Let $A$ be a $C^{*}$-algebra with identity and abelian condition.
If $\varphi\colon A \to \mathbb{C}$ is a linear map so that for all $a\in A$, $\varphi (a^{*} a ) \geq 0$.
How can I show that $ \varphi$ is continuous?
Please help me.
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There is an answer to a more general question here: http://math.stackexchange.com/questions/426487/why-is-every-positive-linear-map-between-c-algebras-bounded – Ulrik Dec 20 '16 at 10:30
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We know that every positive linear map between C∗-algebras is bounded.Does it matter whether $ a$ is self-adjoint or not ? I mean in question $ \varphi ( a^{*}a) \geq 0$ – m.t Dec 20 '16 at 11:40
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Whether or not $a$ is self-adjoint, $a^a$ is always positive, and further every positive element of $A$ can be written as $c^c$ for some $c$. – Josh Keneda Dec 20 '16 at 14:40
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Possible duplicate of positive linear functionals are bounded in $C^*$-algebras – Guy Fsone Dec 04 '17 at 09:35
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With the assumption that $A$ is unital, the argument is much simpler.
For $a\geq0$, we have $$ \varphi(a)\leq\varphi(\|a\|\,I)=\varphi(I)\,\|a\|. $$ Since every $a\in A$ is a linear combination of four positives (with coefficients that have absolute value $1$, and each positive with norm at most the norm of $a$), we get $$ |\varphi(a)|\leq 4\|a\|\,\varphi(I). $$ So $\varphi$ is bounded and $\|\varphi\|\leq4\varphi(I)$.
The above estimates are extremely crude, though: in reality, we always have $\|\varphi\|=\varphi(I)$.
Martin Argerami
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