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

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.

• Hint: Consider any open $S$ such that $S\supset i\Bbb R= \{ir:r\in \Bbb R\}.$ If $r\in \Bbb R$ then $\sin (nir)=\sinh (nr).$ If $n\in \Bbb N$ is large and $r>0$ then $\sinh (nr)\approx e^{nr}/2$. – DanielWainfleet Dec 9 '18 at 18:48
• @DanielWainfleet what about open sets that don't contains the imaginary axis? – Renato Faraone Dec 10 '18 at 8:20
• @RenatoFaraone the question seems to be to show that it doesn't converge uniformly on at least one open set in $\mathbb{C},$ so we are done – Brevan Ellefsen Dec 10 '18 at 12:11
• @BrevanEllefsen maybe this is a case of linguistic confusion, I want to prove: suppose $A$ is a set with non-empty interior, moreover such interior is not contained in the real axis, then such sequence doesn't converge uniformly over $A$. – Renato Faraone Dec 10 '18 at 12:14
• @BrevanEllefsen That's makes a lot more sense. Probably the exercise were not well written or is my fault that I didn't quite understand it. – Renato Faraone Dec 10 '18 at 12:26

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 $$\frac {1}{|y|} we have $$\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|} 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 $$n\to \infty.$$
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.