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‎let ‎‎$‎H‎$ ‎be a‎ ‎Hilbert ‎space.‎ ‎the ‎space ‎of ‎all ‎ ‎operator with finite rank ‎on ‎‎$‎H‎$ ‎is ‎shown‎ ‎by ‎‎$‎F ( H ) ‎$‎.‎

we ‎know ‎that‎ ‎$ F(H) ‎\subseteq ‎K(H) ‎\subseteq‎ B(H) $‎ so that ‎‎$‎K(H‎) $ ‎is ‎compact ‎operator ‎on‎$‎H‎$ ‎and ‎‎$‎B(H)‎$ ‎is ‎bounded ‎operator ‎on ‎‎$‎H‎$‎. ‎

Is ‎it ‎right ‎to ‎say, ‎ if ‎‎$ T ‎\in F( H )‎‎$ then $ ‎‎T‎^{*} \in F (H)‎‎‎‎$‎?

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    Yes, the finite rank operators form an algebraic ideal in $B(H)$ closed under the involution (norm dense in $K(H)$ even ). – 2017-01-08
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    is there a simple prof for $ F(H) ‎\subseteq ‎K(H) ‎\subseteq‎ B(H) ? – 2017-01-13
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    Do you just mean ordinary inclusions or as (algebraic) ideals? Also, I need to know your definition of compact operators to answer you (I gladly will). – 2017-01-14
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    ‎let‎ ‎$ X, ‎\parallel .‎ ‎‎\parallel ‎) ‎‎$‎. ‎the ‎map ‎‎$‎‎T : X ‎‎\longrightarrow X ‎$‎‎‎‎‎‎‎is ‎compact ‎when ‎$ ‎\overline{‎T( ‎B_{‎x} )‎}‎ $‎ ‎is ‎compact ‎in ‎$ ‎X‎ $‎ ,‎‎ ‎so ‎that‎$‎B_{x}= \{ x \in X : ‎\Vert ‎x‎ ‎\Vert‎\leq ‎1‎ ‎‎$. – 2017-01-17

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