Sequence of entire function that converges uniformly over on sets with empty interior

Question

I have to prove that the sequence of entire functions:

$$f_n(z)=\frac 1n \sin(nz)$$

converges uniformly over $\mathbb{R}$ (and this I managed to verify) but doesn't on every set with non-empty interior of the complex plane $\mathbb{C}$. I guess it has to do with Picard theorem but I'm not sure on how to proceed.

Answer 1

We have $\sin n(x+iy)=\sin nz \cosh ny +i\cos nz \sinh ny.$

For convenience let $f(ny)=\frac {e^{|ny|}-1}{2}.$

For $0\ne y\in \Bbb R$ and $\frac {1}{|y|}<n\in \Bbb N$ we have $$\min(|\sinh ny|,|\cosh ny|)=|\sinh ny|=\frac {e^{|ny|}-e^{-|ny|}}{2}>\frac {e^{|ny|}-1}{2}=f(ny).$$ We have $\max (|\sin nx|,|\cos nx|)\geq \frac {1}{\sqrt 2}$ because $|\sin nx|^2+|\cos nx|^2\geq |\sin^2 nx+\cos^2 nx|=1.$

So if $z=x+iy$ with $x,y\in \Bbb R$ and $y\ne 0$, and if $\frac {1}{|y|}<n\in \Bbb N$ then $$n^{-1}|\sin nz|\geq n^{-1} \max (|Re (\sin nz)|,|Im(\sin nz|)=$$ $$=n^{-1}\max (|\sin nx\cosh ny|, |\cos nx \sinh ny|)\geq$$ $$> n^{-1}\max (|\sin nx|\cdot f(ny),|\cos nx|\cdot f(ny))\geq$$ $$\geq n^{-1} \frac {1}{\sqrt 2}f(ny)=\frac {e^{|ny|}-1}{2n \sqrt 2}$$ which $\to \infty$ as $n\to \infty.$

Answer 2

If $A$ is a subset of $\Bbb C$ with nonempty interior, then there exists $z\in A$ with $z=x+i\,y$, $y\ne0$. Then $$ \sin(n\,z)=\sin(n\,x)\cosh(n\,y)+i\cos(n\,x)\sinh(n\,y). $$ It is now easy to see that this sequence is unbounded.