On analytic function differentiable on the circle of convergence of its Taylor series - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-20T11:55:58Z http://mathoverflow.net/feeds/question/119998 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/119998/on-analytic-function-differentiable-on-the-circle-of-convergence-of-its-taylor-se On analytic function differentiable on the circle of convergence of its Taylor series woodbass 2013-01-27T08:20:15Z 2013-01-31T20:11:47Z <p>For simplicity, we assume that $f(z)$ is analytic in the unit disk $\Delta: |z|&lt;1$ 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.<br> Let $f(z)=(1-z)^2\log (1-z)$.</p> <p>Question 1. Is the Taylor series $\sum a_nz^n$ convergent at every point of $S^1$? </p> <p>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".)</p> <p>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): <a href="http://math.stackexchange.com/questions/286119/continuity-of-analytic-function-implies-convergence-of-power-series" rel="nofollow">http://math.stackexchange.com/questions/286119/continuity-of-analytic-function-implies-convergence-of-power-series</a></p> http://mathoverflow.net/questions/119998/on-analytic-function-differentiable-on-the-circle-of-convergence-of-its-taylor-se/120108#120108 Answer by David Speyer for On analytic function differentiable on the circle of convergence of its Taylor series David Speyer 2013-01-28T13:37:49Z 2013-01-29T15:19:50Z <p>Here is my attempt:</p> <p>Set $g(\theta) = f(e^{i \theta})$. </p> <p><b>Lemma</b> The Fourier coefficients of $g$ are $(\ldots, 0,0,0,a_0,a_1,a_2,\ldots)$. </p> <p><b>Proof:</b> For $r&lt;1$, we have <code>$\frac{1}{2 \pi} \int_{\theta=0}^{2 \pi} f(r e^{i \theta} )e^{- n i \theta} d \theta = a_n$</code>. 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 <code>$$\lim_{r \to 1^{-}} \int_{\theta=0}^{2 \pi} f(r e^{i \theta} e^{- n i \theta} ) d \theta$$</code> and conclude that <code>$$\int_{\theta=0}^{2 \pi} f(e^{i \theta} e^{- n i \theta} ) d \theta = a_n.$$</code> $\square$</p> <p>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.</p> <p>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.</p> <p>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 $\Delta$.</p> <hr> <p>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:</p> <p>(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 $\Delta$.</p> <p>We could also give the "real differentiable" analogue of this definition:</p> <p>(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 $\Delta$.</p> <p>I don't know if there is a difference between these.</p> <p>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:</p> <p>(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 <code>$g|_{\Delta \cap U} = f|_{\Delta \cap U}$</code>. </p> <p>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".</p> <p>(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 <code>$g|_{\Delta \cap U} = f|_{\Delta \cap U}$</code>. </p> <p>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.</p>