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Let $A$ be a $C^\ast$-algebra and let $\widetilde{A}$ denote its unitisation. Define a norm on $\widetilde{A}$ as $\|(a,\lambda)\| = \sup_{\|b\|=1} \|ab + \lambda b\|$.

I could show that $\|(a,\lambda)^\ast (a,\lambda)\| = \|(a,\lambda)\|^2$ using the submultiplicativity of the norm but now I'm having trouble showing it is submultiplicative.

Please could someone help me and show me how to show

$\displaystyle \|(a,\lambda)(b,\mu)\| \le \|(a,\lambda)\| \|(b,\mu)\|$?

1 Answers1

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What you are doing here is that you identify the element $(a, \lambda)$ with the linear operator $T_{(a,\lambda)} : x\mapsto ax + \lambda x$.

The norm $\Vert (a,\lambda) \Vert$ is then the operator norm $\Vert T_{(a,\lambda)} \Vert$.

It is easy to see that the operator norm in general is submultiplicative.

First show that $\Vert Tx \Vert \leq \Vert T\Vert \cdot \Vert x \Vert$ for all $x$ (using homogeneity of the norm).

This implies $\Vert TS x \Vert \leq \Vert T\Vert \Vert Sx\Vert \leq \Vert T\Vert \cdot \Vert S\Vert \cdot \Vert x \Vert$, which yields the submultiplicativity.

PhoemueX
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  • Great answer, thank you! There where you mention to use the homogeneity of the norm I am wondering if it doesn't follow straight from the definition: $|T|= \sup_{|x|\neq 0} {|Tx|\over |x|}$. –  Sep 17 '14 at 03:01
  • Oh, yes. I thought you were using the definition $\Vert T \Vert = \sup_{\Vert x \Vert = 1} \Vert Tx\Vert$, which of course is the same in the end. – PhoemueX Sep 17 '14 at 06:33