# The inequality $a^*ca \le \|c\| a^*a$ in a pre-$C^*$-algebra

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?

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.