# 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