Vector addition and scalar multiplication are continuous in a normed space

For addition, $\mathrm{\parallel }(x+h)+(y+k)-(x+y)\mathrm{\parallel }=\mathrm{\parallel }h+k\mathrm{\parallel }\le \mathrm{\parallel }h\mathrm{\parallel }+\mathrm{\parallel }k\mathrm{\parallel }$ by [L2].

For scalar multiplication, $(\lambda +\alpha )(x+h)-\lambda x=\lambda h+\alpha x+\alpha h$, so $\mathrm{\parallel }(\lambda +\alpha )(x+h)-\lambda x\mathrm{\parallel }\le \mathrm{\mid }\lambda \mathrm{\mid }\mathrm{\parallel }h\mathrm{\parallel }+\mathrm{\mid }\alpha \mathrm{\mid }\mathrm{\parallel }x\mathrm{\parallel }+\mathrm{\mid }\alpha \mathrm{\mid }\mathrm{\parallel }h\mathrm{\parallel }$ by the triangle inequality and absolute homogeneity from [L1] and [L2].

Given $\epsilon >0$, step 1.1 shows that if $\mathrm{\parallel }h\mathrm{\parallel }<\epsilon \mathrm{/}2$ and $\mathrm{\parallel }k\mathrm{\parallel }<\epsilon \mathrm{/}2$, then $(x+h)+(y+k)$ lies in the $\epsilon$-ball around $x+y$; this is continuity of addition at $(x,y)$.

For scalar multiplication at $(\lambda ,x)$, first require $\mathrm{\mid }\alpha \mathrm{\mid }<1$ and then choose $\delta >0$ so that $\mathrm{\mid }\lambda \mathrm{\mid }\delta +\delta (\mathrm{\parallel }x\mathrm{\parallel }+\delta )<\epsilon$; step 1.2 then gives $\mathrm{\parallel }(\lambda +\alpha )(x+h)-\lambda x\mathrm{\parallel }<\epsilon$ whenever $\mathrm{\mid }\alpha \mathrm{\mid }<\delta$ and $\mathrm{\parallel }h\mathrm{\parallel }<\delta$.

Since the point $(x,y)$ was arbitrary, step 2.1 proves continuity of addition everywhere, and step 2.2 proves continuity of scalar multiplication everywhere.