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Specifically, let $B$ be a unbounded self-adjoint positive-definite operator on a complex Hilbert space $H$. Denote by $B^{1/2}$ the square root operator of $B$. I am interested in showing that there exists the bounded inverse operator of $B^{1/2}$. I would appreciate if anyone can give me a hint to start with. My intention here is to prove the result without the knowledge of how $B^{1/2}$ is obtained if possible.........

If it helps at all, I am reading this book Operator Approach to Linear Problems of Hydrodynamics., in particular Section 1.4.2.

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    @Aweygan yes, if possible. I know nothing about $C^*$ algebra, I was looking for proofs involving existence of square root operator for an unbounded self-adjoint positive operator, but they seem beyond my level of understanding. – 2017-02-15
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    can you point us to some of the proofs you were looking at? Or, could you state which result was used that seemed beyond your level? – 2017-02-15
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    @Omnomnomnom Quoting the author: By uniform positiveness of $B$, hence of $B^{1/2}$, there exists the bounded inverse operator $B^{-1/2}$. About the existence of square root operator for the unbounded case, I was curious on my own so I decided to look it up myself and found this paper https://www.cambridge.org/core/services/aop-cambridge-core/content/view/S1446788700004560. There are a few immediate links when I type square root operator on MSE, but they seem to be based on the "general spectral theorem", which is way beyond my current background. – 2017-02-15
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    Great, thanks. I'll say more if I think of anything – 2017-02-15
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    @Omnomnomnom Thanks, I appreciate it! I have no problem at all isolating for a few hours and read about square root operators for the unbounded case and hopefully gain some insight, but I thought I would at least ask here if there is a more elementary approach (: – 2017-02-15
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    Of course. My feeling is that since we can't take advantage of analytic functions the way we would in the unbounded case, we're stuck with something like the spectral theorem. My other thought is that with Stone's theorem, we might be able to say something like $B^{1/2} = \exp(\log(B))$; seems ambitious though. – 2017-02-15
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    I meant $\exp(\log(B)/2)$ – 2017-02-15

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