Linear Transformation from C*-Algebra is Bounded if Bounded on Positive Elements
Theorem
Let $\struct {A, \ast, \norm {\, \cdot \,}_A}$ be a $\text C^\ast$-algebra.
Let $\struct {B, \norm {\, \cdot \,}_B}$ be a normed vector space over $\C$.
Let $T : A \to B$ be a linear transformation such that there exists $M > 0$ with:
- $\norm {T x}_B \le M \norm x_A$ for each $x \in A$ positive.
Then $T$ is bounded.
Proof
From Linear Transformation from Banach *-Algebra is Bounded if Bounded on Hermitian Elements, it is enough to show that there exists $M' > 0$ such that:
- $\norm {T x}_B \le M' \norm x_A$ for each $x \in A$ Hermitian.
Let $x \in A$ be Hermitian.
From Hermitian Element of C*-Algebra Decomposes into Positive Elements, there exists positive $x^+, x^-$ such that:
- $x = x^+ - x^-$
with:
- $\norm {x^+}_A \le \norm x_A$ and $\norm {x^-}_A \le \norm x_A$
Then we have:
\(\ds \norm {T x}_B\) | \(=\) | \(\ds \norm {T x^+ - T x^-}_B\) | Definition of Linear Transformation | |||||||||||
\(\ds \) | \(\le\) | \(\ds \norm {T x^+}_B + \norm {T x^-}_B\) | Norm Axiom $\text N 3$: Triangle Inequality | |||||||||||
\(\ds \) | \(\le\) | \(\ds M \norm {x^+}_A + M \norm {x^-}_A\) | ||||||||||||
\(\ds \) | \(\le\) | \(\ds M \norm x_A + M \norm x_A\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds 2 M \norm x_A\) |
for all $x \in A$.
Hence $M' = 2 M$ satisfies:
- $\norm {T x}_B \le M' \norm x_A$ for each $x \in A$ Hermitian.
So by Linear Transformation from Banach *-Algebra is Bounded if Bounded on Hermitian Elements, $T$ is bounded.
$\blacksquare$