How do I show that in a normed space $| (\|x\|-\|y\|) | \leq \|x-y\|$
I need to use normed space axioms but I'm unable to figure it out. I think the scalar axiom and triangle inequality are helpful.
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2Observe that $x=(x-y)+y$ and apply the triangle inequality. – SamM Jun 17 '15 at 15:41
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This is known as the "reverse triangle inequality". – Batman Jun 17 '15 at 15:48
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You can find the proof for absolute value here; Reverse Triangle Inequality Proof. Proof for normed spaces is very similar. – Martin Sleziak Jun 17 '15 at 16:35
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On an unrelated note, this inequality can be used to prove that $f: \mathbb{R^n}\to \mathbb{R}$, defined by $f(x)=||x||$ is uniformly continuous which can be used to prove the equivalence of all norms on $\mathbb{R^n}$. – Jun 17 '15 at 16:47
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To complement @user96343's comment I will add links to this question and this question. – Martin Sleziak Jun 17 '15 at 17:40
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By triangle inequality: $$\|x\| = \|x - y + y\| \leq \|x - y\| + \|y\|$$ implies $\|x\| - \|y\| \leq \|x - y\|$. And $$\|y\| = \|y - x + x\| \leq \|y - x\| + \|x\|$$ implies $\|y\| - \|x\| \leq \|y - x\| = \|x - y\|$.
Therefore $$\left|\|x\| - \|y\|\right| \leq \|x - y\|.$$
Zhanxiong
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$\left \| x \right \|\leq \left \| x-y \right \|+\left \| y \right \|$
$\left \| y \right \|\leq \left \| x-y \right \|+\left \| x \right \|$
so
$\left \| x \right \|-\left \| y \right \|\leq \left \| x-y \right \|$
and
$\left \| y \right \|-\left \| x \right \|\leq \left \| x-y \right \|$
which is what you want.
Matematleta
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