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Sometimes one finds the following definition of a normed algebra: This is an algebra with a norm on the underlying vector space such that there is a constant $K \geq 0$ such that $|x \cdot y| \leq K |x| \cdot |y|$ holds for all $x,y$. Can someone give a specific and natural example of a normed algebra in this sense, appearing in practice, which does not satisfy the inequality $|x \cdot y| \leq |x| \cdot |y|$ for all $x,y$?

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  • $\begingroup$ Maybe the space $C^1 ([0,1]) $ with norm $\|f\|=\max\{\|f\|_\sup, \|f'\|_\sup\} $ does what you want? $\endgroup$ – PhoemueX Dec 5 '15 at 22:25
  • $\begingroup$ Or consider the space of matrices $\Bbb {R}^{n\times n } $ with a non-standard but natural norm like $\|A\|=\max \{|A_{i, j}| \,\mid\, i, j =1,\dots,n\} $. $\endgroup$ – PhoemueX Dec 5 '15 at 23:42
  • $\begingroup$ Thank you. But why do you write answers in comments? $\endgroup$ – Martin Brandenburg Dec 6 '15 at 8:35
  • $\begingroup$ You didn't like my previous answer, so I wanted to verify if this is what you are looking for. I will make an answer out of it. $\endgroup$ – PhoemueX Dec 6 '15 at 10:04
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Here are two examples. In both cases, it is easy to see $|x\cdot y| \leq K \cdot |x|\cdot|y|$ for some $K>0$. Thus, we just have to show that we cannot choose $K=1$.

First example: Take $C^1 ([0,1])$ with the norm $\|f\| := \max \{\|f\|_\sup , \|f'\|_\sup\}$. In this case, we can take $f(x) = x = g(x)$, for which $\|f\| = 1 = \|g\|$, but $(f\cdot g)'(x) = 2x$, so that $\|f \cdot g \| \geq 2 > \|f\| \cdot \|g\|$.

Second example: Consider the space of matrices $\Bbb{R}^{n \times n}$ with the non-standard but naturl norm $\|A\| = \max \{|A_{i,j} \,\mid\, i,j =1,\dots,n\}$. To show that this is not submultiplicative (in general), take $n=2$ and $$x = \left(\begin{array}{cc} 1 & 1 \\ 0 & 1\end{array}\right)=y.$$ Then $\|x\|=1=\|y\|$ but $$ \|xy\| = \left\| \left(\begin{array}{cc} 1 & 2 \\ 0 & 1\end{array}\right) \right\| = 2 > \|x\| \cdot \|y\|. $$


Closing remark (mainly for the sake of future readers of this question): In fact, for a normed algebra as in the question, one can always find an equivalent norm $\|\cdot\|$ which is genuinely submultiplicative. Indeed, define $\|x\|= K \cdot |x| $ with $|\cdot|$, $K $ as in your question (here I assume w.l.o.g. that $K>0$). Then $$ \|xy\| = K |xy|\leq K \cdot K |x||y| = \|x\| \|y\|. $$

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  • $\begingroup$ @MartinBrandenburg: You are right, of course, fixed. $\endgroup$ – PhoemueX Dec 6 '15 at 12:53
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Let $X$ be a metric space and consider the family ${\rm Lip}(X)$ of all real-valued, bounded Lipschitz functions on $X$. One usually considers the norm $$\|f\|=\max\{\|f\|_\infty, \text{Lipschitz constant of }f\}\quad (f\in {\rm Lip}(X))$$ on ${\rm Lip}(X)$. This norm fails to be submultiplicative but multiplication is still continuous (bounded) with respect to this norm. There is a whole book devoted to such algebras:

N. Weaver, Lipschitz Algebras, World Scientific, 1999.

Weaver proposed the term Gelfand algebra for such objects.

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