Let $A$ be a pre-$C^*$-algebra, i.e. $A$ satisfies all axioms for a $C^*$-algebra except completeness. In other words, $A$ is an involutive algebra with a $C^*$-norm.

We say that $x \in A$ is positive (notation: $x \ge 0$) if there exists $a\in A$ with $x = a^*a$ and on self-adjoint elements we define the usual relation $x \le y \iff y-x \ge 0$.

Given a positive element $c \ge 0$ in $A$, and $a \in A$, does the inequality $$a^*ca \le \|c\| a^*a$$ hold? For $C^*$-algebras, this result is well-known. A quick proof is given by faithfully representing $A \subseteq B(H)$ and using $c \le \|c\|1$. Does the same result continue to hold for pre $C^*$-algebras?

Context question: In the book "Hilbert $C^*$-modules" by Lance on p4, it is claimed that all results proved so far continue to hold for inner product modules defined over a pre-$C^*$-algebra. The above inequality is used in the proof of proposition 1.1 and I wanted to check that it still works for pre $C^*$-algebras.

Bonus question: Is the sum of two positive elements again positive?


1 Answer 1


Given your definitions, the first one is an easy no. If $a = 1$ then it says $c \leq \|c\|$, which fails when $A$ is the polynomials in $C[0,1]$: let $c$ be the polynomial $x$, then $\|c\|= 1$ and $1-x$ is not of the form $p\bar{p}$ for any polynomial $p$.

For the second question, take $a = x^2 + y^2$ and $b = z^2$ in the polynomials on $[0,1]^3$. Then $a = (x + iy)(x- iy)$ and $b = z\cdot z$ are both positive, but their sum is not. (Suppose you had a polynomial $p$ with $p\bar{p} = x^2 + y^2 + z^2$. Then $p$ has to be linear in $x$, $y$, and $z$ and the lack of $xy$ and $xz$ cross terms forces the arguments of the coefficients of $y$ and $z$ to be the same or differ by $\pi$, which in either case creates a nonzero $yz$ cross term.)

But I think the usual definition of positivity in a pre C${}^*$- algebra is that it is positive in the completion, which would make both answers trivially yes.

  • 1
    $\begingroup$ Ah okay. Your last paragraph mzans that I interpreted the definition of positivity in the wrong way and it solves all problems. Thanks for the answer! $\endgroup$
    – Andromeda
    Jul 27, 2021 at 7:06
  • $\begingroup$ You are welcome! $\endgroup$
    – Nik Weaver
    Jul 27, 2021 at 11:52

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