On analytic function differentiable on the circle of convergence of its Taylor series

2013-01-27T08:20:15Z

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

Answer by David Speyer for On analytic function differentiable on the circle of convergence of its Taylor series

2013-01-28T13:37:49Z

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.

On analytic function differentiable on the circle of convergence of its Taylor series - MathOverflow

On analytic function differentiable on the circle of convergence of its Taylor series

Answer by David Speyer for On analytic function differentiable on the circle of convergence of its Taylor series