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Let X be a vector space over $\mathbb{R}$ or $\mathbb{C}$, $\|\cdot\|: X\rightarrow [0,\infty)$ is called a quasi-norm if

i) $\|x\|=0 \Rightarrow x=0$

ii) $\|\lambda x\|=|\lambda|\|x\|, \forall \lambda, x$

iii) $\exists K\ge 1$, s.t. $\|x+y\|\le K(\|x\|+\|y\|), \forall x,y$

My question is:

If $\|x_k-x\|\rightarrow 0$, can we conclude that $\|x_k\|\rightarrow \|x\|$?

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1 Answer 1

up vote 5 down vote accepted

No. A counterexample for $\mathbb R^2$ is: $$\|\langle x,y\rangle\|=\cases{K|x|&\text{for } y=0\\|x|+|y|&\text{otherwise}}$$ for some $K>1$.

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Thanks! Could I ask an irrelevant question? Are computer scientists also using things like quasi-Banach spaces? –  Syang Chen Oct 8 '11 at 6:18
    
Not that I know of, but there could easily be applications I'm not aware of. Perhaps in computer vision -- they have some ties to functional analysis. –  Henning Makholm Oct 8 '11 at 6:48
    
Just to spell an example out: take $x_k = \langle 1, \frac{1}{k}\rangle$ and $x = \langle 1, 0\rangle$. Then $$ \|x_k - x\| = \|\langle 1, \frac{1}{k}\rangle - \langle 1, 0\rangle \| = \frac{1}{k} \to 0,$$ while $\|x_k\| = 1 + \frac{1}{k} \to 1$ and $\|x\| = K \gt 1$, so $\|x_k\| \not\to \|x\|$. –  t.b. Oct 8 '11 at 9:09

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