This is an old qual problem I'm working on: Show that there is no entire function $f(z)$ satisfying $|f(z)-e^{\overline{z}}|\leq 3|z|$ for all $z\in \mathbb{C}$.

I tried to use Liouville's theorem by dividing both side by $|z|$ ,but it doesn't quite work because $e^{\overline{z}}$ is not analytic. I tried a few more things, but none gave a result worth to mention here. I would appreciate any kind of help.


2 Answers 2


Suppose the inequality is true. Let $f(z) = \sum_{n=0}^{\infty}a_nz^n.$ Let $g(z) = \exp (\bar z ).$ Note that

$$g(re^{it}) = \sum_{n=0}^{\infty}r^ne^{-int}/n!.$$

Using the orthogonality of the exponentials, we then get

$$(1/2\pi)\int_0^{2\pi}|f(re^{it}) - g(re^{it})|^2\, dt = |a_0-1|^2+\sum_{n=1}^{\infty}|a_n|^2r^{2n} + \sum_{n=1}^{\infty}r^{2n}/(n!)^2 \le 9r^2.$$

Because of the second series, this inequality fails for large $r,$ contradiction.

  • $\begingroup$ Nice answer. I would suggest cleaning it up just a little bit by removing the "suppose the inequality is true" at the start, and just showing that the quantity you calculate grows larger than $9r^2$ for large $r$. $\endgroup$
    – Rhys
    Jul 28, 2015 at 22:25
  • $\begingroup$ @BrianFitzpatrick: because it's not really a proof by contradiction. Trying to frame every proof as a contradiction is an easy trap to fall into, but one that should be avoided. $\endgroup$
    – Rhys
    Jul 28, 2015 at 22:34
  • 3
    $\begingroup$ @Rhys Well, contradiction is the way I started thinking about it. True, the proof could be improved. But what is it we want to do here at MSE? I think showing the audience a proof that works, although one that could use some hedge-trimming, is more valuable than pulling a shiny polished jewel out of a hat. So what I'm saying is: I think my proof, together with your comments, is a more honest presentation of what happens in our discipline. So I'll leave the proof as is, and your comment is most certainly a welcome addition. $\endgroup$
    – zhw.
    Jul 28, 2015 at 23:25

By putting $z = 0$, we see that $f(0) = 1$. Thus $g(z) = \frac{f(z) - e^z}{z}$ is analytic on $\mathbb{C}$ as $0$ is a removable singularity. Notice that for $z = x + \pi n i$, $e^z = e^{\bar{z}}$, so we have that $|g(x + \pi n i)| \leq 3$ for all $x \in \mathbb{R}, n \in \mathbb{Z}$.

If we consider horizontal half-strips $S_{N,t} = \{ z: Re(z) > -t, \pi n \leq \Im(z) \leq \pi(n+1) \}$ separately, we see that $g$ is bounded on the three boundaries. We have $|g| \leq 3$ on the horizontal sides, and on the vertical side, $|g(z)| \leq | \frac{f(z) - e^z}{z} | + |\frac{e^z - e^{\bar{z}}}{3}| \leq 3 + \frac{e^{-t}}{z}$.

But the problem statement tells us $g$ satisfies the hypotheses of the Phragment Lindelof principle, as $|g| \leq Ce^{|z|}$ for large $z$. This allows to conclude $|g(x + iy)| \leq 4$ on the interior of each $S_{N,t}$. But by varying $N$ and letting $t \to +\infty$, we see that $|g| \leq 4$ on $\mathbb{C}$.

By Liouville, $g$ is constant, which forces $f(z) = e^z + az$. But by plugging in $z = x + i\pi/2$ for large positive $x$, this contradicts the original inequality.


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