Prove an entire function under certain conditions is constant.

[mahler1]

Homework Statement

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.

 

[mfb]

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).

 

[mahler1]

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).

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.

 

[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).

 

[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.

 

[mahler1]

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.

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 $f(z+z_0)=f(z)=f(z+z_1)$?

 

[pasmith]

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.

(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.

 

[mfb]

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 $z$.

$n z_0 + m z_1$ 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].