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For simplicity, we assume that $f(z)$ is analytic in the unit disk $\Delta: |z|<1$ and continous in the closed disk $\overline \Delta$. Let $f(z)=\sum a_nz^n$ be its Taylor series in $\Delta$ which has $R=1$ as its radius of convergence. We further assume that $f(z)$ is (complex-)differentiable at every point of $S^1$. (where the differentiability is global at a point instead of only along certain directions in $\overline\Delta$.) Such a function $f(z)$ exists. Please see a simple example.
Let $f(z)=(1-z)^2\log (1-z)$.

Question 1. Is the Taylor series $\sum a_nz^n$ convergent at every point of $S^1$?

Question 2. If the answer to Question 1 is yes, is the convergence of $\sum a_nz^n$ uniform or absolute on $\overline\Delta$? (Note."uniform" does not imply "absolute".)

Notice. There is a function $f$ analytic in $\Delta$ and continous in $\overline\Delta$ such that its Taylor series $\sum a_nz^n$ is divergent at some pont in $S^1$. Please see (who can teach me how to cite a simple super-link to this): http://math.stackexchange.com/questions/286119/continuity-of-analytic-function-implies-convergence-of-power-series

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12 revisions in under two hours. A record? Has it settled down yet? –  Gerry Myerson Jan 27 '13 at 11:05
    
For the "dual" question: mathoverflow.net/questions/110345/… –  Andres Caicedo Jan 27 '13 at 17:12
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@njguliyev: John Baez once said that broken English is the language of science, and he's absolutely right. While everyone should strive for good language, we should also try to make some allowances for those who do not speak it natively, and focus on the mathematics, provided that it is understandable. –  Todd Trimble Jan 29 '13 at 0:49
    
@Todd Trimble:Thank you for your understanding. I am a Chinese while I am trying to improve my English. –  woodbass Jan 29 '13 at 6:11
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1 Answer 1

Here is my attempt:

Set $g(\theta) = f(e^{i \theta})$.

Lemma The Fourier coefficients of $g$ are $(\ldots, 0,0,0,a_0,a_1,a_2,\ldots)$.

Proof: For $r<1$, we have $\frac{1}{2 \pi} \int_{\theta=0}^{2 \pi} f(r e^{i \theta} )e^{- n i \theta} d \theta = a_n$. Since $f$ is differentiable on $\bar{\Delta}$, it is continuous; since $\bar{\Delta}$ is compact, $f$ is uniformly continuous on $\bar{\Delta}$. So it is valid to interchange integration and limit in $$\lim_{r \to 1^{-}} \int_{\theta=0}^{2 \pi} f(r e^{i \theta} e^{- n i \theta} ) d \theta $$ and conclude that $$\int_{\theta=0}^{2 \pi} f(e^{i \theta} e^{- n i \theta} ) d \theta = a_n.$$ $\square$

Now, the missing part. I want to claim that $g$ is not just differentiable. but $C^1$. If so, then a theorem of Dirichlet states that $g$ is the sum of its Fourier series, as desired.

But now that I think about it, $g$ doesn't need to be $C^1$. Unless I have missed something, $f(z) = (z-1)^2 e^{1/(z-1)}$, as a function on $\bar{\Delta}$, is differentiable at $z=1$ (with derivative $0$) but not continuously so. You might think about whether "differentiable" or "$C^1$" is the condition you really want.

If you really want just differentiable, I'd start by looking up the counterexample of a function which is differentiable but whose Fourier series doesn't converge. My guess is that it won't be too hard to extend that example to the interior of $\Delta$.


There is a discussion going on in the comments about what the right definition is of a function $f$ on $\bar{\Delta}$ being differentiable at a point $z_0$ of the boundary. We should also keep track of the distinction between definitions which are the analogue of "complex differentiable" and definitions which are the analogue of "real differentiable". The definition I wanted to give was:

(1) There is a complex number $a$ such that $f(z) = f(z_0) + a (z-z_0) + o(|z-z_0|)$ for $z$ in $\Delta$.

We could also give the "real differentiable" analogue of this definition:

(1') There are complex numbers $a$ and $\bar{a}$ such that $f(z) = f(z_0) + a (z-z_0) + \bar{a} (\overline{z-z_0}) + o(|z-z_0|)$ for $z$ in $\Delta$.

I don't know if there is a difference between these.

I haven't fully understood the definition the OP wants to give; he would like to require there to be a differentiable extension of $f$ to an open neighborhood of $z_0$ in $\mathbb{C}$. Again, I can imagine a complex analytic or a real differentiable version of this question. Here are some things I could think of:

(2) There is an open neighborhood $U$ of $z_0$ in $\mathbb{C}$ and a complex differentiable function $g:U \to \mathbb{C}$ so that $g|_{\Delta \cap U} = f|_{\Delta \cap U}$.

But complex differentiable function are $C^{\infty}$ and, as I've already pointed out, the answer is yes for $C^1$ functions. So this definition makes the answer to the original question be "yes".

(2') There is an open neighborhood $U$ of $z_0$ in $\mathbb{C}$ and a real differentiable function $g:U \to \mathbb{C}$ so that $g|_{\Delta \cap U} = f|_{\Delta \cap U}$.

My gut is that this is the same as (1') and, in particular, that my example $(z-1)^2 e^{1/(z-1)}$ probably has a differentiable extension to a neighborhood of $z=1$. But I haven't found one.

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$C^1$ is not necessary for the Dirichlet Theorem to hold. The classical Dirichlet-Dini test implies that Hölder $C^\alpha$ (for $\alpha>0$) is enough (see en.wikipedia.org/wiki/…). –  ACL Jan 28 '13 at 14:50
    
@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 $z=1$. –  woodbass Jan 28 '13 at 15:52
    
@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. –  woodbass Jan 28 '13 at 16:12
    
@woodbass: On the disc $\bar{\Delta}$, we have $\Re(z-1) \leq 0$. So $\Re(1/(z-1)) \leq 0$. So $|e^{1/(z-1)}| \leq 1$. A bounded function times $(z-1)^2$ is differentiable at $1$, right? –  David Speyer Jan 28 '13 at 16:19
    
I am assuming that, for $K$ a subset of $\mathbb{C}$, with $z_0$ a point of $K$ and $f: K \to \mathbb{C}$ a function, the definition of "$f$ is differentiable at $z_0$ is that there exists a complex number $a$ such that $f(z) = f(z_0) + a (z-z_0) + o(|z-z_0|)$ as $z \to z_0$ through $K$. If this isn't what you mean, then I ask what you think it means for a function to be differentiable at the boundary of its domain. –  David Speyer Jan 28 '13 at 16:27
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