Can the fact "If $\phi$ is a Hermitian linear functional on a $C^*$-algebra $A$ then $||\phi|| = \sup\{\phi(x):x = x^*, ||x|| = 1\}$" be generalized?

148 Views Asked by At

This is Proposition 13.3 on Page 78 in An Introduction to Operator Algebras by Kehe Zhu. The statement is as follows:

Proposition 13.3. If $\varphi$ is a Hermitian linear functional on a $C^*$-algebra $A$, then $\Vert \varphi \Vert = \sup\{\varphi(x):x = x^*, \Vert x\Vert \leq 1\}.$

where a linear functional $\varphi: A \rightarrow \mathbb{C}$ on a $C^*$-algebra $A$ is Hermitian if $\varphi(x^*) = \overline{\varphi(x)}$.

I wonder whether this can be generalized to the case that the "target" of the linear map $\varphi$ is a general $C^*$-algebra, say, $B$? I give the definition below:

  • A linear map $\varphi: A \rightarrow B$ between two $C^*$-algebras $A$ and $B$ is said to be Hermitian if $\varphi(x^*) = \varphi(x)^*$.

and I guess Proposition 13.3 can be generalized as the following statement:

If $\varphi: A \rightarrow B$ is a Hermitian linear map between two $C^*$-algebras $A$ and $B$, then $\Vert \varphi \Vert = \sup\{\Vert\varphi(x)\Vert:x = x^*, \Vert x\Vert \leq 1\}$.

It is evident that in the above statement, the right-hand side is less than or equal to the left-hand side, because by definition $\Vert\varphi\Vert = \sup\{\Vert\varphi(x)\Vert: \Vert x\Vert \leq 1\} $ (without the extra limitation $x = x^*$). However, I have no idea how to obtain the inequality in the reverse direction, because in Zhu's book he involved the properties of complex numbers such as rotations, etc. enter image description here

Could anyone present a proof of this statement or a counterexample to disprove this statement (although I may feel a bit disappointed)? Any help is appreciated.