Ar Ve Son Spectral Solutions
Short Description
Download Ar Ve Son Spectral Solutions...
Description
SOLUTIONS TO
A SHOR SHORT COURSE COURSE ON SPECTR SPECTRAL AL THEOR THEORY Y
BY WILLIAM WILLIAM ARVESON ARVESON
Contents
1. Spectral Spectral Theory Theory and Banac Banach h Algebras Algebras 1.1.. Origin 1.1 Originss of Spect Spectral ral Theo Theory ry 1.2.. The Spect 1.2 Spectrum rum of an Operato Operatorr 1.3. Banach Banach Algebras: Algebras: Example Exampless 1.4. The Regular Regular Represen Representatio tation n 1.5.. The Gene 1.5 General ral Line Linear ar Group Group of A of A 1.6. Spectrum Spectrum of an an Element Element of a Banach Banach Algebra Algebra 1.7.. Spectra 1.7 Spectrall Radius Radius 1.8.. Ideals 1.8 Ideals and Quotie Quotient ntss 1.9. Commutat Commutative ive Banach Banach Algebras Algebras 1.10. 1.1 0. Exampl Examples: es: C (X ) and the Wiener Algebra 1.11. Spectral Spectral Perma Permanence nence Theorem Theorem 1.12. Brief on on the Analytic Analytic Functional unctional Calculus Calculus 2. Operato Operators rs on on Hilber Hilbertt Space Space 2.1.. Operato 2.1 Operators rs and Their Their C ∗ -Algebras 2.2.. Commu 2.2 Commutat tativ ivee C ∗ -Algebras 2.3. Continu Continuous ous Functio Functions ns of Normal Normal Operators Operators 2.4. The Spectral Spectral Theorem Theorem and and Diagonaliz Diagonalization ationss 2.5. Represen Representatio tations ns of Banach Banach *-Algebra *-Algebrass 2.6. Borel Functions unctions of Normal Normal Operators Operators 2.7.. Spectra 2.7 Spectrall Mea Measur sures es 2.8.. Compac 2.8 Compactt Operato Operators rs 2.9.. Adjoin 2.9 Adjoining ing a Unit Unit to to a C ∗ -Algebra 2.10. 2.1 0. Quotie Quotient ntss of C of C ∗ -Algebras 3. Asymptotic Asymptotics: s: Compact Compact Perturb Perturbations ations and and Fredholm Fredholm Theory Theory 3.1.. The Calkin 3.1 Calkin Alg Algebr ebraa 3.2. Riesz Theory Theory of Compact Compact Operators Operators 3.3. Fredholm redholm Operators Operators 3.4.. The Fredhol 3.4 redholm m IInde ndex x 4. Methods Methods and Applica Applicatio tions ns 4.1. Maximal Maximal Abelian Abelian von von Neumann Neumann Algebras Algebras 4.2. Toeplitz Matrices Matrices and and Toeplitz Toeplitz Operato Operators rs ∗ 4.3.. The Toeplitz 4.3 oeplitz C -Algebra 4.4. Index Index Theorem Theorem for for Contin Continuous uous Symbo Symbols ls 4.5. 4.5. Some Some H 2 Function Theory 4.6. Spectra Spectra of Toeplitz Toeplitz Operator Operatorss with Continu Continuous ous Symbol Symbol 4.7. States States and and the the GNS GNS Constru Construction ction 4.8. Existence Existence of of States: States: The Gelfand-Naim Gelfand-Naimark ark Theorem Theorem References 1
2 2 4 5 8 9 11 12 14 15 16 17 18 21 21 23 26 27 31 33 34 37 39 40 42 42 43 43 43 43 43 43 43 43 43 43 43 43 43
2
ARVESON
1.
SPECTRAL THEORY
SOLUTIONS
Spectral Spectral Theory Theory and Banach Banach Algebras
1.1. Origins of Spectral Theory. (1) Fix a sequence sequence an of numbers with 0 < an M , M , and let A : 2 2 be the operator determined by (Ax ( Ax))n = an xn .Show that A is a bounded 2 operator on , and exhibit a bounded operator B on 2 such that AB = BA = 1.
{ }
≤ ≤
Solution. A is bounded because
Ax2 =
(Ax) Ax)n (Ax) Ax)n =
n
an xn an xn =
n
→
| | | | ≤ an
2
2
xn
n
M 2 xn
| |2 = M 2x2
n
1 an xn ;
The inverse operator B is determined by (Bx (Bx))n = bounded by 1/ 1/,, by the same argument.
this operator is
{ }
(2) Let an be a bounded increasing sequence of positive numbers and let Dn = a1 a2 . . . an . Show that the sequence Dn converges to a nonzero limit D(A) iff
∞
(1
n=1
− an) < ∞.
Solution. To clarify, when we say
−
(1 an ) < , we mean to say that the sum converges to a finite number. In particular, the case where it converges to (for example, if all the an are equal to 2) is disallowed. It is clear that either of the conditions whose equivalence we seek to establish implies that an 1, so we will assume this to be the case. The condition that Dn approaches a nonzero limit is equivalent to j ( ln aj ) < . We want to show this is equivalent to j (1 aj ) < . One direction is trivial, trivial, becaus b ecausee ln x 1 x for all positive x (just compare the derivatives to the right and left of the intersection point x = 1). For the other other direction, we note that ln x < 2(1 x) in a neighborhood of x = 1 (again, this follows by comparing the derivatives at x = 1); since an 1, one has ln aj < 2(1 aj ) for sufficiently large j , so that the result follows by the comparison test. We note that this argument is standard in complex analysis in the context of the Weierstra Weierstrass ss factoriza factorization tion theorem; see for instance instance ([ Rud87]) Rud87]) Theorem 15.5 or ([Con78 ([ Con78]) ]) Proposition 5.4.
−∞
∞
↑
∞
−
−
≥ − −
−
−
∞
−
→
−
(3) Let k (x, y) be a continuous function on (x, y) : 0 y x C defined by f C ([0, ([0, 1]). Show that the function g : [0, [0, 1]
{
∈
x
g (x) =
→
≤ ≤ ≤ 1} and let
k (x, y)f ( f (y) dy
0
is contin continuou uous, s, and that the linear map K : f operator operator on C ([0, ([0, 1]).
→
g defines defines a bounded bounded
sup and L = max |k(x, y)|. Fix any x ∈ [0, [0, 1]. Let > 0. Since k is uniformly continuous (its domain is compact), one can choose δ 1 > 0 such that |k (x1 , y1 ) − k (x2 , y2 )| < whenever (x1 , y1 ) −
Solution. Let M = f
ARVESON
SOLUTIONS
SPECTRAL THEORY
≤ h < δ one has
(x2 , y2 ) < δ 1 . Let δ = min(δ,/L min(δ,/L). ). For 0
| ≤ | ≤ | x+h
|g(x + h) − g(x)
=
−
x
k(x + h, y )f ( f (y) dy
0
x+h
[k (x + h, y )
0
− k(x, y)]f )]f ((y) dy +
k(x + h, y )f ( f (y) dy
x
x+h
x
k (x + h, y)
0
− k(x, y)||f ( f (y)| dy +
k(x + h, y ) f ( f (y ) dy
x
|
|
x+h
x
M
| |
k (x, y)f ( f (y) dy
0
x
=
3
k (x + h, y )
0
− k(x, y)| dy + M
k (x + h, y) dy
x
|
≤ M + MLh < 2M . A similar calculation holds for −δ < h ≤ 0 (the second integral is reversed). Thus g (x + h) → g (x) as h → 0, so that g is continuous at x. For the boundedness of K , note that for each x ∈ [0, [0, 1], x |g(x)| = k(x, y)f ( f (y) dy
≤ | ≤ | 0 x
k (x, y) f ( f (y ) dy
0
||
|
x
M
k(x, y) dy
0
|
≤ M L = Lf sup Taking the supremum over x ∈ [0, [0, 1], we have g sup ≤ Lf sup . Thus Thus K op ≤ L. (4) For the kernel kernel k(x, y) = 1 for 0 ≤ y ≤ x ≤ 1 consider the corresponding Volterra operator V : C ([0, ([0, 1]) → C ([0, ([0, 1]), i.e. x (V f )( f )(x x) = f ( f (y) dy , f ∈ C ([0, ([0, 1]). 1]). 0 Given Given a function function g ∈ C ([0, ([0, 1]), 1]), show show that that the equation equation V f = g has a solution f ∈ C ([0, ([0, 1]) iff g is continuously differentiable and g(0) = 0.
Solution. Clearly every element V f of the range is continuously differen-
tiable and has (V ( V f )(0) f )(0) = 0. Conversely, suppose g satisfies those hypothe ses; let f = g . Then (V (V f ) f ) = f = g , so that V f and g differ by an additive constant; but (V (V f )(0) f )(0) = 0 = g(0), so that V f = g . (5) Let k (x, y), 0 x, y 1 be a continuous function on the unit square, and consider consider the bounded bounded operator operator K defined on C ([0, ([0, 1]) by
≤
≤
1
(Kf )( Kf )(x x) =
k (x, y)f ( f (y) dy,
0
{ ∈
≤ }
0
≤ x ≤ 1.
Let B1 = f C ([0, ([0, 1]) : f 1 be the closed unit ball in C ([0, ([0, 1]). Show that K is a compact operator in the sense that the norm closure of the image KB 1 under K is a compact subset of C ([0, ([0, 1]).
4
ARVESON
SPECTRAL THEORY
SOLUTIONS
Solution. Clearly KB 1 is a pointwise bounded family of functions (since K
is a bounded operator, operator, by a straight straightforw forward ard generalization generalization of exercise exercise 3). Given > 0, use the uniform continuity of k to choose δ > 0 such that k(x1 , y1 ) k (x2 , y2 ) < whenever (x1 , y1 ) (x2 , y2 ) < δ . δ . Then for any f B1 and any x1 , x2 [0, [0, 1] satisfying x1 x2 < δ , δ ,
|
−
∈
|
∈
|(Kf )( Kf )(x x1 ) − (Kf )( Kf )(x x2 )
− | − | 1 [k (x1 , y ) − k(x2 , y)]f )]f ((y) dy
| ≤ | ≤ · =
0 1
k (x1 , y)
0
− k(x2, y)||f ( f (y)| dy
1
1 dy = .
0
Thus, KB 1 is uniformly equicontinuous; by the Arzela-Ascoli theorem, it has compact closure. Note: Note: The hint hint is mislea misleadin ding, g, as KB 1 is not unifor uniformly mly Lipsc Lipschit hitzz in genera general; l; indeed indeed,, it ma may y contai contain n functi functions ons which which are not Lipsc Lipschit hitz. z. For example, example, suppose k (x, y) = x and f is the constant function f ( f (y) = 1; then (Kf (Kf )( )(x x) = x is not Lipschitz. Lipschitz. One needs to use uniform rather rather than Lipschitz continuity.
√
√
1.2. The Spectrum of an Operator. (1) Give explicit examples of bounded operators A, B on 2 (N) such that AB = 1 and BA is the projection onto a closed infinite-dimensional subspace of infinite codimension. Solution. We define A and B on the standard basis vectors by Be k = e2k
and Aek =
ek/2 k/2 0
k even else.
Then AB = 1 and BA is the projection onto spane span e2 , e4 , e6 , . . .. ..
