Thread: Complex Analysis -- convergence

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

I'm not sure where to start with this one. Clearly, since the functions are continuous, $\displaystyle \lim f(z_n) = f(z)$. I need a way to make sense of the $\displaystyle \rho$ 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.

What is the definition of $\displaystyle \rho$?

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

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

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.

We can find a $\displaystyle k_0$ such that $\displaystyle z_n\in K_{k_0}$ for all $\displaystyle n$, since the map $\displaystyle z\mapsto d(z,\mathbb C\setminus U)$ is continuous. We get $\displaystyle \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 $\displaystyle \varepsilon>0$, we can find $\displaystyle N(\varepsilon)$ such that for $\displaystyle n\geq N(\varepsilon)$ we have $\displaystyle \frac{d(f_n(z_n),f(z_n))}{1+d(f_n(z_n),f(z_n))}\le \frac{\varepsilon}{1+\varepsilon}$, hence $\displaystyle d(f_n(z_n),f(z_n))\le \varepsilon$.

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