The WOT on the unit ball of $\mathcal{B}(\mathcal{H})$ is determined by total subsets of $\mathcal{H}$.

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The following picture is taken from the Chapter 5 (page-305) of Kadison-Ringrose I. In this picture the authors have described an argument of the fact that the WOT on unit ball of $\mathcal{B}(\mathcal{H})$ is determined by a subset of $\mathcal{H}$ whose closed linear span is equal to $\mathcal{H}$:

enter image description here

Here, I could not verify this last line:

For this, note that $|\langle Ax, x \rangle|$ is small (with $A$ in $(\mathcal{B}(\mathcal{H}))_1$) provided $|\langle Ay, y \rangle|$ is small with $y$ (in the span of $(x_j)$) sufficiently near $x$.

What I have done so far is as follows:

$|\langle Ax, x \rangle|=|\langle Ax-Ay, x-y \rangle- \langle Ay, y \rangle + \langle Ax, y \rangle +\langle Ay, x \rangle|\le||x-y||^2+ |\langle Ay, y \rangle|+|\langle Ax, y \rangle| +|\langle Ay, x \rangle|$

Now, $|\langle Ay, y \rangle|$ and $||x-y||$ are small but how do I manage the last two terms in this last inequality?

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You keep going: $$ |\langle Ax,y\rangle|\leq|\langle A(x-y),y\rangle|+|\langle Ay,y\rangle| \leq \|x-y\|\,\|y\|+|\langle Ay,y\rangle|. $$