# Completely Bounded Maps and Operator Algebras by Vern Paulsen

By Vern Paulsen

This booklet excursions the imperative effects and concepts within the theories of thoroughly confident maps, thoroughly bounded maps, dilation concept, operator areas and operator algebras, besides a few of their major functions. It calls for just a easy historical past in sensible research. The presentation is self-contained and paced adequately for graduate scholars new to the topic. specialists will get pleasure from how the writer illustrates the ability of tools he has built with new and less complicated proofs of a few of the main ends up in the world, lots of that have now not seemed past within the literature. An indispensible creation to the speculation of operator areas.

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Furthermore, if S is an operator system and f is positive, then f is completely positive. Proof. Let (ai, j ) be in Mn (S), and let x = (x1 , . . , xn ), y = (y1 , . . , yn ) be unit vectors in Cn . We have that | f n ((ai, j ))x, y | = f (ai, j )x j y¯ i = f i, j ai, j x j y¯ i i, j ai, j x j y¯ i . ≤ f · i, j Thus, we must show that this latter element has norm less than (ai, j ) . To see this, note that the above sum is the (1,1) entry of the product y¯ 1 · 1 . . y¯ n · 1 x1 · 1 0 .

Iii) Let φn : C([0, 1]) → C([0, 1]) be a sequence of positive maps. Prove that if φn ( f i ) − f i → 0 as n → ∞ for f i (t) = t i , i = 0, 1, and 2, then φn ( f ) − f → 0 as n → ∞ for all f ∈ C([0, 1]). 21 The Bernstein maps φn : C([0, 1]) → C([0, 1]) are deﬁned by n φn ( f )(t) = k=0 n f k k n t k (1 − t)n−k . (i) Verify that the Bernstein maps are positive maps with φn (1) = 1, 2 φn (t) = t, φn (t 2 ) = t 2 + t−t . , prove that the polynomials are dense in C([0, 1]). 22 A sequence {an }∞ n=0 of complex numbers is called a Hausdorff moment sequence if there exists a positive (ﬁnite) Borel measure µ on [0,1] such 1 that an = 0 t n dµ(t) for all n.

Clearly, (i) implies (ii), and since (E i, j )i,n j=1 is positive, (ii) implies (iii). Thus, we shall prove that (iii) implies (i). For this it is sufﬁcient to assume that B = B(H). Fix k, and let x1 , . . , xk belong to H and B1 , . . , Bk belong to Mn . By the above lemma, it is sufﬁcient to prove that i, j φ(Bi∗ B j )x j , xi is positive. n br,s, Er,s so that Write B = r,s=1 Bi∗ B j = n b¯ r,s,i br,t, j E s,t . r,s,t=1 Set yt,r = k j=1 br,t, j x j ; then φ(Bi∗ B j )x j , xi = n n φ(E s,t ) r =1 s,t=1 i, j b¯ r,s,i br,t, j x j , xi i, j n = φ(E s,t )yt,r , ys,r .