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I have a perhaps stupid question. When having a finite-dimensional Vectorspace $X$ (f.e. n-dimensional) and when knowing a basis $V=\left\{v_1,...,v_n\right\}$ of it, so any $x\in X$ can be written as $x=\sum\limits_{i=1}^n \alpha_i v_i$.

What is then the maximumnorm of x? And what is the supremumnorm?

Greetings, math12

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Are you asking, whether the $\sup$-norm of $x$ is independent from the basis? –  Ilya Jan 29 '13 at 14:10
    
I simply do not know how it is defined. :-) But your question interests me, too, of course. –  math12 Jan 29 '13 at 14:12
    
To put it another way: I do not know what the maximumnorm of a vector is, which is written as the linear combination of the basis vectors. And I have the same problem for the supremumnorm. –  math12 Jan 29 '13 at 14:47
    
As far as I know, $\|x\|_\sup = \max_i\alpha_i$ and this certainly depends on the basis. –  Ilya Jan 29 '13 at 14:53
    
Not $\lVert x\rVert_{\mbox{sup}}=\max_i \lvert\alpha_i\rvert$? –  math12 Jan 29 '13 at 15:04
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1 Answer

up vote 1 down vote accepted

Briefly, the maximum norm of a vector $(\alpha_i)_i$ (coordinated in a fixed basis), as Ilya commented, is $$\|x\|_{\max}= \max_i{|\alpha_i|}\ .$$

I would say, that the sup norm arises rather on infinite dimensional vector spaces, where, in most cases, the elements are some functions $X\to\Bbb R$. (If $X=\Bbb N$, we get the real sequences.)

For those, $\|f\|_{\sup} =\displaystyle\sup_{x\in X}|f(x)|$. In $\Bbb R$, the supremum of any set exists. In case when it is guaranteed that the supremum value is taken by $f$ (i.e., $f$ has a maximum -- for example if $X$ is compact and $f$ is continuous), then it can also be called a maximum norm, and then $\|f\|_{\max}=\|f\|_{\sup}$.

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Thank you very much. –  math12 Jan 29 '13 at 15:09
    
You're welcome :) –  Berci Jan 29 '13 at 15:09
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