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Funktionentheorie » Holomorphie » Uniqueness Theorem for Holomorphic Functions
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Universität/Hochschule J Uniqueness Theorem for Holomorphic Functions
Neymar
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Dabei seit: 03.01.2019
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  Themenstart: 2019-03-04

Hi. I would like to ask you how we can obtain an important corollary by the Theorem. "$\mathbf{\text{10.18 Theorem}}$ Suppose $\Omega$ is a region, $f\in H(\Omega)$, and $Z(f) = \{a \in \Omega: f(a) = 0\}$. Then either $Z(f) = \Omega$, or $Z(f)$ has no limit point in $\Omega$. In the latter case there corresponds to each $a\in Z(f)$ a unique positive integer $m = m(a)$ such that $f(z) = (z-a)^m g(z)$, $z \in \Omega$, where $g\in H(\Omega)$ and $g(a) \ne 0$; furthermore, $Z(f)$ is at most countable. (We recall that regions are $\textit{connected}$ open sets.)" "$\mathbf{\text{Corollary}}$ If $f$ and $g$ are holomorphic functions in a region $\Omega$ and if $f(z) = g(z)$ for all $z$ in some set which has a limit point in $\Omega$, then $f(z) = g(z)$ for all $z \in \Omega$." (Walter Rudin, "Real and Complex Analysis", page 208f.) Okay, I have a question: How can we prove this? Usually, Rudin proves corollaries, but here he does not.


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Vercassivelaunos
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Dabei seit: 28.02.2019
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  Beitrag No.1, eingetragen 2019-03-04

Hi Neymar, consider the function \(h(z)=f(z)-g(z)\). Its zeros are the points where \(f(z)=g(z)\), and it is holomorphic on the region \(\Omega\), so the theorem applies. The set of zeros either has no limit point or is all of \(\Omega\). Since by assumption it does have a limit point, it must be \(h(z)=0\forall z\in\Omega\Rightarrow f(z)=g(z)\forall z\in\Omega\).


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Neymar
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  Beitrag No.2, vom Themenstarter, eingetragen 2019-03-05

Wow, what a nice proof! ;-) Yeah, it makes sense. I assume that one could say (in the $\mathbf{\text{Corollary}}$): "If $f$ and $g$ are holomorphic functions in a region $\Omega$ and if $f(z) = g(z)$ for all $z$ in some set which has $\mathbf{\text{at least}}$ a limit point in $\Omega$, then $f(z) = g(z)$ for all $z \in \Omega$." Anyway, how we can check whether or not a set has a limit point in $\Omega$? Honestly, I would like it a lot if we could go through the proof of theorem 10.11 as well. Here it is: "Let $A$ be the set of all limit points of $Z(f)$ in $\Omega$. Since $f$ is continuous, $A \subset Z(f)$. Fix $a \in Z(f)$, and choose $r>0$ so that $D(a; r) \subset \Omega$. By Theorem 10.16 [$\textit{For every open set $\Omega$ in the plane, every $f \in H(\Omega)$ is representable by a power series in $\Omega$.}$], $f(z) = \sum_{n = 0}^{\infty}c_n (z-a)^n (z \in D(a; r))$. There are now two possibilities. Either all $c_n$ are $0$, in which case $D(a;r) \subset A$ and $a$ is an interior point of $A$, or there is a smallest integer $m$ [necessarily positive, since $f(a) = 0$] such that $c_m \ne 0$. [...]" This is of course not the whole proof, but this is the first part of it. I should mention that $D(a;r) := \{z : |z-a| < r\}$ (i) Why does from the continuity of $f$ follow that $A \subset Z(f)$? (ii) Do I understand it correctly that when Rudin writes "There are now two possibilities", he means "there are two possibilites for $f(a) = 0$ given $f(z)$ is representable by a power series"? I think that the first one is clear ("all $c_n$ are $0$"), but I am trying to figure out the second one. Could it - theoretically, as a trivial case - be that for instance $c_3 = 2$ and $c_4 = -2$ and all the other $c_n$'s are $0$, since I would then obtain: $f(z) = \sum_{n = 0}^{\infty}c_n (z-a)^n = 2 \cdot (z-a)^n - 2\cdot (z-a)^n = 0 \forall z \in D(a;r).$ What I do not understand yet is that if we only take $f(z) = \sum_{z = 0}^{\infty}\dots$ for granted, I obtain for $f(z)|_{z = a} = f(a) = c_0 \cdot 0^0 + ...$, which is undefined due to "0^0" ... Vercassivelaunos, many thanks in advance! Best regards, Neymar


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Vercassivelaunos
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  Beitrag No.3, eingetragen 2019-03-05

