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Math Help - Complex Analysis -- convergence

  1. #1
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    Complex Analysis -- convergence

    Let M be a complete metric space. Suppose that f_nis a sequence of
    continuous functions in (\mathcal{C}(U, M), \rho) which converges to f and z_n is a
    sequence in U which converges to a point z in U. Show that \lim f_n(z_n) = f(z).

    I'm not sure where to start with this one. Clearly, since the functions are continuous, \lim f(z_n) = f(z). I need a way to make sense of the \rho metric so that I can pass the correct limit, but I can't seem to be able to do it.

    Can anyone help?

    Thank you in advance.
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  2. #2
    Super Member girdav's Avatar
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    Re: Complex Analysis -- convergence

    What is the definition of \rho?
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  3. #3
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    Re: Complex Analysis -- convergence

    \rho(f,g) = \sum_{n=0}^\infty\frac{1}{2^n}\cdot{\frac{\rho_n(f  ,g)}{1+\rho_n(f,g)}.

    Here, \rho_n(f,g) = \sup_{z\in K_n} d(f(z),g(z)) where K_n = \{z : |z| \leq n\} \cap \{ z : d(z,\mathbb{C}\backslash U) \geq \frac{1}{n}\}

    f,g are defined on an open set U in the complex plane and take values in a complete metric space M with metric d.
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  4. #4
    Super Member girdav's Avatar
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    Re: Complex Analysis -- convergence

    We can find a k_0 such that z_n\in K_{k_0} for all n, since the map z\mapsto d(z,\mathbb C\setminus U) is continuous. We get \frac 1{2^{k_0}}\frac{d(f_n(z_n),f(z_n))}{1+d(f_n(z_n),f  (z_n))}\leq \frac 1{2^{k_0}}\frac{\rho_{k_0}(f_n,f)}{1+\rho_{k_0}(f_  n,f)}\leq \rho(f,f_n) and given \varepsilon>0, we can find N(\varepsilon) such that for n\geq N(\varepsilon) we have \frac{d(f_n(z_n),f(z_n))}{1+d(f_n(z_n),f(z_n))}\le \frac{\varepsilon}{1+\varepsilon}, hence d(f_n(z_n),f(z_n))\le \varepsilon.
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  5. #5
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    Re: Complex Analysis -- convergence

    Thanks very much for this! The first inequality is the key one I was missing.
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