$f$ is entire and maps a rectangle to a rectangle, then $f$ is linear

Question

The problem is in the subject line, I have it for homework. $f$ is a complex valued function. For completeness:

Prove that if $f$ is an entire function and for some rectangle $R$, the image $f(R)$ is also a rectangle, then $f$ is linear.

The composition of linear maps is linear, so we can choose the two rectangles to have two edges coinciding with the real and imaginary axes, and their common vertex at the origin. So as a portion of the real line gets mapped to itself, we can take $f$ to be the analytic continuation of a real function. Then I'm stuck.

This question is confusing me. It's well known from the Riemann mapping theorem that there exists many holomorphic functions taking any rectangle to any other given rectangle. But if $f$ is linear, then the two rectangles must be similar. So the added restraint of $f$ being entire seems to be causing this hassle. Thoughts?

The condition that $f$ maps $[0,\epsilon]$ into $\mathbb{R}$ and $i[0,\epsilon]$ into $i\mathbb{R}$ forces the power series of $f$ to only have odd terms. Perhaps the equivalent conditions at the other 3 corners can say something about the odd terms? (I have assumed that $f(0)=0$ here.) — copper.hat, Nov 13 '13 at 06:22
@MichaelHoppe I don't believe I defined an affine function. The questions specifies a specific rectangle $R$, not a general one. — Kieran Cooney, Nov 13 '13 at 19:20
I can see that you can assume that $f$ maps the real line into the real line. That is straightforward. However, I fail to see why $|f(z)| \le A|z|+B$. Where did that come from? I mean that is tantamount to assuming that $f$ is a degree one polynomial? — copper.hat, Dec 19 '13 at 07:43
I did not make this assumption in my proof, which I did not post. In my solution I used the swarchz reflection principle. — Kieran Cooney, Dec 19 '13 at 07:52
Can you elaborate a little please? The function is entire, so I don't see why you would need the Schwartz reflection principle. — copper.hat, Dec 19 '13 at 08:01

Malik Younsi · Answer 1 · 2013-11-14T19:51:32.427

2

As you already mentioned, we can choose the two rectangles to have two edges coinciding with the real and imaginary axes, and their common vertex at the origin.

Now, $f$ restricted to $R$ can be extended to the whole plane by Schwarz reflections, and this extension must be equal to $f$ since $f$ is assumed to be entire.

By construction, it is easy to see that $f$ must have linear growth, i.e. $$|f(z)| \leq A|z|+B$$ for all $z$, for some constants $A$ and $B$. This implies that $f$ is linear, as is easily seen by Cauchy's formula for $f^{(n)}(0)$.

edited Nov 14 '13 at 19:51

answered Nov 14 '13 at 14:01

Malik Younsi

3,868

6

Why must $f$ have linear growth? Also, I understand that angles are preserved when the derivative is non-zero, and it is intuitively appealing that corners should map corners to corners, but it is not clear to me that being homomorphic implies that this is true. – copper.hat Nov 14 '13 at 16:02
@copper.hat : yes, you are right, I assumed that the derivative was non-zero at the corners. For the general case, I have to give it more thought. $f$ has linear growth because the value of $f$ at a point $z$ is given by the value of $f$ at a point of $R$ plus a quantity of the form $m+ni$ where $m,n$ are less than $|z|$. – Malik Younsi Nov 14 '13 at 17:32
@MalikYounsi I answered the question myself very similarly. I didn't encounter the corner problem though. Instead I argued that since the function is entire $f$ must map the whole line to a whole line, ie we can extend the sides of the rectangles. The easiest way to see this is to expand a real valued function in a power series; a tiny line segment will do. If it is real on this line segment, every coefficient in the series is real, so indeed $f$ maps the real line to the real line. This can clearly be extended to general lines. – Kieran Cooney Nov 14 '13 at 19:50
@MalikYounsi: Could you elaborate a little on why the growth is linear please? I do not follow why the value of $f$ is bounded as you described in your last comment. – copper.hat Dec 19 '13 at 08:43

score 0 · Answer 2 · edited Jun 12 '20 at 10:38

0

Note that, if your map takes rectangles to rectangles, then you should notice the following effects of the map

i) it rotates the sides with angle $90^{o}$ and rescale them and this means using the map $az$ in general (you can think of the sides of the rectangle as vectors).

ii) the other effect of the map should be the translation.

edited Jun 12 '20 at 10:38

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answered Nov 13 '13 at 06:32

Mhenni Benghorbal

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$f$ is entire and maps a rectangle to a rectangle, then $f$ is linear

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