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I want to prove the following theorem (no idea whether it has a name):

Let $V = \mathbb{R}^n$ or $\mathbb{C}^n$ and $\|\cdot\|$ be a norm on $V$. Then, there exist $C_1, C_2 > 0$ such that for all $x \in V$:

$$C_1 \|x\|_\infty \leq \|x\| \leq C_2 \|x\|_\infty$$

I first let $x \neq 0$ (otherwise it would be trivial). Then, I divided by $\|x\|_\infty$ and normalized the vector $x$ such that $\|x\|_\infty = 1$. That left me with

$$C_1 \leq \|x\| \leq C_2$$

but I don't see how this could help me or how I could possibly limit an unknown norm. How can I proceed? Or is this the wrong way anyway? Thanks for any answers.

Srivatsan
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Huy
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    You question is about the equivalence of norm $||\cdot||$ to $||\cdot||_\infty$. In fact all norms on a finite dimensional real (or complex) vector space are equivalent. So search for "equivalence of norms on a finite dimensional real vector space" ... E.g. this page. – Fabian Mar 05 '11 at 13:48
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    It might help to draw a picture, for instance in $\mathbf R^2$. The unit ball in the $||\cdot||$ norm looks like a circle with radius 1, and the unit ball in the $||\cdot||_\infty$ norm looks like a square with sidelength $2$. What you are trying to show is that by rescaling the square, it is both possible to make the square be contained in the circle and the circle contained in the square. – Dan Petersen Mar 05 '11 at 18:46
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    Strengthened version of this result: http://math.stackexchange.com/questions/1669515/equivalent-norms-without-cauchy-schwarz-inequality – Henricus V. Dec 26 '16 at 06:26

3 Answers3

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I try to make a self-contained answer to the question (more info can be found on this page). The short answer is: yes you are on the right track!

What you are missing:

  1. $\|x\|$ is a continuous function with respect to the $\|\cdot\|_\infty$ norm. You can write $x= \sum_m x_m e_m$ with $e_m$ a basis with $\|e_m\|=1$ (similarly, $y= \sum_m y_m e_m$). We are interested in $$\|x-y\| = \left\| \sum_m (x_m - y_m) e_m \right\| \leq \sum_m | (x_m - y_m)| \leq n \| x - y\|_\infty$$ using the triangle inequality. So for every $\epsilon>0$, we can choose a $\delta =\epsilon/n$ such that with $\| x - y\|_\infty < \delta$ it follows that $\|x-y\| < \epsilon$. Therefore, $\|x\|$ is $\infty$-continuous.

  2. The unit sphere $S= \{x :\|x\|_\infty = 1\}$ is a compact set thereby $\|x\|$ assumes a minimum $C_1$ and a maximum $C_2$ on it.

Calvin Khor
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Fabian
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  • This might be somewhat stupid, but is it trivial that the unit sphere is a compact set? I know Heine-Borel's theorem, but as far as I know it is only valid for subsets of $\mathbb{R}^n$. – Huy Mar 05 '11 at 16:37
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    @Huy: If you pick a basis for your space, then you've constructed a linear isomorphism between your space and $\mathbb R^n$. These are continuous, so the Heine-Borel theorem holds on your space. Alternatively, the proof of the Heine-Borel theorem should go through mostly unchanged for a finite-dimensional normed vector space. – Gunnar Þór Magnússon Mar 05 '11 at 17:00
  • @Fabian: Assuming that $| x |\infty := \max{i \in {1, \dots , n} } x_i$ I think you are missing an $n$ in the following line: $$ | x - y | = \dots \leq n | x - y|_\infty$$ – Rudy the Reindeer Jan 05 '12 at 14:42
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    I personally find it slightly confusing to use $n$ as an index variable to sum over if $n$ also denotes the dimension of the space. – Rudy the Reindeer Jan 05 '12 at 14:43
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    @Matt: thank you for the remarks. I changed the answers accordingly. – Fabian Jan 06 '12 at 22:53
  • @Fabian Nice : ) Thank you! – Rudy the Reindeer Jan 07 '12 at 07:13
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If you are interested in an elementary proof, try $C_1 = \min_j \| e_j \|$ and $C_2 = \sum_j \| e_j \|$.

x_y_z
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The absolute value of every component is smaller than the absolute value of the max of all components. So the sum of the absolute values of components is smaller than $n \cdot max$ of components. This gives $C_2=n$. On the other hand, the absolute value of the max of components is smaller than the sum of the absolute values of all components, which is nothing but the first norm, $C_1=1.$

janindunethmal
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Zakaria Oussaad
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