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Thread: square root lemma

  1. #1
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    square root lemma

    Hello,

    in the book functional analysis by reed, there is a lemma which says:
    If $\displaystyle A \in L(H)$ is a bounded linear operator on a Hilbert Space, and$\displaystyle A\ge0$. Then there is a unique $\displaystyle \( B \in L(H) \) B\ge0$ s.t. $\displaystyle B^2 = A.$


    I have a question about the proof. The author argues:
    "It is sufficient to consider the case where$\displaystyle \|A\| \ge 1$."

    Why is it sufficient to proof just this case?

    I couldn't find the right argument.

    Regards
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  2. #2
    Super Member girdav's Avatar
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    Re: square root lemma

    If the result has been shown for the operators which has a norm $\displaystyle \geq 1$, let $\displaystyle T\in\mathcal{B}(H)$, $\displaystyle T\geq 0$. If $\displaystyle T=0$ take $\displaystyle B=0$ and if $\displaystyle \lVert T\rVert \neq 0$ then $\displaystyle S:=\frac{T}{\lVert T\rVert}\in\mathcal{B}(H)$ and $\displaystyle S\geq 0$. We can find an unique $\displaystyle B\geq 0$ such that $\displaystyle B^2=S$ i.e. $\displaystyle \lVert T\rVert^2B^2 =T$. We take $\displaystyle B' := \lVert T\rVert B$.
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  3. #3
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    Re: square root lemma

    Thank you very much!
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