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Why is every positive linear functional bounded in $C^*$-algebras?

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See page 11: math.berkeley.edu/~theojf/CstarAlgebras.pdf – 1015 Feb 6 '13 at 21:27
    
In general, every positive $ \ast $-homomorphism from a C$ ^{\ast} $-algebra to another is necessarily bounded. – Haskell Curry Feb 6 '13 at 21:39
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@Haskell: but functionals are not homomorphisms, so why "in general"? Also, note that every $*$-homomorphism is positive. – Martin Argerami Feb 16 '13 at 14:20
up vote 5 down vote accepted

For self-adjoint elements $a$, we have the inequality $-\lVert a\rVert e\le a\le\lVert a\rVert e$, where $e$ is the identity. So if $f$ is a positive linear functional, $-\lVert a\rVert f(e)\le f(a)\le\lVert a\rVert f(e)$ follows; i.e., $\lvert f(a)\rvert\le\lVert a\rVert f(e)$. For non-selfadjoint $a$, write $a=b+ic$ with $b$ and $c$ selfadjoint and use the result just shown.

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How do you reduce to the selfadjoint case exactly? – Alexander Frei Mar 27 at 22:09
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@Freeze_S With notation as in the answer, $b=\frac12(a+a^*)$, so $\|b\|\le\|a\|$ by the triangle inequality. Similarly, $\|c\|\le\|a\|$. So $|f(a)|^2=|f(b)+if(c)|^2=f(b)^2+f(c)^2\le C^2(\|b\|^2+\|c\|^2)\le 2C^2\|a\|^2$, where $C=f(e)$. – Harald Hanche-Olsen Mar 28 at 15:47
    
Thanks alot!!! :) – Alexander Frei Mar 29 at 10:12

The C*-algebra may not has identity. Your answer is in the following reference: C*algebras and operator theory by Gerard Murphy, page 88, thm 3.3.1

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