I'm having trouble with the following exercise from Ahlfors' text (not homework)

"If $f(z)$ is analytic in a neighborhood of $\infty$ and if $z^{-1} \Re f(z) \to0$ as $z \to \infty$, show that $\lim_{z \to \infty} f(z)$ exists. (In other words, the isolated singularity at $\infty$ is removable)

Hint: Show first, by use of Cauchy's integral formula, that $f = f_1 +f_2$ where $f_1(z) \to 0$ for $z \to \infty$ and $f_2(z)$ is analytic in the whole plane.

My attempt:

$f$ is analytic in the exterior of some disk $\{|z|\geq R_0 \}$. For large enough $|z|$ the circle $C_1(z)$ around $z$ with radius one lies in the domain of analyticity, and we can apply Cauchy's integral formula: $$f(z)=\frac{1}{2\pi i}\oint_{C_1(z)}\frac{f(\zeta)}{\zeta-z} \mathrm{d} \zeta=\frac{1}{2 \pi i} \oint_{C_1(z)}\frac{\Re f(\zeta)}{\zeta-z} \mathrm{d} \zeta+\frac{1}{2\pi} \oint_{C_1(z)}\frac{\Im f(\zeta)}{\zeta-z} \mathrm{d} \zeta. $$ I want to say that the last expression is exactly the decomposition $f_1+f_2$. Indeed, using Cauchy's estimate (and the fact about the real part) one can show that the first integral tends to zero for large $|z|$, but I can't understand how to define the other one for all $z \in \mathbb C$.

In fact, even if I have such a decomposition, I don't think that it will suffice: For instance $f(z)=f_1(z)+f_2(z)$ with $f_1(z)=0,f_2(z)=\exp(z)$ has an essential singularity at $\infty$.

Thank you for any directions.


The existence of the decomposition does not depend on the assumption on the real part of $f$. The shortest way to get it is to expand $f$ in Laurent series in some neighbourhood of infinity $W=\{ \vert z\vert >a\}$ : $$f(z)=\sum_{n=-\infty}^{-1} c_n z^n + \sum_{n=0}^\infty c_n z^n\, .$$ The first sum is $f_1(z)$. The second series is a power series converging in $W$, hence in fact in the whole complex plane, so it defines an entire function $f_2(z)$.

If you want to follow the hint to get the decomposition, you can proceed as follows. Fix $R_0$ such that $f$ is holomorphic in an open set containing $\{\vert z\vert\geq R_0\}$. For each $R\geq R_0$, denote by $\Gamma_R$ the circle $\{\vert z\vert=R\}$ and define $$f_R(z)=\frac{1}{2i\pi}\int_{\Gamma_R} \frac{f(\xi)}{\xi -z}d\xi\, ,$$ which is holomorphic in $\mathbb C\setminus\Gamma_R$. By Cauchy's formula, we have $$f(z)=f_R(z)-f_{R_0}(z)$$ whenever $R_0<\vert z\vert <R$. This shows that if $R_0<R< R'$, then the functions $f_R$ and $f_{R'}$ agree in $\{ R_0<\vert z\vert<R\}$, and hence on the disk $D(0,R)$ by analytic continuation. In other words (letting $R\to\infty$), we get a single entire function $f_2$ equal to $F_R$ on $D(0,R)$ for any $R>R_0$. Finally, put $f_1=-f_{R_0}$.

Having done this, observe that the function $f_2$ satisfies the same assumption as $f$, since $f$ and obviously $f_1$ do. Moreover, $f_1$ has a limit at $\infty$ (!) Thus, to prove that $f$ has a limit at $\infty$, it is in fact enough to assume that $f$ is an $entire$ function.

