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Suppose an entire function $f$ exists with the properties: $$f(z) = f(z+1)$$

$$f(z) = f(z+i)$$ For all $z \in \mathbb{C}$. Show that $f$ is a constant.

I think I must inspect the unit square and show that it characterizes the whole function (rigorously i suppose), and show that $f$ is bounded in that unit square and hence $f$ is bounded, and by Liouville's theorem $f$ is then constant.

I am not sure how to show f is bounded in the unit square, is it possible for an analytic function to be unbounded/have its absolute value get arbitrarily large at any point?

Uq'''12wn1F12u2x3uW31H1JBk9m
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1 Answers1

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What you say is correct.

Just focus on the closed rectangle with vertices $0,1,i,1+i$. Let the rectangle be denoted by $R$. Then $R$ is closed and bounded (hence compact).

Then as $f$ is continuous (entire $\Rightarrow$ continuous) $f(R)$ is compact (hence closed and bounded). ($\because$ continuous image of a compact set is compact)

Now $f(\Bbb C)=f(R)$ is bounded. Hence Liouville's theorem is applicable.

Here is another approach for you which avoids compactness and other topological stuff :

Given $f$ is entire. This implies that given a point $z_0 \in \Bbb C$, $\lim_{z \to z_0} |f(z)| \neq \infty$. (Reason : For if $\lim_{z \to z_0} |f(z)|=\infty$ for some point $z_0$, then $z_0$ has a pole which contradicts with $f$ being analytic at $z_0$).

Let our rectangle be as above. Suppose $f(R)$ is unbounded. That means there exists a point $z_1 \in R$ such that $\lim_{z \to z_1} |f(z)|=\infty.$ Which is a contradiction for our first statement.

Error 404
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  • So is any continuous function $f$ on a bounded set like our rectangle hence bounded? – Uq'''12wn1F12u2x3uW31H1JBk9m Mar 28 '17 at 01:13
  • @Uq'''12wn1F12u2x3uW31H1JBk9m No! Any continuous function on a bounded set might not be bounded. For example, take $\tan x$ on $(-\frac {\pi}2,\frac {\pi}2)$. The key is both closed and bounded in this case. Our rectangle is both closed and bounded and hence it becomes compact in $\Bbb C$ by Heine-Borel theorem. However, in an arbitrary metric space, a closed and bounded set may not be compact. Hope this helps. – Error 404 Mar 28 '17 at 04:10