Continuity of vector space operations in a normed space

You can use the same idea that works for $+$: $$\|(\alpha,x)\|_{K\times X}=|\alpha|+\|x\|.$$ In any case, you can endow $K\times X$ with the product topology and no explicit norm is required.

EDIT:

$$\eqalign{\|\alpha x - \alpha_0 x_0\|_X & = \|\alpha x - \alpha_0 x + \alpha_0 x - \alpha_0 x_0\|_X\cr &\le\|\alpha x - \alpha_0 x\|_X + \|\alpha_0 x - \alpha_0 x_0\|_X\cr & = |\alpha - \alpha_0|\cdot\|x\|_X + |\alpha_0|\cdot\|x-x_0\|_X\cr &\le(1+\|x_0\|_X)\cdot|\alpha - \alpha_0|+ |\alpha_0|\cdot\|x-x_0\|_X\cr &\le(1+\|x_0\|_X+|\alpha_o|)\|(\alpha-\alpha_0,x-x_0)\|_{K\times X}\cr & = (1+\|x_0\|_X+|\alpha_o|)\|(\alpha,x)-(\alpha_0,x_0)\|_{K\times X}.}$$ (Why $\|x\|_X\le 1+\|x_0\|_X$?)


The fields $\mathbb R$ and $\mathbb C$ are equipped with a standard topology (derived from the standard metric and the standard absolute value).

Tags:

Analysis

Metric Spaces

Real Analysis

Functional Analysis

Normed Spaces

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