We say that a linear functional $f$ on a $C^*$-algebra $A$ is positive if $f(a^*a)\geq 0$ for all $a\in A$. Why must it be the case that every positive linear functional on a $C^*$-algebra is bounded?
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1See page 11: http://math.berkeley.edu/~theojf/CstarAlgebras.pdf – Julien Feb 06 '13 at 21:27
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In general, every positive $ \ast $-homomorphism from a C$ ^{\ast} $-algebra to another is necessarily bounded. – Haskell Curry Feb 06 '13 at 21:39
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4@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
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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.
Harald Hanche-Olsen
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3@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 '16 at 15:47
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The following proof can be found in $\S$4.1 of the course notes A (Very) Short Course on $C^*$-Algebras by D.P. Williams.
Let $f$ be a positive linear functional on a $C^*$-algebra $A$. Suppose that $$\{f(a):a\in A^+,\,\|a\|\leq1\}$$ is unbounded. Then there exists a sequence $a_n\in A^+$ with $\|a_n\|\leq1$ and $f(a_n)\geq n$. Letting $a:=\sum_{k=1}^\infty\frac{a_k}{k^2}$ and $b_n:=\sum_{k=1}^n\frac{a_k}{k^2}$ for all $n$, we have that $a-b_n=\sum_{k=n+1}^\infty\frac{a_k}{k^2}\in A^+$ (since $A^+$ is closed under addition and scalar multiplication by non-negative reals, and is norm-closed).
Since $f$ is positive, this gives that $f(a)\geq f(b_n)\geq\sum_{k=1}^n\frac{1}{k}$ for all $n$, a clear contradiction. We conclude that $$m:=\sup\{f(a):a\in A^+,\,\|a\|\leq1\}<\infty.$$ Since every element of a $C^*$-algebra can be expressed in the form $x+iy$, where $x$ and $y$ are self-adjoint and have norm no greater than $\|a\|$, and each self-adjoint element is a difference of positive elements with norm no greater than $\|a\|$, we have that each $a\in A$ with $\|a\|\leq 1$ can be expressed in the form $$a=b_1-b_2+i(b_3-b_4),$$ where each $b_i\in A^+$ has norm no greater than $1$. Then $$|f(a)|\leq f(b_1)+f(b_2)+f(b_3)+f(b_4)\leq 4m$$ shows that $\|f\|\leq 4m<\infty$ and completes the proof.
user829347
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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
wikihowwhy
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