POLONICI MATHEMATICI LXVI (1997)
On the joint spectral radius
by Vladim´ır M¨ uller (Praha)
Abstract. We prove the `
p-spectral radius formula for n-tuples of commuting Banach algebra elements. This generalizes results of some earlier papers.
Let A be a Banach algebra with the unit element denoted by 1. Let a = (a
1, . . . , a
n) be an n-tuple of elements of A. Denote by σ(a) the Harte spectrum of a, i.e. λ = (λ
1, . . . , λ
n) 6∈ σ(a) if and only if there exist u
1, . . . , u
n, v
1, . . . , v
n∈ A such that
n
X
j=1
(a
j− λ
j)u
j=
n
X
j=1
v
j(a
j− λ
j) = 1.
Let 1 ≤ p ≤ ∞. The (geometric) spectral radius of a is defined by r
p(a) = max{kλk
p: λ ∈ σ(a)},
where
kλk
p= max
1≤j≤n|λ
j| (p = ∞), ( P
nj=1
|λ
j|
p)
1/p(1 ≤ p < ∞);
see [10], cf. also [4].
If σ(a) is empty we put formally r
p(a) = −∞.
Clearly, r
p(a) depends on p. On the other hand, instead of the Harte spectrum we can take any other reasonable spectrum (e.g. the left, right, approximate point, defect, Taylor etc.) without changing the value of r
p(a);
see [4], [9].
For a single Banach algebra element the just defined spectral radius r
p(a) does not depend on p and coincides with the ordinary spectral radius r(a
1) = max{|λ
1| : λ
1∈ σ(a
1)}. By the well-known spectral radius formula
1991 Mathematics Subject Classification: Primary 46H05, 46J05.
Key words and phrases: Banach algebra, spectrum, spectral radius.
The research was supported by the grant No. 119106 of the Academy of Sciences of the Czech Republic.
[173]
we have in this case
r(a
1) = lim
k→∞
ka
k1k
1/k= inf
k
ka
k1k
1/k.
The spectral radius formula for n-tuples of Banach algebra elements was studied by a number of authors, see e.g. [1], [2], [6], [7], [8]. In this paper we generalize results of [6], [7] and [10].
Let a = (a
1, . . . , a
n) be an n-tuple of elements of a Banach algebra A.
Instead of powers of a single element it is natural to consider all possible products of a
1, . . . , a
n.
Denote by F (k, n) the set of all functions from {1, . . . , k} to {1, . . . , n}.
Let further
s
k,p(a) =
X
f ∈F (k,n)
ka
f (1). . . a
f (k)k
p1/p(1 ≤ p < ∞) and
s
k,∞(a) = max
f ∈F (k,n)
ka
f (1). . . a
f (k)k.
Lemma 1. s
k+l,p≤ s
k,p(a) · s
l,p(a).
P r o o f. The statement is obvious for p = ∞. For p < ∞ we have [s
k,p(a) · s
l,p(a)]
p= X
f ∈F (k,n)
ka
f (1). . . a
f (k)k
p· X
g∈F (l,n)
ka
g(1). . . a
g(l)k
p≥ X
f,g
ka
f (1). . . a
f (k)a
g(1). . . a
g(l)k
p= [s
k+l,p(a)]
p. It is well known that the above lemma implies that lim
k→∞(s
k,p(a))
1/kexists and it is equal to inf
k(s
k,p(a))
1/k.
Thus we may define r
00p(a) = lim
k→∞
X
f ∈F (k,n)
ka
f (1). . . a
f (k)k
p1/(pk). Similarly we define
(1) r
p0(a) = lim sup
k→∞
X
f ∈F (k,n)
r
p(a
f (1). . . a
f (k))
1/(pk)(we write briefly r
p(x) instead of (r(x))
p).
In general, the limit in (1) does not exist. The limit exists if a
1, . . . , a
nare mutually commuting. This can be proved analogously as in Lemma 1 by using the submultiplicativity of the spectral radius.
Theorem 2. Let a = (a
1, . . . , a
n) be an n-tuple of elements of a Banach algebra A. Let 1 ≤ p ≤ ∞. Then
r
p(a) ≤ r
p0(a) ≤ r
p00(a).
P r o o f. The case p = ∞ was proved in [7], Theorem 1.
Let p < ∞. The second inequality is clear.
Let λ = (λ
1, . . . , λ
n) ∈ σ(a). Denote by A
0the closed subalgebra of A generated by the unit 1 and the elements a
1, . . . , a
n. By [5], Proposition 2, there exists a multiplicative functional h : A
0→ C such that h(a
j) = λ
jfor j = 1, . . . , n. Then
X
f ∈F (k,n)
r
p(a
f (1). . . a
f (k)) ≥ X
f ∈F (k,n)
|h(a
f (1). . . a
f (k))|
p= X
f ∈F (k,n)
|λ
f (1)|
p. . . |λ
f (k)|
p= (|λ
1|
p+ . . . + |λ
n|
p)
k= kλk
pkp. Thus
X
f ∈F (k,n)
r
p(a
f (1). . . a
f (k)) ≥ r
ppk(a) and r
p0(a) ≥ r
p(a).
