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We know that if $\mathcal{A}$ is a unital $C^*$-algebra and if $f:\mathcal{A}\to\mathbb{C}$ is a positive linear functional then it is Hermitian. It simply follows from the fact that in $\mathcal{A}$ every Hermitian element is difference of two positive elements. But what will happen if $\mathcal{A}$ is not a $C^*$-algebra to begin with? Is the result still valid?

My attempt : I have considered a unital abelian $*$-algebra which has no complete $C^*$-norm. More precisely, I was considering $\mathcal{A}$ to be the space of all complex polynomials defined on $\mathbb{C}$ with $*$ given by "coefficient wise" conjugation. The only non-trivial positive linear functional on $\mathcal{A}$ which I can find is $p\mapsto p(0)$ but it is Hermitian.

Can we find any positive linear functional on this $\mathcal{A}$ which is not Hermitian? If we can please give me an example or proof of existence. Do we have any general result regarding this?

Definitions : If $\mathcal{A}$ is a unital $*$-algebra then an element $x\in\mathcal{A}$ is Hermitian (resp. positive) if $x^*=x$ (resp. $x=y^*y$ for some $y\in\mathcal{A}$).

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  • $\begingroup$ With a little modification of algebra you are considering the answer is negative. Instead of polynomial algebra let's consider the algebra of lorentz polynomials with $*$ operation as follows: $(az^n)^*=\bar{a}z^{-n}$ then this is the group algebra of all function on $\mathbb{Z}$ with compact support so according to its pre $C^*$ algebra structure every positive functional must be Heremitian. $\endgroup$
    – Ali Taghavi
    Commented Jul 24 at 15:40
  • $\begingroup$ Can one modify this idea to answer your original question? $\endgroup$
    – Ali Taghavi
    Commented Jul 24 at 15:42

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In fact, any positive linear functional $\varphi$ on a unital $\ast$-algebra $A$ is Hermitian. Indeed, for any Hermitian $x \in A$ and any real number $r$, we have,

$$0 \leq \varphi((1 + rx)^\ast(1 + rx)) = \varphi(1 + 2rx + r^2x^2) = \varphi(x^2)r^2 + 2\varphi(x)r + \varphi(1)$$

This is a quadratic polynomial in $r$ taking real values for all real inputs, thus, by taking derivatives, we see that all coefficients of this polynomial must be real. In particular, $\varphi(x) \in \mathbb{R}$, so $\varphi$ is Hermitian.

Tangentially, let me also note that the result is not true without assuming unitality. Indeed, a somewhat silly example is $A = \mathbb{C}$ equipped with complex conjugate as its $\ast$-operation but with trivial product (i.e., $xy = 0$ for any $x, y$). Then all linear functionals are positive, as the only positive element is $0$, but clearly not all linear functionals are Hermitian.

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  • $\begingroup$ Exactly the answer I was looking for. So silly of me that I have missed this. Upvoted and thank you. $\endgroup$
    – UtsabrajSarkar
    Commented Jul 26 at 23:33
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    $\begingroup$ @UtsabrajSarkar No problem! I was surprised myself to find this proof. I actually thought at first there would be counterexamples before realizing every “counterexample” I came up with was all non-unital. I was quite shocked to eventually find one can prove this just purely algebraically, and that functional calculus was not necessary. $\endgroup$
    – David Gao
    Commented Jul 27 at 3:32
  • $\begingroup$ It was an elementary proof but very enlightening. So it's a purely algebraic thing and doesn't depend upon the analytic/topological properties of the algebra. Beautiful. $\endgroup$
    – UtsabrajSarkar
    Commented Jul 28 at 11:58

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