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 [itex]f[/itex] into components.

If [itex]z_0[/itex] and [itex]z_1[/itex] are [itex]\mathbb{R}[/itex]-linearly independent, then for every [itex]z \in \mathbb{C}[/itex] there exist unique integers [itex]n[/itex], [itex]m[/itex] and a unique [itex](s,t) \in [0,1)^2[/itex] such that [itex]z = (n + s)z_0 + (m + t)z_1[/itex].

This suggests looking at a suitable continuous [itex]g: [0,1]^2 \to \mathbb{C}[/itex] and showing that [itex]f(\mathbb{C}) = g([0,1]^2)[/itex].

If [itex]f(\mathbb{C})[/itex] is compact then it is (closed and) bounded, which by definition requires that [itex]|f(z)|[/itex] is bounded.

 

[mahler1]

It's not necessary to split [itex]f[/itex] into components.

If [itex]z_0[/itex] and [itex]z_1[/itex] are [itex]\mathbb{R}[/itex]-linearly independent, then for every [itex]z \in \mathbb{C}[/itex] there exist unique integers [itex]n[/itex], [itex]m[/itex] and a unique [itex](s,t) \in [0,1)^2[/itex] such that [itex]z = (n + s)z_0 + (m + t)z_1[/itex].

This suggests looking at a suitable continuous [itex]g: [0,1]^2 \to \mathbb{C}[/itex] and showing that [itex]f(\mathbb{C}) = g([0,1]^2)[/itex].

If [itex]f(\mathbb{C})[/itex] is compact then it is (closed and) bounded, which by definition requires that [itex]|f(z)|[/itex] 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 [itex]g[/itex] I was thinking of, but nevermind. For this [itex]g[/itex] you need to show that [itex]f(\mathbb{C}) = f(g([0,1]^2))[/itex] (and also choose values for [itex]n[/itex] and [itex]m[/itex]; simplicity suggests [itex]n = m = 0[/itex]).

Let [itex]\Omega = g([0,1]^2)[/itex]. By continuity of [itex]g[/itex] we have that [itex]\Omega[/itex] is compact. A consequence of this is that [itex]\mathbb{C} \setminus \Omega[/itex] is not empty.

You need to show that [itex]f(\mathbb{C}) = f(\Omega)[/itex]. That reduces to showing that for every [itex]z \in \mathbb{C} \setminus \Omega[/itex] there is a [itex]w \in \Omega[/itex] such that [itex]f(z) = f(w)[/itex]. This is where you need the periodicity of [itex]f[/itex].

The first step is to show that if [itex]f(z + z_0) = f(z) = f(z + z_1)[/itex] then [tex]f(z) = f(z + pz_0 + qz_1)[/tex] for all integers [itex]p[/itex] and [itex]q[/itex].

 

[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].