If a = (a
1, . . . , a
n) is an n-tuple of mutually commuting elements then a better result can be proved.
We use the standard multiindex notation. Denote by Z
+the set of all non-negative integers. For α = (α
1, . . . , α
n) ∈ Z
n+and m ∈ Z
+define |α| = α
1+ . . . + α
n, α! = α
1! . . . α
n!, a
α= a
α11. . . a
αnnand mα = (mα
1, . . . , mα
n).
If k is an integer, k ≥ |α|, then let
k α
= k!
α!(k − |α|)!
(for n = 1 this definition coincides with the classical binomial coefficients).
We shall use frequently the following formula (for commuting variables x
i):
(x
1+ . . . + x
n)
k= X
|α|=k
k α
x
α. In particular, for x
1= . . . = x
n= 1 we have P
|α|=k k
α
= n
k.
If a = (a
1, . . . , a
n) is a commuting n-tuple of elements of a Banach algebra A, then the definitions of r
p0(a) and r
00p(a) assume a simpler form (for 1 ≤ p < ∞):
r
0p(a) = lim
k→∞
X
|α|=k
k α
r
p(a
α)
1/(pk),
r
p00(a) = lim
k→∞
X
|α|=k
k α
ka
αk
p 1/(pk).
Theorem 3. Let a = (a
1, . . . , a
n) be an n-tuple of mutually commuting elements of a Banach algebra A. Let 1 ≤ p ≤ ∞. Then
r
p(a) = r
p0(a) = r
p00(a).
P r o o f. For p = ∞ the first equality was proved in [10] and the second in [7], Theorem 2.
We assume in the following p < ∞.
Recall that the number of all partitions of the set {1, . . . , k} into n parts is equal to
k+n−1n−1≤ (k + n − 1)
n−1.
We have
|α|=k
max
k α
ka
αk
p≤ X
|α|=k
k α
ka
αk
p≤ k + n − 1 n − 1
|α|=k
max
k α
ka
αk
p. Note that
k→∞
lim
k + n − 1 n − 1
1/k= 1.
Thus
r
00p(a) = lim
k→∞
X
|α|=k
k α
ka
αk
p 1/(kp)= lim
k→∞
max
|α|=k
k α
ka
αk
p 1/(kp). Similarly,
r
0p(a) = lim
k→∞
max
|α|=k
k α
r
p(a
α)
1/(kp). We now prove the inequality r
0p(a) ≤ r
p(a):
Choose k and α ∈ Z
n+, |α| = k. Let µ ∈ σ(a
α) satisfy |µ| = r(a
α). By the spectral mapping property there exists λ = (λ
1, . . . , λ
n) ∈ σ(a) such that µ = λ
α11. . . λ
αnn. Then
k α
r
pp(a
α) = k α
|µ|
p= k α
|λ
1|
α1p. . . |λ
n|
αnp≤ X
|β|=k
k β
|λ
1|
β1p. . . |λ
n|
βnp= (|λ
1|
p+ . . . + |λ
n|
p)
k= kλk
pkp≤ r
pkp(a).
Thus
r
0p(a) = lim
k→∞
max
|α|=k
k α
r
p(a
α)
1/(kp)≤ r
p(a).
The remaining inequality r
p00(a) ≤ r
0p(a) will be proved by induction on n.
For n = 1, Theorem 3 reduces to the well-known spectral radius formula
for a single element.
Let n ≥ 2 and suppose that the inequality r
p00≤ r
0pis true for all com- muting (n − 1)-tuples.
For each k there is α ∈ Z
n+, |α| = k, such that
k α
ka
αk
p= max
|β|=k
k β
ka
βk
p. Using the compactness of [0, 1]
nwe can choose a sequence
{α(i)}
∞i=1= {(α
1(i), . . . , α
n(i))}
∞i=1⊂ Z
n+such that lim
i→∞|α(i)| = ∞, (2) |α(i)|
α(i)
ka
α(i)k
p= max
|β|=|α(i)|
|α(i)|
β
ka
βk
p(i = 1, 2, . . .)
and the sequences {α
j(i)/|α(i)|}
∞i=1are convergent for j = 1, . . . , n. Define k(i) = |α(i)| and
t
j= lim
i→∞
α
j(i)
k(i) ∈ [0, 1] (j = 1, . . . , n).
