Dynamics of a compact operator

0.18 MB·

by  Teck-Cheong Lim

Dynamics of a Compact Operator
1
1 Teck-Cheong Lim
0 Department of Mathematical Sciences
2
George Mason University

n
4400, University Drive
a
J Fairfax, VA 22030

6 U.S.A.
1 e-mail address: [email protected]

]
S January 18, 2011
D
.
h
Abstract
t
a LetT :X →X beacompactlinear(ormoregenerallyaffine)operator
m from a Banach space into itself. For each x∈X, thesequence of iterates
[ Tnx,n = 0,1,··· and its averages 1Pn Tk−1x,n = 0,1,··· are either
k k=0
bounded or approach infinity.
2
v Keywords: compact operator, dynamics of linear operator, average,iterate
5
7
2
Let X be a set and f : X X a map from X into X. For an x X,
0 → ∈
. the sequence of iterations x,f(x),f2(x), ,fk(x), can be considered as a
2 ··· ···
trajectoryofadynamicalsystemwheretimeisthediscretenonnegativeintegers:
1
starting with the initial (time t = 0) state x, the state at time t = k is fk(x).
0
1 Suppose now X =Cn and f is the transformationdefined by ann n complex
×
: matrix A. What can one say about the general behavior of trajectories of A?
v
More generally, we consider affine maps on X, i.e. maps of the form Ax+c,
i
X where A is linear and c is a constant vector, and we also allow X to be infinite
r dimensional. Moreover,we study the behavior of the sequence of averages:
a
1
Ave f(x)= (x+f(x)+ +fk−1(x)),k =1,2,
k
k ··· ···
Note that the method of averaging was used in [2] in approximating solutions
of a system of linear equations. In case of linear operators,problems about the
linear span of the iterates Tkx,n=0,1, can be found in [4].
···
RecallthatasquarematrixN iscallednilpotentifNs =0forsomenonnegative
integer s.
Throughout this paper, for nonnegative integers k,j,k j, C(k,j) denotes the
≥
binomial coefficient
k!
j!(k j)!
−
1
By convention, C(k,0)=1 for k 0 and C(k,j)=0 if k <j.
≥
k
If k <j, the sum u is considered as an empty sum and its value is 0.
i=j i
Theorem 1 Let TP:Cn Cn be an affine map defined by Tx=Ax+c, where
A is an n n complex ma→trix and c a constant vector in Cn. Let be a norm
on Cn. Th×en for any vector x Cn, the sequence k·k
∈
Tkx,k =0,1,2,
···
is either bounded or lim Tkx = .
k
k k ∞
Proof.
By Jordancanonicaldecompositiontheorem, Cn =V V V for some
1 2 m
⊕ ⊕···⊕
subspaces V ,i = 1,2, ,m with the following properties: (a) each V is an
i i
invariant subspace of A·,··i.e. Av V for all v V , and (b) there exists λ C
i i i
∈ ∈ ∈
and a nilpotent matrix N such that Av =λ v+N v for all v V .
i i i i
Let P be the algebraic projection of Cn onto V associated wi∈th the decompo-
i i
sition Cn =V V V .
1 2 m
Define a new no⊕rm ⊕·o·n·⊕Cn by
|·|
v = P v + P v + + P v .
1 2 m
| | k k k k ··· k k
Let x be a vector in Cn. Let
x =Tkx=Akx+c+Ac+ +Ak−1c,k=1,2, .
k
··· ···
For any vector x, we have the following equalities:
x = P x+ +P x
1 m
···
Akx = AkP x+ +AkP x
1 m
···
Akx = P Akx+ +P Akx
1 m
···
The last equality follows from the commutative property of A and P ,i =
i
1, ,m since A is invariant in each V .
j
···
Fix an i and write v = P x,d = P c,λ = λ ,V = V ,N = N . Let s be the
i i i i i
smallest nonnegative integer such that Nsv =0 and t the smallest nonnegative
integer such that Ntd = 0. Then for k > max s,t and s,t 1 (if s = 0 or
{ } ≥
t=0, the corresponding sum below is defined as 0), one has
P x = (λI +N)kv+d+(λI +N)d+ +(λI +N)k−1d (1)
i k
···
s−1 t−1
= C(k,j)λk−jNjv+ S(j,k)Njd (2)
j=0 j=0
X X
where
S(j,k)=C(j,j)+C(j+1,j)λ+ +C(k 1,j)λk−1−j.
