundary. The dissicussion above actually has given such example. |

Jan
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On analytic function differentiable on the circle of convergence of its Taylor series
@David Speyer: $f(z)$ is (complex-)differentiable at $z=1$, which does not imply that $f'(z)$ is continuous at $z=1$. Of Course, if $f(z)$ is differentiable on a domain, then it is analytic and is $C^\infty$. In my OP, we say that $f(z)$ is differentiable at $z=1$ which does not imply that $f(z)$ is differentiable in a $\mathbb{C}$-neighborhood of $z=1$ although we require that $f(z)$ can be defined on a larger neighborhood of $\overline\Delta$. |

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On analytic function differentiable on the circle of convergence of its Taylor series
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Jan
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On analytic function differentiable on the circle of convergence of its Taylor series
@Todd Trimble:Thank you for your understanding. I am a Chinese while I am trying to improve my English. |

Jan
28 |
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On analytic function differentiable on the circle of convergence of its Taylor series
Now, The quesition can be considered under your and my understanding of the differentiability respectively. I think both of them are interesting. |

Jan
28 |
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On analytic function differentiable on the circle of convergence of its Taylor series
@David Speyer:With your definition on $K=\overline\Delta$, you are right. However, if you consider your example at $K$ which is a ball-neighborhood of $z=1$, $f$ is not differentiable at $z=1$. This is my point. |

Jan
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On analytic function differentiable on the circle of convergence of its Taylor series
We only require that $f$ is differentiable on every point of $S^1$ although we need to define the differentiability at $z=1$, say, on a greater test neighborhood. |

Jan
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On analytic function differentiable on the circle of convergence of its Taylor series
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Jan
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On analytic function differentiable on the circle of convergence of its Taylor series
If $f$ is continuous on $\overline\Delta$, we can extend $f$ continously to a neighborhood of the closed unit disk. But the differentiability cannot be done so. Of course, it is also of great interest to restrict the consideration with your supposition. |

Jan
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On analytic function differentiable on the circle of convergence of its Taylor series
@David Speyer: "where the differentiability is global at some point instead of only along certain directions in $\overline\Delta$". So the differentiability is defined in a full neighborhood of $z=1$, say. |

Jan
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On analytic function differentiable on the circle of convergence of its Taylor series
@David Speyer: "differentiable" is a suitable intermeium contdition compared to $C^1$ and "continuous", isn't it? You have solved the question under "continuous" assumption just as I cited from mathstackexchange. The $C^1$ condition is too strong. |

Jan
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On analytic function differentiable on the circle of convergence of its Taylor series
@David Speyer: I am interested in your attempt. But It seems that $f(z)=(z-1)^2e^{\frac{1}{z-1}}$ is not differentiable at $z=1$. |

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