Let $(X,||·||)$ be a normed space over a field $\mathbb{K}$, where $$||x||=\max\{||x_1||,||x_2||\}$$ for any $x=(x_1,x_2)\in\mathbb{K}\times X$. Show that the mapping \begin{equation*} \begin{split} f : \mathbb{K}\times X & \to X \\ (\alpha, x) & \mapsto \alpha x \end{split} \end{equation*} is continuous.
Firstly, I have shown that $||·||$ is in fact a well defined norm. Also, in this case $||·||=|·|$ for any $\alpha\in\mathbb{K}$.
Let $\varepsilon>0$, we choose $\delta>0$ such that if $$||(\alpha,x)-(\beta,y)||=\max\{|\alpha-\beta|,||x-y||\}<\delta$$ then $$||f(\alpha,x)-f(\beta,y)||=||\alpha x-\beta y||=||\alpha x-\alpha y +\alpha y-\beta y||\leq |\alpha|||x-y||+|\alpha-\beta|||y||\leq (|\alpha|+||y||)\max\{|\alpha-\beta|,||x-y||\}$$
However, I don't know how to conclude it, because of the term $|\alpha|+||y||$ which depends on the point I choose.