If $f$ is entire and $f(z)/z$ is bounded, then $z = 0$ is a removable singularity of $f(z)/z$.

Question

Let $f$ be an entire function with $\sup_{z\in\mathbb{C}}|f(z)/z|<\infty$. Show that $z=0$ is a removable singularity of $g(z):=f(z)/z$.

To prove the claim, I need to show that $0 = \lim_{z\to 0}(z-0)g(z) = \lim_{z\to 0}(f(z)$, that seems really obvious to me because if a fraction where the denominator goes to zero is bounded, then the nominator has to go to zero?

I would like to do that more formally. So what can I write for a proof?

Answer 1

In order to prove that $\lim_{z\to0}zg(z)=0$, take $\varepsilon>0$. Let $S=\sup_{z\in\mathbb{C}\setminus\{0\}}\left|\frac{f(z)}z\right|$. If $S=0$, then $f$ is the null function and the statment is trivial. Otherwise, let $\delta=\frac\varepsilon S$. Then$$|z|<\delta\iff|z|<\frac\varepsilon S\implies\bigl|f(z)\bigr|=|z|.\left|\frac{f(z)}z\right|<\frac\varepsilon S.S=\varepsilon.$$Therefore, by the definition of limit, $\lim_{z\to0}zg(z)=0$.

Answer 2

Since $g(z)$ is analytic in $0 < \left| z \right| < \infty $ by laurent's theorem we can write: $$g(z) = \sum\limits_{ - \infty }^\infty {{a_n}{z^n}} $$ for all $0 < \left| z \right| < \infty $,
and $${a_n} = {1 \over {2\pi i}}\oint\limits_\gamma {{{g\left( z \right)} \over {{z^{n + 1}}}}dz} $$ for any simple-closed curve $\gamma $ that lies in the annulus $0 < \left| z \right| < \infty $ and circles $z=0$.

Notice that for all $R>0$ by the estimation lemma: $$\eqalign{ & \left| {{a_n}} \right| = \left| {\frac{1}{{2\pi i}}\oint\limits_{\left| z \right| = R} {\frac{{g\left( z \right)}}{{{z^{n + 1}}}}dz} } \right| \leqslant \frac{1}{{2\pi }}\mathop {\sup }\limits_{\left| z \right| = R} \left| {\frac{{g\left( z \right)}}{{{z^{n + 1}}}}} \right| \cdot 2\pi R = R \cdot \mathop {\sup }\limits_{\left| z \right| = R} \left| {\frac{{\frac{{f\left( z \right)}}{z}}}{{{z^{n + 1}}}}} \right| = \frac{1}{{{R^n}}}\mathop {\sup }\limits_{\left| z \right| = R} \left| {\frac{{f\left( z \right)}}{z}} \right| \cr & {\text{ }} \leqslant \frac{1}{{{R^n}}} \cdot \mathop {\sup }\limits_{z \in {\Bbb C}} \left| {\frac{{f\left( z \right)}}{z}} \right| \cr} $$ Let $M \equiv \mathop {\sup }\limits_{z \in {\Bbb C}} \left| {\frac{{f\left( z \right)}}{z}} \right|$.

We now have $\left| {{a_n}} \right| \le {M \over {{R^{n}}}}$ for all $R>0$

Since ${{a_n}}$ does not depend on $R$ (which was arbitrary) we can take the limit $R \to {0^ + }$:

Thus, if $n \leqslant - 1$ we have

$$\left| {{a_n}} \right| = \mathop {\lim }\limits_{R \to {0^ + }} \left| {{a_n}} \right| \leqslant \mathop {\lim }\limits_{R \to {0^ + }} \frac{M}{{{R^{n}}}} = 0$$

Since ${a_n} = 0$ for all $n \leqslant - 1$ we can conclude that $z=0$ is a removable singularity for $g(z)$.

But we can actually show more than that!

We can also take the limit ${R \to \infty }$:

Thus, if $n \geqslant 0$ we have $$\left| {{a_n}} \right| = \mathop {\lim }\limits_{R \to \infty } \left| {{a_n}} \right| \leqslant \mathop {\lim }\limits_{R \to \infty } \frac{M}{{{R^{n}}}} = 0$$

In conclusion, $g\left( z \right) = \sum\limits_{ - \infty }^\infty {{a_n}{z_n}} = {a_0}$ (a constant)