I am reading a nice proof that there is only one square root of a positive self-adjoint operator using spectral measures in the book of Konrad Schmudgen: Unbounded Self-adjoint Operators on Hilbert Space.
I was confused only in the two parts of red that I marked, in the first, wouldn't it be $ = $ instead of the inclusion? Is the second part really necessary? Doesn't equality follow from before? The figure below also follows the Theorem used
Best Answer
I think the argument Schmüdgen had in mind was the following. First note that $T^\ast T$ is self-adjoint for every closed densely defined $T$ (Proposition 3.18). Thus $B^2$ in the question is self-adjoint. Moreover, self-adjoint operators cannot have proper self-adjoint extensions: If $S$ and $T$ are self-adjoint and $S\subseteq T$, then $T=T^\ast\subseteq S^\ast=S$, hence $S=T$. I'm pretty sure this is also mentioned somewhere in the book. In the situation of the question that means that $B^2$ must be equal to $A$, since the latter is a self-adjoint extension of the former.