Complex Analysis -- convergence

**harbottle** · September 15th 2011, 07:46 AM

Let M be a complete metric space. Suppose that $f_n$ is 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.

**girdav** · September 15th 2011, 10:47 AM

What is the definition of $\rho$ ?

**harbottle** · September 15th 2011, 11:30 AM

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

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.

**girdav** · September 15th 2011, 12:52 PM

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 $d(f_n(z_n),f(z_n))\le \varepsilon$ .

**harbottle** · September 15th 2011, 01:02 PM

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

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