Ollie
• Member for 11 years, 2 months
• Last seen more than 4 years ago
• Lancaster, UK

Here's a mistake I've seen from students taking a first course in linear analysis. For a vector $g$ in a Hilbert space $H$, it is true that $\langle f,g\rangle=0$ for every $f\in H$ implies $g=0$. ...

A low tech (naive?) piece of intuition comes straight from the definition of the modular operator and what happens if one tries to carry it over to finite fields. The nontrivial automorphism $z\... View answer 12 votes Since you've answered part of the question, let me elaborate on the Dunford integral. If$f:\Omega\to E$is weakly measurable and satisfies$\langle x^* ,f\rangle\in L^1(\Omega)$for all$x^*\in E^*$... View answer Accepted answer 8 votes I highly recommend Segal's original paper Equivalences of Measure Spaces [American Journal of Mathematics Vol. 73, No. 2 (1951), pp. 275-313], where he introduced localizable spaces, since this was ... View answer Accepted answer 6 votes Essentially I think weighted shifts should be a sufficiently rich class of operators. Consider, for instance, the following example. Take the doubly infinite sequence$$w_k=\left\{\begin{array}{ll} ... View answer 5 votes Yes, Araki-Woods showed they're always ITPFI factors and ITPFI factors are hyperfinite. See the following: Araki-Woods, A classification of factors It's a bit of a monster paper, the stuff on CCR ... View answer 4 votes I'm reasonably convinced the algebra generates all of$B(\mathcal{F}(H))$. Note that$1-\sum_{i=1}^\infty s_is_i^*$converges strongly to the projection$P\_0$onto$\mathbb{C}$and similarly the sum$...

For any contraction $X$ on $H$, the operator $\Gamma(X)$ (in Alain's notation S(X)) is a contraction. It is essentially algebraic to check that a pair of contractions $X,Y$ on $H$ will satisfy $\Gamma(... View answer 2 votes Here is another proof. The elements of$C(\mathbb{R};U(n)) $satisfying$f^2=1$are functions whose values (under the standard representation of U(n)) are self-adjoint unitaries. There are$n+1\$ ...