THE SPECTRAL SEMIDISTANCE IN BANACH ALGEBRAS R BRITS Let A be a - - PowerPoint PPT Presentation

THE SPECTRAL SEMIDISTANCE IN BANACH ALGEBRAS

R BRITS

Let A be a Banach complex algebra with identity 1. For elements a, b ∈ A the spectral semidistance between a and b is defined as follows: Denote the commutator Ca,b := La − Rb, and then consider powers evaluated at 1: C n

a,b1 = n

• k=0

(−1)k n k

• an−kbk.

Writing ρ(a, b) := lim sup

n

• C n

a,b1

• 1/n

we define the spectral semidistance d(a, b) := sup{ρ(a, b), ρ(b, a)}. The spectral semidistance is a semimetric and could be viewed as the noncommutative generalization of the distance induced by the spectral radius in the commutative case. If d(a, b) = 0 then a and b are called quasinilpotent equivalent.

The spectral semidistance between decomposable operators S, T ∈ L(X) can be formulated in terms of spectra via Vasilescu’s geometric formula: d(S, T) = sup{∆(σT(x), σS(x)) : 0 = x ∈ X} where σS(x) and σT(x) are, respectively, the local spectra of S and T at x ∈ X.

THEOREM

Suppose σ(a) and σ(b) are finite with σ(a) = {λ1, . . . , λn}, σ(b) = {β1, . . . , βk}. If {p1, . . . , pn} and {q1, . . . , qk} are the corresponding Riesz projections then ρ(a, b) = sup{|λi − βj| : piqj = 0}. (1)

THEOREM

Suppose σ′(a) and σ′(b) are discrete sets which cluster at 0 ∈ C, if

• anywhere. If σ′(a) = {λ1, λ2, . . . } and σ′(b) = {β1, β2, . . . }

denote the nonzero spectral points of a and b, and if {p1, p2, . . . } and {q1, q2 . . . } are the corresponding Riesz projections, then ̺ takes at least one of the following values: (i) ρ(a, b) = sup{|λi − βj| : piqj = 0}, or (ii) ρ(a, b) = |λi| for some i ∈ N, or (iii) ρ(a, b) = |βi| for some i ∈ N. Moreover, ρ(a, b) = 0 if and only if the spectra and the corresponding Riesz projections of a and b coincide.

Let f be an entire function from C into a Banach algebra A. Then f has an everywhere convergent power series expansion f (λ) =

∞

• n=0

anλn, with coefficients an belonging to A. Define a function Mf (r) = sup

|λ|≤r

f (λ), r > 0.

DEFINITION

The function f is said to be of finite order if there exists K > 0 and R > 0 such that Mf (r) < erK holds for all r > R. The infimum of the set of positive real numbers, K, such that the preceding inequality holds is called the order of f , denoted by ωf . If ωf = 1 then f is said to be of exponential order.

DEFINITION

Suppose f is entire, and of finite order ω := ωf . Then f said to be

• f finite type if there exists L > 0 and R > 0 such that

Mf (r) < eLrω holds for all r > R. The infimum of the set of positive real numbers, L, such that the preceding inequality holds is called the type of f , denoted by τf . It is well-known that the order and type of A-valued entire functions are given by the formulae ωf = lim sup

n

• n log n

log an−1

• and τf =

1 eωf lim sup

n

• n n
• anωf
• .

Let a, b ∈ A, and define f : λ → eλae−λb, λ ∈ C. The corresponding series expansion, valid for all λ ∈ C, is given by f (λ) = eλae−λb =

∞

• n=0

λnC n

a,b1

n! . The important observations for us are the following:

◮ f is of order at most one ◮ If f is of order precisely one, then the type of f is given by

ρ(a, b).

THEOREM

Suppose d(a, b) = 0, and suppose f is analytic on an open set U containing σ(a) = σ(b). Then d(f (a), f (b)) = 0.

THEOREM (P´

• lya)

Let f : C → A be an entire function from the field C into a Banach algebra A. If f is (norm) bounded over Z and if lim sup

r→∞

log Mf (r) r ≤ 0, then f is constant.