By (2) we have
r
00pp(a) = lim
i→∞
k(i) α(i)
ka
α(i)k
p 1/(k(i)p). We distinguish two cases:
(a) t
j= 0 for some j, 1 ≤ j ≤ n. Without loss of generality we may assume that t
n= 0. Define a
0= (a
1, . . . , a
n−1), α
0(i) = (α
1(i), . . . , α
n−1(i))
∈ Z
n−1+and k
0(i) = |α
0(i)| = k(i) − α
n(i). Clearly lim
i→∞k
0(i)/k(i) = 1.
We have ka
α(i)k ≤ ka
0α0(i)k · ka
nk
αn(i). Then r
p00p(a
0) ≥ lim sup
i→∞
k
0(i) α
0(i)
ka
0α0(i)k
p 1/k0(i)≥ L
1· L
2· L
3, where
L
1= lim sup
i→∞
k
0(i) α
0(i)
k(i) α(i)
1/k0(i),
L
2= lim
i→∞
k(i) α(i)
ka
α(i)k
p 1/k0(i)and
L
3= lim
i→∞
ka
nk
−αn(i)p/k0(i).
Since lim
i→∞α
n(i)/k
0(i) = 0, we have L
3= 1.
Further,
L
2= lim
i→∞
k(i) α(i)
ka
α(i)k
p 1/k(i)k(i)/k0(i)= r
00pp(a).
Finally, L
1= lim sup
i→∞
k
0(i)! · α
n(i)!
k(i)!
1/k0(i)≥ lim sup
i→∞
(α
n(i)/3)
αn(i)k(i)
αn(i) 1/k0(i)= lim sup
i→∞
α
n(i) 3k(i)
(αn(i)/k(i))·(k(i)/k0(i))= 1 since lim
i→∞k(i)/k
0(i) = 1 and
i→∞
lim
α
n(i) 3k(i)
αn(i)/k(i)= lim
x→0+
x 3
x= lim
x→0+
x
x= lim
x→0+
e
x ln x= 1.
Thus r
p00(a
0) ≥ r
p00(a).
By the induction assumption r
00p(a
0) = r
0p(a
0) = r
p(a
0) and by the defini- tion r
p(a
0) ≤ r
p(a) = r
p0(a). Hence r
00p(a) ≤ r
p0(a).
(b) There remains the case t
j> 0 (j = 1, . . . , n), with t
j= lim
i→∞α
j(i)/k(i). Choose ε > 0, ε < min
1≤j≤nt
j/n. For i sufficiently large we have
t
j− ε
4 ≤ α
j(i)
k(i) ≤ t
j+ ε 4 .
We approximate t
1, . . . , t
nby rational numbers. Fix positive integers c
1, . . . , c
n, d such that
t
j− ε 2 ≤ c
jd ≤ t
j− ε
4 (j = 1, . . . , n).
Let γ = (c
1, . . . , c
n) ∈ Z
n+and u = a
γ= a
c11. . . a
cnn. For each i write k(i) = m(i)d + z(i), where 0 ≤ z(i) ≤ d − 1. So, for i sufficiently large, we have
c
jd ≤ α
j(i)
k(i) , α
j(i) k(i) − c
jd ≤ 3ε 4 and
α
j(i) − m(i)c
j= α
j(i) − k(i) − z(i)
d · c
j= k(i) α
j(i) k(i) − c
jd
+ z(i)c
jd . Thus α
j(i) − m(i)c
j≥ 0 (1 ≤ j ≤ n) and
k(i) − m(i)|γ| =
n
X
j=1
(α
j(i) − m(i)c
j) ≤ k(i) · 3εn 4 +
n
X
j=1
z(i)c
jd ≤ εnk(i)
for i large enough. We have
ka
α(i)k ≤ ka
m(i)c1 1. . . a
m(i)cn nk · ka
1k
α1(i)−m(i)c1. . . ka
nk
αn(i)−m(i)cn≤ ku
m(i)k · K
nεk(i),
where K = max{1, ka
1k, . . . , ka
nk}. Then, since
m(i)|γ|m(i)γ1/(m(i)|γ|)≤ n, we have
r
0pp(a) ≥ lim sup
i→∞
m(i)|γ|
m(i)γ
r
p(a
m(i)γ)
1/(m(i)|γ|)= lim sup
i→∞
m(i)|γ|
m(i)γ
1/(m(i)|γ|)· r(u)
p/|γ|= lim sup
i→∞
m(i)|γ|
m(i)γ
ku
m(i)k
p 1/(m(i)|γ|)≥ L
1· L
2· L
3, where
L
1= lim inf
i→∞
m(i)|γ|
m(i)γ
k(i) α(i)
1/(m(i)|γ|),
L
2= lim inf
i→∞
k(i) α(i)
ka
α(i)k
p 1/(m(i)|γ|)and
L
3= lim inf
i→∞
K
−nεpk(i)/(m(i)|γ|). Since
1 ≤ k(i)
m(i)|γ| ≤ 1 1 − nε for i sufficiently large, we have L
3≥ K
−nεp/(1−nε).