(2) Let A and B be the operators defined on 2 (N) be A(x1 , x2 , . . . ) = (0, (0, x1 , x2 , . . . ) B (x2 , x2 , . . . ) = (x2 , x3 , x4 , . . . )
Show that A = B = 1, and compute both BA and AB. AB . Deduce Deduce that that A is injective but not surjective, B is surjective but not injective, and that σ (AB) AB ) = σ (BA). BA ).
Solution. Clearly A is isometric and B is contractive; since B (0, (0, 1, 0, 0, . . . ) =
(0, (0, 1, 0, 0, . . . ) we see that B = 1.
BA = 1, whereas AB is projection onto the complement of the span of e1 . The former former of these implies implies the injectivity injectivity of A and surjectivity of B ; if A were surjective or B injective, the composition AB would have to be as well, but it’s not. The non-injectivity of AB implies that its spectrum includes 0, which σ (BA) BA ) does not.
ARVESON
SOLUTIONS
SPECTRAL THEORY
5
(3) Let E be a Banach space and let A and B be bounded operators on E . Show that 1 AB is invertible iff 1 BA is invertible.
−
−
Solution. Suppose 1
1 + BC A. Then
− AB is invertible, and let C be its inverse. Let D =
D(1 BA) = 1+BC A BA BCABA = 1+B(C CAB )A BA = 1+BA BA = 1
−
− −
−
−
−
and
−
−
−
−
−
−
(1 BA)D = 1 BA+BC A BABCA = 1 BA+B(C ABC )A = 1 BA+BA = 1 so that 1 BA is invertible. Interchanging A and B yields the other direction of implication. We note that norm properties enter nowhere into this calculation; the result holds in any unital algebra.
−
(4) Use the result of the preceding exercise to show that for any two bounded operators A, B acting on a Banach space, σ(AB) and σ(BA) agree except perhaps for 0: σ(AB) 0 = σ(BA) 0 .
\{ }
\{ }
λ∈ / σ(AB) ⇔ λ − AB invertible ⇔ λ 1 − Aλ B invertible
Solution. For λ = 0,
⇔ −
A invertible λ BA invertible
λ 1
B
⇔λ− ⇔ λ ∈/ σ(BA).
1.3. Banach Algebras: Examples. (1) Let E be a normed linear space. Show that E is a Banach space iff for every sequence of elements xn X satisfying n xn < , there is an element y X such that
∈
∈
∞
→∞ y − (x1 + ·· · + xn ) = 0.
lim
n
{ } be an absolutely summable sebe the nth partial sum. Then for n ≥ m,
Solution. Suppose E is Banach. Let xn
n i=1 xi
quence. Let yn =
n
yn − ym
=
i=m+1
≤ ≤ n
xi
∞
xi
i=m+1
xi .
i=m+1
−
Since the tails of a convergent sequence tend to zero, this shows that yn ym 0 as n, m , so that yn is Cauchy. By completeness, it converges to a limit y. Thus, every absolutely summable sequence in E is summable. Conversely, suppose E is a normed space in which every absolutely summable sequence is summable. Let xn be a Cauchy sequence in E . Let xnk be a subsequence such that xnk+1 xnk < 2−k . Let yk =
→
{ }
→∞
{ }
{ } −
6
ARVESON
−
SPECTRAL THEORY
SOLUTIONS
{ }
xnk+1 xnk . Then yk is absolutely summable, hence summable. Let y be its sum. For each k, k
xnk+1 = xn1 +
yk ;
i=1
since the right-hand side approaches a limit as k , the left-hand side must as well. Hence the sequence xnk converges. But then xn is a Cauchy sequence with a convergent subsequence, so it converges as well.
→∞
{ }
(2) Prove that the convolution algebra L1 (R) does not have an identity. Solution. Suppose g is a convolution identity. Let f n = 1[0,1/n] . Then
0
∗
1 = f n (0) = (f n g)(0) =
g(y) dy.
−1/n
Since L1 functions generate absolutely continuous measures ( [Fol99] Theorem 3.5), this is impossible. Another way to prove impossibility is to cite the fact that the convolution of L1 functions with L∞ functions is continuous ([Fol99] Proposition 8.8). But then the convolution of g with a characteristic function would have to be a continuous characteristic function, which is impossible. (3) For every n = 1, 2, . . . let φn be a nonnegative function in L1 (R) such that φn vanishes outside the interval [ 1/n, 1/n] and
∞
−
φn (t) dt = 1.
−∞
Show that φ1 , φ2 , . . . is an approximate identity for the convolution algebra L1 (R) in the sense that
∗ − f 1 = 0
lim f φn
→∞
n
for every f
∈ L1(R). L1 (R),
Solution. For any f
f ∗ φn
∈ − | ∗ − | − − ≤ | − − || | | − | − f =
(f φn )(x)
f (x) dx
R
=
f (x
R
y)φn (y) dy
f (x)
R
φn (y) dy dx
R
1/n
f (x
R
y)
f (x) φn (y) dydx
−1/n
1/n
=
φn (y)
−1/n
f y (x)
f (x) dxdy
R
1/n
=
φn (y) f y
f
1
dy.
−1/n
By the uniform continuity of translation on L1 , this tends to 0 as n .
∞
→
ARVESON
(4) Let f
SPECTRAL THEORY
SOLUTIONS
7
∈ L1(R). The Fourier transform of f is defined as follows: ˆ )= f (ξ
∞
eitξ f (t) dt,
ξ R.
∈
−∞
Show that f ˆ belongs to the algebra C ∞ (R) of all continuous functions on R that vanish at .
∞ Solution. Fix ξ ∈ R. Let > 0 be given. Let K ⊆ R be a compact subset such that \K |f | < . Now for any η ∈ R,
R
|
ˆ ) f (ξ
−
| ≤
ˆ f (η) =
e
R
− − − | | − | |
iξt
f (t) dt
R
eiξt
R
≤ 2 +
e
f (t) dt
R
eiξt
=
iηt
eiηt f (t) dt
eiηt f (t) dt
eiξt
eiηt f (t) dt.
K
As η ξ , eiηt eiξt uniformly on K , so that the above expression is less than 3 for η sufficiently close to ξ . This proves that f ˆ is continuous. For the Riemann-Lebesgue lemma, one could use the identification of R with the maximal ideal space of L1 (R), and the corresponding identification of the Fourier transform with the Gelfand transform; my presentation in Jorgensen’s class in spring 2011 goes into this. A more elementary proof, however (thank to [SS05]), is as follows: For any ξ = 0, the change of variables x = x + πξ yields
→
→
− − − −
f ˆ(ξ ) =
f x
R
π eix ξ dx . ξ
Averaging this with the original expression for f ˆ(ξ ) yields ˆ )= 1 f (ξ 2 from which f ˆ(ξ ) of translation.
|
f (x)
f x
R
| ≤ 12 f − f π/ξ .
π ξ
eixξ dx
The result follows by the L1 continuity
(5) Show that the Fourier transform is a homomorphism of the convolution algebra L1 (R) onto a subalgebra of C ∞ (R) which is closed under complex conjugation and separates points of R.
A
Solution. Linearity is obvious. Multiplicativity is a well-known consequence
of Fubini’s theorem. For self-adjointness, transform the function f ∗ (x) = f ( x). For separation of points, suppose ξ < η and consider the Fourier transform of the characteristic function of [0 , |2π η| ].
−
|| ||
8
ARVESON
SPECTRAL THEORY
SOLUTIONS
1.4. The Regular Representation. (1) Let E and F be normed linear spaces with E = 0 . Show that B(E, F ) is a Banach space iff F is a Banach space.
{}
Solution. Suppose F is Banach. Let T n be a Cauchy sequence in B(E, F ). For each e E and each m, n N,
{ }
∈ ∈ T me − T ne = (T m − T n)e ≤ T m − T ne so that {T n e} is a Cauchy sequence in F . Define the function T : E → F by T e = lim T n e. Clearly this is linear; it is bounded because
T e = lim T ne = lim T ne ≤ (lim T n)e so that T ≤ lim T n (which limit exists because {T n } is a Cauchy sequence in R). Finally, we must prove that T n → T in norm, not just pointwise. Given > 0, choose N ∈ N such that T m − T n < for m, n ≥ N . Then for n ≥ N and e ∈ E , T e − T ne = lim T m e − T n e = lim T m e − T n e ≤ e m m since T m e − T n e < e for each m. Hence T − T n → 0. Conversely, suppose B(E, F ) is Banach. Let {f n } be a Cauchy sequence in F . Let e ∈ E be some nonzero vector. Define linear maps T n : E → F by T n e = f n and T n = 0 on the (algebraic) complement of the span of e. Clearly T n is linear with norm f en . Moreover, since (T m
− T n)x =
α(f m 0
− f n)
x = αe else,
f n we see that T m T n = f m− , so that T n is Cauchy in B(E, F ). Let e T be its limit, and let f = T e. Then
−
{ }
f − f n = T e − T ne ≤ T − T ne → 0. Thus F is Banach.
∈
(2) Let A B(E ) be an operator with the property that there is a sequence Am of finite-rank operators such that A An 0. Show that A is compact.
{ }
− →
{ } be a sequence of points in the unit ball of E .
Solution. Let xn
Define x as follows: = xn , while is a (i+1) Ai+1 xn converges. (This is possible by the compactness of the unit ball for any finite-dimensional Banach space, (n) such as the range of each Am .) Let yn = xn be the diagonal subsequence, so that Ak yn converges for each fixed k; call the limit zk . Then (i) subsequences xn recursively (i) subsequence of xn such that
{
(0) xn
{
(i+1) xn
}
}
zj − zk = lim Ak yn − lim Aj yn = lim (Ak − Aj )yn ≤ Ak − Aj n n n since (Ak − Aj )yn ≤ Ak − Aj for each n. This proves that {zk } is Cauchy; let its limit be z. Then for each k,
Ayn − z = Ak yn − zk +(A − Ak )yn +(zk − z) ≤ Ak yn − zk +A− Ak +zk − z.
ARVESON
SPECTRAL THEORY
SOLUTIONS
9
→∞ { }
Since all three terms on the right approach 0 as k , we see that Ayn z. Thus, we have found a subsequence yn such that Ayn converges. So A is compact. By the way, this diagonal subsequence trick is used in general topology (nets) to show that a cluster point of the cluster points of A is a cluster point of A; see 11.5 of my putative functional HW. It’s also used to prove Arzela-Ascoli, etc.
→
{
}
(3) Let a1 , a2 , . . . be a bounded sequence of complex numbers and A the corresponding multiplication operator on 2 . Show that A is compact iff an 0.
→
Solution. Suppose an
→ 0. Let An be the truncated multiplication operator with 0 for k > n. Then A − An = supk>n |ak | →
which replaces ak 0. Since each An has finite rank, the previous exercise implies that A is compact. Conversely, suppose an 0. Let > 0 and ank a subsequence with ank > for each k. Then enk is a sequence of unit vectors such that Aenk has no convergent subsequence (indeed, Aenk Aenj > 2), so that A is not compact.
→ { }
| | { } (4) Let k
−
√
∈ C ([0, 1] × [0, 1]) and define A : C ([0, 1]) → C ([0, 1]) by
1
(Af )(x) =
k(x, y)f (y) dy.
0
Show that A is bounded with A
≤ ksup.