··· −
2
Using the identity C(j,i+1)+C(j,i) = C(j +1,i+1), we have for λ = 1,
S(j,k)=C(k,j+1), and for λ=1, S(j,k) is given recursively by
6
1 λk
S(0,k)= −
1 λ
−
and (by subtracting λS(j,k) from S(j,k))
(1 λ)S(j,k)=S(j 1,k) λk−jC(k,j),j =1,2, ,t 1,( for t 2),
− − − ··· − ≥
from which we get an alternate formula for S(j,k):
1 λk j−1 C(k,i+1)λk−i−1
S(j,k)= − ,j =1, ,t 1 (3)
(1 λ)j+1 − (1 λ)j−i ··· −
− i=0 −
X
Note that (3) is also valid for j = 0 for −1 is an empty sum. We shall show
i=0
that for λ=1,
6 w−1P
P x =B+ λk−jC(k,j)A (4)
i k j
j=0
X
where w =max s,t ,and A ,B,j =0, ,w 1 are constantvectors indepen-
j
{ } ··· −
dent of k; and for λ=1,
l
P x =v+ C(k,j)B (5)
i k j
j=1
X
where l = max s 1,t , and B ,j = 1, ,l are constant vectors independent
j
{ − } ···
of k. More precisely, for k >max s,t ,
{ }
t−1
1
A =Njv Nid,j =0, ,min s,t 1,
j − (1 λ)i−j+1 ··· { }−
i=j −
X
t−1
1
A =ǫ(s t)Njv ǫ(t s) Nid,j =min s,t , ,w 1
j − − − (1 λ)i−j+1 { } ··· −
i=j −
X
and
t−1
1
B = Nid;
(1 λ)i+1
i=0 −
X
and
B =Njv+Nj−1d,j =1, ,min s 1,t ,
j
··· { − }
B =ǫ(s 1 t)Njv+ǫ(t s+1)Nj−1d,j =min s 1,t +1, ,l,
j
− − − { − } ···
3
where ǫ(r)=1 for r 0 and ǫ(r)=0 for r <0.
≥
Indeed substituting (3) into (2), we get
s−1 t−1 1 λk j−1 C(k,i+1)λk−i−1
P x = C(k,j)λk−jNjv+ − Njd
i k (1 λ)j+1 − (1 λ)j−i !
j=0 j=0 − i=0 −
X X X
s−1 t−1 1 t−1 j C(k,i)λk−i
= C(k,j)λk−jNjv+ Njd Njd
(1 λ)j+1 − (1 λ)j−i+1
j=0 j=0 − j=0i=0 −
X X XX
s−1 t−1t−1 C(k,i)λk−i
= C(k,j)λk−jNjv+B Njd
− (1 λ)j−i+1
j=0 i=0 j=i −
X XX
s−1 t−1t−1 C(k,j)λk−j
= C(k,j)λk−jNjv+B Nid
− (1 λ)i−j+1
j=0 j=0i=j −
X XX
s−1 t−1 t−1
1
= C(k,j)λk−jNjv+B C(k,j)λk−j Nid
− (1 λ)i−j+1
j=0 j=0 i=j −
X X X
Now (4) follows by considering cases s>t,s=t or s<t.
Nextconsiderthe caseλ=1. Substituting S(j,k)=C(k,j+1)into(2), weget
s−1 t−1
P x = C(k,j)Njv+ C(k,j+1)Njd
i k
j=0 j=0
X X
s−1 t
= v+ C(k,j)Njv+ C(k,j)Nj−1d
j=1 j=1
X X
Now (5) follows by considering cases s 1>t,s 1=t or s 1>t.
