# Banach Alg. Lecture 3

Results/Exercises

1. Prove that in a unital banach algebra the left invertible (reps. right invertible) elements form open sets.
2. Let B be a unital banach algebra, $a \in B$ be on the boundry of the invertible (meaning left and right invertible) elements.  Prove $a$ is a topological zero divisor in $B$, meaning $\exists (b_n)_1^\infty \subset B$ with $|b_n| = 1$ and $|ab_n|,|b_na$ both tend to zero.

Examples of spectrum

1. If $\mathscr{X}$ is cmpt and Hausdorff, set $B = C(\mathscr{X})$, then the spectrum of any $f \in B$ is its range: $f(\mathscr{X})$.
2. If $\mathscr{X}$ is loc. compt and not compt, then set $B = C_0(\mathscr{X})$ (cont. functions that vanish at infinity).  Then for $f \in B$, $sp(f) = f(\mathscr{X}) \cup 0 = \overline{f(\mathscr{X})}$.
3. For a top. sp. $\mathscr{X}$ set $B = C_b(\mathscr{X})$.  Then $sp(f) = \overline{f(\mathscr{X})}$
4. If $B = L^\infty(\mu)$ for a measure $\mu$ then $sp(f) =$ essential range of $f$ = $\{ z \in range(f): \mu(f^{-1}U) > 0\}$ where $U$ ranges over all nonempty open sets containing $z$.
5. For $f \in C^n[0,1]$ the spectrum is $sp(f) = f([0,1])$.
6. The spectrum of an element of $L^1(\mathbb{R})$? Later ….

Lemma I.5.1 Let $B$ be a banach algebra, $a \in B$.  Then $\lim_{n\to \infty} |a^n|^{1/n}$ exists and is equal to $\inf \{|a^n|^{1/n}: n \in \mathbb{N}\}$.

proof: The case $a = 0$ is trivial so assume $a \ne 0$ and fix $m \in \mathbb{N}$.  For all $n > m$ use the Euclidean algorithm to write $n = km + r$ where $0\le r .  Then using the product inequality

$|a^n| \le |a^m|^k|a|^r$

Take the $nth$ root and notice that as $n\to \infty$ the terms on the rhs approach to $|a^m|^{1/m}$ and 1.  It follows that

$\limsup_{n\to \infty} |a^n|^{1/n} \le |a^m|^{1/m}$

This holds for all m from which the following inequality follows

$\limsup_n |a^n|^{1/n} \le \inf_m |a^m|^{1/m} \le \liminf_n |a^n|^{1/n}$

The opposite inequality is true which shows they are all in fact equalities.QED.

In view of the lemma, define the spectral radius of convergence of $a \in B$ as $\lim_{n\to \infty} |a^n|^{1/n}$.

Thm I.5.1 For $a \in B$ have $sp(a) \subset \{|z|\le r(a)\} \subset \mathbb{C}\cdot 1$.  Further, $\exists z_0$ such that $|z_0| = r(a)$ and $z \in sp(a)$.

proof: WLOG $B$ is unital.  Set $r = r(a)$ and fix $e>0$. From $\lim_n \frac{|a^n|^{1/n}}{r+e} < 1$ it can be concluded that $\exists n_0 \in \mathbb{N}, \rho \in (0,1)$ such that for $n\ge n_0$

$\frac{|a^n|}{(r+e)^n} \le \rho^n$

Recall the von neumann series associated to an element $a \in B$: $vn_a(z) = -\sum_{n=0}^\infty \frac{a^n}{z^{n+1}}$.  Recall in lecture 2 it was proved that locally in the resolvent set $(a - z)\cdot vn_a(z) = 1$.  It follows that $vn_a(z)$ converges aboslutely and uniformly for $|z| \ge r+e$.  This shows the desired inclusion.

Note if $r = 0$ then $sp(a) = 0$ and the second claim follows.  Now assume $r > 0$.  Proceeding by contradiction assume $sp(a) \cap \{|z| = r\} = \emptyset$.  The resolvent function $r_a(z) = (a - z)^{-1}$ is holomorhpic off the spectrum thus holomophic in an open set containing $\{ |z|\ge r\}$.  Now choose $r_0 < r$ such that $vn_a(z)$ is holomorphic in an openset containing $\{|z| \ge r_0\}$.  Choose a bnd. linear functional $\phi \in B^*$ apply it to $vn_a(z)$ to get that

$\sum_n \frac{\phi(a^n)}{z^{n+1}} = \sum_n \frac{\langle \phi, a^n \rangle}{z^{n+1}}$

Converges in an openset containing $\{ |z| \ge r_0\}$; in particular $\sup_n \frac{\phi(a^n)}{r_0^n}$ is bounded. By the principle of uniform boundedness $\sup_n \frac{a^n}{r_0^n}$ is bounded say by $m < infty$ then

$\lim_{n \to \infty} \frac{|a^n|^{1/n}} \le \lim_{n\to \infty} m^{1/n} \cdot r_o$

implying that $r \le r_0$ which is a contradiction.QED.

The principal of uniform boundedness says (according to wikipedia, a short proof is also on the wiki page):

Theorem. Let X be a Banach space and Y be a normed vector space. Suppose that F is a collection of continuous linear operators from X to Y. The uniform boundedness principle states that if for all x in X we have

$\sup_{T \in F} |T(x)| < \infty$

then

$\sup_{T \in F} |T| < \infty$.