Suppose $a,b>0$ are contants and $F$ is a non-constant function such that $F(z+a)=F(z)$ and $F(z+ib)=F(z)$. Prove $F$ is not analytic in the rectangle $0\leq a \leq b$ and $0\leq y \leq b$

I don't see how I'd apply Liouville (requirements are bounded+analytic).

I know bounded+analytic imply constant, but does non-constant imply unbounded+non-analytic?

  • $\begingroup$ Hint: you can cover the whole plane with translates of the rectangle. And a holomorphic function on a compact set is bounded. $\endgroup$ – carmichael561 Apr 11 '16 at 16:45
  • $\begingroup$ @carmichael561 I don't see what you mean by translates. Do you mean that $b \rightarrow \infty$? $\endgroup$ – GRS Apr 11 '16 at 16:50
  • $\begingroup$ @carmichael561 Assume it is analytic, then it should attain max on the boundary, implying that $F$ is constant, but it's non constant, so contradictions? $\endgroup$ – GRS Apr 11 '16 at 16:52
  • $\begingroup$ By translates I mean translating the rectangle so that it covers the whole plane. It would help to draw a picture. $\endgroup$ – carmichael561 Apr 11 '16 at 17:03
  • $\begingroup$ @carmichael561 is my proof incorrect? $\endgroup$ – GRS Apr 11 '16 at 17:03

Since there is still some confusion after the comment of @carmichael561. Suppose $F$ is analytic for all $z=x+iy$ with $0\leq x \leq a$ and $0\leq y \leq b$. Since $F$ is analytic it is also continous so $\vert F(z) \vert$ will obtain it's maximum on $z_0$ in this rectangle. Say $\vert F(z_0) \vert=M$. Now for arbitrary $z \in \mathbb{C}$ you find $m,n \in \mathbb{Z}$, s.t. $z=x+iy=x'+an+(y'+mb)i$ with $0\leq x'\leq a$ and $0\leq y'\leq b$.

Now by the functional equation that is satisfied by $F$ you get

$$ \vert F(x+iy)\vert=\vert F(x'+iy')\vert\leq M $$

so $F$ is bounded. Now use Liouville.

  • $\begingroup$ Could you explain what are $y'$ and $x'$? are they the coordinated of $z_0$?If so, are you saying that if there is a point inside a boundary, equal to the maximum on the boundary, then its constant? $\endgroup$ – GRS Apr 11 '16 at 17:28
  • $\begingroup$ @GRS I will give you an example, hopefully you can figure out what I mean from there: Let $a=1=b$ now take $z=x+iy=4.7+2.5i$. This is an complex number. Now what I'm trying to say is that you can find $x',y'$ s.t. $0\leq x'\leq a$ and $0\leq y'\leq b$ and some integers $m,n$ with $x+iy=x'+an+(y'+bm)i$. In this case it's very simple: pick $x'=0.7$ and $y'=0.5$ and let $n=4$ and $m=2$. Then you have $$ 4.7+2.5i=0.7+4\cdot 1+(0.5 +2 \cdot 1)i $$ $\endgroup$ – Maik Pickl Apr 11 '16 at 18:43

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