Completely positive map is $*$-homomorphism

Question

Suppose $A$ is a $C^*$ algebra and we have a completely positive contractive map $f \colon A\rightarrow B(H)$ such that $sup_{a,b \in A}\lVert f(ab)-f(a)f(b)\rVert =0$. Can we conclude that $f$ is a $*$-homomorphism?

Answer 1

The conditions immediately imply that $f$ is an algebra homomorphism. To show that it preserves adjoints, let $a\in A$ be self-adjoint. Then $a=a_+-a_-$, where $a_+,a_-$ are positive elements of $A$. Then $f(a)=f(a_+)-f(a_-)$ is a difference of positive elements (as $f$ is a positive map), hence self-adjoint. If now $b\in A$ is arbitrary, write $b=b_1+ib_2$, where $b_1,b_2$ are self-adjoint. Then $$f(b)^*=(f(b_1)+if(b_2))^*=f(b_1)-if(b_2)=f(b_1-ib_2)=f(b^*),$$ and therefore $f$ is a $*$-homomorphism.

Indeed, the above work proves the following proposition:

Let $A,B$ be $C^*$-algebras, and let $f:A\to B$ be a positive linear map. Then $f$ is $*$-preserving, i.e., $f(a)^*=f(a^*)$ for all $a\in A$.

Answer 2

Assuming that the sup is take over all $a,b\in A$, then the condition $\sup_{a,b}\|f(ab)-f(a)f(b)\|=0$ is exactly the same as $f(ab)=f(a)f(b)$ for all $a,b$. Neither contractive not completely positive are relevant in that case. All you need is $f$ to be linear and preserve adjoints.