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Let $A$ be a $C^*$-algebra, and $\ f:A \to \mathbb C$ a positive linear functional on $A$. Where $\tilde A$ is the unitization of $A$, is there a nice way to extend $f$ to a positive linear functional on $\tilde A$? Is this always possible?

Two approaches I took to this were to set $f(a + \lambda I) = f(a) + \lambda$ (I identified $a \in A$ with its counterpart in $\tilde A$), and $f(a + \lambda I) = f(a)$. I tried to show that $\|f\|_{\tilde A} = 1$ in the first approach, but I got stuck. The second thing that I tried was just computational: $$ f((a + \lambda I)^*(a + \lambda I)) = f(a^*a + \bar\lambda a + \lambda a^*) + |\lambda|^2 $$ This seemed promising, but I'm not sure how to proceed. It may be my naivety thinking, but I seem to doubt there's any hope for the second approach.

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  • $\begingroup$ A positive linear functional defined on a subalgebra can always be extended to the whole algebra, regardless of whether or not the whole algebra is the unitization. This is proven, for example, in Theorem 3.3.8 of Murphy's book C$^*$-Algebras and Operator Theory. $\endgroup$ – Aweygan Nov 28 '17 at 13:48
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Your second approach cannot succeed, because $f(I)=0$; this guarantees that $f$ is not positive unless its zero (you have $|f(a)|\leq f(\|a\|\,I)=0$).

In your first approach, the right extension is $$ f(a+\lambda)=f(a)+\lambda\,\|f\|, $$ and then we may assume without loss of generality that $\|f\|=1$ (that is, $f$ is a state).

Now we can check positivity directly: if $b\in \bar A$ is positive, the $b=(a+\lambda)^*(a+\lambda)$ for some $a\in A$, $\lambda\in\mathbb C$. Then, using Kadison's inequality, \begin{align} f(b)&=f((a+\lambda)^*(a+\lambda))=f(a^*a+2\text{Re}\,\bar\lambda a+|\lambda|^2)\\ \ \\ &=f(a^*a)+2\text{Re}\,\bar\lambda f(a)+|\lambda|^2\\ \ \\ &\geq |f(a)|^2+2\text{Re}\,\bar\lambda f(a)+|\lambda|^2\\ \ \\ &=|f(a)+\lambda|^2\geq0. \end{align}

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  • $\begingroup$ Ah! Very nice. Thank you. Perhaps I need to spend some more time playing with the inequalities associated with plfs. Where can I find a proof of Kadison's inequality? Arveson? $\endgroup$ – fauxefox Nov 28 '17 at 17:15
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    $\begingroup$ You will find proofs in wherever they talk about completely positive maps. But, for a state, it is simply Cauchy-Schwarz: if $\{b_j\}$ is an approximation of the unity, then $$|f(a)|^2=\lim_j|f(b_j^*a)|^2,$$ and $$|f(b_j^*a)|^2\leq f(a^*a) f(b_j^*b_j)\leq f(a^*a)\,\|b_j\|^2\leq f(a^*a).$$ $\endgroup$ – Martin Argerami Nov 28 '17 at 19:45

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