Solution. Already done in exercise 1.1.5, where we showed A is in fact
compact.
−
(5) Show that there is a sequence of finite-rank operators An such that An A 0.
→
Solution. As can easily be checked, if A is another integral operator with
kernel k , then A A k k sup . Now by Stone-Weierstrass, there (n) (n) n exists a sequence kn of kernels of the form kn (x, y) = N j=1 pj (x)q j (y),
− ≤ −
(n)
(n)
− →
with pj and q j polynomials, such that kn k 0, and therefore A An 0, where An is the operator with kernel kn . But each An has (n) finite rank; in particular, its range is spanned by the pj .
− →
1.5. The General Linear Group of A. Let A be a unital Banach algebra with 1 = 1, and let G = A−1 . (1) Show that for every x A with x < 1, there exists a continuous function f : [0, 1] G with f (0) = 1 and f (1) = (1 x)−1 .
∈
→
Solution. We define f (t) = (1
−
− tx)−1. This is well-defined since tx < t.
It is continuous because it is the composition of the continuous functions 1 tx with y t y −1 .
→ −
→
∈
(2) Show that for every element x G there is an > 0 with the following property: For every element y G satisfying y x < there is an arc in G connecting y to x.
∈
−
10
ARVESON
SPECTRAL THEORY
SOLUTIONS
Solution. Let = x−1
−1. If h ∈ A−is1 such that x + h ∈ G and h < ,
then let f be a path from 1 to 1 + x h as in the previous problem; then t xf (t) is a path in G from x to x + h. In words, what we’ve just shown is that G is locally path connected.
→
(3) Let G0 be the set of all finite products of elements of G of the form 1 x or (1 x)−1 , where x A satisfies x < 1. Show that G0 is the connected component of 1 in G.
−
∈
Solution. Let
−
I denote the connected component of the identity. Note that
this is the same as the path component of the identity, because G is locally path connected by problem 2. By problem 1 above, each element of the form ( 1 x)−1 is in . is closed under products because, if f is a path from 1 to x and g a path from 1 to y, then
• • I
−
h(t) =
f (2t) xg(2t
−
I
≤ ≤ ≤ ≤
0 t 1/2 1) 1/2 t 1
defines a path from 1 to xy.
• I is closed−under inverses because, if f is a path from 1 to x, then 1 g(t) = f (t) is a path from 1 to x−1 . • Thus, G0 ⊆ I . • G0 is open: Let a = a1 . . . an be an element of G0 , where ai = 1 − xi or − 1 (1 − xi ) for each i, with xi < 1. Then for h ∈ A with h < a−1 , one has a − h = a(1 − a−1 h) and since both a and 1 − a−1 h are elements of G0 , this product is again in G0 . • G0 is clopen, because open subgroups of topological groups are always clopen: Suppose G is a topological group and H a subgroup. Then
\
G H =
•
xH
x G H
∈ \
is a union of open sets (because left multiplication y omorphism), hence open, so that H is closed. Thus, G0 , completing the proof.
→ xy is a home-
I⊆
(4) Deduce that G0 is a normal subgroup of G and that the quotient topology on G/G0 makes it into a discrete group. Solution. The connected component of the identity is a subgroup, as shown
above; it is normal because, if f is a path from 1 to x and y G, then t yf (t)y −1 is a path from 1 to yxy −1 , so that G0 is closed under conjugation by elements of G. The cosets of G0 in G are precisely the connected components of G; to see this, let Gx denote the component of some x G, and note that xG0 Gx and x−1 Gx G0 . As a result, the preimage of any subset of G/G0 will be a union of connected components of G, and therefore open.
∈
⊆
∈
→
⊆
ARVESON
SPECTRAL THEORY
SOLUTIONS
11
1.6. Spectrum of an Element of a Banach Algebra. (1) Give an example of a one-dimensional Banach algebra that is not isomorphic to the algebra of complex numbers. Solution. Such an algebra would have to have the same linear structure
but a different multiplication. The simplest way to do this is to define all products to be zero. Note that it is impossible to give a unital counterexample, because the multiplicative structure of a unital one-dimensional algebra is determined by (λ1)(µ1) = λµ1. (2) Let X be a compact Hausdorff space and let A = C (X ) be the Banach algebra of all complex-valued continuous functions on X . Show that for every f C (X ), σ(f ) = f (X ).
∈
Solution. The invertible functions in C (X ) are precisely the functions which
are never zero. Thus, for a given λ, f
− λ1 is invertible iff λ ∈/ f (X ).
(3) Let T be the operator defined on L2 ([0, 1]) by (T f )(x) = xf (x). What is the spectrum of T ? Does T have point spectrum? Solution. T has no point spectrum, because if f is a function such that
xf (x) = f (x) then f (x) = 0 for x = 1, so that f = 0 in L2 . The spectrum is [0, 1]. It is at most this, because for λ / [0, 1], the inverse of T λ1 is the operator (Sf )(x) = x−1 λ f (x). On the other hand, if λ [0, 1] then such an inverse S cannot exist; when f (x) is the constant function f (x) = 1, we must have (Sf )(x) = x−1 λ a.e., but this is not square integrable. We recall that, more generally, the spectrum of a multiplication operator M f on L2 is the essential range of f .
∈
−
∈
For the remaining exercises, let an be a bounded sequence of complex numbers and H a Hilbert space with ONB en . (4) Show that there is a (necessarily unique) bounded operator A B(H ) satisfying Aen = an en+1 for every n. Such an operator A is called a unilateral weighted shift .
{ }
{ }
∈
Solution. This operator is just the composition of the multiplication oper-
ator A determined by an with the unilateral shift S . These are both bounded by previous work.
{ }
(5) Let A B(H ) be a weighted shift as above. Show that for every complex number λ with λ = 1 there is a unitary operator U λ B(H ) such that U AU −1 = λA.
∈
||
∈
Solution. Define U by U en = λn en . It’s easy to see that this is unitary and
that the desired conjugation relation holds.
(6) Deduce that the spectrum of a weighted shift must be the union of (possibly degenerate) concentric circles about z = 0. Solution. For any µ, λ
∈ C with |λ| = 1, an operator T is the inverse for
A − µ1 iff U T U −1 is the inverse for λA − µ1. Hence µ ∈ ρ(A) iff µ/λ ∈ ρ(A). It follows that ρ(A), and therefore σ(A), is radially symmetric.
12
ARVESON
SPECTRAL THEORY
SOLUTIONS
{ } ∈ ∞ .
(7) Let A be the weighted shift associated with a sequence an (a) Calculate A in terms of an . (b) Assuming that an 0, show that An 1/n 0.
{ }
→
→
Solution.
| |
(a) We have A = sup an . Clearly it is at least this, by considering only its action on the ONB; on the other hand, it’s at most this, because of the factorization A = SA mentioned above. (b) Because each An maps the ONB onto an orthonormal set, Parseval’s identity implies that the norm may be calculated as
An = sup An ek . k Now
k+n 1
n
A ek =
−
aj
ek+n
j=k
so that
k+n 1
−
Anek 1/n = sup k
1/n
aj
.
j=k
| |
This approaches 0. Explicitly, given > 0, choose N such that an < /2 for n N ; let M be the product of all a1 , . . . aN with absolute value greater than 1, if there are any; choose N 2 such that M (/2)N 2 −N < ; then the geometric mean of any N 2 consecutive terms of the sequence an is less than . For amusement, one could apply AM-GM and do the above epsilontics with arithmetic rather than geometric means.
≥
{ }
1.7. Spectral Radius. (1) Let an be a sequence with an shift operator has spectrum 0 .
{ }
{}
→ 0.
Show that the associated weighted
Solution. This follows from exercise 1.6#7(b) and the spectral radius for-
mula.
⊆ [0, 1]n defined by ∆n = {(x1 , . . . , xn ) ∈ [0, 1]n : x1 ≤ x2 ≤ · · · ≤ xn }.
(2) Consider the simplex ∆ n
1 n! .
Show that the volume of ∆ n is
First Solution. We have the explicit expression
1
|∆n| =
xn
xn
1
−
x2
2
−
...
0
0
0
1dx1 dx2 . . . d xn .
0
One can see inductively that performing the first k integrals yields the xk
integrand k+1 k! , so that one ends up integrating 1 an answer of n! .
1 xn n (n 1)! −
−
from 0 to 1, yielding
ARVESON
SPECTRAL THEORY
SOLUTIONS
13
Second Solution. We proceed by induction. First, let us define more general
regions
∈ [0, r]n : x1 ≤ ·· · ≤ xn ≤ r} for each r ≥ 0. By homothety, |∆n,r | = r n |∆n |. 1 Now the base case |∆1 | = 1! is clear; for the induction, we note that 1 1 |∆n| . |∆n+1| = 1dV n+1 = 1dV n dxn+1 = xnn+1 |∆n | dxn+1 = n+1 ∆n,r = (x1 , . . . , xn )
{
∆n+1
0
∆n,xn+1
0
Third Solution. We note that the permutation group S n acts isometrically
on the cube [0, 1]n . Also,
π(∆n ) = [0, 1]n .
π S n
∈
Finally, the images π(∆n ) are almost disjoint. For this, it suffices to show ∆n is almost disjoint from π(∆n ) for π = id. Referring to the cycle decomposition of π, if (a1 a2 . . . ) is a cycle in standard form (i.e. smallest element first), then
∆n
∩ π(∆n) ⊆ {(x1, . . . , xn) ∈ [0, 1]n : xa
1
= xa2
}
which has volume zero. Thus, [0, 1]n is the union of n! almost-disjoint regions of equal volume, 1 so each must have volume n! . (3) Let k(x, y) be a Volterra kernel as in Example 1.1.4, and let K be its corresponding integral operator on the Banach space C ([0, 1]). Estimate the norms K n by showing that there is a positive constant M such that for every f C ([0, 1]) and every n = 1, 2, . . . , M n n K f f . n! Solution. By induction,
∈
n
(K f )(t) =
≤
k(t, xn )k(xn , xn+1 ) . . . k(x2 , x1 )f (x1 )dx1 . . . d xn
∆n,t
so that n
|(K f )(t)
|≤
|k(t, xn)| . . . |k(x2, x1)||f (x1)|dx1 . . . d xn ∆ where M = k C ([0,1] ) . Hence, K n ≤ M n! . n,t
2 n
≤
∆n,t
n
(4) Let K be a Volterra operator as in the preceding exercise. Show that for every complex number λ = 0 and every g C ([0, 1]), the Volterra equation of the second kind Kf λf = g has a unique solution f C ([0, 1]).
∈
− ∈ Solution. Since (n!)1/n → ∞ (one can compare n! to any fixed an to get this result, or use the fact that (n!)n → e), the previous exercise implies that the spectral radius of K is 0, so that the spectrum is {0}. Hence 0. K − λ1 is invertible for every λ = 1/n
n
n
M f t M f ≤ M f dx1 . . . d xn = n! n! n
14
ARVESON
SPECTRAL THEORY
SOLUTIONS
1.8. Ideals and Quotients. (1) Let V and W be finite-dimensional vector spaces over C and let T : V W be a linear map satisfying T V = W , and having kernel K = x V : T x = 0 . Then we have a short exact sequence of vector spaces
{ ∈
}
0
→
−→ K −→ V −→ W −→ 0.
Show that dim V = dim K + dim W .
