Linear Transformation from C*-Algebra is Bounded if Bounded on Positive Elements

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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.

$\blacksquare$