− − −
We now proceed to finish the proof of the theorem:
Case1: λ >1. IfoneoftheseA ,j =0, ,w 1isnonzero,thenweseefrom
j
| | ··· −
(4) that P x as k ; otherwise P x =B is a constant vector.
i k i k
k k→∞ →∞
Case 2: λ = 1 and λ = 1. If one of these A ,j = 1, ,w 1 is nonzero,
j
| | 6 ··· −
then we see from (4) that P x as k ; otherwise the sequence
i k
k k → ∞ → ∞
P x =B+λkA ,k =1, , is bounded.
i k 0
···
Case 3: λ = 1. If one of these B ,j = 1, ,l is nonzero, then we see from
j
···
(5) that P x as k ; otherwise P x is the constant vector v.
i k i k
k k→∞ →∞
Case 4: λ < 1. We see from (4) that the sequence P x ,k = 1, , con-
i k
| | ···
verges to the constant vector B.
We conclude that for each i, the sequence P x ,k = 1, is either bounded
i k
···
or tends to infinity. If one of these sequences tends to infinity, then since
x P x , the sequence x ,k = 1, tends to infinity in norm and
k i k k
| | ≥ k k ··· |·|
4
henceinnorm ,sincethetwonormsareequivalent. Otherwiseallsequences
k·k
P x ,k =1, are bounded for all i=1, ,m, and from which it follows that
i k
··· ···
x ,k =1, is bounded. This completes the proof. Q.E.D.
k
···
The following example shows that in infinite dimensional spaces, Theorem
1 is false.
Example 1 For nonnegative integers n, let c = 1n(n+1). Define λ = 1
n 2 i 2
for c2n i c2n+1 1,n = 0,1, and λi = 2 for c2n−1 i c2n 1,n =
≤ ≤ − ··· ≤ ≤ −
1,2, . Define a linear operator A : l l such that Ae = λ e ,i =
2 2 i i i+1
··· →
0,1, . Then the sequence of iterates Ake0 =λ0λ1 λk−1ek,k=1,2, con-
··· ··· ···
tains subsequences that converge to 0, (e.g. Ac2n−1e = 1/2n,n = 1,2, ),
0
···
subsequences that approach infinity (e.g.Ac2ne = 2n,n = 1,2, ), and in-
0
···
finitely many bounded nonconvergent subsequences (e.g. Ane = e for n =
0 n
2,4,8,12,18,24,32,40, ). Clearly A is bounded but not compact.
···
Remark 1 For T linear, Theorem 1 appeared in a 1997 unpublished article
“On the behavior of the iterates of a matrix” of the author.
In the case that the mapping T in Theorem 1 is linear, we can actually say
more:
Theorem 2 Let A : Cn Cn be a linear map. Let be a norm on Cn.
Then for any vector x C→n, either limk→∞Tkx=0 orkli·mkk→∞ Tkx = or
∈ k k ∞
F Tkx G for sufficiently large k’s, where F,G are positive numbers with
≤k k≤
F G.
≤
Proof.
If c=0 in Theorem 1, then the constant vectors B and d in its proof are 0. It
follows that in all cases where P x ,k=1,2,... is bounded and not convergent
i k
to 0, it is bounded away from 0. If P x is bounded away from 0 for some i,
i k
then so is x = Tkx since x P x . Otherwise P x 0 converges to 0
k k i k i k
| | ≥ k k →
for all i and hence so does x . Q.E.D.
k
Example 2 The following simple example shows that linearity is needed in
Theorem 2: Let T : C C be the map Tx = ix+c, where c C is nonzero
→ ∈
and i = √ 1. Then T4n(0) = 0 and T4n+1(0) = c for all positive integers n,
−
showing that Tk(0),k = 1,2, is neither convergent to 0 nor bounded away
···
from zero.
Definition 1 Let X be a Banachspace and A:X X a linear operator. We
→
say A has property (P) if X is a direct sum of two closedsubspaces V ,V such
1 2
that (1) eachV is invariantunder A, (2) V is finite dimensional, and (3) there
i 1
5
exists 0 r < 1 and a positive integer N such that Akx rk x for all
≤ k k ≤ k k
x V and all k N.