{
Solution. Let k1 , . . . , kn
}
{
}
be a basis for K and w1 , . . . , wm a basis for W . For each j = 1, . . . , m let vj V be a vector such that T vj = wj . We will show that B = k1 , . . . , kn , v1 , . . . , vm is a basis for V . It is linearly independent because, if
∈
{
}
· ·· + ankn + b1v1 + ·· · + bmvm = 0, then, applying T , we have b1 w1 + · ·· + bm wm = 0, which by the independence of w1 , . . . , wm implies that b1 = · · · = bm = 0; we therefore have a1 k1 + · · · + an kn = 0, which by the independence of k1 , . . . , kn implies a1 = · ·· = an = 0. Furthermore, B spans V , because if v ∈ V , there exist scalars c1 , . . . , cm such that T v = c1 w1 + · ·· + cm wm ; but then v − (c1 v1 + ·· · +cm vm ) ∈ K , so that v − (c1 v1 + ·· · + cm vm ) = d1 k1 + ·· · + dn kn for some scalars d1 , . . . , dn . a1 k1 +
We thus have
v = d1 k1 +
· ·· + dnkn + c1v1 + · · · + cmvm.
(2) For n = 1, 2, . . . let V 1 , V 2 , . . . , Vn be finite-dimensional vector spaces and set V 0 = V n+1 = 0. Suppose that for k = 0, . . . , n we have a linear map T k : V k V k+1 such that
→
T T T T T → V 1 → V 2 → . . . → V n → 0 n is exact. Show that k=1 (−1)k dim V k = 0. Solution. By exactness, R(T k−1 ) = ker T k , so that dim R(T k−1 ) = dim ker T k . By the Rank Theorem, dim ker T k +dim R(T k ) = dim V k . Substituting the
0
0
1
2
n−1
n
former equation into the latter,
dim V k = dim (T k−1 ) + dim
R
R(T k ).
Thus n
−
n
k
( 1) dim V k =
k=1
−
( 1)k [dim
k=1
R(T k−1)+dim R(T k )] = − dim R(T 0)+(−1)n dim R(T n)
since the sum telescopes. But both T 0 and T n have 0-dimensional range. Hence the sum is zero. (3) Show that every normed linear space E has a basis consisting of unit vectors, and deduce that every infinite-dimensional normed linear space has a discontinuous linear functional f : E C.
→
ARVESON
SPECTRAL THEORY
SOLUTIONS
15
Solution. Any basis can be scaled to a unit basis. Alternatively, one can
adapt the standard proof of the existence of a basis to find a maximal set of independent unit vectors. If eα is a unit basis of the infinite-dimensional normed space E , let en be a countable subset. Define f (en ) = n and f (eα ) = 0 for all other α. Then f is discontinuous.
{ }
{ }
(4) Let A be a complex algebra and let I be a proper ideal of A. Show that I is a maximal ideal iff the quotient algebra A/I is simple. Solution. Suppose A/I is not simple, so that it has a nontrivial ideal J .
Then π −1 (J ) is an ideal strictly between A and I , where π : A A/I is the canonical quotient map, so that I is not maximal. Conversely, suppose I is not maximal, and let J be a nontrivial ideal of A containing I . Then π(J ) is a nontrivial ideal of A/I , so that A/I is not simple.
→
(5) Let A be a unital Banach algebra, let n be a positive integer, and let ω :A M n be a homomorphism of complex algebras such that ω(A) = M n . Show that ω is continuous. Deduce that every multiplicative linear C is continuous. functional f : A
→
→
Solution. Since M n is simple, the previous problem implies that the kernel
of ω would be a maximal ideal in A, and therefore closed. But a homomorphism is continuous precisely when its kernel is closed (indeed, when it is, then the homomorphism factors into the contractive quotient map followed by an isomorphism of finite-dimensional algebras), so ω is continuous. The final comment is simply the case n = 1. 1.9. Commutative Banach Algebras. (1) Show that if A is nontrivial in the sense that A = 0 (equivalently, 1 = 0), one has sp(A) = 0 .
{}
{}
Solution. Apply Theorem 1.9.5 to the element 1.
→
(2) Show that ω ker ω is a bijection of the Gelfand spectrum onto the set of maximal ideals in A. Solution. This maps into the set of maximal ideals as claimed, because
ker ω has codimension one. It is injective because if ker ω1 = ker ω2 then ker(ω1 ω2 ) includes both ker ω1 and 1 and hence is all of A. Finally, it is surjective, because if I is a maximal ideal, then A/I is a division algebra and hence isomorphic to C by Gelfand-Mazur; hence the quotient map A A/I has the form a ω(a) + I for some complex homomorphism ω, from which it follows that I = ker ω.
−
→
→
(3) Show that the Gelfand map is an isometry iff x2 = x
2 for every x ∈ A.
2xˆ2 = 2xˆ2; hence, if the Gelfand map is an isometry, one must have x = x . Conversely, suppose x2 = x2 . By induction, this implies that x2 = x2 for each n ∈ N. The spectral radius then implies that x = r(x). But one always has r(x) = x ˆ as well, so that x = x ˆ . Solution. One always has
n
n
16
ARVESON
SPECTRAL THEORY
SOLUTIONS
(4) The radical of A is the set of quasinilpotent elements rad(A) =
∈ x
1/n = 0 n→∞
A : lim xn
.
Show that rad(A) is a closed ideal in A with the property that A/rad(A) has no nonzero quasinilpotents (such a commutative Banach algebra is called semisimple ). Solution. The quasinilpotent elements are precisely those with spectrum
{0}, by the spectral radius formula. The set of all such is a closed ideal because it is the kernel of the contractive homomorphism Γ : A → C (sp(A)). Hence the induced map A/rad(A) → C (sp(A)) is injective; a quasinilpo-
tent element of A/rad(A) would be in the kernel and therefore would be zero. (5) Let A and B be commutative unital Banach algebras and let θ : A B be a unital algebra homomorphism. ˆ (a) Show that θ induces a continuous map θˆ : sp(B) sp(A) by θ(ω) = ω θ. (b) Assuming that B is semisimple, show that θ is bounded. (c) Deduce that every automorphism of a commutative unital semisimple Banach algebra is a topological automorphism.
→
→
◦
Solution.
(a) Clearly θˆ maps sp(B) to sp(A) as claimed. For continuity, suppose ων ω in sp(B), which means ων (b) ω(b) for every b B. Then
→
→
ˆ ν )(a) = ων (θ(a)) θ(ω
∈
ˆ → ω(θ(a)) = θ(ω)(a) ˆ ν ) → θ(ω) ˆ for every a ∈ A, so that θ(ω in sp(A). (b) Suppose (aν , θ(aν )) → (a, b). Then for every ω ∈ sp(B), ω(θ(aν )) → ω(b). On the other hand, because θˆ is continuous, we have ˆ ω(θ(aν )) = θ(ω)(a ν )
ˆ → θ(ω)(a) = ω(θ(a)). It follows that ω(θ(a)) = ω(b) for every ω ∈ sp(B); by Theorem 1.9.5, this means θ(a) − b has spectrum {0}, hence is quasinilpotent. By
semisimplicity, θ(a) = b. By the Closed Graph Theorem, θ is continuous. (c) This is a special case of the previous statement with A = B.
1.10. Examples: C (X ) and the Wiener Algebra. ¯ ) which are represented by convergent (1) Let be the space of functions in C (D power series
B
f (z) =
| |
∞
an z n
n=0
∞
with an < . Prove the following analogue of Wiener’s theorem: If f ¯ , then 1 f (z) = 0 for every z D . f
∈
∈ B
∈ B satisfies
ARVESON
SPECTRAL THEORY
SOLUTIONS
17
B
Solution. First, we show that the maximal ideal space of (also known as the disk algebra A(D)) is homeomorphic to D. Clearly point evaluations
are MLF’s on . For the converse, suppose ω sp( ) ; let z0 = ω(id). D. By linearity and multiplicativity, ω agrees with Since ω = 1, z0 evaluation at z0 on all polynomials; but the latter are dense in , so ω is evaluation at z0 . ¯ Also, Now the Gelfand transform is just the inclusion map into C (D). the Gelfand transform preserves invertibility, by Theorem 1.9.5. If f , ¯ then Γ(f ) has an inverse in C (D), and therefore f has an inverse in .
B
∈
∈
B
B
∈ B B
(2) Let Z+ denote the semigroup of nonnegative integers. Let T be the right shift on 1 (Z+ ). Let a = (a0 , a1 , . . . ) 1 (Z) . Show that the set of translates a,Ta,T 2 a , . . . spans 1 (Z+ ) iff the power series
{
∈
}
f (z) =
∞
an z n ,
n=0
|z| ≤ 1
has no zeros in the closed unit disk. Solution. Note that 1 (Z+ ) is a Banach algebra; indeed, it is a closed subalgebra of 1 (Z).
A calculation like that on page 30 shows that the maximal ideal space ¯ , with the Gelfand transform being the of 1 (Z+ ) is homeomorphic to D n Fourier transform a an z . By the previous exercise, f has no zeros ¯ which happens iff a is invertible in in the disk iff f is invertible in C (D), 1 (Z+ ) since the Gelfand transform preserves invertibility. Finally, note that the convolution of a with another element of 1 consists of a (convergent infinite) sum of translates of a. Hence the span of such translates includes the delta function; but, being translation-invariant, it therefore includes translates of the delta function, hence everything.
→
1.11. Spectral Permanence Theorem. (1) Let A be a unital Banach algebra, let x A, and let Ω∞ be the unbounded component of C σA (x). Show that for every λ Ω∞ there is a sequence of polynomials p1 , p2 , . . . such that
∈
\
∈
−1 →∞ (x − λ1) − pn (x) = 0.
lim
n
Solution. Let B be the unital subalgebra generated by x. Then σB (x) has
∈
the same unbounded component as σA (x); in particular, λ / σB (x), so that x λ1 has an inverse in B; but this inverse must be a norm limit of polynomials, because everything in B is.
−
(2) Let A be a unital Banach algebra that is generated by x A. Show that σA (x) has no holes.
∈
{1, x} for some
Solution. Let K denote the “filling in” of σA (x); that is, K is the comple-
ment of the unbounded component of the complement. We wish to show that K = σA (x). By Theorem 1.9.5, it suffices to show that for each ζ K there is a multiplicative linear function ω on A with ω(x) = ζ . Given ζ , define ω on polynomials p(x) by ω( p(x)) = p(ζ ); this is contractive because, given a polynomial p, the Maximum Modulus Principle implies that there
∈
18
ARVESON
∈
SPECTRAL THEORY
SOLUTIONS
|
| ≤| | | |≤ ≤ |ω( p(x))| = | p(ζ )| ≤ | p(y)| ≤ p(x)
exists y ∂K with p(ζ ) p(y) . But by the Spectral Mapping Theorem, p(y) σA ( p(x)), so that p(y) r( p(x)) p(x) . We thus have
∈
so that ω is contractive. We can therefore extend ω to a multiplicative linear functional on the closure of the set of all polynomials in x; but this is all of A. C be a compact set (3) Deduce the following theorem of Runge: Let X whose complement is connected. Show that if f (z) = p(z)/q (z) is a rational function with q (z) = 0 for z X , then there is a sequence of polynomials f 1 , f 2 , . . . such that
⊆
∈
|
sup f (z) z X
∈
− f n(z)| → 0.
