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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
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
up vote 18 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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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
@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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