Example of continuous function that is analytic on the interior but cannot be analytically continued? - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-23T00:56:10Z http://mathoverflow.net/feeds/question/10831 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/10831/example-of-continuous-function-that-is-analytic-on-the-interior-but-cannot-be-ana Example of continuous function that is analytic on the interior but cannot be analytically continued? Johan 2010-01-05T18:39:32Z 2010-03-27T02:51:59Z <p>I am looking for an example of a function $f$ that is 1) continuous on the closed unit disk, 2) analytic in the interior and 3) cannot be extended analytically to any larger set. A concrete example would be the best but just a proof that some exist would also be nice. (In fact I am not sure they do.)</p> <p>I know of examples of analytic functions that cannot be extended from the unit disk. Take a lacuanary power series for example with radius of convergence 1. But I am not sure if any of them define a continuous function on the closed unit disk. </p> http://mathoverflow.net/questions/10831/example-of-continuous-function-that-is-analytic-on-the-interior-but-cannot-be-ana/10837#10837 Answer by David Speyer for Example of continuous function that is analytic on the interior but cannot be analytically continued? David Speyer 2010-01-05T19:10:43Z 2010-01-05T19:10:43Z <p>Let $f(z) = \sum z^n/n^2$, which is continuous and bounded on the closed unit disc but not analytic near $1$. Then consider </p> <p>$$\sum f(z^n)/n^2.$$</p> <p>This should have a singularity at every root of unity; and should be analytic in the interior because it is uniformly convergent.</p> http://mathoverflow.net/questions/10831/example-of-continuous-function-that-is-analytic-on-the-interior-but-cannot-be-ana/10838#10838 Answer by Harald Hanche-Olsen for Example of continuous function that is analytic on the interior but cannot be analytically continued? Harald Hanche-Olsen 2010-01-05T19:41:11Z 2010-01-05T19:41:11Z <p>I suggest this function: <code>$$f(z)=\sum_{n=1}^\infty \frac{z^{n!}}{n^2}.$$</code> It converges uniformly on the closed unit disk, and the derivatives blow up as you approach any root of unity radially.</p> http://mathoverflow.net/questions/10831/example-of-continuous-function-that-is-analytic-on-the-interior-but-cannot-be-ana/10839#10839 Answer by jvp for Example of continuous function that is analytic on the interior but cannot be analytically continued? jvp 2010-01-05T19:42:57Z 2010-03-27T02:51:59Z <p>One can invoke <a href="http://en.wikipedia.org/wiki/Carath%25C3%25A9odory%27s_theorem_%28conformal_mapping%29" rel="nofollow">Carathéodory's theorem</a>. </p> <blockquote> <p>If $U$ is a simply connected open subset of the complex plane with a <a href="http://en.wikipedia.org/wiki/Jordan_curve_theorem" rel="nofollow">Jordan curve</a> as boundary then the <a href="http://en.wikipedia.org/wiki/Riemann_mapping_theorem" rel="nofollow">Riemann map</a> $f : U \to \mathbb D$ extends continuosly to the boundary and the extension is a homeomorphism $\partial U \to S^1$ at the boundary.</p> </blockquote> <p>To obtain the sougth function, it suffices to consider a simply connected open set $U\subset \mathbb C$ having a nowhere analytic Jordan curve as boundary and take the inverse of the Riemann map of $U$. </p> http://mathoverflow.net/questions/10831/example-of-continuous-function-that-is-analytic-on-the-interior-but-cannot-be-ana/10845#10845 Answer by engelbrekt for Example of continuous function that is analytic on the interior but cannot be analytically continued? engelbrekt 2010-01-05T21:11:25Z 2010-01-05T21:11:25Z <p>Here is a very concrete example:</p> <p>$g(z) = \sum_{n=0}^{\infty}\frac{z^{2^n + 1}}{2^n + 1}.