THEOREM

Let A be a Banach algebra, and let a, b ∈ A. If 0 / ∈ σ∗(a), then a = b if and only if a and b are quasinilpotent equivalent, and {anb−n : n ∈ Z} is bounded. More generally, two elements, a and b, in a Banach algebra coincide if and only if they are are quasinilpotent equivalent, and there exists α / ∈ σ∗(a) such that {(α + a)n(α + b)−n : n ∈ Z} is bounded.

COROLLARY

Let A be a C ∗-algebra, and let a, b ∈ A be both self-adjoint or both be unitary . Then a = b if and only if a and b are quasinilpotent equivalent.

THEOREM (Brits & Raubenheimer)

Let A be a C*-algebra. If a and b are normal elements of A, and if 0 is the only possible accumulation point of σ(a), then d(a, b) = 0 if and only if a = b. The above results can be improved to hold for arbitrary normal elements using the fact that the spectral semidistance is related to the growth characteristics of an entire function from C to A. For this we shall need a typical version of the Phragm´ en-Lindel¨

• f

device.

THEOREM (Phragm´ en-Lindel¨

• f)

Let u be a subharmonic function on the half-plane H := {λ ∈ C : Re λ > 0}, such that for some constants A, B < ∞ u(λ) ≤ A + B|λ|, λ ∈ H. (2) If lim sup

λ→ζ

u(λ) ≤ 0 for all ζ ∈ ∂H\{∞} (3) and if lim sup

t→∞

u(t) t = L, (t ∈ R+) (4) then u(λ) ≤ L Re λ, λ ∈ H. (5)

THEOREM

Let A be a C ∗-algebra and let a, b ∈ A be normal elements. Then a = b if and only if a and b are quasinilpotent equivalent.

SKETCH OF PROOF

◮ Define entire functions, f and g, from C into A by respectively

f (λ) = eλiae−λi(b−b∗)e−λia∗ and g(λ) = eλi(a−a∗)e−λi(b−b∗), and notice that rσ(f (λ)) = rσ(g(λ)) for all λ ∈ C.

◮ Then define

C : λ → log rσ(f (λ)) which is subharmonic on C.

◮ Applying the growth characteristics it follows that, given ǫ > 0

arbitrary, there exists R(ǫ) > 0 such that for all r > R(ǫ) log rσ(f (λ)) ≤ log

• eλiae−λib eλib∗e−λia∗
• ≤ ǫr

whenever |λ| ≤ r, hence (2) holds.

◮ To establish (3): if ζ lies on the imaginary axis, then

lim sup

λ→ζ

log rσ(f (λ)) = lim sup

λ→ζ

log rσ(g(λ)) ≤ 0

◮ Finally, one also obtains

lim sup

t→∞ t>0

log rσ(f (t)) t = 0, whence it follows that rσ(f (λ)) = rσ(g(λ)) ≤ 1 for all λ ∈ {λ ∈ C : Re λ ≥ 0}.

◮ Using a symmetric argument we can prove the same result for

−H, and hence that rσ(f (λ)) = rσ(g(λ)) ≤ 1 for all λ ∈ C.

◮ Now define an entire function

h(λ) = eλi(a−a∗)e−λi(b−b∗) − 1

• /λ

if λ = 0 i(a − a∗) − i(b − b∗) if λ = 0.

◮ Since rσ(g(λ)) is bounded on C it follows that

lim sup|λ|→∞ rσ(h(λ)) = 0. But rσ(h(λ)) is subharmonic on C and therefore, by Liouville’s Theorem (for subharmonic functions), it must be constantly zero on C

◮ In particular, we see that rσ(i(a − a∗) − i(b − b∗)) = 0. But

i(a − a∗) − i(b − b∗) being self-adjoint it follows that a − a∗ = b − b∗.

◮ Writing c := a − a∗ = b − b∗ we see that c commutes with

both a and b from which we then obtain d a + a∗ 2 , b + b∗ 2

• = d
• a − c

2, b − c 2

• = d(a, b) = 0.

So, using a preceding corollary, we get a + a∗ = b + b∗ and hence that a = b as required.