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Thread: complex numbers- hurwitz theorem

  1. #1
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    complex numbers- hurwitz theorem

    Suppose that D is a connected set open set , $\displaystyle f_{n} \in H(D)$ and
    $\displaystyle f_{n} -> f $ uniformly on compact subsets of D. If f is nonconstant and
    $\displaystyle z \in D $ then there exists N and a sequence $\displaystyle z_{n} -> z$ such that $\displaystyle f_{n} ( z_{n} ) = f(z)$ for all $\displaystyle n \geq N$
    Hint: Assume that f(z) = 0. Apply the hurwitz theorem in the disk $\displaystyle D(z, r_{j}) $ for a suitable sequence $\displaystyle r_{j}->0 $

    I didn't even understand the hint.hurwitz theorem states that $\displaystyle f_{n}$
    and f have the same number of the zeroes in the disk. Can anyone help, thanx.
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  2. #2
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    Quote Originally Posted by hermanni View Post
    Suppose that D is a connected set open set , $\displaystyle f_{n} \in H(D)$ and
    $\displaystyle f_{n} -> f $ uniformly on compact subsets of D. If f is nonconstant and
    $\displaystyle z \in D $ then there exists N and a sequence $\displaystyle z_{n} -> z$ such that $\displaystyle f_{n} ( z_{n} ) = f(z)$ for all $\displaystyle n \geq N$
    Hint: Assume that f(z) = 0. Apply the hurwitz theorem in the disk $\displaystyle D(z, r_{j}) $ for a suitable sequence $\displaystyle r_{j}->0 $

    I didn't even understand the hint.hurwitz theorem states that $\displaystyle f_{n}$
    and f have the same number of the zeroes in the disk. Can anyone help, thanx.
    I remember doing this exercise once and using a Cantor's diagonal argument somewhere, but I can't remember why I did all that. Take a look at this and let me know if something's off (It's pretty simple, which makes me wary):

    Pick $\displaystyle z_0\in D$ with $\displaystyle f(z_0)=0$. There exists an $\displaystyle r>0$ such that $\displaystyle f(z)\neq f(z_0)=0$ for all $\displaystyle z\in \overline{D}_{r}(z_0)\setminus \{ z_0 \} \subset D$ (the closed punctured disk or radius $\displaystyle r$ and center $\displaystyle z_0$), and by Hurwiz theorem there exists $\displaystyle N$ such that for all $\displaystyle n\geq N$ the functions $\displaystyle f_n$ have the same zeroes as $\displaystyle f$ in the disk, so there exists $\displaystyle z_n\in D_{r}(z_0)$ with $\displaystyle f_n(z_n)=0$. It now suffices to prove that $\displaystyle z_n \rightarrow z_0$, but this follows since $\displaystyle \overline{D}_r(z_0)$ is compact (therefore sequentially compact) and the fact that the sequence $\displaystyle z_n$ can only have one accumulation point by how we picked our disk and the fact that $\displaystyle |f_n(y_n)-f(y)| \rightarrow 0$ whenever $\displaystyle y_n,y\in D$ and $\displaystyle y_n\rightarrow y$.

    Check it carefully, maybe I overlooked something.
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  3. #3
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    Everything seems OK to me , thank u very much )
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