A be the unital subalgebra of C (K ) generated by the identity function—that is, the algebra of norm limits of polynomials. We wish to show that f ∈ A. Clearly it suffices to show that q is invertible in A. Solution. Let
Now σC (K ) (id) = K , which has no holes, so that σA (id) = K as well. By Spectral Mapping, σA (q ) = σA (q (id)) = q (K ), which by hypothesis does not contain 0. Hence q is invertible in . I’m not sure why the previous problem is needed for this; Corollary 2 is enough.
A
1.12. Brief on the Analytic Functional Calculus. Exercise 1.12.1: Let C be an oriented curve in C, f a continuous function from C to a Banach space E , and the set of finite ordered partitions of C . (1) Show that for every > 0 there is a δ > 0 with the property that for every pair of oriented partitions 1 , 2 satisfying δ for k k = 1, 2, one has R(f, 1 ) R(f, 2 ) . (2) Verify the estimate
P
C
P P P ≤
P −
≤
f (λ) dλ
P ≤
| | ≤ λsup f (λ)(C ). ∈C
f (λ) d λ
C
Solution: (1) Since f is a continuous function with compact domain C , it is uniformly continuous. Thus, given > 0, we may choose δ > 0 such that z w < δ implies f (z) f (w) < 2(C C . ) for any z, w Suppose 1 and 2 are two partitions with 1 < δ and 2 < δ . Let 3 = 1 denote 2 be the common refinement. Let γ 0 , . . . , γN the points in 1 . Let k 0, . . . , N 1 . Suppose 3 contains the additional points η1 , . . . , ηm between γ k and γ k+1 . (It is possible that m = 0, i.e. that there are no additional points.) We consider the partial Riemann sums
| − | | − P P P P ∪ P P ∈{ − }
|
P
∈
P
P
P 1) := f (γ k+1)(γ k+1 − γ k ) m−1 Rk (f, P 3 ) := f (η1 )(η1 − γ k ) + f (ηj+1 )(ηj+1 − ηj ) + f (γ k+1 )(γ k+1 − ηm ) Rk (f,
j=1
ARVESON
SOLUTIONS
SPECTRAL THEORY
19
for the portion C k of C between γ k and γ k+1 . (In the case m = 0 we have instead Rk (f, 3 ) = Rk (f, 1 ).) Note that we can expand
P
Rk (f,
P 1) = f (γ k+1)
P
m 1
(η1
−
− γ k ) +
(ηj+1
j=1
− ηj ) + (γ k+1 − ηm)
.
Then
|Rk (f, P 3) − Rk (f,
P |
1 ) = f (η1 )
− f (γ k )
| m 1
(η1
− γ k )+
−
f (ηj+1 )
j=1
+ f (γ k+1 )
− f (γ k )
− f (γ k )
m 1
≤|f (η1) − f (γ k )||η1 − γ k | ≤ ≤
−
(ηj+1
(γ k+1
− ηj
− ηm)
− f (γ k )||ηj+1 − ηj | j=1 +|f (γ k+1 ) − f (γ k )||γ k+1 − ηm | m−1 |η1 − γ k | + |ηj+1 − ηj | + |γ k+1 − γ k | +
f (ηj+1 )
2(C ) (C k ). 2(C )
j=1
Summing over k, we have
|R(f, P 3) − R(f,
P | ≤ | 1)
=
Rk (f,
k
≤ 2(C )
P − 3)
Rk (f,
k
Rk (f,
k
P 1)
P 3) − Rk (f, P 1)| (C k ) =
k
The same calculation holds with Inequality, it follows that
. 2
P 2 in place of P 1.
By the Triangle
|R(f, P 1) − R(f, P 2)| ≤ . (2) For every partition P one has by the Triangle Inequality R(f, P ) ≤ f (γ k+1)|γ k+1 − γ k | = R(f , P )
k
where the right-hand side denotes the Riemann sum of the function R with respect to d λ . Taking the limit of both sides as f : C 0, we have C f (λ) dλ f (λ) d λ . For the second half, C if M = supλ∈C f (λ) , then f (λ) M at each point λ, so that
→ P →
C
|| ≤ | | ≤ f (λ)d|λ| ≤ M d|λ| = M (C ).
C
Notation: For exercise 2-4, let E be a Banach space and T
∈ B (E ).
20
ARVESON
SPECTRAL THEORY
SOLUTIONS
C : z < R be an open disc containing Exercise 1.12.2: Let D = z C be an analytic function defined on D with power σ(T ). Let f : D series
{ ∈
→
∞
||
f (z) =
cn z n ,
}
∈ D.
z
n=0
Show that the infinite series of operators
∞
cn T n
n=0
| | | |
converges absolutely in the sense that n cn T n < . ˜ with r(T ) < R ˜ < R. The spectral radius formula Solution: Choose some R ˜ n for sufficiently large n, say n N , so implies that T n < R
| |
˜n cn R
cn T n <
n N
≥
∞ ≥
n N
≥
which converges because power series converge absolutely on proper subdiscs. Exercise 1.12.3: Give a definition of sin T and cos T using power series. Solution: We define sin T =
∞
k=0
( 1)k T 2k+1 , (2k + 1)!
−
cos T =
∞ ( 1)k
− k=0
(2k)!
T 2k .
By the previous exercise, these series are absolutely convergent for any T , since the complex-valued functions sin and cos are entire. Exercise 1.12.4: Use your definitions in the preceding exercise to show that (sin T )2 + (cos T )2 = 1. Solution: This is immediate from the holomorphic functional calculus, which sends the complex function sin 2 z + cos2 z 1 to 0. Alternatively, one can prove it directly by manipulating the above series: Since
−
2
∞
j,k=0
and 2
∞
− − − − − − − −− −− − −
sin T =
cos T =
( 1)j+k T 2(j+k+1) = (2k + 1)!(2 j + 1)!
−
∞ ( 1)j+k
j,k=0
(2k)!(2 j)!
2(j+k)
T
= 1+
( 1)−1 T
s odd
=1
∞
( 1) T 2
=1
1 s!(2 s)!
−
1 , s!(2 s)! s even
−
we have
sin2 T + cos2 T = 1 +
∞
1 s!(2 s)! s even
( 1) T 2
=1
= 1+
∞
2
2
( 1) T
s=0
=1
s odd
1 s!(2 s)!
−
( 1)s . s!(2 s)!
But, by the Binomial Theorem, 2
s=0
( 1)s 1 = s!(2 s)! (2)!
2
( 1)s
s=0
2 s
=
1 (1 (2)!
1)2 = 0,
ARVESON
SPECTRAL THEORY
SOLUTIONS
21
so that sin2 T + cos2 T = 1. 2.
Operators on Hilbert Space
2.1. Operators and Their C ∗ -Algebras. C be a sesquilinear form defined on a Exercise 2.1.1: Let [ , ] : H H Hilbert space H . Show that [ , ] satisfies the polarization formula
··
3
4[ξ, η] =
× → ··
ik ξ + ik η, ξ + ik η .
k=0
Solution: We expand the right-hand side using sesquilinearity:
− −
3
3
ik ξ + ik η, ξ + ik η =
k=0
ik [ξ, ξ ] + ik [η, ξ ] + ( i)k [ξ, η] + [η, η]
k=0 3
3
k
=
[ξ, ξ ] +
i
k=0
3
( 1)
k
[η, ξ ] +
k=0
3
ik
[ξ, η] +
k=0
k=0
= 4[ξ, η]
Exercise 2.1.2: Let A (H ) be a Hilbert space operator. The quadratic C defined by q A (ξ ) = Aξ,ξ . The form of A is the function q A : H numerical range and numerical radius of A are defined, respectively, by
∈ B
→
} ⊆ C, w(A) = sup{|q A (ξ )| : ξ = 1 }. {
W (A) = q A (ξ ) : ξ = 1
(a) Show that A is self-adjoint iff q A is real-valued. (b) Show that w(A) A 2w(A) and deduce that q A = q B only when A = B. Solution: (a) We have
≤ ≤
∀ξ : Aξ,ξ ∈ R ⇔ ∀ξ : Aξ,ξ = Aξ,ξ ⇔ ∀ξ : Aξ,ξ = ξ,Aξ ⇔ ∀ξ : Aξ,ξ = A∗ ξ, ξ ⇔ ∀ξ, η : Aξ,η = A∗ξ, η ⇔ A = A∗
The penultimate equivalence uses the polarization identity from the previous exercise, which implies that the values of the sesquilinear form (ξ, η) Aξ,η are determined by the values of the quadratic form ξ Aξ,ξ . (b) By Cauchy-Schwarz, q A (ξ ) Aξ ξ A ξ 2 for all ξ , so that w(A) A . For the other direction, the polarization formula implies that
→ → ≤
A =
sup
η≤1 ξ≤1
|Aξ,η| ≤
|
1 sup η≤1 4
ξ≤1
| ≤ ≤
3
|
q A (ξ + ik η)
k=0
|≤
sup
χ≤√ 2
|q A(χ)| = 2w(A).
Finally, observe that q A+B = q A + q B . Then
A − B = 0 ⇔ w(A − B) = 0 ⇔ q A−B = 0 ⇔ q A − q B = 0 ⇔ q A = q B .
[η, η]
22
ARVESON
SPECTRAL THEORY
SOLUTIONS
Exercise 2.1.3: Show that the adjoint operator A A∗ in continuous but not strongly continuous. Solution: If T ν 0 in WOT, then for any ξ, η H ,
→
B (H ) is weakly
→ ∈ T ν ∗ ξ, η = ξ, T ν η = T ν η, ξ → 0,
so that T ν ∗ 0 in WOT. Hence adjunction is WOT-continuous. However, if S is the right shift on 2 , then S ∗n 0 in SOT, whereas n S 0 in SOT, so that adjunction is not SOT-continuous. Exercise 2.1.4: Show that the only operators that commute with all operators in (H ) are the scalar multiples of the identity. Solution: For each ξ, η H let ξ η ∗ denote the rank-one operator given by (ξ η ∗ )(y) = y, η ξ . If an operator T commutes with all rank-one operators, then for any ξ and η,
→
→
→
B ⊗
∈
⊗
η2(T ξ ) = T ◦ (ξ ⊗ η∗ )(η) = [(ξ ⊗ η∗ ) ◦ T ](η) = T η , ηξ. Note that this implies that every ξ is an eigenvector of T . Let cξ denote the corresponding eigenvalue. Returning to the above equation with T ξ = cξ ξ and T η = cη η, we have
η2cξ ξ = cη η2ξ, so that cξ = cη . That is, all vectors have the same eigenvalue. Thus T is a scalar times the identity. Exercise 2.1.5: Let be the closure in the strong operator topology of the set of all unitary operators in (H ). Show that consists of isometries. Solution: Suppose U ν are unitaries and S an operator such that U ν S in SOT. Then for any ξ H ,
C
B
C
→
∈ Sξ = lim U ν ξ = lim U ν ξ = ξ ν ν since U ν ξ = ξ for all ν . Thus S is an isometry.
Note that we never used the fact that U ν are actually unitaries, not isometries; this proof shows that the set of isometries is SOT-closed. Exercise 2.1.6: Show that the unilateral shift S belongs to be exhibiting a sequence of unitary operators U 1 , U 2 , . . . that converges to S in the strong operator topology. Solution: Define unitaries U n on the standard ONB by
C
U n ek =
ek+1 e1 ek
1 k n k=n k > n.
