Banach Lecture 7

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

Excerpts

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.