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I'm working on some practice exams and in one I am looking at the following question:

Let $f$ be a function holomorphic on $\mathbb{C}$. Suppose that there exist [real] constants $A$ and $B$ such that:

$$\lvert f(z) \rvert \leq A + B \lvert z \rvert ^{1/2} ~ ~ ~ (z \in > \mathbb{C})$$

Prove $f$ is constant on $\mathbb{C}$.

This is what I did, can someone check if it is correct and the steps follow properly? If not, what needs to be fixed? Thanks!


We have:

$$\lvert f(z) \rvert \leq A + B \lvert z \rvert ^{1/2} ~ ~ ~ \iff ~ ~ ~ \frac{\lvert f(z) \rvert - A}{\lvert z \rvert^{1/2}} \leq B$$

And by the triangle inequality and properties of the complex absolute value:

$$\left \lvert \frac{f(z) - A}{z^{1/2}} \right \lvert \leq \frac{\lvert f(z) \rvert - A}{\lvert z \rvert^{1/2}} \leq B$$

By Liouville's theorem, it follows the function:

$$\frac{f(z) - A}{z^{1/2}}$$

is constant. Therefore, for some constant $M \in \mathbb{C}$ we have:

$$\frac{f(z) - A}{z^{1/2}} = M ~ ~ ~ \implies ~ ~ ~ f(z) = A + M z^{1/2}$$

But $f$ is holomorphic on $\mathbb{C}$, whereas $z^{1/2}$ cannot be holomorphic on $\mathbb{C}$. Hence $M = 0$ and so:

$$f(z) = A$$

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  • $\begingroup$ The triangle inequality doesn't imply that $|f(z) - A| \le |f(z)| - A$, though. $\endgroup$ – user61527 Jun 24 '14 at 4:54
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No, this is not correct, The inequality

$$|f(z) - A| \le |f(z)| - A$$

is quite false, and does not follow from the triangle inequality.


For a slightly different way to proceed, try using the Cauchy integral formula to $f'$, since $f$ is entire. For any $z$, we have

\begin{align*} |f'(z)| &= \left|\frac{1}{2\pi i} \int_{\gamma} \frac{f(s)}{(s - z)^2} ds\right| \end{align*}

where $\gamma = z + R e^{it}$, $0 \le t \le 2\pi$ is a large circle with center $z$. Now the numerator can be bounded away in terms of $R^{1/2}$, the path length is $R$, but the denominator contributes a factor of $R^2$. So

$$|f'(z)| \lesssim \frac{1}{R^{1/2}}$$

for all large $R$, so....

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  • $\begingroup$ Thank you! $\left . \right. $ $\endgroup$ – Thomas Jun 24 '14 at 9:29

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