Since
i→∞
lim
k(i) α(i)
ka
α(i)k
p 1/k(i)= r
p00p(a), we have L
2≥ min{r
p00p(a), (r
p00p(a))
1/(1−nε)}.
To estimate L
1, we use the well-known Stirling formula l! = l
le
−l√
2πl(1 + o(l)).
We have
(1 − ε) α
j(i) e
αj(i)/(m(i)|γ|)≤ (α
j(i)!)
1/(m(i)|γ|)≤ (1 + ε) α
j(i) e
αj(i)/(m(i)|γ|)for j = 1, . . . , n and for i sufficiently large. Similar estimates can be used for
(m(i)c
j)!, (m(i)|γ|)! and |α(i)|!. Thus, for i sufficiently large, we have (to
simplify the expressions we write m, k and α instead of m(i), k(i) and α(i))
m|γ|
mγ
k α
1/(m|γ|)=
(m|γ|)!α
1! . . . α
n! k!(mc
1)! . . . (mc
n)!
1/(m|γ|)≥ 1 − ε 1 + ε
n+1× m|γ| · α
α11/(m|γ|). . . α
αnn/(m|γ|)· e
k/(m|γ|)· e
c1/|γ|. . . e
cn/|γ|e · e
α1/(m|γ|). . . e
αn/(m|γ|)· k
k/(m|γ|)· (mc
1)
c1/|γ|. . . (mc
n)
cn/|γ|= 1 − ε 1 + ε
n+1α
1mc
1 c1/|γ|. . .
α
nmc
n cn/|γ|× α
(α1 1−mc1)/(m|γ|). . . α
(αn n−mcn)/(m|γ|)· m|γ|
k
k/(m|γ|)≥ 1 − ε 1 + ε
n+1· α
1k
(α1−mc1)/(m|γ|). . . α
nk
(αn−mcn)/(m|γ|)· m|γ|
k . Then
L
1≥ 1 − ε 1 + ε
n+1(1 − nε)(t
1. . . t
n)
ε/(1−nε). Hence
r
p0p(a) ≥ 1 − ε 1 + ε
n+1(1 − nε)(t
1. . . t
n)
ε/(1−nε)× K
−nεp/(1−nε)· min{r
p00p(a), (r
p00p(a))
1/(1−nε)}.
Since ε was an arbitrary positive number, we conclude that r
0p(a) ≥ r
00p(a).
Theorem 3 is proved.
We now apply the previous result to the case of n-tuples of operators.
Let T = (T
1, . . . , T
n) be an n-tuple of bounded operators in a Banach space X. Define
kT k
p= sup
x∈X kxk=1
X
nj=1
kT
jxk
p 1/p.
Equivalently, kT k
pis the norm of the operator e T : X → X
pn, where X
pnis the direct sum of n copies of X endowed with the `
p-norm, kx
1⊕ . . . ⊕ x
nk = ( P
nj=1
kx
jk
p)
1/p, and e T x = T
1x ⊕ . . . ⊕ T
nx (for p = ∞ the definitions are changed in the obvious way). Let T = (T
1, . . . , T
n) ∈ B(X)
nand S = (S
1, . . . , S
m) ∈ B(X)
m. Denote by T S the mn-tuple
T S = (T
1S
1, . . . , T
1S
m, T
2S
1, . . . , T
2S
m, . . . , T
nS
1, . . . , T
nS
m).
Further, let T
2= T T and T
k+1= T · T
k. With this notation we can state the spectral radius formula in the familiar way:
Theorem 4. Let T = (T
1, . . . , T
n) be an n-tuple of mutually commuting operators in a Banach space X, and let 1 ≤ p ≤ ∞. Then
r
p(T ) = lim
k→∞
kT
kk
1/kp. P r o o f. We have
kT
kk
p= sup
kxk=1
X
|α|=k
k α
kT
αxk
p 1/pand
r
p(T ) = lim
k→∞
X
|α|=k
k α
kT
αk
p 1/(kp)= lim
k→∞
max
|α|=k
k α
kT
αk
p 1/(kp)= lim
k→∞
max
|α|=k
sup
kxk=1
k α
kT
αxk
p 1/(kp)= lim
k→∞
sup
kxk=1
max
|α|=k
k α
kT
αxk
p 1/(kp)= lim
k→∞
sup
kxk=1
X
|α|=k
k α
kT
αxk
p 1/(kp)= lim
k→∞
