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I can't solve this exercise from the book, can anyone give me a hint?

Show that if $f$ is holomorphic in the unit disc, is bounded, and converges uniformly to zero in the sector $\theta < \arg z < \varphi$ as $|z| \to 1$, then $f = 0$.

(Use the Cauchy inequalities or the maximum modulus principle)

My idea was to extend $f$ continuously to the border of the domain : $θ < \arg z < \varphi$ as $|z| = 1$ then since $f=0$ on the border, $f=0$ in the whole domain. However I can't show that $f$ is continuously extendable.

thank you!

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  • $\begingroup$ The hypothesis gives you the continuous extensibility to the arc $\{ e^{it} : 0 < t < \varphi\}$, with values $0$ there. Then you can reflect in that arc, and the identity theorem asserts that $f \equiv 0$. Note that the boundedness of $f$ is not required. But the exercise is probably before the reflection principle is introduced. $\endgroup$ Jan 3, 2016 at 19:03
  • $\begingroup$ Hi Daniel, what do you mean by "reflect in that arc"? An idea that I had to do this problem, was to first rotate the sector then apply a power map $z^{\alpha}$ so that the image of the sector the whole disk and then apply the maximum modulus principle. $\endgroup$
    – user135520
    Jun 18, 2018 at 20:46

1 Answer 1

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Here's one way to do it. Let $M$ be a bound for $|f|$ on the unit disc and let $t$ be slightly smaller than $\varphi-\theta$ and define $$ g(z) = f(z)f(ze^{it})f(ze^{2it})\cdots f(ze^{nit}) $$ where $n$ is chosen so large that $nt > 2\pi$. Let $\varepsilon > 0$. By assumption there is an $r < 1$ such that $|f(z)| < \varepsilon$ for $r < |z| < 1$ and $\theta < \arg z < \phi$. Hence $$ |g(z)| < M^n \varepsilon $$ for all $z$ with $r < |z| < 1$. (One factor has modulus less than $\varepsilon$ and the other factors less than $M$.) By the maximum modulus principle, $|g| < M^n\varepsilon$ on the whole disc, and since $\varepsilon$ was arbitrary, we must have that $g(z) = 0$ for all $z$ in the unit disc.

Hence one of the factors, and consequently all factors, of $g$ vanishes identically (otherwise $g$ would have at most countably many zeros).

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    $\begingroup$ wow! just wow! $\phantom{ }$ $\endgroup$ Sep 26, 2016 at 18:02
  • $\begingroup$ Sorry, what was the significance of choosing $n$ such that $nt> 2\pi$? $\endgroup$
    – user135520
    Jun 18, 2018 at 21:00
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    $\begingroup$ @user135520 You need to have enough rotated copies of $f$ to cover the whole circle. $\endgroup$
    – mrf
    Jun 19, 2018 at 7:32
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    $\begingroup$ How to go from the maximum modulus principle to get $|g|<M^n\epsilon$ for the whole disc? Why there cannot exist $|z|\leq r$ such that $|g(z)|\geq M^n \epsilon$? $\endgroup$
    – SHBaoS
    Oct 16, 2018 at 1:01
  • $\begingroup$ Nice Proof ,SIr I had question AS on boundary of arc f is uniformly convegeging to 0 . So why we have consider M factor for that . ? It is working But still is it necessary to bound g? $\endgroup$ Nov 18, 2018 at 17:16

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