# Consistency of Square Root in C*-Algebras (Functional Calculus)

Suppose that $$A$$ is a unital C*-algebra and $$a\in A$$ is positive, that is, $$a$$ is normal and $$\sigma(a)\subset[0,\infty)$$.

Then we can define the element $$a^{1/2}\in A$$ to be the unique element satisfying $$(a^{1/2})^{2}=a$$.

On the other hand, we can consider the continuous map $$f\colon\sigma(a)\to\mathbb{C}$$ defined by $$f(x):=\sqrt{x}$$ and apply the (continuous) functional calculus. Then we get an element $$f(a)\in A$$.

I don't understand why the definitions of $$f(a)$$ and $$a^{1/2}$$ coincide. Can someone explain what's going on? Any help will be greatly appreciated!

By the way, I use the definitions in Murphy's book on C*-algebras and Operator Theory.

• Note that $a^{1/2}$ is the unique positive element such that $(a^{1/2})^2=a$. indeed, one can cook up many different $b$ such that $b^2=a$, but only one of them can be positive. – Aweygan Mar 5 at 19:31

Suppose that $$b$$ is positive and $$b^2=a$$. Let $$g(t)=t^2$$. The functional calculus respects composition: that is, since $$t=f (g (t))$$ on $$\sigma (b)$$, then $$b=f (g (b))$$.
Then $$b=f (g (b))=f (b^2)=f (a).$$