2
∈ ≥
NotethatbyGelfand’sspectralradiustheorem,condition(3)aboveisequivalent
to that A, as an operator on V , has spectral radius less than 1.
2
Itiswell-knownthateverycompactoperator,ormoregenerally,Rieszoperator,
has property (P), see e.g. [1].
Theorem 3 Let(X, )beaBanachspaceandA:X X aboundedoperator
k·k →
having property (P). Let c be a constant vector in X, and let T(x) = Ax+c
for x X. Then for any vector x X, either Tkx,k = 0,1, is bounded
∈ ∈ { ···}
or limk→∞ Tkx = . Moreover, if c = 0, then either limk→∞Tkx = 0
k k ∞
or limk→∞ Tkx = or F Tkx G for sufficiently large k’s, where
k k ∞ ≤ k k ≤
0<F G< .
≤ ∞
Proof.
Let V ,i = 1,2,N,r be as in Definition 1. Let P ,i = 1,2 be the projections of
i i
X onto V ,i=1,2 respectively. Define a norm on X as
i
|·|
x = P x + P x .
1 2
| | k k k k
is equivalent to and T commutes with P ,i=1,2.
i
|·| k·k
For any v X, write v =P v,i=1,2. Then for x X, and k N one has
i i
∈ ∈ ≥
Tkx = Akx +c +Ac + +Ak−1c
2 2 2 2 2
k k k ··· k
rk x + c + +AN−1c +(rN + +rk−1) c
2 2 2 2
≤ k k k ··· k ··· k k
1
rk x + c + +AN−1c + c ,
2 2 2 2
≤ k k k ··· k 1 rk k
−
showingthatTkx ,k =1,2, isbounded. ByTheorem1, Tkx ,k =1,2,
2 1
··· k k ···
iseitherboundedorapproachinginfinity. Hence Tkx = Tkx + Tkx ,k=
1 2
| | k k k k
1,2, is either bounded or approaching infinity. Since norms and are
··· |·| k·k
equivalent, the same is true for Tkx ,k=1,2, .
k k ···
If c=0, thenc =0 andTkx 0 as k , andthe lastpartofthe theorem
2 2
→ →∞
follows readily from Theorem 2.
Corollary 1 Let (X, )be a Banach space and A:X X a Riesz operator.
k·k →
Let c be a constant vector in X, and let T(x) = Ax+c for x X. Then for
∈
any vector x X, either Tkx,k =0,1, is bounded or limk→∞ Tkx = .
∈ { ···} k k ∞
Moreover, if c = 0, then either limk→∞Tkx = 0 or limk→∞ Tkx = or
k k ∞
F Tkx G for sufficiently large k’s, where 0<F G< .
≤k k≤ ≤ ∞
Corollary 2 Let (X, ) be a Banach space and A : X X a compact
k · k →
operator. LetcbeaconstantvectorinX,andletT(x)=Ax+cforx X. Then
∈
for any vector x X, either Tkx,k =0,1, is bounded or limk→∞ Tkx =
∈ { ···} k k
. Moreover, if c=0, then either limk→∞Tkx=0 or limk→∞ Tkx = or
∞ k k ∞
F Tkx G for sufficiently large k’s, where 0<F G< .
≤k k≤ ≤ ∞
6
Let us consider now the behavior of the sequence of averagesof T:
1
Ave T(x)= (x+Tx+ +Tk−1x),k =1,2,
k
k ··· ···
We have
1 1
Ave T(x)= (x+Ax+ +Ak−1x)+ ((k 1)c+(k 2)Ac+ +Ak−2c)
k
k ··· k − − ···
As in the proof of Theorem 1, we may assume that A = λI +N, and that s,t
are defined as in there. By replacing c by x, x by 0, and hence s by 0 and t by
s in the proof of Theorem 1, we obtain:
For λ=1,
6
s−1
1 1 C(k,j)
(x+Ax+ +Ak−1x)= B+ λk−j A
j
k ··· k k
j=0
X
where
s−1
1
A = Nix,j =0, ,s 1
j − (1 λ)i−j+1 ··· −
i=j −
X
s−1
1
B = Njx.
