Banach Alg. Lecture 3


  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 ….

I.5 Spectral Radius

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 <m.  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


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


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