Lecture 2

I.3 Adjunction of a unit

If B is a nonunital banach algebra.  Make it unital as follows B_1 = B \oplus \mathbb{C} with norm

|a \oplus c| := |a| + |c|

and multiplication

a \oplus c \cdot b \oplus c' = (ab + cb + c'a)\oplus cc'

Clear that 0\oplus 1 is a unit.  Also use smart factoring and triangle inquality to show

|(x\oplus c)(y\oplus c')| \le |x \oplus c|\cdot |y \oplus c'|

so we get a bonafied banach algebra.

 I.4 

Let B be a unital banach algebra, and a \in B.  The spectrum of a, sp_B(a) = sp(a) = \{z \in \mathbb{C}| a - z\cdot 1 \mbox{ not invertible}\}.  If B is nonunit set sp(a) = sp_{B_1}(a \oplus 0).  The complement of the spectrum of a is the resolvent set.  Define a function r_a \colon \mathbb{C} - sp(a) \to B (in the nonunital case its a map to B_1) by

r_a(z) = (a - z)^{-1}

Thm: The spectrum is nonempty and sp(a) \subset \{ z \in \mathbb{C}| |z| le ||a||\}.  Further, r_a is holmorphic in the resolvent set. 

Note, r_a has values in B; for such a function to be holomorphic means for each point in its domain there is an open nbd for which it is represented by a convergent power series.  Another definition would be that after applying any functional B \to \mathbb{C}, the resulting composition is holomorphic.  The first definition implies the second, the opposite implication is a hard thm.

proof:  WLOG B is unital.  (Step 1) let |z| > ||a|| then formally

(a - z)^{-1} = -1/z \cdot (1 - a/z)^{-1} = - \sum_{n = 0}^\infty \frac{a^n}{z^{n+1}}

the latter series (as a function in z) converges locally uniformly in |z|>||a|| consequently the product of the series with (a - z) can be taken term by term (it be regarded as the limit of the taking the product with all finite subsums); doing this shows indeed this series is the inverse.  This gives the second claim of the thm.  Note (a - z)^{inv} tends to 0 as z tends to infinity.

(Step 2) fix z_0 \in \mathbb{C} - sp(a).  Write

a - z = (a - z_0) - (z - z_0) = (a-z_0)(1 - (z-z_0)(a - z_0)^{-1})

For |z - z_0||a-z_0| < 1

\sum_{n = 0}(z-z_0)^n(a-z_0)^{-n}

conv. abs. hence get local uniform convergence inside |z - z_0| < 1/|a - z_0|.  Check again this series is actually an inverse by multiplying things out.  The conclusion is that (a - z)^{-1} is holomorphic in the claimed region. 

(Step 3) By step 2, the resolvent set is open and r_a is holomorhphic in it.  Its also a bnd function that tends to 0, so it the spectrum were empty it would be an entire function (using the Banach Louville thm; can reduce to ususal complex case by applying a functional to B) hence identically 0, this is an obvious contradiction.  QED

 

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