By Shilov G.

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**Extra info for An introduction to the theory of linear spaces**

**Example text**

Now suppose (a), that C∗ (G) = C∗r (G). Let a ∈ C∗ (G) have ϕ(a) = 0 for each ultraweakly continuous linear functional ϕ on B( 2 (G)). Then λ(a) = 0 and thus a = 0 since λ is faithful on C∗ (G). It follows that the space of ultraweakly continuous functionals on the regular representation is weak-∗ dense in C∗ (G)∗ . 14, this implies that A+ (G) is dense in B+ (G) in the topology of pointwise convergence. Now the constant function 1 belongs to B+ (G) (it corresponds to the trivial representation of G) and thus there is a net in A+ (G) converging pointwise to 1.

The Hardy space H2 (S1 ) is the closed subspace of L2 (S1 ) spanned by the functions zn , for n 0. A Toeplitz operator on H2 (S1 ) is a bounded operator Tg of the form (f ∈ H2 (S1 )), Tg (f) = P(gf) where H2 (S1 ) and P is the orthogonal projection from . The function g is called the symbol of Tg . g ∈ L∞ (S1 ) L2 (S1 ) onto Lemma: The C∗ -algebra generated by the Toeplitz operators is isomorphic to T, and the map g → Tg is a linear splitting for the quotient map T → C(S1 ) that appears in the Toeplitz extension.

Proof: Suppose that K ⊆ H is a closed invariant subspace for T. 42) K = 0, then K contains a nonzero vector, and by applying a suitable power of V ∗ if necessary we may assume that K contains a vector v = (x0 , x1 , x2 , . . ) with x0 = 0. Since P ∈ T we nd that Pv = v0 e0 ∈ K and so e0 ∈ K. Now applying powers of V we nd that each en = V n e0 ∈ K, so K = H. Thus T is irreducible. 39. The unilateral shift operator is unitary modulo the compacts, and so its essential spectrum, that is the spectrum of its image in the Calkin algebra Q(H) = B(H)/K(H), is a subset of the unit circle S1 .