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Given $(\eta_i), (\xi_i) \in \ell^p$, for some $p \geq 1$ in $\mathbb{R}$, is there a graphically way to see the inequality $$\left|(\sum_{i=1}^{\infty}|\eta_i|^p)^{1/p} - (\sum_{i=1}^{\infty}|\xi_i|^p)^{1/p}\right| \leq (\sum_{j=1}^{\infty}|\xi_i-\eta_i|^p)^{1/p}\:\:?$$

John Mars
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1 Answers1

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Triangle inequality?
Indeed, $$\big|\|y\| - \|x\| \big| \le \|x-y\|$$ in any normed space. This follows from the usual triangle inequality.

Proof.
First, $(y) + (x-y) = x$, so $$ \|y\| + \|x-y\| \le \|x\| \tag1$$ Next, $x + (y-x) = x$, so $$ \|x\| + \|y-x\| \le \|y\| \tag2$$

Now from $(1)$ we get $$ \|x\|-\|y\| \le \|x-y\| $$ and from $(2)$ we get $$ \|y\| - \|x\| \le \|y-x\| \\ -\big(\|x\|-\|y\|\big)) \le \|x-y\| $$ But from $$ \|x\|-\|y\| \le \|x-y\| \le \|x-y\|\quad\text{and}\quad -\big(\|x\|-\|y\|\big) \le \|x-y\| $$ we conclude $$ \big|\|x\|-\|y\|\big| \le \|x-y\| $$

GEdgar
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  • I'm looking for specifics of it in the ℓp spaces. Can it only seen through the general sense within a normed space? – John Mars Mar 26 '21 at 17:21