Hahn-Banach Theorem in the C*-algebra

Question

What is the Hahn-Banach Theorem in the C*-algebra(or W*-algebra maybe)?

If B is an nondense subalgebra of C*-algebra(or W*-algebra maybe), can we get an state f of A which is always zero at the subalgebra B from the version of Hahn-Banach Theorem?

Answer 1

Every state is a completely positive map. Then use Arveson extension theorem. For details see this notes.

Answer 2

I will assume that $A$ is unital (otherwise we can consider $B$ inside the unitization of $A$).

The answer is obviously no when $A$ is unital and $1\in B$: in that case, any $x\in B^+$ will satisfy $0\leq f(x)\leq f(\|x\|\,1)=0$, so $f(x)=0$ and by linearity $f(x)=0$ for all $x\in B$.

When $1\not\in B$, the answer is yes. Since $1\not\in B$, we can define (by linear independence) a linear functional $\varphi$ with $\varphi|_B=0$ and $\varphi(1)=1$. The space $B_1=B+\mathbb C\,1$ is closed (this is a consequence of the fact that $\mbox{dist}(1,b)>\delta$ for all $b\in B$ and some fixed $\delta>0$) and it is $*$-algebra, so it is a C$^*$-subalgebra of $A$. Note also that $\varphi$ is positive: $$ \varphi((b+\mu\,1)^*(b+\mu 1))=\varphi(b^*b+2\mbox{Re}\bar\mu b+|\mu|^2\,1)=|\mu|^2\geq0. $$ Being positive, we have that $\|\varphi\|=\varphi(1)=1$ (every positive functional achieves its norm on the identity).

Now we can extend $\varphi$, by Hahn-Banach, to a functional $\tilde\varphi$ on all of $A$ with the same norm. So $\tilde\varphi$ satisfies $\|\tilde\varphi\|=\|\varphi\|=\varphi(1)=\tilde\varphi(1)$, which implies that it is positive.

In the end, $\tilde\varphi$ is a state on $A$ that annihilates $B$.