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Let $f(z)$ be an entire function such that $|f(z)-f(2z)|\leq 10$.

Prove that $f(z)$ is a constant.

Let $g(z)=f(z)-f(2z)$, $g(z)$ is an entire function as a sum of two entire functions.

Also $|g(z)|\leq 10$ so $g(z)$ is bounded and therefore constant.

Moreover $||f(z)|-|f(2z)||\leq|f(z)-f(2z)|\leq 10$ or $||f(z)|-|f(2z)||\leq 10$.

How can I continue from here?

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  • $\begingroup$ @ChristianF typo, edited $\endgroup$
    – gbox
    May 2, 2018 at 12:39

3 Answers 3

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With $f(z)-f(2z)=c$, we find $c=0$ from plugging in $z=0$. So $f(2z)=f(z)$ for all $z$. For all $z\in\Bbb C^\times$, we find $n\in\Bbb Z$ such that $1\le |2^nz|< 2$, hence the value of $f(z)$ occurs already in that annulus. It follows that $f$ is bounded, hence constant.

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  • $\begingroup$ Can you please elaborate on" we find $n\in\Bbb Z$ such that $1\le |2^nz|< 2$" $\endgroup$
    – gbox
    May 2, 2018 at 19:02
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As you've shown, the function $g(z)=f(z)-f(2z)$ is constant.

Differentiating iteratively, \begin{align*} &f'(z)=2f'(2z) \implies f'(0)=0\\[4pt] &f''(z)=4f''(2z) \implies f''(0)=0\\[4pt] &f'''(z)=8f'''(2z)\implies f'''(0)=0\\[4pt] &\;\;\;\cdots \end{align*} hence, all coefficients of the power series of $f$, centered at zero, are zero, except possibly for the constant term.

It follows that $f$ is constant.

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Can't comment because I have low rep, but elaboration on Hagen's answer is as follows:

So we know $f(z)=f(2z)$ for all $z$. Let's look at the range of $f$. Letting $A:=\lbrace z\in\mathbb{C}:1\leq|z|\leq 2\rbrace$, we see that $f(\mathbb{C})=f(A)$, since we may multiply/divide any complex number by 2 repeatedly to yield a number in $A$.

Now, $A$ is compact, hence $f(A)$ too, so $f$ is bounded and constant by Liouville.

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  • $\begingroup$ Why have we choose $1\leq |z| \leq 2$? $\endgroup$
    – gbox
    May 9, 2018 at 16:53
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    $\begingroup$ It doesn't matter, $2\leq|z|\leq4$ works too, or anything sensible. We just need a compact set. $\endgroup$ May 10, 2018 at 6:50

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