Banach Lecture 7

This lecture ( banach7 ) was given on 9/11/09


Recall for a \in B, I get a functional on \mathscr{M}(B) via \hat a (\phi) = \phi(a).  Recalling Thm I.7.1 from lecture 5, if \hat a =0 then sp(a) = 0, meaning a is quasi nilpotent.  Suppose \hat a \ne 0, then \phi(a) = c \ne 0 for some \phi.  It follows that \phi(a - c) = 0.  Hence a - c cannot be invertible, meaning \hat a is not quasinilpotent.  

The Gelfond transformation gives a representation of B, it has no quasi nilpotents we say B is semisimple.

Thm I.7.2.  Let B be unital, a \in B.  and h holomophic in a nbd of sp(a).  Then h\circ \hat a = \widehat{h(a)} where this is all meant in the sense of HFC (lecture 4)

a \in B is a generator of B if polynomial in a are dense in B.  It is a rational generator if the functions h(a) (HFC) where h is a rational function whose poles don’t lie in sp(a) are dense in B.

Prop. I.7.1 Let B be unital, a a generator then the map \phi \mapsto \hat a(\phi) of \mathscr{M}(B) \to \mathbb{C} is a homemorphism to sp(a).

Prop. I.7.2 Same statement for rational generators.

I.8 Gelfond Theory Examples

Consider B = l^1(\mathbb{Z}) with convolution as the multiplication.  Define e_k \in B by e_k(i) = \delta_{ik}.  Then e_o is a unit, e_k * e_j = e_{k+j} and e_1 is a rational generator.  From ||e_1|| = 1, conclude sp(e_1) \subset \overline{\mathbb{D}} the closed unit disk.  On the other hand, if z \subset \mathbb{D}, then

(e_1 - ze_0)(e_{-1} + ze_{-2} + z^2e_{-3} + ...)

= e_0 - ze_{-1} + ze_{-1} + ... = e_0

since |z^n| tends to 0, so in fact sp(e_1) \subset \mathbb{T}, the unit circle.


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