Sign up ×
MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

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).

My question is:

Assuming further that $\| \varphi \| \leq 1$, can we choose $H$, $\pi$ and $\psi, \eta$ as above, satisfying the additional requirement that $\| \psi \| \leq 1$ and $\| \eta \| \leq 1$, such that (again) $\varphi(a) = \langle \pi(a) \psi, \eta \rangle$ for all $a \in A$?

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 $$ 1 \geq \| \varphi \| = \sup_{\| a \| \leq 1} |\langle \pi(a) \xi, \xi \rangle | \geq |\langle \pi(1) \xi, \xi \rangle| = \| \xi \|^2$$

share|cite|improve this question

2 Answers 2

up vote 8 down vote accepted

Yes, you can do this using polar decomposition. We can also consider $\phi$ to be a normal linear functional on $A^{**}$, and there is a positive $\omega \in A^*$ and a partial isometry $v \in A^{**}$ such that $\phi(a) = \omega(va)$ for all $x \in A$. (I'm sure this is in volume 1 of Takesaki, probably also in Pedersen.) We have $\|\omega\| = \|\phi\| \leq 1$, so we can apply GNS to $\omega$ and get $\phi(a) = \omega(va) = \langle \pi(va)\xi,\xi\rangle = \langle \pi(a)\psi,\eta\rangle$ with $\psi = \xi$ and $\eta = \pi(v^*)\xi$. As you note, $\|\xi\|^2 = \langle \pi(1)\xi,\xi\rangle = \omega(\xi) \leq 1$.

share|cite|improve this answer
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 $\| \varphi \| \geq \omega(1) = \| \xi\|^2$). – Martin Wanvik Jun 7 '12 at 13:10

Another point of view is that this result can be seen as a special case of its "completely contractive" version (in the same way that Stinespring's dilation theorem for completely positive maps generalizes the GNS representation of states). In particular, the following is true: if $\Phi:A\to B(H)$ is any completely contractive map, then there exists another Hilbert space $K$, a representation $\pi:A\to B(K)$, and operators $V_1, V_2:H\to K$ with $\|V_i\|\leq 1$ such that $$ \Phi(a) =V_2^*\pi(a)V_1 $$ for all $a\in A$. This result may be found e.g. in Chapter 8 of Paulsen's Completely Bounded Maps and Operator Algebras. Essentially the idea is to use Arveson's extension theorem to embed the original $\Phi$ as the "upper right corner" of a completely positive map into the $2\times 2$ matrices over $B(H)$, and then use the Stinespring theorem.

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.