I'm reading the book "Theory of operator algebras" by Takesaki. In this book, the $\sigma$-strong$^*$ topology on the space $B(H)$ (bounded operators on the Hilbert space $H$) is defined (see def 2.3 p68) to be the locally convex topology generated by the seminorms

$$B(H)\ni x \mapsto \left(\sum_{n=1}^\infty \|x\xi_n\|^2 + \|x^*\xi_n\|^2\right)^{1/2}.$$I am trying to understand the proof of lemma 2.4:

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Why does it suffice to use only one seminorm? Shouldn't we have something like $$\sum_{k=1}^m \left(\sum_{n=1}^\infty (\|x(\xi_k)_n\|^2+ \|x^*(\xi_k)_n\|^2\right)^{1/2} \le 1 \implies |\omega(x)| \le 1$$ where $\xi_1, \dots, \xi_m$ are sequences in $H$.


1 Answer 1


So, we are dealing with locally convex topological vector spaces. I think, in general, given a family of seminorms, you would need to consider the finite intersections of the open balls they form, see wikipedia article. So a basic open set about $0$ is of the form $$ \{ x : p(x)<r \ (p\in F) \} $$ where $F$ is a finite subset of the set $\mathcal P$ of generating seminorms, and $r>0$. However, in the special case of the $\sigma$-strong$^*$-toplogy, this isn't necessary.

In our case, the seminorms are of the form $$ p_{\xi}(x) = \Big( \sum_{n=1}^\infty \|x\xi_n\|^2 + \|x^*\xi_n\|^2 \Big)^{1/2}, $$ for any sequence $\xi=(\xi_n)$ with $\sum_n \|\xi_n\|^2<\infty$. We can always rescale, so it suffices to take $r=1$ above. Now notice the special property: given $\xi,\eta$, with $\gamma$ the union of the sequences $\xi,\eta$ (say $\gamma_{2n} = \xi_n, \gamma_{2n-1} = \eta_n$) then if $p_\gamma(x) < 1$ then certainly $p_\xi(x)<1$ and $p_\eta(x)<1$. The same argument applies to finite families.

We conclude that for our $\mathcal P$, for any basic open set $U$ about $0$ there is some $p\in\mathcal P$ so that $\{ x : p(x)<1 \}$ is contained in $U$.

A linear functional $f$ is continuous if and only if $f$ is bounded on some open set about $0$. So, with $\mathcal P$ having our property, f is continuous if and only if there is some $p\in\mathcal P$ with $\{ f(x) : p(x)<1 \}$ bounded. Again by rescaling, equivalently, there is some $p\in\mathcal P$ with $|f(x)|\leq 1$ for all $x$ with $p(x)<1$.

Finally, a further standard rescaling argument shows that you can use $p(x)\leq 1$ instead of $p(x)<1$.


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