Apparent Contradiction to Liouville's Theorem

Question

The function of a real variable defined by $f(x)=\sin(x)$ is analytic everywhere and bounded because $|\sin(x)|\leqslant 1$ for all $x$ but it is certainly not a constant. Should this contradict Liouville's theorem? Why is it not a contradiction?

It is not bounded as a complex variable function. Let $x=-1000i$. — André Nicolas, Dec 23 '15 at 15:51
No purely real, non-constant function is analytic in the complex plane. — Mark Viola, Dec 23 '15 at 16:42

score 11 · Accepted Answer · edited Jul 06 '16 at 18:31

11

The inequality $|\sin(x)|\leq 1$ is only true for real values of $x$. In fact, $\sin$ is unbounded when considered as a complex function. Note that we have $$ \sin(z)=\frac{e^{iz}-e^{-iz}}{2i} $$ so $\sin$ grows unbounded along the imaginary axis.

edited Jul 06 '16 at 18:31

Zain Patel

16,802

answered Dec 23 '15 at 15:52

TomGrubb

12,909

score 1 · Answer 2 · edited Apr 13 '17 at 12:21

1

The function $\sin$ isn't bounded in $\Bbb C$.

edited Apr 13 '17 at 12:21

Community

1

answered Dec 23 '15 at 15:52

Martín-Blas Pérez Pinilla

41,546
4
46
89

Apparent Contradiction to Liouville's Theorem

2 Answers2

Linked

Related