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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)

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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.

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  • $\begingroup$ Thank you, that was really easy, I meant to use the double barred norm, is there a difference? Also I see (but with the double norm notation) $\lvert . \rvert_\infty$ and other numbers as sub-script, are they arbitrary do they mean something? I ask because I'd get it if they were natural numbers, then they would be arbitrary, but infinity? Seems a bit cocky to call one norm the infinitth(?) norm, also I see the subscript infinity used to denote other things, some supremum, what was it! $\endgroup$ – Alec Teal Nov 17 '13 at 1:48
  • $\begingroup$ bump for curiousity $\endgroup$ – Alec Teal Nov 17 '13 at 2:02
  • $\begingroup$ I'm not sure what you are asking, do you mean the $p$ in $\|\cdot\|_p$? This is just the standard notation for the $L^p$ (or $l_p$ for sequences) norm $\|f\|_p = \sqrt[p]{\int |f|^p}$ (for $1 \le p < \infty$) and $\|f\|_\infty = \operatorname{esssup} |f|$. $\endgroup$ – copper.hat Nov 17 '13 at 6:40

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