≤ ≤ −1
This is unitary because it maps an ONB to an ONB (it permutes the standard basis). Fix x 2 , and for each n, let xn be the projection of x onto the span of (e1 , . . . , en−1 ). Then U n xn = Sx n , so that
∈
U nx − Sx = U n[xn + (x − xn)] − S [xn + (x − xn)] = U nxn − Sx n + (U n − S )(x − xn) = (U n − S )(x − xn ) ≤ U n − S x − xn ≤ 2x − xn → 0. Hence U n → S in SOT.
ARVESON
SPECTRAL THEORY
SOLUTIONS
23
Exercise 2.1.7: Let (X, µ) be a σ-finite measure space and let f : X C be a bounded complex-valued Borel function. Show that the essential range of f can be characterized as the intersection
→
{
g(X ) : g
∼ f }
of all the closed ranges of all bounded Borel functions g : X C that agree with f almost everywhere (dµ). Solution: Suppose λ is in the essential range of f , and g = f a.e. Then for every n N, g −1 (B(λ, 1/n)) has positive measure; in particular, it is nonempty, so that g(X ) contains points in B(λ, 1/n) for every n. Thus λ is in the closure of g(X ). Conversely, suppose λ is not in the essential range of f . Let r > 0 such that f −1 (B(λ, r)) has measure zero. Let µ C B(λ, r). Define g : X C by
→
∈
∈ \
g(x) =
→
f (x) f (x) / B(λ, r) µ else.
∈
∈
Then g agrees with f a.e., and λ / g(X ). 2.2. Commutative C ∗ -Algebras. Exercise 2.2.1: Let A be a unital Banach algebra and a0 , a1 , . . . and b0 , b1 , . . . be two sequences of elements of A such that an < and bn < , and let x = an , y = bn . Prove that the product xy is given by the series xy = cn , where
cn = a0 bn + a1 bn−1 +
∞
· · · + an b 0 ,
the series cn being absolutely convergent in the sense that Solution: Let us introduce the notation
∞
∞.
cn <
xn = a0 + yn = b0 +
· · · + an · ·· + bn X n = a0 + ·· · + an Y n = b0 + · · · + bn n
S n =
ck
k=0
n
An =
aj
bk =
j+k n
ck .
k=0
≤
Then An X n Y n , so that An is bounded above and hence converges, i.e. ck converges absolutely. Now
≤
{ }
xnyn − S n =
j,k n j+k>n
≤
aj bk
≤
A2n
− An
24
ARVESON
SPECTRAL THEORY
SOLUTIONS
{ }
which tends to zero since An is a convergent sequence. Thus,
∞
ck = lim S n = lim xn yn = xy. n
k=0
n
→∞
→∞
We note that the key estimate above, using A2n An as an upper bound, can be pictured as a relationship between squares and right triangles on a quarter-infinite lattice. If S n indicates an n n square and T n an n n right triangle, then T n S n T 2n whereas S n T 2n S 2n . Here the points of the lattice represent pairs ( j, k) over which one sums aj bk . Exercise 2.2.2: Let A be a C ∗ -algebra. (a) Show that the involution in A satisfies x∗ = x . (b) Show that if A contains a unit 1, then 1 = 1. Solution: (a) Obviously this is true if x = 0; for nonzero x, divide the inequality x 2 = x∗ x x∗ x by x to obtain x x∗ . Applying the same result to x∗ yields x∗ x , from which it follows that ∗ x = x . (b) For any x ,
−
⊆
×
⊆
⊆
×
⊆
≤ ≤ ∈A
≤
1∗ x∗ = (x1)∗ = x∗ = (1x)∗ = x∗ 1∗
so that 1∗ is a multiplicative identity; hence 1∗ = 1. Then 1 = 1∗ 1 = 1 2 and, since 1 = 0, we can divide both sides by 1 to obtain 1 = 1. Notation: In the following exercises, X and Y denote compact Hausdorff spaces, and θ : C (X ) C (Y ) denotes an isomorphism of complex algebras. We do not assume continuity of θ. Exercise 2.2.3: Let p Y . Show that there is a unique point q X such that (θf )( p) = f (q ) for all f C (X ). Solution: Let = f C (Y ) : f (q ) = 0 , which is a maximal ideal in C (Y ) corresponding to the MLF of evaluation at q . Since θ is an isomorphism, = θ −1 is a maximal ideal in C (X ). As proved in section 1.10, every maximal ideal in C (X ) is the zero set of a unique point p X . Given f C (Y ), let λ = f (q ). Then f λ1 , so that θ−1 (f ) λ1 . Hence (θ−1 f )( p) λ = 0, so (θ−1 f )( p) = λ. Thus p has the desired property. If x X with x = p, then there is some g with g(x) = 0, by the uniqueness result from chapter 1.10 (or by Urysohn’s lemma). Since (θg)(q ) = 0, this x does not have the desired property. Hence p is unique. Exercise 2.2.4: Show that there is a homeomorphism φ : Y X such that θf = f φ. Solution: The previous exercise gives us a bijection φ : Y X with this property, so we need only show φ is continuous. (This will imply it’s a homeomorphism, since it’s a continuous bijection from a compact space to a Hausdorff space.) Suppose not; then there is some closed F X such ¯ G and let x = φ(y). that G = φ−1 (F ) is not closed in Y . Let y G By Urysohn’s lemma, there exists f C (X ) such that f (x) = 1 and f ¯ while also (θf )(y) = 1, vanishes on F . But then θf = 0 on G, hence on G, a contradiction. Hence φ is continuous.
→ ∈ I { ∈
J
∈
∈
}
I
•
∈
J
•
∈
− ∈ I
−
∈
∈ J
− ∈
→ →
◦
∈
∈ \
⊆
ARVESON
SPECTRAL THEORY
SOLUTIONS
25
Exercise 2.2.5: Conclude that θ is necessarily a self-adjoint linear map in the sense that θ(f ∗ ) = θ(f )∗ , f C (X ). Solution: This is immediate since (f φ)∗ = f ∗ φ. Exercise 2.2.6: Formulate and prove a theorem that characterizes unital algebra homomorphisms θ : C (X ) C (Y ) in terms of certain maps φ : Y X . Which maps φ give rise to isomorphisms? Solution: Theorem: For every unital algebra homomorphism θ : C (X ) C (Y ) there exists a unique continuous function Θ : Y X such that θ = Θ (that is, for all φ C (X ), θ(φ) = φ Θ). Moreover, θ is (injective, surjective) iff Θ is (surjective, injective). In particular, θ is an isomorphism iff Θ is a homeomorphism. Proof: Let Ψ X : X Σ(C (X )) be the canonical homeomorphism x (φ φ(x)), and similarly with ΨY . Define θˆ : Σ(C (Y )) Σ(C (X )) by ˆ ˆ θ= θ, that is, for all ω Σ(C (Y )) we have θ(ω) = ω θ, which in turn ˆ means that for all φ C (X ) we have θ(ω)(φ) = ω(θ(φ)). Define Θ : Y X − 1 ˆ by Θ = ΨX θ ΨY . 1 ˆ Let φ C (X ). By definition of Θ, we have φ Θ = φ Ψ− θ ΨY . In X case it’s not immediately obvious that this equals θ(φ), let’s slowly unwind the definitions: Let y Y . ΨY (y) is the multiplicative linear functional on C (Y ) given by ψ ψ(y). ˆ Y (y)) is the multiplicative linear functional on C (X ) given by ψ θ(Ψ θ(ψ)(y). 1 ˆ Ψ− X such that, for all ψ C (X ), X (θ(ΨY (y))) is the unique point x ˆ ψ(x) = [θ(ΨY (y))](ψ). 1 ˆ ˆ φ(Ψ− X (θ(ΨY (y)))) is, therefore, equal to [ θ(ΨY (y))](φ). ˆ Y (y)), this is θ(φ)(y). By the definition of θ(Ψ Thus, θ(φ) = φ Θ as claimed. This proves the existence of Θ; for uniqueness, suppose there were a different map Ξ with the same properties. Let y Y such that Θ(y) = Ξ(y). By Urysohn’s lemma, there exists φ C (X ) with φ(Θ(y)) = φ(Ξ(y)). Hence Θ is unique. The last assertions are a consequence of the fact that surjectivity and injectivity correspond to left and right cancellations, and that the correspondence between θ and Θ is given by a contravariant functor. Thus, if Z is any other compact Hausdorff space and λ1 , λ2 : C (Z ) C (X ) unital homomorphisms, corresponding to continuous maps Λ 1 , Λ2 : X Z via λi = Λi , then (λi θ) = (Θ Λi ). Therefore, if Θ is right-cancellative, in that Θ Λ1 = Θ Λ2 implies Λ1 = Λ2 , then θ is left-cancellative, meaning that λ1 θ = λ2 θ implies λ1 = λ2 . For the other direction (leftcancellativity of Θ corresponds to right-cancellativity of θ) we note that Θ ˆ and repeat the above argument. has the same cancellativity properties as θ, In more concrete terms, however, suppose Θ is not surjective. Let x be in the complement of the range, which is open; by Urysohn, there exists φ C (X ) which is zero on the range of Θ and such that φ(x) = 1. Then θ(φ) = φ Θ = 0, proving that θ is not injective. Conversely, if θ is not
∈
◦
→
→
·◦
∈
◦
◦ ◦
∈
∈
∈
◦
◦
→
→
→
◦
◦ ◦
∈
→ →
∈
• •
∈
◦
∈
∈
→
·◦
∈
→
→
→
→ ·◦
• • • •
◦
◦
◦
◦
◦ ◦
·◦ ◦
◦
→
26
ARVESON
SPECTRAL THEORY
SOLUTIONS
∈
◦
injective, let φ C (X ) be nonzero such that φ Θ = 0; then φ is zero on the range of Θ, but is not zero everywhere, so that Θ is not surjective. Similarly, if Θ is not injective, let y1 = y2 such that Θ(y1 ) = Θ(y2 ). Choose ψ C (Y ) with ψ(y1 ) = 1 and ψ(y2 ) = 0. If θ were injective, there would exist φ C (X ) with θ(φ) = ψ. But then
∈
∈
φ(Θ(y1 )) = θ(φ)(y1 ) = ψ(y1 ) = 1 = 0 = ψ(y2 ) = θ(φ)(y2 ) = φ(Θ(y2 ))
which is impossible since Θ(y1 ) = Θ(y2 ). Hence θ is not surjective. Conversely, suppose θ is injective. Then it is a homeomorphism onto its range ˙ , so that θ factors as ι χ where χ : Y ˙ is a homeomorphism and X X ˙ ι : X X is the inclusion map. Correspondingly, we get the factorization ˙ ˙ ) is the map θ˙ = ι, and η : C (X ˙) θ = η θ, where θ˙ : C (X ) C (X C (Y ) is the isomorphism η = χ. Thus θ will be surjective iff θ˙ is; but this is the restriction map, which is surjective by Urysohn’s lemma again (or, if you like, by the Tietze extension theorem). Notation: In the remaining exercises, let H be a Hilbert space and let T B(H )−1 be an invertible operator. Define θ : B(H ) B(H ) by
◦ → ·◦
→ ◦
Θ(A) = T AT −1 ,
→
·◦
A
∈ B(H ).
