2

Let $a,b \in \mathbb{C}$ linearly independent over $\mathbb{R}$ and $f: \mathbb{C} \to \mathbb{C}$ analytic with $f(z)=f(z+a)=f(z+b)$ for all $z \in \mathbb{C}$. Show that f is constant.

WLOG, I'll assume that $a=1;b=i$. Now, we have that $f(\mathbb{C})=f([0,1]^2)$, because given $w=x+yi; \ x, y \in \mathbb{R}$; $w'=(x-\lfloor{x}\rfloor)+(y-\lfloor y \rfloor)i \in [0,1]^2$, so: \begin{align*} &f(w')=f((x-\lfloor{x}\rfloor)+\lfloor{x}\rfloor+(y-\lfloor y \rfloor)i) \\[7pt] =& f(x+(y-\lfloor y \rfloor)i+(\lfloor y \rfloor)i)\\[7 pt] =& f(x+yi)=f(w) \end{align*}

But as $[0,1]^2$ is compact, $|f|$ attains a maximum for some $z_0 \in [0,1]^2$. But by the property above, $|f(z)| \leq |f(z_0)|; \forall z \in \mathbb{C}$ and as $f$ is analytic, by Maximum Modulus Principle, $f$ is constant.

For other $a,b$, the procedure is analogous, considering $w'=xa+yb; \ x, y \in \mathbb{R}$.

Is this a correct approach; any suggestions?

J P
  • 1,152
  • That's a good approach. Your WLOG argument is really dealing with a special typical case only. For general $a,b$ there is still a small technical gap: what replaces $[0,1]^2$ in that case?? – Hans Engler Apr 13 '24 at 15:06
  • 1
    That is the standard proof, compare e.g. https://math.stackexchange.com/a/1244421/42969 – Martin R Apr 13 '24 at 15:08
  • 1
    @HansEngler, it'd be the paralellogram defined by the complex numbers $0,a,b,a+b$ as vertices – J P Apr 13 '24 at 16:17

0 Answers0