# Does Liouville's Theorem work in real analysis?

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.

• What examples of analytic real functions do you know? Jul 19, 2011 at 3:35
• $X^2 + y^2 =r^2$ Jul 19, 2011 at 3:39
• That is not a function, but an implicit equation. Jul 19, 2011 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 $e^{-x^2}$. Jul 19, 2011 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.) Jul 19, 2011 at 4:42

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?

• How, so ? I mean how can we prove it ?
– Our
Oct 27, 2018 at 11:36

Take $f(x)=\sin{x}$. clearly $|f| \leq 1$ is bounded and entire but is not constant

• What do you mean with entire here? Jul 19, 2011 at 10:24
• @wildidildlife-it means it is analytic everywhere Jul 19, 2011 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. Jul 20, 2011 at 11:25