Pure states on commutative C* algebra are exactly the characters – elementary proof

banach-algebrasc-star-algebrasfunctional-analysis

I'm trying to come up with an 'elementary proof' that if $$A$$ is a commutative $$C^*$$ algebra, then the pure states of $$A$$ are exactly its characters.

I have already come up with a proof which uses properties of the GNS representation in order to show this. One can also identify the algebra with $$C(X)$$ and it's probably easier to prove this on this specific algebra.
But I was wondering if the is a more 'algebraic' proof which uses relatively elementary results and does not require much more than the basic definitions and properties of pure states.

I can show by simple arguments that all characters are pure states. But I'm having difficulty with the other direction – showing that in the unital and commutative case, pure states must be multiplicative.

Does anyone have an idea?

There is no need for machinery. Properly, one needs a tiny bit of ad-hoc functional calculus to use square roots; concretely, if $$0\leq x\leq 1\$$ then one can define $$x^{1/2}=1-\sum_{n=1}^\infty \frac{2(2n-2)!}{4^nn!(n-1)!}\,(1-x)^n$$ and properties of series show that $$x^{1/2}≥0$$ and $$(x^{1/2})^2=x$$.
Suppose that $$\phi$$ is pure. Fix $$a\in A_+$$ with $$a\leq1$$. If $$\varphi(a)\ne0$$, $$\varphi(1-a)\ne0$$, define states $$\phi_a(x)=\frac1{\phi(a)}\,\phi(ax),\qquad \phi_{1-a}(x)=\frac1{\phi(1-a)}\,\phi((1-a)x).$$ The fact that $$A$$ is commutative guarantees that $$\phi_a$$ is a state, because if $$x\geq0$$ then $$\phi_a(x)=\frac1{\phi(a)}\,\phi(a^{1/2}xa^{1/2})\geq0.$$ Since $$\phi(x)=\phi(a)\,\phi_a(x)+\phi(1-a)\,\phi_{1-a}(x),$$ the fact that $$\phi$$ is pure gives us $$\phi_a=\phi$$. This is $$\tag1 \phi(ax)=\phi(a)\phi(x).$$ If $$\phi(a)=0$$, then for $$x\geq0$$ $$\phi(ax)=\phi(a^{1/2}xa^{1/2})\leq \|x\|\,\phi(a)=0.$$ So again $$\tag2 \phi(ax)=\phi(a)\phi(x),\qquad x\geq0.$$ If $$\phi(1-a)=0$$ instead, the same argument applied to $$1-a$$ gives us $$\phi((1-a)x)=\phi(1-a)\phi(x)$$, which is again $$(2)$$.
As the positive elements span the algebra, $$(2)$$ becomes $$\tag3 \phi(ax)=\phi(a)\phi(x),\qquad x\in A.$$ Now for arbitrary $$a\geq0$$ we apply the above to $$a/\|a\|$$, and then use again that positive elements span $$A$$ to get $$\tag4 \phi(ax)=\phi(a)\phi(x),\qquad a,x\in A.$$