How do you show that a doubly-periodic continuous function $f:\Bbb C→\Bbb C$ is bounded?
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Mike Pierce
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J.doe
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3What about the exponential function $z \to e^z$? Even for doubly periodic you need the entire function. – Arpit Kansal Aug 31 '16 at 07:04
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3Perhaps you mean doubly-periodic? Compare http://math.stackexchange.com/questions/1244416/proving-that-a-doubly-periodic-entire-function-f-is-constant. – Martin R Aug 31 '16 at 07:05
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It is not, unless it is constant. – H. H. Rugh Aug 31 '16 at 07:05
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1@ArpitKansal If the function is doubly periodic and continuous, then it is necessarily bounded. No need for it to be entire, although you do need that it is defined on the whole of $\Bbb C$. – Arthur Aug 31 '16 at 07:16
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@Arthur Ah thanks..it will be constant if its entire. – Arpit Kansal Aug 31 '16 at 07:20
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1See this. – Ivo Terek Sep 01 '16 at 00:13
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The result is not true unless you mean doubly-periodic instead of periodic.
If that is what you mean, then here is a hint:
A continuous function on a compact subset of $\mathbb{C}$ is bounded (and attains its minimum and maximum).
The values of a doubly-periodic function are all equal to the values on some initial parallelogram containing the origin.
Caleb Stanford
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@mrf But the result is not true without the doubly-periodic assumption. Even $z \mapsto e^z$ will be a counterexample. – Caleb Stanford Aug 31 '16 at 07:49
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1Of course it's not true, but it's just as likely that the OP has misunderstood this (after all, it's true for $\mathbb{R}$) as that (s)he mistook periodic and doubly periodic. Surely, it's better to wait for clarification from the only one who actually knows what they want to ask. – mrf Aug 31 '16 at 07:52
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I don't understand the part 2 of the answer. Would you explain it a little more? Thanks! – J.doe Sep 01 '16 at 21:47
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@J.doe let's say the two periods are x and y. Just thinking of them as vectors, all complex numbers z can be written as ax+by for a and b real numbers. Then subtract or add x and y until you get a'x + b'y where a', b' are between 0 and 1. The value of the function is the same since subtracting x or y didn't affect it. The set of z such that z = a'x + b'y and a', b' are between 0 and 1 is a parallelogram. – Caleb Stanford Sep 01 '16 at 21:56
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So 0 and 1 are just some random numbers right? We just need them to be bounded. – J.doe Sep 01 '16 at 22:04
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@J.doe Yeah! That's right. Of course, you can only subtract or add an integer number of $x$s and $y$s, so you can't get it between $0$ and $\frac12$, for example. But instead of between $0$ or $1$, you could get between $-10$ and $10$ or something. – Caleb Stanford Sep 01 '16 at 22:30
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