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I'm preparing for a PhD prelim in Complex Analysis, and I encountered this question from an old PhD prelim:

Suppose $f(z)$ is an entire function such that $|f(z)| \leq \log(1+|z|) \forall z$. Show that $f \equiv 0$.

Well, for $z=0$, $|f(0)| \leq 0$. On the other hand, for $z \neq 0$, $\log(1+|z|) > 0$, a positive constant. I'm guessing this would mean that $f$ turns out to be a bounded entire function, so then by Liouville's theorem, $f$ is constant, but this doesn't necessarily mean that $f \equiv 0$, does it? Am I wrong somewhere? Some guidance would be much appreciated!

Libertron
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    Certainly, if $f$ is constant, then it's 0. (proof: $|f(0)| \leq 0 \Rightarrow f(0) = 0$.) –  Jul 16 '12 at 18:11
  • @countinghaus: So essentially, my proof is correct? – Libertron Jul 16 '12 at 18:13
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    @countinghaus is right. But $|f(z)| \leq \log(1+|z|)$ doesn't imply that $f$ is bounded. – joriki Jul 16 '12 at 18:13
  • yeah, we're not done yet, I was just saying that if you can get $f$ to be constant, then you're done. –  Jul 16 '12 at 18:14
  • @joriki: So what should be done with the issue where $|f(z)| \leq \log(1+|z|)$? Would exponentiating both sides of the inequality be of any help? – Libertron Jul 16 '12 at 18:16

3 Answers3

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Since $f$ is entire it can be expressed as a power series that converges everywhere: $f(z) = \sum_{n=0}^\infty a_n z^n$. From $|f(0)| \leq 0$ we know that $f(0)=0$, hence $a_0 = 0$. So $g(z) := f(z)/z$ can be continued to an entire function that satisfies $$|g(z)| \leq \frac{\log(1+|z|)}{|z|} \quad \text{for all } z\neq 0.$$ The right side converges to zero for $|z| \to \infty$, in particular it is bounded. By Liouville's theorem, $g$ is constantly zero and so is $f$.

marlu
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  • So from the above it follows that $$\log(1+|z|)=kz,,,,k\in\Bbb C,,,\text{a constant}?$$ – DonAntonio Jul 16 '12 at 18:55
  • @DonAntonio I don't see how that follows from the above. – marlu Jul 16 '12 at 19:07
  • Perhaps I misunderstood something @Marlu, but isn't this what you wrote in the last two lines of your answer? – DonAntonio Jul 16 '12 at 19:11
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    Do you mean that the same reasoning as above would show that $\log(1+|z|)/z$ is constant? It doesn't because $g$ needs to be entire to apply Liouville's theorem. – marlu Jul 16 '12 at 19:29
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We will prove a slightly more genaral fact. The proof is based on this answer

Theorem. Let $f\in\mathcal{O}(\mathbb{C})$ and for all $z\in\mathbb{C}$ we have $|f(z)|\leq\varphi(|z|)$. Assume that $$ \lim\limits_{R\to+\infty}\frac{\varphi(R)}{R^{p+1}}=0 $$ then $f$ is a polynimial with $\deg (f)\leq p$.

Proof. Since $f\in\mathcal{O}(\mathbb{C})$, then $$ f(z)=\sum\limits_{n=0}^\infty c_n z^n $$ for all $z\in \mathbb{C}$. Moreover, for all $R>0$ we have integral representation for coefficients $$ c_n=\int\limits_{\partial B(0,R)}\frac{f(z)}{z^{n+1}}dz $$ Then, we get an estiamtion $$ |c_n|\leq \oint\limits_{\partial B(0,R)}\frac{|f(z)|}{|z|^{n+1}}|dz|\leq \oint\limits_{\partial B(0,R)}\frac{\varphi(|z|)}{|z|^{n+1}}|dz|= \frac{2\pi R\varphi(R)}{R^{n+1}}= \frac{2\pi \varphi(R)}{R^{n}} $$ Hence for $n>p$ we obtain $$ |c_n|\leq\lim\limits_{R\to+\infty}\frac{2\pi\varphi(R)}{R^n}= 2\pi\lim\limits_{R\to+\infty}\frac{1}{R^{n-p-1}}\lim\limits_{R\to+\infty}\frac{ R\varphi(R)}{R^{p+1}}=0 $$ which implies $c_n=0$ for $n>p$. Finally we get $$ f(z)=\sum\limits_{n=0}^p c_n z^n+\sum\limits_{n=p+1}^\infty c_n z^n=\sum\limits_{n=0}^p c_n z^n $$ This means that $f$ is a polynimial with $\deg(f)\leq p$.

For your particular problem $\varphi(R)=\log(1+R)$, and it is easy to check that $$ \lim\limits_{R\to+\infty}\frac{\varphi(R)}{R}=0 $$ Hence, $f(z)=c_0$ is a constant function. Moreover, $$ |c_0|=|f(0)|\leq\log(1+|0|)=0 $$ so $c_0=0$ and $f(z)=0$ for all $z\in\mathbb{C}$.

Community
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Norbert
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We already know that $f(0) = 0$. Now from $|f(z)| \leq \log(1 + |z|)$ and we get that on every circle of radius $R$ about the origin, $$|f^{(n)}(0)| \leq \frac{\log(1 + |R|) \times n!}{R^n} \hspace{5mm} \forall n \geq 1$$ by the Cauchy estimate. Since $R^n$ grows faster than $\log (1+ |R|)$ when $n \geq 1$, we see that $f^{(n)}(0) = 0$ for all $n \geq 1$. But we also know $f(0) = 0$ so that $f^{(n)}(0) = 0$ for all $n \geq 0$. Since $f$ is entire the identity principle implies $f \equiv 0$.