→
∈
→
Exercise 2.2.7: Show that θ is an automorphism of the Banach algebra structure of B(H ). Solution: Clearly Θ is an algebraic automorphism; it is bounded because
Θ(A) ≤ T AT −1 ⇒ Θ ≤ T T −1, and its inverse A → T −1 AT is bounded in the same way. Exercise 2.2.8: Show that θ(A∗ ) = θ(A)∗ for all A ∈ B(H ) iff T is a scalar multiple of the identity operator. Solution: The “if” direction is clear. Conversely, suppose Θ(A∗ ) = Θ(A)∗ for all A. That is, T A∗ T −1 = (T AT −1 )∗ = T −∗ A∗ T ∗
for all A
This implies T ∗ T A∗ = A∗ T ∗ T
for all A
∈ B(H ).
∈ B(H ),
i.e. that T ∗ T B(H ) = CI . Let r > 0 such that T ∗ T = r2 I . The ∗ equation Θ(T ) = Θ(T )∗ implies T ∗ T = T T ∗ , so that T is normal. Then U = 1r T satisfies
∈
U ∗ U =
1 ∗ 1 T T = I = 2 T T ∗ = U U ∗ 2 r r
and hence is unitary. 2.3. Continuous Functions of Normal Operators. Exercise 2.3.1: Show that the spectrum of a normal operator T (H ) is ∗ 1 connected if and only if the C -algebra generated by T and contains no projections other than 0 and 1. Solution: By the spectral theorem, C ∗ (T ) contains a nontrivial projection iff C (σ(T )) does. But a nontrivial projection in C (σ(T )) is the characteristic function of a nontrivial union of connected components, which can only exist of σ(T ) is disconnected.
∈ B
ARVESON
SPECTRAL THEORY
SOLUTIONS
27
C. Notation: Consider the algebra of all continuous functions f : C There is no natural norm on , but for every compact subset X C there is a seminorm f X = sup f (z) .
C
C
⊆
z X
∈
|
→
|
Exercise 2.3.2: Given a normal operator T (H ), show that there is a natural extension of the functional calculus to a *-homomorphism f f (T ) (H ) that satisfies f (T ) = f σ(T ) . Solution: Given T , we obtain a restriction homomorphism ρT : C (σ(T )) which satisfies ρT (f ) = f σ(T ) . Composing this with the functional calculus yields the result. Exercise 2.3.3 (Continuity of the functional calculus): Fix a function and let T 1 , T 2 , . . . be a sequence of normal operators that converges f in norm to an operator T . Show that f (T n ) converges in norm to f (T ). Solution: First note that T is normal, since norm convergence implies T n∗ T ∗ and therefore T T ∗ T T ∗ = lim T n T n∗ T n∗ T n = 0. Let K C be a compact set containing all the σ(T n ). Since one eventually has r(T n ) = T n T + 1, this is possible. Let P k be a sequence of polynomials converging uniformly to f in C (K ). Given > 0, choose k such that P k f K < . Since P k (T n ) P k (T ) in norm (by norm-continuity of products and sums), there is N N such that P k (T n ) P k (T ) < for n > N . Then for n > N one has
∈ B
∈ B
∈C →
C→
∈C
− ≤
⊆
→
−
−
−
→ ∈
f (T n) − f (T ) = f (T n) − P k (T n) + P k (T n) − P k (T ) + P k (T ) − f (T ) ≤ f (T n) − P k (T n) + P k (T n) − P k (T ) + P k (T ) − f (T ) < 3. Thus f (T n ) → f (T ). 2.4. The Spectral Theorem and Diagonalizations. Exercise 2.4.1: Let X be a Borel space, f a bounded complex-valued Borel function on X , and µ and ν two σ-finite measures on X . The multiplication operator M f defines bounded operators A on L2 (X, µ) and B on L2 (X, ν ). Assuming that µ and ν are mutually absolutely continuous, show that there is a unitary operator W : L2 (X, µ) L2 (X, ν ) such that W A = BW . Solution: By Radon-Nikodym, dν = hdµ for some nonnegative function h. (Nonnegative because L2 spaces are defined with respect to positive measures.) Define W : L2 (X, µ) L2 (X, ν ) by (W φ)(x) = 1 φ(x). Then
→
√ h(x)
→
W A and BW are both M f/ √ h , i.e. (W Aφ)(x) =
f (x) φ(x) = (BW φ)(x) h(x)
for every φ L2 (X, µ). Exercise 2.4.2: Show that every diagonalizable operator on a separable Hilbert space is unitarily equivalent to a multiplication operator M f acting on L2 (X, µ) where (X, µ) is a probability space, that is, a measure space for which µ(X ) = 1. Solution: By definition, if T B( ) is diagonalizable, there is a σ-finite measure space (X, ν ), a unitary U : L2 (X, ν ), and a function φ L∞ (X, ν ) such that such that U T = M φ U . Let X = X i where ν (X i )
∈
∈ H
H→
∈
28
ARVESON
SPECTRAL THEORY
SOLUTIONS
∩
is finite and the X i are disjoint. Define a measure µ on X by µ(E ∩Xi ) X i ) = ν 2(iE ν (Xi ) . Then µ is a probability on X which is mutually absolutely continuous with respect to ν . By the previous exercise, M φ is unitarily equivalent to a multiplication operator on L2 (X, µ), so T is as well. Notation: The following exercises concern the self-adjoint operator A defined on the Hilbert space 2 (Z) by Aξ n = ξ n+1 + ξ n−1 . Exercise 2.4.3: Show that A is diagonalizable by exhibiting an explicit unitary operator W : L2 (T,dθ/2π) H for which W M f = AW , where iθ R is the function f (e ) = 2cos θ. Deduce that the spectrum of A f : T is the interval [ 2, 2] and that the point spectrum of A is empty. Solution: The unitary is just the Plancherel transform, i.e. the operator given by
→
→
−
(W g)(n) = n ˆ=
− 1
f (ζ )ζ −n dζ =
0
T
For any g L2 (T) one then has W g = dard basis for 2 (Z). Hence
∈
AW g =
n
whereas
gˆ(n)[ξ n+1 + ξ n−1 ] =
W M f g =
f (e2πiθ )e−2πinθ dθ.
[ˆg(m
f g(n)ξ n =
{ }
gˆ(n)ξ n where ξ n is the stan1) + gˆ(m + 1)]ξ m
m
n
n
[ˆg(n
n
− 1) + gˆ(n + 1)]ξ n.
Exercise 2.4.4: Let U be the operator defined on L2 (T,dθ/2π) by (U f )(eiθ ) = f (e−θ ). Show that U is a unitary operator that satisfies U 2 = 1 and which commutes with M f . [Typo in the book: says “commutes with A”.] Solution: Clearly U 2 = 1. It’s also clear that U φ = φ , so that U is an isometry; since it’s also invertible, it’s unitary. Finally, for any φ L2 (T),
∈
(U M f φ)(eiθ ) = (M f φ)(e−iθ ) = 2 cos( θ)φ(e−iθ ) = 2 cos(θ)φ(e−iθ ) = (M f U φ)(eiθ )
−
so that U commutes with M f . Exercise 2.4.5: Let be the set of all operators on L2 (T,dθ/2π) that have the form M f + M g U where f, g L∞ (T,dθ/2π) and U is the unitary operator of the preceding exercise. Show that is *-isomorphic to the C ∗ algebra of all 2 2 matrices of functions M 2 ( 0 ), where 0 is the abelian C ∗ -algebra L∞ (X, µ), X being the upper half of the unit circle X = z T : im z 0 and µ being the restriction of the measure dσ = dθ/2π to X . Solution: First, we note that one can recover f and g from the operator M (f ) + M (g)U ; that is, the linear map Φ : L∞ (T) L∞ (T) given by Φ(f g) = M (f ) +M (g)U is injective. This is clear from a consideration of this operator applied to the characteristic functions of the upper and lower semicircles. Next, we introduce notation for some of the relevant maps between spaces:
B
×
∈
B B
B
{ ∈
≥ }
⊕
⊕
→ B
ARVESON
SPECTRAL THEORY
SOLUTIONS
29
• M : L∞(T) → B(L2(T)) is the *-algebra homomorphism that takes f to the multiplication operator by f • P u : L∞(T) → L∞ (X ) and P : L∞(T) → L∞(Y ) are the *-homomorphisms of projection onto the upper and lower semicircles, which we denote by X and Y ; we view L∞ (X ) and L∞ (Y ) as subalgebras of L∞ (T) : L∞ (T) L∞ (T) is the *-homomorphism ( f )(eiθ ) = f (e−iθ ), which satisfies 2 = id. Note that (L∞ (X )) = L∞ (Y ) and conversely. Ψ: M 2 ( 0 ) is the map
• F •
→
B→
F
F
F
B
P u (f ) P u (g) Ψ(M (f ) + M (g)U ) = , P u ( (g)) P u ( (f ))
F
F
which we will prove to be a *-algebra isomorphism. Here are some elementary relations between these maps, all easy to verify: U M (f ) = M ( (f ))U P = P u and Pu = P It is clear that the map Ψ is bijective and linear. It is self-adjoint because
• • F
F
F
F
F
F F F
Ψ [M (f ) + M (g)U ]∗ = Ψ M (f )∗ + U M (g)∗
F F
¯ + M ( (¯g))U = Ψ M (f )
F
=
¯ P u (f ) P u ( (g¯)) ¯ P u ( ( (¯g))) P u ( (f ))
=
P u (f ) P u (g) P u ( (g)) P u ( (f ))
F
∗
and is multiplicative because
Ψ [M (f ) + M (g)U ][M (h) + M (k)U ]
= Ψ M (f )M (h) + M (f )M (k)U + M (g)U M (h) + M (g)U M (k)U
= Ψ M (f )M (h) + M (f )M (k)U + M (g)M ( (h))U + M (g)M ( (k))
F P u (f h + g F (k)) P u (F (f k + g F (h)))
F
F
F
= Ψ M (f h + g (k)) + M (f k + g (h))U = =
F F
P u (f k + g (h)) P u ( (f h + g (k)))
F P u (f )P u (h) + P u (g)P u (F (k)) P u (F (f ))P u (F (k)) + P u (F (g))P u (h)
F
P u (f ) P u (g) P u (h) P u (k) P u ( (g)) P u ( (f )) P u ( (k)) P u ( (h)) = Ψ(M (f ) + M (g)U )Ψ(M (h) + M (k)U ). =
F
F
F
P u (f )P u (k) + P u (g)P u ( (h)) P u ( (f ))P u ( (h)) + P u ( (g))P u (k)
F
F F
F
Notation: The following exercises ask you to compare the operator A to a related operator B that acts on the Hilbert space L2 ([ 2, 2], ν ) where ν is Lebesgue measure on the interval [ 2, 2]. The operator B is defined by (Bf )(x) = xf (x).
