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.
Does Liouville's Theorem work in real analysis?
1
$\begingroup$
real-analysis
complex-analysis
-
0entire functions are imposed the Cauchy–Riemann condition. However there's no such a constraint to real functions. – 2011-07-19
2 Answers
22
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?
-
0How, so ? I mean how can we prove it ? – 2018-10-27
6
Take $f(x)=\sin{x}$. clearly $|f| \leq 1$ is bounded and entire but is not constant
-
1@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. – 2011-07-20