How to prove the closedness of \sqrt{A}(H)?

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SUMMARY

The discussion focuses on proving that the range of the square root of a symmetric operator, denoted as \(\sqrt{A}(H)\), is a closed set in a separable Hilbert space \(H\). It is established that if \(A\) is a symmetric operator satisfying \(A \geq 0\) (where \((Ax, x) \geq 0\) for any \(x\)), then \(\sqrt{A}(H)\) inherits the closedness property from \(A(H)\). The key facts include that the square root of a symmetric operator remains symmetric and that the non-negativity condition applies to \(\sqrt{A}(H)\).

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Homework Statement


If A is an symmetric operator in separable hilbert space (H) and
1)A>=0 (which means that (Ax, x)>=0 for any x)
2)A(H) is a closed set

How do you proove that \sqrt{A}(H) is a closed set


Homework Equations





The Attempt at a Solution


Facts that i know that might help..
1)Square root of a symmetric operator is also symmetric
2)\sqrt{A}(H)>=0
 
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Is there another section that I should ask about the homework problems I have? I don't think anyone of them was yet answered?
 

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