(1 λ)j+1
j=0 −
X
For λ=1,
s
1 1
(x+Ax+ +Ak−1x)= C(k,j)Nj−1x.
k ··· k
j=1
X
By expanding Aj we have
(k 1)c+(k 2)Ac+ +Ak−2c=T0c+T1Nc+ +Tt−1Nt−1c
− − ··· ···
where
k−2
T = (k i 1)C(i,j)λi−j,j =0, ,t 1.
j
− − ··· −
i=j
X
Assume that λ=1. By subtracting λT fromT and using the geometric series
0 0
6
formula, one gets
k 1 λk
T = −
0 1 λ − (1 λ)2
− −
Using the relation (i j)C(i,j)=(j+1)C(i,j+1), we see that
−
1 d
T = T
j+1 j
j+1dλ
We shall prove in the Appendix that
k j+1 C(k,j)λk−j
T = + D(k,j,λ) (6)
j (1 λ)j+1 − (1 λ)j+2 (1 λ)2
− − −
7
for j =0, ,t 1, where
··· −
1
D(k,j,λ)= (1 λ)j[B0(k,j)λj +B1(k,j)λj−1+···+Bj−1(k,j)λ+1]
−
and
C(k j,2)
B (k,j)=( 1)j−iC(j,i) − , i=0, ,j
i
− C(k i,2) ···
−
Note that
B (k,j) ( 1)j−iC(j,i)
i
→ −
so that D(k,j,λ) approaches1 as k . Substituting the formulae we obtain
→∞
thus far into Ave T(x), with w =max s,t , we get for λ=1,
k
{ } 6
1
Ave T(x)=E+ F +G(k,λ) (7)
k
k
where
w−1
1 D(k,j,λ)
G(k,λ)= C(k,j)λk−j ǫ(s 1 j)A +ǫ(t 1 j) Njc
k − − j − − (1 λ)2
j=0 (cid:18) − (cid:19)
X
(8)
t−1
1
E = Njc,
(1 λ)j+1
j=0 −
X
s−1 t−1
1 j+1
F = Njx Njc,
(1 λ)j+1 − (1 λ)j+2
j=0 − j=0 −
X X
and ǫ(r)=1 for r 0 and ǫ(r)=0 for r <0.
≥
If λ=1, then T is equal to
j
S =C(k 2,j)+2C(k 3,j)+ +(k j 1)C(j,j)
j,k
− − ··· − −
We have
Sj,k Sj,k−1 =C(k 2,j)+C(k 3,j)+ +C(j,j)=C(k 1,j+1)
− − − ··· −
where the last equality follows from repeatedly applying the identity
C(j,i+1)+C(j,i)=C(j+1,i+1). From this it follows that
Sj,k = Sj,k−1+C(k 1,j+1)
−
= Sj,k−2+C(k 2,j+1)+C(k 1,j+1)
− −
=
···
= S +C(j+2,j+1)+ +C(k 1,j+1)
j,j+2
··· −
= C(j+1,j+1)+C(j+2,j+1)+ +C(k 1,j+1)
··· −
= C(k,j+2)
8
Therefore for λ=1,
1
Ave T(x) = [C(k,1)x+C(k,2)Nx+ +C(k,s)Ns−1x] (9)
k
k ···
1
+ [C(k,2)c+C(k,3)Nc+ +C(k,t+1)Nt−1c] (10)
k ···
Case 1: λ >1. Suppose that t>s. Then from (7) and (8), 1λkC(k,t 1)
dominatesa|ll|othercoefficients. SinceD(k,t 1,λ) 1ask kandNt−1−c=
− → →∞ 6
0, we see that Ave T(x) as k . The same is true if s > t since
k
k k → ∞ → ∞
As−1 = 0. So assume that s = t. In the trivial case s = t = 0, we have
6
Ave T(x) = 0 for all k. So assume that s = t 1. Direct checking (see
k
Appendix, item 1) shows that C(k,j)C(j,i)C(k−j,2)≥=C(k,i)C(k i 2,j i)
C(k−i,2) − − −
so that C(k,j)D(k,j,λ) is a polynomial in k. Then from (8), we have
t−1
1 D(k,j,λ)
G(k,λ) = C(k,j)λk−j A + Njc
k j (1 λ)2
j=0 (cid:18) − (cid:19)
X
λk t−1
= p (k,j,λ)A +p (k,j,λ)Njc
1 j 2
k
j=0
X
where for fixed λ, p1,p2 are polynomials in k. Since limk→∞ λkkp(k) = ∞ for
any polynomial p(k) we see that if
t−1
H(k,λ)= p (k,j,λ)A +p (k,j,λ)Njc,
1 j 2
j=0
X
asapolynomialink,isidenticallyzero,thenclearlyfrom(7)wehaveAve T(x)
k
→
E, otherwise Ave T(x) .
