Let $G \subseteq \mathbb C$ be an open set and $K \subseteq G$ be compact. Let $(\Omega,d)$ be a metric space and $C(G,\Omega)$ be the space of all continuous functions from $G$ to $\Omega.$ Define a function $\rho : C(G,\Omega) \times C(G,\Omega) \longrightarrow \mathbb R$ by $$\rho (f,g) = \sup\limits \{d(f(z),g(z))\ |\ z \in K\}$$ $f,g \in C(G,\Omega).$ Show that $\rho$ satisfies triangle inequality i.e. for any $f,g,h \in C(G,\Omega)$ we have $$\rho (f,g) + \rho (g,h) \geq \rho (f,h).$$
I have tried to approach in the following way $:$
We have for any $f,g,h \in C(G,\Omega)$ $$\begin{align*} \rho (f,g) + \rho (g,h) & = \sup\limits \{d(f(z),g(z))\ |\ z \in K\} + \sup\limits \{d(g(z),h(z))\ |\ z \in K\} \\ & = \sup\limits \{d(f(z),g(z)) + d(g(z'),h(z')) \ |\ z,z' \in K\} \\ & \geq \sup\limits \{d(f(z),g(z)) + d(g(z),h(z))\ |\ z \in K\} \\ & \geq \sup\limits \{d(f(z),h(z))\ |\ z \in K\} \\ & = \rho (f,h) \end{align*}$$
Is it fine what I did or should I add something more to it? Please verify it.
Thanks for your time.
Your proof is correct. However, it can be simplified. For example, the equality $$ \sup\{d(f(z),g(z))\ |\ z \in K\} + \sup\{d(g(z),h(z))\ |\ z \in K\} \\ = \sup\limits \{d(f(z),g(z)) + d(g(z'),h(z')) \ |\ z,z' \in K\} $$ is true but requires some thinking. On the other hand, you need only $\ge$, which is easier to verify.
I would write it as follows: For all $z \in K$ is $$ d(f(z), h(z)) \le d(f(z), g(z)) + d(g(z), h(z)) \le \rho(f, g) + \rho(g, h) $$ and since the right-hand side does not depend on $z$, $\rho(f, g) \le \rho(f, g) + \rho(g, h)$ follows. This is more or less equivalent to your calculation, but easier to read in my opinion.