Yes, one could also demand that there must be at least one limit point. It's really just a more emphasized way of saying "has a limit point". As to your question of how to check wether a set has a limit point, there's different characterizations which can be useful for different sets. For instance, the following are equivalent: \[1.\quad z\in\Omega \textrm{ is a limit point of }A\subset\Omega\\ 2.\quad B_\epsilon(z)\backslash\{z\}\cap A\neq\emptyset \textrm{ for all }\epsilon>0\\ 3.\quad \textrm{There exists a sequence $a_n\in A$ with $a_n\to z$ and $a_i\neq a_j\forall i\neq j$}\] Now we could define the set \(A:=\{\frac{1}{n}~\vert~n\in\mathbb N\}\). The third characterization is useful here, since the set is basically defined as the elements of the sequence \(\frac{1}{n}\), which converges to 0. So 0 is a limit point of this set. Another set might be defined as \(B:=\{e^{in}~\vert~n\in\mathbb N\}\). This set is also essentially defined as the elements of a sequence, but this one doesn't converge. We can still show that -1 is a limit point of this sequence using the second characterization: It's obvious that \(B\) is a subset of the unit circle. The distance between two numbers on the unit circle is always less than the difference between their arguments (since that's their distance along the circle arc, which is longer than the direct distance). So if for any number on the unit circle we find a number in \(B\) whose argument is within \(\epsilon\) of \(\pi\) (the argument of -1) we have proven that \(B_\epsilon(-1)\cap B\neq \emptyset\). So we want to find an \(n\) where \(k(\pi-\epsilon)\frac{1}{\epsilon}\) we get \[k(\pi-\epsilon)=k\pi-k\epsilon


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Neymar
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  Beitrag No.4, vom Themenstarter, eingetragen 2019-03-06

I hope that you are okay when we continue. Just tell me please when I should open another thread. "$\mathbf{\text{10.19 Definition}}$ If $a \in \Omega$ and $f\in H(\Omega - \{a\})$, then $f$ is said to have an $\mathbf{\textit{isolated}}$ singularity at the point $a$. If $f$ can be so defined at $a$ that the extended function is holomorphic in $\Omega$, the singularity is said to be $\mathbf{\textit{removable}}$. $\mathbf{\text{10.20 Theorem}}$ Suppose $f \in H(\Omega - \{a\})$ and $f$ is bounded in $D'(a;r)$, for some $r>0$. Then $f$ has a removable singularity at $a$. Recall that $D'(a;r) = \{z: 0< |z-a| < r\}$. PROOF Define $h(a) = 0$ and $h(z) = (z-a)^2f(z)$ in $\Omega - \{a\}$. Our boundedness assumption shows that $h'(a) = 0$. Since $h$ is evidently differentiable at every other point of $\Omega$, we have $h \in H(\Omega)$, so $h(z) = \sum_{n = 2}^{\infty}c_n (z-a)^n (z \in D(a;r))$. We obtain the desired holomorphic extension of $f$ by setting $f(a) = c_2$, for then $f(z) = \sum_{n = 0}^{\infty}c_{n+2}(z-a)^n (z \in D(a;r))$." _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ (i) Why did Rudin not define $h(z) := (z-a)^2 f(z) \forall z \in \Omega$ straightforwardly, since we would then obtain $h(z)|_{z = a} = 0$? (ii) "Our boundedness assumption shows that $h'(a) = 0$." $>$ I do not understand yet why this follows from $f$ being bounded in $D'$? Is it not obvious that when $h(a) = 0 \Rightarrow h'(a) = 0$? (iii) We know that $h \in (\Omega)$ and there is a theorem stating that every holomorphic function is representable by a power series, so why is it not $h(z) = \sum_{n = 0}^{\infty}c_n (z-a)^n$ ? (iv) I think that the gist of the proof is unclear to me. We define another function that is holomorphic in $\Omega$. And then we just extend our function $f$ by adding one value to it!? Hmm ... Neymar


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Vercassivelaunos
Senior Letzter Besuch: in der letzten Woche
Dabei seit: 28.02.2019
Mitteilungen: 1239
  Beitrag No.5, eingetragen 2019-03-06

It's fine, we can go on for now. (i) \((x-a)^2f(z)\) isn't well-defined in \(z=a\), since \(f\) is undefined at $z=a$, so $h$ has to be defined separately at $z=a$. (ii) It's not so obvious without boundedness. Consider this counterexample: Let $a=0$ and $f(z)=\frac{1}{z^3}$. This function is holomorphic on $\mathbb C-\{0\}$, and it is unbounded, since $\vert\frac{1}{z^3}\vert\to\infty$ for $z\to0$. Now we have $h(z)=(z-a)^2f(z)=\frac{z^2}{z^3}=\frac{1}{z}$ if $z\neq 0$, and $h(0)=0$. This is not even continuous, let alone differentiable. The boundedness allows the factor $(z-a)^2$ to dominate the derivative as we go to $a$. (iii) It's because both $h(a)$ and $h'(a)$ are zero, so the first two terms of the power series have to be zero as well, since they correspond to $h$ and $h'$, respectively. (iv) The gist is this (with a not so rigorous notation): \(h=z^2f\) is holomorphic, so it has a power series. So $z^2f=\textrm{power series}\Rightarrow f=\frac{\textrm{power series}}{z^2}=\textrm{other power series}$, where the last equality is possible since the power series of $h$ starts at the quadratic term. And if $f$ is representable as a power series around $a$, then it must be continuously extendable to $a$, since power series (with a nonzero convergence radius) always converge and are continuous at their center ($a$ in this case).


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Neymar
Wenig Aktiv Letzter Besuch: vor mehr als 3 Monaten
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  Beitrag No.6, vom Themenstarter, eingetragen 2019-03-07

Okay, thanks for your answer. For now, I do not have any urgent questions which is why I close the thread. Many thanks for your commitment, Neymar


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