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$$\|x\|_1=\sum_{i=1}^{n} |x_i|$$ and $$\|x\|_2=(\sum_{i=1}^{n}|x_i|^2)^{1\over 2}$$ these two norm induce topologies on $\mathbb{R}^n$, I want to know whether they are comparable?

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It's a standard result that all norms on a finite-dimensional real vector space are equivalent. –  kahen Sep 10 '13 at 10:38
    
I found this picture from Wikipedia: commons.wikimedia.org/wiki/File:Vector_norms2.svg simply by Googling for equivalent norms circle square. I think it nicely illustrates the relation between various norms in $\mathbb R^2$; and it can help with your intuition for $n$-dimensional case. –  Martin Sleziak Sep 10 '13 at 13:55

2 Answers 2

up vote 3 down vote accepted

As @Kahen said, it is a standard result that all norms on a finite-dimensional real vector space are equivalent, however, in this particular case, it can be deduced with some ease (the following inequalities are immediate):

$$|x_i|\leq \left(\sum_{i=1}^n |x_i|^2\right)^{1/2}$$

and

$$\sum_{i=1}^n |x_i|^2\leq \left(\sum_{i=1}^n |x_i|\right)^2$$

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Neighbourhoods in the first topology are (unions of) arbitrarily small N-spheres, those in the second are (unions of) arbitrarily small N-cubes. It's obvious that every sphere contains a cube, which is enough for comparability; but also that every cube contains a sphere, so we have equivalence.

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