$</p> <p>The power series converges uniformly to a continuous function on the closed unit disk. Differentiating we obtain $g'(z) = f(z)$ with</p> <p>$f(z) = \sum_{n=0}^{\infty}z^{2^n}.$</p> <p>This is <em>the</em> standard example of a function with a natural boundary. Clearly $f(x) \rightarrow +\infty$ as $x \rightarrow 1^{-}$ on the real axis. The functional equation </p> <p>$f(z) = z + f(z^2)$</p> <p>shows that $f(x) \rightarrow + \infty$ as $x \rightarrow (-1)^{+}$ on the real axis, then $|f(z)| \rightarrow \infty$ as $z$ tends radially to ${\pm}i$, and so on, so that $|f(z)|$ tends to $\infty$ as $z$ tends radially to any root of unity of order $2^m$. Hence $f(z)$ has a dense set of singularities on the unit circle, and so does $g(z)$, thus $g(z)$ has the unit circle as natural boundary.</p> http://mathoverflow.net/questions/10831/example-of-continuous-function-that-is-analytic-on-the-interior-but-cannot-be-ana/19483#19483 Answer by Jonas Meyer for Example of continuous function that is analytic on the interior but cannot be analytically continued? Jonas Meyer 2010-03-27T01:47:34Z 2010-03-27T01:47:34Z <p>I found some neat stuff in Remmert's <em>Classical topics in complex function theory</em>.</p> <p><a href="http://books.google.com/books?id=BHc2b0iCoy8C&amp;lpg=PP1&amp;ots=9hKxbNioim&amp;dq=remmert%252520classical%252520topics&amp;pg=PA256#v=onepage&amp;q=&amp;f=false" rel="nofollow">Fabry's gap theorem</a> gives a way to construct many examples including some already mentioned. Stated for the unit disk, it says:</p> <blockquote> <p>If <code>$m_1,m_2,\ldots$</code> is a sequence of positive integers such that <code>$\displaystyle{\lim_{n\to\infty}}\frac{m_n}{n}=\infty$</code> and if $\displaystyle{f(z)=\sum_{n=1}^\infty a_nz^{m_n}}$ has radius of convergence 1, then the unit disk is the domain of holomorphy of $f$.</p> </blockquote> <p>For example, if $p_n$ is the $n^{th}$ prime, then $$f(z)=\sum_{n=1}^\infty \frac{z^{p_n}}{n^2}$$ converges uniformly on the closed disk and is therefore continuous. It is not analytically extendable to any larger set because it satisfies the hypotheses of Fabry's theorem. </p> <hr> <p>An interesting result that yields many such functions in a nonconstructive way is a theorem of Fatou-Hurwitz-Pólya:</p> <blockquote> <p>If $\displaystyle{f(z)=\sum_{n=0}^\infty a_n z^n}$ has radius of convergence 1, then the set of functions $$f_\epsilon(z)=\sum_{n=0}^\infty \epsilon_na_nz^n$$ for <code>$\epsilon_n\in\{\pm1\}$</code> whose domain of holomorphy is the unit disk has cardinality $2^{\aleph_0}$.</p> </blockquote> <p>Hausdorff showed further that if $\displaystyle{\lim_{n\to\infty} |a_n|^{1/n}}$ exists (and equals 1) then the set of such functions whose domain of holomorphy is <em>not</em> the unit disk is at most countable. This applies in particular to the function $\displaystyle{f(z)=\sum_{n=1}^\infty \frac{z^n}{n^2}}$, which therefore yields examples by changing the signs of the coefficients in all but countably many ways.</p> <hr> <p>One more, this time an explicit example from Remmert: The series $$f(z)=1+2z+\sum_{n=1}^\infty\frac{z^{2^n}}{2^{n^2}}$$ is one-to-one and has real derivatives of all orders on the closed disk, and has the open disk as domain of holomorphy. </p> <p>Reference: Remmert's <em><a href="http://books.google.com/books?id=BHc2b0iCoy8C&amp;lpg=PP1&amp;ots=9hKxbNioim&amp;dq=remmert%2520classical%2520topics&amp;pg=PP1#v=onepage&amp;q=&amp;f=false" rel="nofollow">Classical topics in complex function theory</a></em>, pages <a href="http://books.google.com/books?id=BHc2b0iCoy8C&amp;lpg=PP1&amp;ots=9hKxbNioim&amp;dq=remmert%2520classical%2520topics&amp;pg=PA252#v=onepage&amp;q=&amp;f=false" rel="nofollow">252</a>-258. (Fatou-Hurwitz-Pólya is stated on a page without preview.)</p>