# Alternative proof of existence of absolute value of a functional on a C*-algebra

The usual proof of the existence of an absolute value of a functional on a C*-algebra $$A$$ uses the polar decomposition of normal functionals on $$A^{**}$$, which relies on the compactness of the unit ball of $$A^{**}$$ in the weak*-topology.

Is it possible to derive the existence of an absolute value of a bounded linear functional on a C*-algebra via a compactness argument in $$A^*$$? For a definition of the absolute value of $$\varphi \in A^*$$, I mean a $$\tau \in A^*_+$$ such that $$\|\tau\| = \|\varphi\|$$ and $$|\varphi(a)|^2 \leq \|\varphi\| \tau(a^* a)$$.

Furthermore, is it possible to use this to show that $$\varphi$$ can be represented as $$\varphi(a) = \langle \pi_\tau(a) \xi \,\vert\, \eta\rangle$$ where $$\pi_\tau$$ is the GNS representation associated with $$\tau$$ and $$\|\varphi\| = \|\xi\| \|\eta\|$$?

• I haven't worked out the details, but another way to think about absolute values is to use a $2\times 2$ matrix construction. What I have in mind is how you move from completely bounded maps to completely positive maps, by putting the cb map on the off diagonal, and using a Hahn-Banach argument to find cp maps on the diagonal which make the whole matrix CP. This allows you to use Stinespring to prove to representation result for CB maps. I think the same works for linear functionals, but I haven't written this out to check there is not a lurking circular argument... Jun 10 '20 at 8:31
• That does work, although you need to embed a proof that the CB norm of a linear functional is its norm, so it's not that much different than the full CB representation theorem. One other approach I tried that doesn't work is proving representation for self-adjoint functionals using Jordan Decomposition, and then trying to lift that to a self-adjoint functional in the 2x2 case. You do better than the naive estimate of $4 \|\varphi\|$, but you end up with a $\sqrt{2}$ factor in the norms of the vectors. Jun 10 '20 at 14:02

This is maybe a "backwards" answer to what you might have been hoping for...

An affirmative answer to the 2nd question would give the 1st question as well. Indeed, if there is a $$*$$-representation $$\pi:A\rightarrow B(H)$$ and $$\xi,\eta\in H$$ with $$\|\varphi\| = \|\xi\| \|\eta\|$$ and $$\varphi(a) = \langle \pi(a)\xi, \eta \rangle$$, then by rescaling we may suppose that $$\|\xi\|=\|\eta\|=\|\varphi\|^{1/2}$$, and then $$\tau:a\mapsto \langle \pi(a)\xi,\xi\rangle$$ is a positive functional with $$\|\tau\| = \|\xi\|^2 = \|\varphi\|$$. Further, $$\|\varphi\| \tau(a^*a) = \|\eta\|^2 \|\pi(a)\xi\|^2 \geq | \langle \pi(a)\xi, \eta \rangle|^2 = |\varphi(a)|^2.$$

I do not know a "compactness" argument which can show this. However, a while ago I wrote up an argument of how to prove Kaplansky Density using Arens products; see Notes on GitHub. To do this in a non-circular way, you need to use Hahn-Banach, and need to find a way to prove exactly this representation result, in a "simple" way. The way I did this was as follows.

Let $$H$$ be a Hilbert space. Let $$B(H)_*$$ be the trace-class operators on $$H$$, the predual of $$B(H)$$. By Hahn-Banach (Goldstine's Theorem) for $$\mu\in B(H)^*$$ there is a net $$(\omega_i)$$ in $$B(H)_*$$ converging weak$$^*$$ to $$\mu$$, and with $$\|\omega_i\|=\|\mu\|$$ for each $$i$$. Thus, for an ultrafilter $$\mathcal U$$ refining the order filter, the natural map $$(B(H)_*)_{\mathcal U} \rightarrow B(H)^*$$ (given by "take weak$$^*$$-limit") is a metric surjection. With $$K = \ell^2(H)$$, and $$\pi_0:B(H)\rightarrow B(K)$$ the "diagonal" map, for each $$\omega\in B(H)_*$$ we can find $$\xi,\eta\in K$$ with $$\omega = \omega_{\xi,\eta}\circ\pi_0$$. Now let $$K = (\ell^2(H))_{\mathcal U}$$ the ultrapower, a Hilbert space. We can find $$\xi = (\xi_i), \eta=(\eta_i)\in K$$ with $$\omega_i = \omega_{\xi_i,\eta_i}\circ\pi_0$$ for each $$i$$. Thus, with $$\Pi:B(H)\rightarrow B(K)$$ the diagonal of $$\pi_0$$, we have that $$\langle \Pi(x)(\xi), \eta \rangle = \lim_{i\rightarrow\mathcal U} \langle \pi_0(x)\xi_i, \eta_i \rangle = \lim_{i\rightarrow\mathcal U} \omega_i(x) = \mu(x).$$

Given a $$C^*$$-algebra $$A$$, by the GNS construction, we can exhibit $$A$$ as a subalgebra of $$B(H)$$ for some $$H$$. For $$\varphi\in A^*$$ take a Hahn-Banach extension to $$\mu\in B(H)^*$$. Then the previous paragraph, with $$\pi$$ the restriction of $$\Pi$$ to $$A$$, gives the required representation.

• I was going to update my question when I fully convinced myself I didn't understand, but there's a proof like the one I suggested in the middle of Theorem 1 in [ousar.lib.okayama-u.ac.jp/en/list/authors/T/Minoru,Tomita/item/… theory of operator algebras I) by Tomita. He says that since every bounded functional is a linear combination of states, you can find such a $\tau$, but maybe not one of the desired norm. Why can you find such a $\tau$ to begin with? It's also strange that he picks one of minimal norm, but never uses minimality... Jun 9 '20 at 20:06
• That aside, your proof seems like the morally correct one, so I'm going to accept it. There's a variant of this idea in Theorem 1.10.8 of Ilijas Farah's book Combinatorial Set Theory of C*-Algebras, which is actually attributed to Ozawa in the acknowledgements. However, it uses a Banach limit of functionals on the diagonal rather than a Hilbert space ultrapower, and hence needs a further step in the form of an inequality that is rather difficult to motivate. Jun 9 '20 at 20:26
• I don't know your motivation for an alternative proof. But maybe Bishop-Phelps theorem helps? Jun 10 '20 at 0:55