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Math Help - Sequence of the positive elements in B(H)

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    Sequence of the positive elements in B(H)

    Hello, could you please help me to solve this problem:

    Let B(H) - the algebra of bounded linear operators on a Hilbert space H, and x\in B(H) is positive. Then sequence x (\frac{1}{n}+x)^{-1} is monotone increasing to the range projection of x:\quad [x].
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    MHF Contributor Drexel28's Avatar
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    Quote Originally Posted by karkusha View Post
    Hello, could you please help me to solve this problem:

    Let B(H) - the algebra of bounded linear operators on a Hilbert space H, and x\in B(H) is positive. Then sequence x (\frac{1}{n}+x)^{-1} is monotone increasing to the range projection of x:\quad [x].
    This makes no sense. Could you rephrase? What sequence? What projection?
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  3. #3
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    What projection?
    For each operator x\in B(H) we define the range projection of x as the projection on the closure of xH.

    What sequence?
    The sequence of operators is x_n=x(\frac{1}{n}+ x)^{-1}.
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    Quote Originally Posted by karkusha View Post
    Let B(H) - the algebra of bounded linear operators on a Hilbert space H, and x\in B(H) is positive. Then sequence x (\frac{1}{n}+x)^{-1} is monotone increasing to the range projection of x:\quad [x].
    Have you come across the functional calculus for selfadjoint operators? If so, you can translate this problem into a problem about functions on the spectrum S of the operator  x. In fact, you just need to show that the sequence of functions f_n(t) = t\bigl(\frac1n+t\bigr)^{-1} is monotone increasing (on the set S) to the function p(t), where p(0)=0 and p(t) = 1 for t>0.

    To do this without the functional calculus would be harder. You could start by observing that the operators  x, \bigl(\frac1m+x\bigr)^{-1} and \bigl(\frac1n+x\bigr)^{-1} all commute with each other. Let x_n = x \bigl(\frac{1}{n}+x\bigr)^{-1}. If n>m then x_n-x_m =  \bigl(\frac1m-\frac1n\bigr)x\bigl(\frac{1}{m}+x\bigr)^{-1}\bigl(\frac{1}{n}+x\bigr)^{-1}. The right side of that is a product of three commuting positive operators and is therefore positive. That shows that the sequence (x_n) is monotone increasing.

    To complete the proof, you need to know that \|x_n\|\leqslant1. This again is easy if you know a bit of spectral theory. In fact, it follows from the fact that |f_n(t)|\leqslant1 on S.

    First, notice that 1-x_n = \bigl((\frac1n+x)-x\bigr)\bigl(\frac{1}{n}+x\bigr)^{-1} = \frac1n\bigl(\frac{1}{n}+x\bigr)^{-1}. It follows that if x\xi is in the range of x (where \xi\in H) then (1-x_n)x\xi = \frac1nx_n\xi\to0 as n\to\infty. Thus x_n\to 1 strongly on the range of x and hence (since \|x_n\|\leqslant1 for all n) on the closure of the range of x.

    The orthogonal complement of the range of x is the null space of x. So if \eta is in that orthogonal complement then x_n\eta = \bigl(\frac{1}{n}+x\bigr)^{-1}x\eta = 0. Thus x_n\to0 strongly on the orthogonal complement of the range of x.

    Put that all together to see that x_n converges strongly to the range projection of x.
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