1. Not finding help here? Sign up for a free 30min tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Prove an entire function under certain conditions is constant.

  1. Jul 2, 2014 #1
    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. jcsd
  3. Jul 2, 2014 #2

    mfb

    User Avatar
    2016 Award

    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).
     
  4. Jul 2, 2014 #3
    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
  5. Jul 2, 2014 #4
    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).
     
  6. Jul 3, 2014 #5

    pasmith

    User Avatar
    Homework Helper

    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.
     
  7. Jul 3, 2014 #6
    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)##?
     
  8. Jul 3, 2014 #7

    pasmith

    User Avatar
    Homework Helper

    (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].
     
  9. Jul 3, 2014 #8

    mfb

    User Avatar
    2016 Award

    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].
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?
Draft saved Draft deleted



Similar Discussions: Prove an entire function under certain conditions is constant.
Loading...