# Homework Help: Prove an entire function under certain conditions is constant.

1. Jul 2, 2014

### mahler1

The problem statement, all variables and given/known data

Let $f$ be an entire function such that there exist $z_0,z_1 \in \mathbb C$, $\mathbb R$-linearly independent, with $f(z+z_0)=f(z)$ and#f(z+z_1)=f(z)$for all$z \in \mathbb C$. Show that$f$is constant. The attempt at a solution From the hypothesis, I know that$f$is not injective and that if$z_0=x_0+iy_0, z_1=x_1+iy_1$and$f(x+iy)=u(x,y)+iv(x,y)$, then$u$and$v$are not injective. I'm under the impression that the idea is to use Liouville's theorem, but in order to use it, I have to show that$f$is bounded. If this is a correct way to solve the problem, I would like suggestions on how could I prove the function is bounded. Last edited: Jul 2, 2014 2. Jul 2, 2014 ### mfb ### Staff: Mentor This is not true. Let$z_0 = i$,$z_1=2\pi + i$. Then the equation reads$f(z+i)=f(z+i+2\pi)$for every$z \in \mathbb C$. "For every z" is equivalent to "for every z+i", so we can simplify the statement to$f(z)=f(z+2\pi)$for every$z \in \mathbb C$. f(z)=sin(z) is a counterexample. I think the equation should be$f(z)=f(z+z_0)=f(z+z_1)$. Then it is a meaningful problem statement and you can use Liouville's theorem. Don't split it into components, first consider what the equation tells you (the sin(z) from above is a hint). 3. Jul 2, 2014 ### mahler1 Sorry, I've edited my post with the statement that you've correctly guessed to be the correct one. So, the function is periodic, so a big part of the problem reduces to prove that continuous functions are bounded. Last edited: Jul 2, 2014 4. Jul 2, 2014 ### mahler1 I know that$[a,b]\times[c,d]$is a compact set in$\mathbb R^2$. If$g:\mathbb R^2 \to \mathbb R$is a continuous function, then$f([a,b]\times[c,d])$is compact, which at the same times implies it is bounded (I've proved these statements some time ago). Now, I want to prove the following: 1)$u(x,y)$and$v(x,y)$are periodic, for example,$u(x,y)=u(x+x_0,y+y_0)$for$(x_0,y_0) \neq (0,0)$(it is here where I guess I must use the hypothesis$z_0,z_1$are$\mathbb R$-linearly independent). 2)$u(\mathbb R^2)$is the same set that$u([a,b]\times[c,d])$for some interval$[a,b]\times[c,d]$(and analogously for$v$). If I could show (1) and (2), then$|f(x+iy)|=|u(x,y)+i(x,y)|\leq |u(x,y)|+|v(x,y)|$, and from here it is immediate that$f$is bounded. I would appreciate suggestions to show (1) and (2). 5. Jul 3, 2014 ### pasmith It's not necessary to split $f$ into components. If $z_0$ and $z_1$ are $\mathbb{R}$-linearly independent, then for every $z \in \mathbb{C}$ there exist unique integers $n$, $m$ and a unique $(s,t) \in [0,1)^2$ such that $z = (n + s)z_0 + (m + t)z_1$. This suggests looking at a suitable continuous $g: [0,1]^2 \to \mathbb{C}$ and showing that $f(\mathbb{C}) = g([0,1]^2)$. If $f(\mathbb{C})$ is compact then it is (closed and) bounded, which by definition requires that $|f(z)|$ is bounded. 6. Jul 3, 2014 ### mahler1 Thanks for that simple answer, as you've suggested, one considers$g(s,t):[0,1]^2 \to \mathbb C$to be$g(s,t)=(n+s)z_0+(m+t)z_1$, since$g$is continuous and$[0,1]^2$is compact, then$g([0,1]^2)$is compact which, in particular, means$Im(g)$is bounded. As$Im(f)=Im(g)$, from here one can apply Liouville's theorem. I have two questions: 1) If$z_0$and$z_1$are$\mathbb R$-linearly independent, then for every$z \in \mathbb C$, there exist unique$k_0(z),k_1(z) \in \mathbb R$such that$z=k_0(z)z_0+k_1(z)z_1$. I don't see how you've deduced from here that there exist unique integers$m,n$and and unique$s,t \in \mathbb[0,1]^2$with$z=(n+s)z_0+(m+t)z_1$for all$z$. By the way, are$m,n$the same for every$z$? If not, I have no idea how to define$g$. 2) Where have you used the hypothesis$f(z+z_0)=f(z)=f(z+z_1)$? 7. Jul 3, 2014 ### pasmith (Best not to use "Im(g)" for the image of a complex-valued function; it may be confused with the imaginary part.) That's not the $g$ I was thinking of, but nevermind. For this $g$ you need to show that $f(\mathbb{C}) = f(g([0,1]^2))$ (and also choose values for $n$ and $m$; simplicity suggests $n = m = 0$). Let $\Omega = g([0,1]^2)$. By continuity of $g$ we have that $\Omega$ is compact. A consequence of this is that $\mathbb{C} \setminus \Omega$ is not empty. You need to show that $f(\mathbb{C}) = f(\Omega)$. That reduces to showing that for every $z \in \mathbb{C} \setminus \Omega$ there is a $w \in \Omega$ such that $f(z) = f(w)$. This is where you need the periodicity of $f$. The first step is to show that if $f(z + z_0) = f(z) = f(z + z_1)$ then $$f(z) = f(z + pz_0 + qz_1)$$ for all integers $p$ and $q$. 8. Jul 3, 2014 ### mfb ### Staff: Mentor$n z_0 + m z_1## is like a grid in the complex plane (as the two complex numbers are R-linearly independent, i. e. have a different complex phase). This allows to split the complex plane into a set of parallelograms, where s and t define the position within the parallelogram.

Note that m,n,s,t are unique only with [0,1) as range for s and t, not with [0,1].