$\Omega$ is an open subset of $\mathbb{R}^{n}$ . Given any function $f \in \mathcal{C}(\Omega)$ and any compact subset $K$ of $\Omega,$ let $$|f|_{K} :=\sup _{x \in K}|f(x)| .$$ Suppose that $K_i$ is a sequence of compact subset of $\Omega$ such that $$ K_{i} \subset \operatorname{int} K_{i+1} \text { for all } i \geq 1 \text { and } \Omega=\bigcup_{i=1}^{\infty} K_{i}$$ Let $\alpha_i$ be a sequence of positive numbers whose sum is finite.For $f,g \in \mathcal{C}(\Omega )$ define $$ d(f, g) :=\sum_{i=1}^{\infty} \alpha_{i} \frac{|f-g|_{K_{i}}}{1+|f-g|_{K_{i}}} $$ The problem asks me to show that $d$ defines a distance, but the corresponding metric space is not normable, i.e. we cannot find a norm which induces the same topology.
I have shown that $d$ is a distance and the metric space is complete.
My trouble is that I do not know what method can be used to show that a metric space is not normable. I have also searched on this site and find some discussions that use results on Rudin, which I do not know.Can anyone help me?
Hint: It is shown in Rudin's Functional Analysis that a space is normable iff there is a bounded convex neighborhood of $0$. It is also shown in it that a subset of $E$ this space is bounded iff $sup_{f \in E} \sup_{x \in K_n} |f(x)| <\infty $ for each $n$. Now take any ball $B(0,\epsilon)$ show that it cannot be bounded. (You can find a continuous function which is 'small' on any $K_n$ but 'large' on $K_{n+1}$.