The sequence of polynomials that generates $\sqrt{ x}$

Question

Let $A$ be a unital C*-algebra, and $x\in A^+$. We know $\sqrt{x}$ is a unique square root of $x$, is limit of polynomials generated by $x$.

My question: What is the representation of these polynomials?

I think about using Functional calculus, and find the polynomials that generated the function $\sqrt{x}$. Thought about Maclaurin series, but I could not find anything.

Please advise me. Thanks a lot.

Answer 1

There are many ways to find a sequence of polynomials that converge to a continuous function (the sequence is not unique). In this case, you can consider the polynomials defined inductively by $$ p_0(t) = 0, p_{n+1}(t) = p_n(t) + \frac{1}{2}[t-p_n^2(t)] $$ This is a sequence that converges pointwise to $\sqrt{t}$. However, they are also a monotone sequence, so by Dini's theorem, they must converge uniformly on compact sets. Hence, $p_n(x)$ converges to $\sqrt{x}$.

Note that $p_n(0) = 0$ for all $n\in \mathbb{N}$, which is also sometimes useful (if you want to show that $\sqrt{x}$ is a compact operator whenever $x$ is compact, for instance)

Answer 2

Arveson's book on Spectral Theory has exercises that guide you through using the binomial expansion to find the square root of such an $x$ where $\|x\|\le 1$. This proof gives you a square root without requiring positivity, which allows the proof to go through in a Banach algebra. And, if you have positivity such as in a Banach $^*$ algebra or in the operators on a Hilbert space, then the sequence of approximates are positive and monotone. So you get something nice in either case.