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Possible Duplicate:
Is norm non-decreasing in each variable?

Let $\| \cdot \|$ be any norm on $\mathbb{R}^{2}$. Let $0 \leq a \leq c$ and $0 \leq b \leq d$. Show that $\|(a,b)\| \leq \|(c,d)\|$.

Letting $A=(a,b)$ and $B=(c-a,d-b)$, this would follow if we could show $\|A-B\| \leq \|A+B\|$, as this would give

$2\|A+B\| \geq \|A-B\| + \|A+B\| \geq \|A-B+A+B\| = 2\|A\|$,

as desired. When $d=2b$, it is easy to show this using the following lemma I already proved: given $a,b,c,d$ as above,

$\|(a,0)\| \leq \|(c,0)\|$ and

$\|(0,b)\| \leq \|(0,d)\|$.

But I don't know how to prove this when $d \neq 2b$.

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    Since this is answered in the other question, I guess it might be closed as duplicate. (The other question is more general, but the counterexample given in the answer there is in $\mathbb R^2$.)2012-10-27

0 Answers 0