User martin wanvik - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-22T20:44:59Z http://mathoverflow.net/feeds/user/19506 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/98955/bounded-linear-functionals-and-representations Bounded linear functionals and representations Martin Wanvik 2012-06-06T14:44:43Z 2012-06-06T17:39:49Z <p>Suppose that $A$ is a unital C$^*$-algebra and that $\varphi: A \to \mathbb{C}$ is a bounded linear functional. Then there exists a Hilbert space $H$, a representation $\pi: A \to B(H)$ and vectors $\psi, \eta \in H$ such that $$\varphi(a) = \langle \pi(a)\psi, \eta \rangle$$ for all $a \in A$ (this can be proved by decomposing the functional as a linear combination of four states and considering the direct sum of the representation spaces associated to the GNS-construction for each state). </p> <p>My question is: </p> <blockquote> <p>Assuming further that $\| \varphi \| \leq 1$, can we choose $H$, $\pi$ and $\psi, \eta$ as above, <strong>satisfying the additional requirement that $\| \psi \| \leq 1$ and $\| \eta \| \leq 1$</strong>, such that (again) $\varphi(a) = \langle \pi(a) \psi, \eta \rangle$ for all $a \in A$?</p> </blockquote> <p>Note that this is clearly true for a positive functional - simply write $\varphi(a) = \langle \pi(a)\xi, \xi \rangle$ (using the GNS-construction) and note that $$1 \geq \| \varphi \| = \sup_{\| a \| \leq 1} |\langle \pi(a) \xi, \xi \rangle | \geq |\langle \pi(1) \xi, \xi \rangle| = \| \xi \|^2$$</p> http://mathoverflow.net/questions/98955/bounded-linear-functionals-and-representations/98963#98963 Comment by Martin Wanvik Martin Wanvik 2012-06-07T13:10:39Z 2012-06-07T13:10:39Z Great! Thanks, Nik. For posterity, I'll add the exact reference to Takesaki's book: theorem 4.2 and proposition 4.6 in chapter III (I actually did ask someone else about this prior to posting here on MO, and was pointed in this direction - what I failed to notice, however, was the equality $\| \omega \| = \| \phi \|$, and ended up with the estimate $\| \varphi \| \geq |\varphi(1)| = |\langle \pi(v)\xi,\xi \rangle|$, rather than $\| \varphi \| \geq \omega(1) = \| \xi\|^2$).