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Liouville's Theorem states that every bounded entire function must be constant. Does it work in real analysis? Justify your answer! I asked it because Liouville's Theorem is proved by complex analysis.

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What examples of analytic real functions do you know? –  Qiaochu Yuan Jul 19 '11 at 3:35
    
$X^2 + y^2 =r^2$ –  Victor Jul 19 '11 at 3:39
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That is not a function, but an implicit equation. –  Qiaochu Yuan Jul 19 '11 at 3:42
    
Entire function is a concept from complex analysis, so one would have to clarify what your question means. But think $\sin x$, or $e^{-x^2}$. –  André Nicolas Jul 19 '11 at 3:44
    
Is the reciprocal of a polynomial with no real zeros real analytic? Is it bounded? (Hint: you may use complex analysis to prove that such a function is real analytic.) –  Amitesh Datta Jul 19 '11 at 4:42
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up vote 16 down vote accepted

Actually it does work in real analysis. The question is only which condition replaces the "entire" because it is certainly not true for all real-valued functions (take $\sin(x)$ as Chandru states). However, if a real-valued function $f$ is harmonic which means that:

$$\frac{\partial^2f}{\partial x_1^2} +\frac{\partial^2f}{\partial x_2^2} +\cdots +\frac{\partial^2f}{\partial x_n^2} = 0$$

It actually has the Liouville Property, isn't that neat?

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+1 An excellent answer. –  Amitesh Datta Jul 19 '11 at 13:13
    
@Amitesh: Thank you. –  Listing Jul 19 '11 at 15:52
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Take $f(x)=\sin{x}$. clearly $|f| \leq 1$ is bounded and entire but is not constant

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What do you mean with entire here? –  wildildildlife Jul 19 '11 at 10:24
    
@wildidildlife-it means it is analytic everywhere –  Victor Jul 19 '11 at 15:22
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@Victor In the context of functions $f:\mathbb{C}\to\mathbb{C}$, "entire" is standard terminology for functions "analytic everywhere". However, in the context of functions $f:\mathbb{R}\to\mathbb{R}$, I think most people would use "analytic everywhere" rather than "entire". I think people prefer to reserve "entire" for complex analysis. –  Amitesh Datta Jul 20 '11 at 11:25
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I will again restate the question as I see it:

True or False: Every bounded, everywhere-differentiable, real-valued function must be constant.

This is false. I'm tempted to give you a counter-example, but that would be against the fundamental principles of patient problem-solving. I will instead follow Qiaochu's lead and note that you should consider some analytic real functions. A simple counter-example exists (simple meaning an elementary function).

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