In a normed vector space $(V,\lvert . \rvert)$ show that $f:V\rightarrow \mathbb{R}$ with $f(v)=\lvert v\rvert$ is uniformly continuous

Question

In a normed vector space $(V,\lvert . \rvert)$ show that $f:V\rightarrow \mathbb{R}$ with $f(v)=\lvert v\rvert$ is uniformly continuous

The first part of the question says to prove the "reverse triangle inequality" which is $\lvert u\rvert -\lvert v\rvert \le \lvert u-v\rvert$ I sense that might be a clue.

Normally I'd start from the definitions, but I'm not sure how it applies to a vector space (Once I consider an actual vector space (say $\mathbb{R}^n$ I loose all confidence)

Answer 1

From the triangle inequality we have $|x|-|y| \le |x-y|$. Reversing the roles of $x,y$ gives $|y|-|x| \le |x-y|$ and combining gives $||x|-|y|| \le |x-y|$, or $|f(x)-f(y)| \le |x-y|$.

Hence $f$ is uniformly continuous, in fact it is Lipschitz continuous with rank one.

Exactly the same analysis (with $| \cdot |$ replaced by $\|\cdot\|$) applies to any norm.