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I was reading the book Elements of noncommutative geometry and in page 160 lemma 4.21 the authors state that in a Hilbert B-A bimodule $E$ the two norms induced by the two inner products coincide. Reading the proof I noticed something missing to justify one of the inequalities. He states that $$\|(\{s|s\}s|\{s|s\}s)\|_A\leq \|t\mapsto \{s|s\}t\|^2\|(s|s)\|_A$$ and that $$\|t\mapsto \{s|s\}t\|=\|\{s|s\}\|_B$$ Here $\{\cdot |\cdot \}$ denotes the inner product over $B$ that makes ${}_B E$ a left Hilbert $B$-module and $(\cdot | \cdot)$ denotes the inner product over $A$ that makes $E_A$ a left Hilbert $A$-module.

However since $t\mapsto \{s|s\}t$ can be seen as a bounded operator on $E_A$ and on ${}_{B}E$, where the norms could be different (in principle) then the norm of $t\mapsto \{s|s\}t$ isn't unique. This means that we can't mix the equation above with the inequality, since those $\|t\mapsto \{s|s\}t\|$ represent different norms in each case. When trying to come up with a fix for this proof I felt some extra hypothesis was missing, for example have both $E_A$ and ${}_{B}E$ be full Hilbert modules. Reading Blackadar's book on operator theory, I found that in page 151 he proves the same result for full Hilbert bimodules as I suspected. Is the fullness a necessary hypothesis?

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  • $\begingroup$ I don't think it matters. You can replace $A$ by the closure of $(E|E)$ and $B$ by the closure of $\{E|E\}$ to obtain a full bimodule with the same norms. $\endgroup$ – MaoWao Nov 21 '18 at 14:28

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