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Let $A$ be a C$^\ast$-algebra and let $S,T: A \to A$ be bounded linear operators such that $\|T\|=\|S\|$.

Is it true that $\|T^2\| = \|ST\|=\|TS\|=\|S^2\|$?

I believe not but if not I don't understand why the last equation holds here:

If $a \in A$ and $\|a\|\le 1$ then $$ \|L(a)\|^2 =\|(L(a))^\ast L(a)\|=\|L^\ast(a^\ast) L(a)\| = \|a^\ast R^\ast L(a)\| \le \|R^\ast L\| = \|T^\ast T\| $$

where $A$ is a C star algebra and $T = (L,R)\in M(A)$ is a double centraliser (element of the multiplier algebra). Please could someone help me understand?

1 Answers1

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Counterexample for the statement: Let's look at a $2 \times 2$ matrix example – define $$ S = \pmatrix{1&1\\0&1}, \quad T= \pmatrix{\phi & 0\\0&0} $$ Where $\phi := \frac 12(1 + \sqrt5)$. Note that $\|S\| = \|T\| = \phi$, where $\|M\| = \sigma_1(M)$.

However:

  • $\|S^2\| = \sqrt{3 + 2\sqrt{2}} = 1 + \sqrt 2$
  • $\|ST\| = \phi$
  • $\|TS\| = \sqrt{2}\phi$
  • $\|T^2\| = \phi^2 = 1 + \phi$
Ben Grossmann
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    Having gone through the trouble, I realize there's an easier example. Namely: $$ S = \pmatrix{0&1\0&0}, \quad T = \pmatrix{1&0\0&0} $$ The failure of the statement in this case, however, is arguably less spectacular. – Ben Grossmann Sep 12 '14 at 04:11
  • Why does $|R^\ast L| = |T^\ast T| $ hold? –  Sep 12 '14 at 04:37
  • @student I'm not sure. I'll update my answer if I figure it out. – Ben Grossmann Sep 12 '14 at 04:50