Fbe9ea99 37e7 4e75 85cf D72fe02a0d86 Pdf 1 let h h be a finite dimensional hilbert space, and u u be its unitary group. let b(h) b (h) be the space of all bounded linear operators on h h. why u u is compact in the sot (strong operator topology) of b(h) b (h)? in heine borel theorem, for a subset s s of rn r n, s s is compact (in norm topology) if and only if it is closed and bounded. Do you know about the continuous functional calculus for a normal element of a c* algebra? one can use this to immediately get that a normal element (in any c* algebra) is positive if and only if it has non negative spectrum.
7da3930e 8e96 4bfa Aa7a 90dd5ce9e47a By Dciprianno On Deviantart Let $h$ be a hilbert space. in murphy's book on operator algebras, the weak topology on $b (h)$ is the locally convex topology determined by the seminorms $u\mapsto\langle ux,y\rangle$ for all $x,y\in h$. Since $b (h)$ is not separable in the norm topology if $h$ is infinite dimensional these spaces will be very large. specifically $b (h)$ contains a subspace isomorphic to $\ell^ {\infty} (\bbb n)$. Let $h$ be a finite dimensional vector space, and $g$ be a compact group. let $b (h)$ be the bounded operators on $h$, let $c (g)$ be the complex valued continuous functions on $g$, and let $c (g;b (h). If h is a hilbert space, is b(h) under the operator norm a hilbert space? if not, is there exists any norm on b(h) that makes it a hilbert space?.
50501749 9 D6 F 4 Aa6 8911 8 D735 A8 Ffdcd Postimages Let $h$ be a finite dimensional vector space, and $g$ be a compact group. let $b (h)$ be the bounded operators on $h$, let $c (g)$ be the complex valued continuous functions on $g$, and let $c (g;b (h). If h is a hilbert space, is b(h) under the operator norm a hilbert space? if not, is there exists any norm on b(h) that makes it a hilbert space?. Careful @arbautjc, i believe you need the derivative of ln to do that, which requires evaluating the limit in question. For convenience, let $s$ denote the closed unit ball of $b (h)$. if $h$ is finite dimensional, then $b (h)$ is a finite dimensional normed space, so its closed unit ball is compact in the norm topology. A bit unexpected for me, but when you think about it knowing the double dual of k(h) k (h) is not that interesting, as is knowing its dual (which gives the weak topology) and the predual of b(h) b (h) (which gives the weak∗ ∗ topology, usually called σ σ weak operator topology). If $h$ has uncountable dimension, then the set of operators whose range is separable forms a proper closed two sided ideal and contains projections onto infinite dimensional subspaces.
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