At this point, I don't see how to conclude without using the $Borel$-$Caratheodory\;inequality$, which allows to control $\vert f\vert$ by ${\rm Re} f$. A special case of this inequality reads as follows : for any $r>0$ $$\sup_{\vert z\vert=r} \vert f(z)\vert\leq 2\sup_{\vert z\vert=2r} {\rm Re}f(z)+3\vert f(0)\vert\, .$$ From this, it follows that $z^{-1} f(z)\to 0$ as $z\to\infty$. Then the usual proof of Liouville's theorem shows that $f$ is constant (recall that we are assuming that $f$ is entire).

  • $\begingroup$ Comment on the last paragraph: as hinted in the preceding exercise of Ahlfors, when $f$ is entire, $f(z)=o(\lvert z\rvert)$ follows from Shwarz integral formula. $\endgroup$ – Yai0Phah Apr 13 '14 at 10:20

Show first, by use of Cauchy's integral formula, that $f=f_1+f_2$ where $f_1(z)\to 0$ for $z \to\infty$ and $f_2(z)$ is analytic in the whole plane.

That is a Laurent decomposition of $f$. Supposing $f$ is defined on $\{\lvert z\rvert > r_0\}$, you get it by applying the Cauchy integral formula for an annulus,

$$f(z) = \frac{1}{2\pi i}\int_{\lvert \zeta \rvert = R} \frac{f(\zeta)}{\zeta-z}\,d\zeta - \frac{1}{2\pi i}\int_{\lvert \zeta \rvert = r} \frac{f(\zeta)}{\zeta-z}\,d\zeta$$

where $r_0 < r < \lvert z\rvert < R$. Neither integral depends on the choice of $r$ resp. $R$, as long as the circle doesn't cross $z$.

So let

$$f_2(z) = \frac{1}{2\pi i}\int_{\lvert\zeta\rvert = R} \frac{f(\zeta)}{\zeta-z}\,d\zeta$$


$$f_1(z) = -\frac{1}{2\pi i}\int_{\lvert\zeta\rvert = r} \frac{f(\zeta)}{\zeta-z}\,d\zeta$$

where $r$ and $R$ are chosen arbitrarily subject to the condition above.

Choosing $r = r_0 + 1$, it is easily seen that $\lim\limits_{z\to\infty} f_1(z) = 0$ by the standard estimate.

It is also easy to see that $f_2$ is an entire function.

So it remains to see that

$$\lim_{z\to\infty} \frac{\Re f_2(z)}{z} = 0 \Rightarrow f_2 \equiv \operatorname{const}.$$

Can you take it from there?

$u = \Re f_2$ is an entire harmonic function that grows slower than $\lvert z\rvert$. Differentiating the Poisson integral and using the standard estimate, we find $$\left\lvert\frac{\partial u}{\partial x}(w)\right\rvert \leqslant C\cdot \sup\limits_{\lvert z\rvert = R} \frac{\lvert u(z)\rvert}{R}$$ for $\lvert w\rvert \leqslant \frac{R}{2}$ (and similarly for $\partial u/\partial y$). Letting $R \to \infty$ shows that $u$ is constant. Hence $f_2$ is constant.

  • $\begingroup$ Isn't $f_1$ identically $0$? The curve $\vert \zeta\vert=r$ winds $0$ times arround $z$. $\endgroup$ – leo Aug 1 '13 at 13:36
  • $\begingroup$ But $f$ isn't (generally) holomorphic in the disk $\lvert z\rvert < r+\varepsilon$. If it is, $f$ is entire, and then $f_1 \equiv 0$ is fine. $\endgroup$ – Daniel Fischer Aug 1 '13 at 13:49
  • $\begingroup$ Indeed. Thanks! $\endgroup$ – leo Aug 1 '13 at 13:54
  • $\begingroup$ Let $u = \Re f_2$. Then $u$ is an entire harmonic function that grows slower than $\lvert z\rvert$. Hence $u$ is constant (Poisson integral). That implies that $f_2$ is constant. $\endgroup$ – Daniel Fischer Aug 1 '13 at 14:26
  • $\begingroup$ Alternatively math.stackexchange.com/q/229312/8271 $\endgroup$ – leo Aug 1 '13 at 14:27

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