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$f$ is an entire function and $|f(z)| < 5 + |z|^{\frac{1}{3}}$ for all $z \in \Bbb C$. Prove $f$ is a constant.

So I want to bound my function in order to use Liouville's theorem, but in order to bound my function, I'm not sure how to do this. I looked at Cauchy's Estimates that says if $f$ is holomorphic on $B(a, R) = \{z : |z - a| < R\}$ and $|f(z)| \leq M$ for some number $M > 0$ and for all $z \in B(a, R)$, then $|f^n (a)| \leq \frac{n! M}{R^n}$ ($f^n (a)$ is the $n$th derivative of $f$ at $a$).

However, in our given inequality we have a variable and not a number. How can I deal with this?

mr eyeglasses
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  • This is similar to http://math.stackexchange.com/questions/326886/entire-fz-le1-sqrtz-implies-f-is-constant – Alex M. Apr 29 '15 at 20:07

2 Answers2

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For all $ z \in B(0, R)$ we have $$ |f(z)| < 5 + |z|^{1/3} < 5 + R^{1/3} \, . $$ Then the Cauchy estimate gives for all $R > 0$ $$ |f^{(n)}(0)| \le \frac {n! (5 + R^{1/3}) }{R^n} $$ With $R \to \infty$ you can conclude that $f^{(n)}(0) = 0$ for all $n \ge 1$. So the power series of $f$ has only the constant term.

Martin R
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  • Could you briefly explain the maximum principle or link a page explaining it? Thank you – mr eyeglasses Apr 29 '15 at 19:35
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    @ᴇʏᴇs: http://en.wikipedia.org/wiki/Maximum_modulus_principle. In this case: If $|f(z)| < 5+R^{1/3}$ for $|z| = R$ then it follows that $|f(z)| < 5+R^{1/3}$ for $|z| < R$. – Martin R Apr 29 '15 at 19:36
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    You don't even need the maximum modulus principle here: For any $z\in B(0, R)$, we have $|f(z)| < 5 + |z|^{1/3} \leq 5 + R^{1/3}$. – anomaly Apr 29 '15 at 19:58
  • @anomaly: You are completely right, thanks! Answer updated. – Martin R Apr 29 '15 at 20:02
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Let $f(z)=a + zg(z)$. Then $|\frac a z + g(z)|= |\frac {f(z)} z| \leq \frac 5 {|z|} + \frac 1 {|z|^{\frac 2 3}}$. For $|z|\to \infty$ this gives $\lim |g(z)| =0$, which shows that $\exists R>0$ such that $g(z)$ is bounded for $|z|>R$. Since $g$ is holomorphic, it will also be bounded for $|z|\leq R$, which shows that $g$ is bounded on the whole $\mathbb{C}$. Then, by Liouville's theorem, $g$ must be constant, so $f=a+zg$ ($g$ constant now!). Now, divide by $|z|^{\frac 2 3}$ and you'll get $|\frac a {|z|^{\frac 2 3}} + \frac z {|z|^{\frac 2 3}} g| \leq \frac 5 {|z|^{\frac 2 3}} + \frac 1 {|z|^{\frac 1 3}}$. For $|z|\to \infty$ this gives $\infty \leq 0$, unless $g=0$ and then $f=a$, a constant.

Alex M.
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  • How is $a$ defined? Why is $g(z) = (f(z)-a)/z$ holomorphic? (I am probably overlooking something :) – Martin R Apr 29 '15 at 19:53
  • @MartinR: In general, if $f$ is holomorphic then $f=\sum \limits _{n=0} ^\infty a_n z^n$. In our case $a=a_0$ and $g(z)=\sum \limits _{n=1} ^\infty a_n z^{n-1}$ (obviously holomorphic). – Alex M. Apr 29 '15 at 19:59
  • OK, I see now what you meant, thanks. – Martin R Apr 29 '15 at 20:04