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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.

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  • $\begingroup$ 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 $\sinh (nr)\approx e^{nr}/2$. $\endgroup$ – DanielWainfleet Dec 9 '18 at 18:48
  • $\begingroup$ @DanielWainfleet what about open sets that don't contains the imaginary axis? $\endgroup$ – Renato Faraone Dec 10 '18 at 8:20
  • $\begingroup$ @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 $\endgroup$ – Brevan Ellefsen Dec 10 '18 at 12:11
  • $\begingroup$ @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 $A$. $\endgroup$ – Renato Faraone Dec 10 '18 at 12:14
  • $\begingroup$ @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. $\endgroup$ – Renato Faraone Dec 10 '18 at 12:26
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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 $n\to \infty.$

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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.

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