Timeline for Must the set of lines through the origin on which a nonconstant entire function is bounded be finite?
Current License: CC BY-SA 2.5
6 events
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| Jun 28, 2010 at 12:20 | comment | added | Willie Wong | Ah! Good points both from Jonas and Andrey. That cleared things up for me. I don't remember my Mittag-Leffler function well, and was under the mis-impression that $\alpha$ has to be integral. Sorry about the noise. | |
| Jun 28, 2010 at 12:03 | comment | added | Andrey Rekalo | Thanks for the comments. @Willie Wong: In any case, $E_{\alpha,1}$ is bounded everywhere except for a small sector when $\alpha$ is small. This allows for a continuum of complete lines where the function is bounded. | |
| Jun 27, 2010 at 23:18 | comment | added | Jonas Meyer | That's OK, just take $z\mapsto e^{z^2}$ and consider $\pi/4<\arg z<3\pi/4$. +1: This answers the cardinality question much more simply. | |
| Jun 27, 2010 at 23:07 | comment | added | Willie Wong | Though Andrey, I think one of the points of the question is that the restriction of the function be bounded on both the positive and negative directions of the angular sector. | |
| Jun 27, 2010 at 22:41 | history | edited | Andrey Rekalo | CC BY-SA 2.5 |
added 101 characters in body; edited body; deleted 5 characters in body
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| Jun 27, 2010 at 22:09 | history | answered | Andrey Rekalo | CC BY-SA 2.5 |