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Suppose that $f$ and $g$ are holomorphic in a domain containing the unit disc $D=\{z| |z| \le 1 \}$. Suppose that $f$ has a simple zero at $z=0$ and vanishes nowhere in the unit disc.

Let $f_{\epsilon}(z)=f(z)+ \epsilon g(z)$. Show that if $\epsilon$ is sufficiently small then

a) $f_{\epsilon}(z)$ has a unique zero in the unit disc.

b) Show that the map $\epsilon \to z_{\epsilon}$ is continuous

My try:

For (a): Since $f$ has a simple zero at $z=0$ and vanishes nowhere in the unit disc, on $|z|=1$ ,$f$ attains the minimum value (say $\delta \gt 0$). Since $g$ is holomorphic on $D$, $g$ attains its maximum (say $M$). Then we have $$|f_{\epsilon}(z)-f(z)|=| \epsilon g(z)| \lt \epsilon M$$

For $\epsilon \lt \frac{\delta}{M}$, on $|z|=1$, we have $$|f_{\epsilon}(z)-f(z)| \lt \delta \le |f(z)|$$ . The conclusion follows by Rouche's theorem.

For (b) I have no idea on how to proceed.

Thanks for the help!!

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1 Answer 1

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If a holomorphic function $h$ has a zero of multiplicity $k$ at $z_0$, then we have $h(z) = (z-z_0)^k\cdot \tilde{h}(z)$ with $\tilde{h}$ holomorphic in a neighbourhood of $z_0$ and $\tilde{h}(z_0) \neq 0$. Thus we have

$$\frac{h'(z)}{h(z)} = \frac{k(z-z_0)^{k-1}\tilde{h}(z) + (z-z_0)^k\tilde{h}'(z)}{(z-z_0)^k\tilde{h}(z)} = \frac{k}{z-z_0} + \frac{\tilde{h}'(z)}{\tilde{h}(z)},$$

and the second summand is holomorphic in a neighbourhood of $z_0$. The residue of $\frac{h'}{h}$ in $z_0$ is therefore $k$. Then

$$z\frac{h'(z)}{h(z)} = \frac{k\cdot z}{z-z_0} + z\frac{\tilde{h}'(z)}{\tilde{h}(z)},$$

so

$$\operatorname{Res} \biggl(z\frac{h'}{h}; z_0\biggr) = \operatorname{Res} \biggl(\frac{k\cdot z}{z-z_0}; z_0\biggr) = k\cdot z_0.$$

Here, $f_\epsilon$ has only one zero in the unit disk (counting multiplicities), and therefore

$$z_\epsilon = \frac{1}{2\pi i} \int_{\lvert z\rvert = 1} z\cdot\frac{f_\epsilon'(z)}{f_\epsilon(z)}\,dz = \frac{1}{2\pi i} \int_{\lvert z\rvert = 1} z\cdot\frac{f'(z) + \epsilon g'(z)}{f(z) + \epsilon g(z)}\,dz,$$

which shows that the function $\epsilon \mapsto z_\epsilon$ is continuous [even holomorphic], since the integrand depends continuously [holomorphically] on $\epsilon$.

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  • 1
    $\begingroup$ @user135520 Yes, one can differentiate under the integral. But it's more idiomatic to use the continuity (by the dominated convergence theorem) and then Morera's theorem (change the order of integration by Fubini) to see that an integral whose integrand depends holomorphically on a parameter depends holomorphically on that parameter [when that is the case, which most of the time - and particularly here - it is, but there are exceptions]. $\endgroup$
    – Daniel Fischer
    Jun 26, 2018 at 18:10
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    $\begingroup$ You need to consider arbitrary sequences converging to $\epsilon$, not only a particular sequence. In cases like this, where the domain of integration is compact and the integrand continuous (jointly, as a function of all involved variables, here $z$ and $\epsilon$), you can also show the continuity in a more elementary way. For each compact set $E$ such that $H_{\epsilon}(z)$ has no singularity on $E \times \{ z : \lvert z\rvert = 1\}$, $(\epsilon, z) \mapsto H_{\epsilon}(z)$ is uniformly continuous there, and then the standard estimate yields the continuity of the integral. $\endgroup$
    – Daniel Fischer
    Jun 26, 2018 at 20:43
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    $\begingroup$ Given $\eta > 0$ there is a $\delta > 0$ such that $\lvert H_{\epsilon_1}(z) - H_{\epsilon_2}(z)\rvert \leqslant \eta$ for all $\epsilon_1, \epsilon_2 \in E$ with $\lvert \epsilon_1 - \epsilon_2\rvert \leqslant \delta$ and all $z$ with $\lvert z\rvert = 1$, and hence $$\Biggl\lvert \int_{\lvert z\rvert = 1} H_{\epsilon_1}(z) - H_{\epsilon_2}(z)\,dz\Biggr\rvert \leqslant \eta\cdot 2\pi\,,$$ which gives you the uniform continuity of $$I(\epsilon) = \frac{1}{2\pi i}\int_{\lvert z\rvert = 1} H_{\epsilon}(z)\,dz$$ on $E$. $\endgroup$
    – Daniel Fischer
    Jun 26, 2018 at 20:43
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    $\begingroup$ @user135520 Once we have the continuity, Morera's theorem asserts the analyticity of $I(\epsilon)$ once we've seen that $$\int_{\partial \Delta} I(\epsilon)\,d\epsilon = 0$$ for all triangles $\Delta$ contained in the domain of $I$. That follows by Cauchy's and Fubini's theorems since $$\int_{\partial \Delta} I(\epsilon)\,d\epsilon = \frac{1}{2\pi i}\int_{\partial \Delta} \int_{\lvert z\rvert = 1} H_{\epsilon}(z)\,dz\,d\epsilon = \frac{1}{2\pi i} \int_{\lvert z\rvert = 1} \int_{\partial \Delta} H_{\epsilon}(z)\,d\epsilon\,dz\,,$$ where the change of order of integration is justified by $\endgroup$
    – Daniel Fischer
    Jun 26, 2018 at 21:03
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    $\begingroup$ Fubini's theorem, and the inner integral $\int_{\partial \Delta} H_{\epsilon}(z)\,d\epsilon$ vanishes for every $z$ by Cauchy's theorem. [Of course one can construct situations where Fubini's theorem isn't applicable, but in most naturally arising situations there's no problem.] $\endgroup$
    – Daniel Fischer
    Jun 26, 2018 at 21:03

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