Proposition 41.3.14.label Let $X$ be a compact Hausdorff space, then:
- (1)
$C(X; M_{n}(\complex))$ equipped with pointwise product, pointwise involution, and the norm
\[\norm{f}_{C(X; M_n(\complex))}= \sup_{x \in X}\norm{f(x)}_{M_n(\complex)}\]is a $C^{*}$-algebra.
- (2)
The mapping
\[M_{n}(C(X; \complex)) \to C(X; M_{n}(\complex)) \quad M \otimes f \mapsto M \cdot f\]is a *-isomorphism.
- (3)
For any $f \in C(X; M_{n}(\complex))$, $f \ge 0$ if and only if $f(x) \ge 0$ in $M_{n}(\complex)$ for all $x \in X$.
Proof. (1): For any $f \in C(X; M_{n}(\complex))$,
(3): If $f \ge 0$, then there exists $g \in C(X; M_{n}(\complex))$ such that $f(x) = g^{*}(x)g(x)$ for all $x \in X$, so $f(x) \ge 0$ in $M_{n}(\complex)$ for all $x \in X$.
On the other hand, if $f(x) \ge 0$ in $M_{n}(\complex)$ for all $x \in X$, then $g(x) = f(x)^{1/2}$ is defined for all $x \in X$. By the continuous functional calculus, $g$ is continuous from $X$ to $M_{n}(\complex)$. Therefore there exists $g \in C(X; M_{n}(\complex))$ such that $f = g^{*}g$.$\square$
Post a Comment