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!

  • $\begingroup$ Certainly, if $f$ is constant, then it's 0. (proof: $|f(0)| \leq 0 \Rightarrow f(0) = 0$.) $\endgroup$ – user29743 Jul 16 '12 at 18:11
  • $\begingroup$ @countinghaus: So essentially, my proof is correct? $\endgroup$ – Libertron Jul 16 '12 at 18:13
  • 3
    $\begingroup$ @countinghaus is right. But $|f(z)| \leq \log(1+|z|)$ doesn't imply that $f$ is bounded. $\endgroup$ – joriki Jul 16 '12 at 18:13
  • $\begingroup$ yeah, we're not done yet, I was just saying that if you can get $f$ to be constant, then you're done. $\endgroup$ – user29743 Jul 16 '12 at 18:14
  • $\begingroup$ @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? $\endgroup$ – Libertron Jul 16 '12 at 18:16

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 $f$.

| cite | improve this answer | |
  • $\begingroup$ So from the above it follows that $$\log(1+|z|)=kz\,\,,\,k\in\Bbb C\,\,\,\text{a constant}?$$ $\endgroup$ – DonAntonio Jul 16 '12 at 18:55
  • $\begingroup$ @DonAntonio I don't see how that follows from the above. $\endgroup$ – marlu Jul 16 '12 at 19:07
  • $\begingroup$ Perhaps I misunderstood something @Marlu, but isn't this what you wrote in the last two lines of your answer? $\endgroup$ – DonAntonio Jul 16 '12 at 19:11
  • 2
    $\begingroup$ 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. $\endgroup$ – marlu Jul 16 '12 at 19:29

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 $z\in\mathbb{C}$.

| cite | improve this answer | |
  • $\begingroup$ shouldn't there in the representation for $\,c_n\,$ be some factorial (I think it is $\,n!\,$ ) somewhere? $\endgroup$ – DonAntonio Jul 16 '12 at 18:54
  • $\begingroup$ No, it shouldn't. $\endgroup$ – Norbert Jul 16 '12 at 19:02

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 $f \equiv 0$.

| cite | improve this answer | |

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.