# Linear functionals can be decomposed as linear combinations of positive ones?

I am reading Arveson's Notes on Extensions of $C^*$-algebras. In proving theorem 1, he needs to establish some results concerning bounded linear functionals. However, he said it suffices to prove for positive linear functionals, which I do not quite understand.

So are bounded linear functionals over a $C^*$-algebra always the linear combinations of 4 positive ones?

Thanks!

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Every hermitian linear functional is a combination of two positive ones. Even in the case of commutative C^* algebras this is non-trivial, it is the Hahn-Jordan decomposition of a finite real-valued measure. –  user16299 Jul 9 '12 at 19:42
@YemonChoi I know the theorem and that it implies the existence of the decomposition when the functionals are defined over measurable functions. And I guess commutative $C^*$ is okay given the correspondence between them and compact hausdorff spaces and the riesz, which connects such spaces to measures. But I wonder how could one prove the general case. –  Hui Yu Jul 9 '12 at 20:20

I seem to remember you saying that you have access to Murphy's book?

If so, then the result you want is Theorem 3.3.10 (see also Theorem 3.3.6). The key trick is that there is a positive, real-linear isometry from the real vector space $A_{sa}$ into $C_{\mathbb R}(\Omega)$ for a suitable $\Omega$.

If you do not have access then I could copy out the details.

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It's been so kind of you. I have the book at my hand. I will jump a few sections and read that part first. Thanks! –  Hui Yu Jul 9 '12 at 23:53

Certainly not, if it's a complex C^* algebra: the difference of positive functionals is real on the hermitian elements.

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Well, I edited my question. I first decompose the linear functional into real and imaginary parts, then ask whether these are difference of positive functionals. –  Hui Yu Jul 9 '12 at 19:11
@JonasMeyer ha! Yes! By 'two' I mean 'four' :) –  Hui Yu Jul 10 '12 at 1:39