k
k k→∞
Case 2: λ = 1,λ = 1. Suppose that t > s. Since D(k,t 1,λ) 1 as
k andN| |t−1c=0,6 weseethat Ave T(x) ask −ift 3→,andit
k
→∞ 6 k k→∞ →∞ ≥
approachesto0orisboundedift 2. Thesameistrueifs>tsinceAs−1 =0.
≤ 6
So assume that s = t. In the trivial case s = t = 0, we have Ave T(x) = 0 for
k
all k. So assume that s = t 1. As in Case 1, we consider the polynomial (in
≥
k)
t−1
D(k,j,λ)
H(k,λ)= C(k,j)λ−j A + Njc
j (1 λ)2
j=0 (cid:18) − (cid:19)
X
so that
λk
G(k,λ)= H(k,λ)
k
If degree of H is one or less, then we see from above that G(k,λ) is bounded
and hence Ave T(x) is bounded by (7). If degree of H is two or more, then
k
G(k,λ) as k and hence so is Ave T(x) .
k
k k→∞ →∞ k k
9
Case 3: λ = 1. We refer to (9) and (10). If s = t = 0, then c = x = 0 and
Ave T(x) = 0. If t = 0 and s = 1, then Ave T(x)= x. If t = s 1,s 2 and
k k
− ≥
Nic+Ni+1x=0 for all i=0, ,t 1, then Ave T(x)=x. In all other cases
k
··· −
lim Ave T(x) = .
k k
k k ∞
The previous discussions yield the proof of the following:
Theorem 4 Let T :Cn Cn be an affine map defined by Tx=Ax+c, where
A is an n n complex ma→trix and c a constant vector in Cn. Let be a norm
on Cn. Fo×r any vector x Cn, define k·k
∈
1
Ave T(x)= (x+Tx+ +Tk−1x)
k
k ···
Then the sequence
Ave T(x),k =0,1,2,
k
···
is either bounded or lim Ave T(x) = .
k k
k k ∞
If T is linear, i.e. if c=0, we can say more:
Theorem 5 Let A : Cn Cn be linear map. Let be a norm on Cn. For
any vector x Cn, define→ k·k
∈
1
Ave A(x)= (x+Ax+ +Ak−1x)
k
k ···
Then the sequence
Ave A(x),k =0,1,2,
k
···
is either (i) convergent to 0, or (ii) lim Ave A(x) = , or (iii) F
k k
k k ∞ ≤
Ave Ax G for sufficiently large k’s, where F,G are positive numbers with
k
k k ≤
F G.
≤
Proof.
Ifc=0inTheorem4,thenE =0,andbyexaminingitsproofweseethatinall
caseswhereTkxisboundedandnotconvergentto0,itisboundedawayfrom0.
(InCase2oftheproof,ifH(k,λ)isofdegreeone,thenG(k,λ)isboundedaway
from 0, and if H(k,λ) is of degree 0, i.e. a constant vector, then G(k,λ) 0.)
→
Q.E.D.
ThefollowingexampleshowsthatTheorem5isfalseifthemapisnotlinear.
Example 3 Let
i 1
A=
0 i
(cid:18) (cid:19)
10