# Proving that $\sqrt{A}^2=A$

Let $A \in \mathfrak{U}$ where $\mathfrak{U}$ is a $C^*$ algebra and let $A=A^*$ with $\sigma(A)\subseteq \mathbb{R}^+$. Let $f := \sqrt{\cdot}\in \mathcal{C}(\sigma(A))$ where $\mathcal{C}(X)$ is the set of continuous functions from $X$ to $\mathbb{R}$

1.Prove that $f(A)$ is well-defined using the continuous functional calculus

1. Show that $\sqrt{A}^2 = A$

These are my questions, I'm studying elements of mathematical theory between quantum mechanics and classical mechanics. I don't quite understand how we can apply the continuous functional calculus usually I'm having trouble to show it.

These are my goals: I really think that this is not something difficult. But still I could not use the correct line of reason here. Using the continuous functional calculus we know that for an $f$ like the one in the question there exists an Gelfand homomorphism $\phi: \mathcal{C}(\sigma(A)) \to \mathfrak{U}$ such that, by the Weierstrass theorem, $\phi(f)$ is the limit of the operator norms of $\phi(p)$ with $p$ the polynomials that approximate to $f$ in the sup norm. So, we define $\phi$ in this way.

Then, we use for the second part of the problem that $\sqrt{A}^2 = \phi(f)\phi(f) = \phi(f\cdot f) = \phi(f^2) = A$

Can someone help me in showing some technicalities behind these types of questions? Thanks in advance.

Another try: 1. Let's use the continuous functional calculus to say that it is properly defined. So, what we have is that, for any function $f \in \mathcal{C}(\sigma(A))$ there is a $\phi: \mathcal{C}(\sigma(A))\to \mathfrak{U}$ such that $\phi(f)$ is defined as the limit of $\phi(p)$ for the $p$ polynomials in Weierstrass approximation theorem. This makes $\phi(f)$ well-defined. Now, for $f := \sqrt{\cdot}$ we define $\phi(f):=\sqrt{A}$. This concludes the first part.

2.With this an the properties of $\phi$ as a $\star$-homomorphism we have that, (see that $f^2 = \text{id}_{\mathcal{C}(\sigma(A))}$) $$\tag{1}\sqrt{A}^2 = \phi(f^2) = \phi(\text{id}_{\mathcal{C}(\sigma(A))}) = \text{id}_{\mathfrak{U}}(A) = A$$ because we have that $\mathfrak{U}$ is a $C^*$ unital algebra.

The $(a)$ line I think that there is a problem in here. Because how can I say the last equality? This seems something that I'm cheeting to get the correct answer.

You arguments are correct. When you consider a selfadjoint (normal is enough) $A$ as you do, the Gelfand transform maps the identity function to $A$. So you don't need to write $\text{id}_{\mathfrak{U}}(A)$, it is enough to write $$\tag{1}\sqrt{A}^2 = \phi(f^2) = \phi(\text{id}_{\mathcal{C}(\sigma(A))})= A.$$ Note, though, that there is no "cheating" in $\text{id}_{\mathfrak{U}}(A)$: you are just applying the polynomial $p(t)=t$ to $A$, so $p(A)=A$.
    Moffat J, Oniga T, Wang CHT (2017) Unitary Representations of the Translational Group Acting as Local Diffeomorphisms of Space-Time. J Phys Math 8: 233. doi: 10.4172/2090-0902.1000233