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location P.R.China
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visits member for 1 year, 8 months
seen Mar 20 at 9:56
Associate Professor in Mathematics

May
22
comment 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
31
comment 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
31
comment 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 $\overline\Delta$.
Jan
31
revised On analytic function differentiable on the circle of convergence of its Taylor series
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Jan
29
comment 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
comment 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
comment 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
28
comment 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
28
revised On analytic function differentiable on the circle of convergence of its Taylor series
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Jan
28
comment 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
28
comment 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
28
comment 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
28
comment 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 $z=1$.
Jan
28
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Jan
27
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Jan
27
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Jan
27
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Jan
27
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Jan
27
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Jan
27
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