−
−
30
ARVESON
SPECTRAL THEORY
SOLUTIONS
−
Exercise 2.4.6: Show that B has spectrum [ 2, 2] and that it has no point spectrum. Deduce that for every f C ([ 2, 2]) we have f (A) = f (B) . Solution: The spectrum of a multiplication operator is its essential range, which in this case is [ 2, 2]. Since A and B are both normal (in fact, selfadjoint) and have the same spectrum, the functional calculi for the two of them together imply that f (A) = f (B) for all f C ([ 2, 2]). I’m not sure why he mentioned the point spectrum, as it’s unnecessary for the above proof. Exercise 2.4.7: Show that A and B are not unitarily equivalent. Solution: First, we show that B = B , where denotes the commutant. Clearly every multiplication operator commutes with B. Conversely, suppose T commutes with B = M x . Let g = T 1. We will show that g L ([ 2, 2]) and that B = M g . For any polynomial p, p(M x ) = M p(x) , so that T commutes with M p(x) as well. It follows that T commutes with M f for any f C ([ 2, 2]). Indeed, since polynomials are dense in C ([ 2, 2]) by the Weierstrass approximation theorem, given any f C ([ 2, 2]) one can find polynomials pn f uniformly; then M pn M f 0, because M pn M f = M pn −f and the norm of a multiplication operator is the essential supremum of its symbol. Then
∈ −
−
∈ −
{ } { }
• • • • •
∈ ∞−
→
− ∈ − − →
∈
−
−
T M f − M f T = T (M f − M p) + T M p − M pT + (M p − M f )T = T (M f − M p) + (M p − M f )T ≤ T M f − M p + M p − M f T → 0. • If we knew that T commuted∞with M f for any f ∈ L∞ ([−2, 2]), we could quickly infer that g ∈ L ([−2, 2]) and that T = M g as follows: For any h ∈ L∞ ([−2, 2]), T h = T M h 1 = M h T 1 = M h g = M g h
so that T and M g agree on the dense subspace L∞ ([ 2, 2]) of L2 ([ 2, 2]). If we know that g L∞ ([ 2, 2]), so that M g is bounded, we can conclude that T = M g . If g / L∞ , let hn be the characteristic function of x : g(x) > n , and it would follow that
∈
{ |
•
|
}
−
− ∈
−
T hn = ghn > nhn
so that T would not be bounded. Unfortunately I don’t know a good way to show that T commutes with M g for all g L∞ , since C ([ 2, 2]) is not dense in L∞ ([ 2, 2]). However, we can still prove that g L∞ and that T = M g as follows: Let φ L2 ([ 2, 2]). Let f n C ([ 2, 2]) such that f n φ in L2 . Let f nj be a subsequence such that f nj φ a.e. (see for instance Theorem 3.12 in ([Rud91]) or Exercise 6.9 in ([Fol99]) for the existence of such).
∈ −
∈
− ∈ ∈ − →
Then gf nj = T f nj a.e. gf njk T φ. Then
→
2
L → T φ.
− →
Let f njk be a sub-subsequence such that
T φ = lim T f njk = lim gf njk = gφ, a.e.
so that T φ = M g φ for all φ g L∞ ([ 2, 2]) as well.
∈
−
a.e.
∈ L2. Since T is bounded, one must have
ARVESON
SPECTRAL THEORY
SOLUTIONS
31
• I have a feeling that there’s a simpler way to prove that {B} consists
only of multiplication operators, but I don’t know what it is. Suggestions welcome! If denotes the algebra of multiplication operators, the fact that B implies that B = . On the other hand, since . Hence B = = is commutative, we also have = B . On the other hand, it is not true that A = A . If it were, it would follow that Aˆ = Aˆ , where Aˆ := M 2 cos θ is the operator shown to be unitarily equivalent to A in Exercise 3. However, U Aˆ as previously noted, so that Aˆ U . But the latter does not contain multiplication operators of non-even functions, so it is strictly smaller than Aˆ , which contains all multiplication operators. Thus, we have found a property which is true of B and false of A, and which is a unitary invariant. It follows that B is not unitarily equivalent to A. Exercise 2.4.8: Show that A is unitarily equivalent to B B. Solution: It is clear that A is unitarily equivalent to the direct sum of M 2 cos θ on L2 ([0, π],dθ/2π) and M 2 cos θ on L2 ([ π, 0],dθ/2π), and that these are unitarily equivalent to each other. Hence it suffices to show that A T 2 where T is M 2 cos θ on L ([0, π],dθ/2π). Define a linear map Φ : L2 ([ 2, 2]) L2 ([0, π],dθ/2π) by (Φf )(θ) = 2 πf (2cos θ) sin θ. Then
• M { } ⊆M M ⊆{ } M M M⊆M { } M M { } { } { } { } { } ∈{ } { } ⊆{ } { } ⊕
−
−
√
√
Φf
2
| | | π
1 = 2π =2
0 π
→
(Φf )(θ) 2 dθ
|
f (2cos θ) 2 sin θ dθ
|
0
=
∼
2 1 f (u) 2 du = f 4π −2
2
|
so that Φ is an isometry. Moreover, Φ is bijective, with (Φ−1 g)(x) =
2
1 π(4
−
x2 )
g(cos−1 (x/2)),
so that Φ is unitary. Finally, T Φ = ΦB, proving that T and B are unitarily equivalent. 2.5. Representations of Banach *-Algebras. Exercise 2.5.1: Let = ∗ on a Hilbert space, and let
A A ⊆ B(H ) be a self-adjoint algebra of operators { ∈ H : Aξ = {0}}
N = ξ
A A
be the null space of . Show that the orthogonal complement of N is the closed linear span of H = T ξ : T , ξ H and that both N and [ H ] are -invariant subspaces.
A
A
{
∈A ∈ }
32
ARVESON
A
SPECTRAL THEORY
SOLUTIONS
A A A h ∈ (AH )⊥ ⇔ ∀ξ 1 , . . . , ξn ∈ H : ∀a1 , . . . , an ∈ A : h, ai ξ i = 0 ⇔ ∀ξ 1, . . . , ξn ∈ H : ∀a1, . . . , an ∈ A : h, ai ξ i = 0 ⇔ ∀ξ 1, . . . , ξn ∈ H : ∀a1, . . . , an ∈ A : a∗i h, ξ i = 0 ⇔ ∀a ∈ A : ah = 0 ⇔ h ∈ N.
Solution: Let ( H ) denote the (non-closed) linear span of H , which has closure [ H ]. Then ( H )⊥ = [ H ]⊥ . But show ( H )⊥ = N , because
A
A
∈
Exercise 2.5.2: Let A be a Banach *-algebra with unit 1, and let π rep(A, H ) be a representation of A. Show that π is nondegenerate iff π(1) = 1H . Solution: Because π(1) = π(1∗ ) = π(1)∗ and π(1) = π(12 ) = π(1)2 , we see that π(1) must be a projection P B(H ). Then for any h H and a ,
∈
∈
∈A
π(a)h = π(1a)h = π(1)π(a)h = P π(a)h so that π(a)h P H ; thus, P H is the essential subspace for π. By definition, π is nondegenerate iff this is all of H , which happens iff P = 1H . Exercise 2.5.3: Let A be a Banach *-algebra. A representation π rep(A, H ) is said to be cyclic if there is a vector ξ H with the property that the set of vectors π(A)ξ is dense in H . Show that a representation π rep(A, H ) is nondegenerate iff it can be decomposed into a direct sum of cyclic subrepresentations in the following sense: There is a family H i H, i I of nonzero subspaces of H that are mutually orthogonal, π(A)-invariant, that sum to H , and such that for each i I there is a vector ξ i H i with π(A)ξ i = H i . Solution: If π is a direct sum of cyclic subrepresentations, then it must be degenerate, as π( )H contains each H i and therefore all of H . Conversely, suppose π is nondegenerate. Let be the family of all subrepresentations of π which are direct sums of cyclic subrepresentations, with the partial order that µ ν if µ is a subrepresentation of ν . Clearly is nonempty, as any vector ξ H gives rise to a cyclic subrepresentation πξ on the subspace π( )ξ . Moreover, every chain in has an upper bound given by direct summation over its members. By Zorn’s lemma, there is a maximal representation ψ . Suppose ψ is a proper subrepresentation of π, on a proper subspace H H . Let ξ (H )⊥ be any nonzero vector; then ψ πξ is an element of strictly greater than ψ, a contradiction. Hence ψ = π and π is a direct sum of cyclic subrepresentations. Exercise 2.5.4: Let A be a Banach *-algebra. A representation π rep(A, H ) is said to be irreducible if the only closed π(A)-invariant subspaces of H are the trivial ones 0 and H . Show that π is irreducible iff the commutant of π(A) consists of scalar multiples of the identity operator. Solution: A closed subspace of H is π( )-invariant iff its projection commutes with π( ). Hence π is irreducible iff π( ) contains no nontrivial projections. This in turn is equivalent to π( ) being one-dimensional because π( ) is a von Neumann algebra, and therefore is generated by its projections. (This fact was mentioned on page 42 of the text.)
∈
∈ ∈ ⊆ ∈
∈
∈
A
Z
Z
A
≤ ∈
Z
∈ Z ⊆ Z
⊕
∈
∈
{}
A
A
A
A
A
∈
ARVESON
SOLUTIONS
SPECTRAL THEORY
33
Exercise 2.5.5: Let X be a compact Hausdorff space and let π be an irreducible representation of C (X ) on the Hilbert space H . Show that H is onedimensional and that there is a unique point p X such that π(f ) = f ( p)1 for all f C (X ). Solution: By the previous exercise, π(C (X )) contains only the scalar operators. But π(C (X )) is commutative, so that π(C (X )) π(C (X )) . Hence π(C (X )) contains only scalar operators, so that π has the form C is a nonzero homomorphism, aka a π(f ) = ω(f )1, where ω : C (X ) multiplicative linear functional. As previously established (chapter 1.10), the only MLF’s on C (X ) are point evaluations. Finally, any closed subspace of H will be invariant under scalar operators, so the irreducibility of π implies that H is one-dimensional.
∈
∈
⊆
→
2.6. Borel Functions of Normal Operators. Exercise 2.6.1: Show that for every f B(X ) and every > 0 there is a finite linear combination of characteristic functions in B(X ) (i.e. a simple function) g = c1 χE1 + + cn χEn
∈
· ··
such that f g . Solution: Let C be tiled by squares with diameter ; a finite collection S 1 , . . . , Sn of them cover f (X ) . Let E i = f −1 (S i ) X , and choose an arbitrary point ci S i for each i. Then g = ci χEi satisfies f g . Exercise 2.6.2: Let (X, ) be a Borel space. For every σ-finite measure µ on X let πµ be the representation of B(X ) on L2 (X, µ) defined by
− ≤ ∈
⊆
B
πµ (f )ξ ( p) = f ( p)ξ ( p),
ξ
− ≤
∈ L2(X, µ).
(a) Show that πµ is a σ-representation of B(X ) on L2 (X, µ). (Notice that the definition of σ-representation makes sense in this more general context.) (b) Given two σ-finite measures µ, ν on (X, ), show that πµ and πν are unitarily equivalent iff µ and ν are mutually absolutely continuous. (c) Deduce that a multiplication operator acting on the L2 space of a σ-finite measure is unitarily equivalent to a multiplication operator acting on the L2 space of a finite measure space. Solution: (a) Let f n be uniformly bounded functions in B(X ) (say f n M for 2 all n) such that f n ( p) 0 for all p X . For any ξ L (X, µ), the 2 2 functions ξ f n are dominated by M 2 ξ 2 , so
B
{ }
lim πµ (f n )ξ
2
→
= lim
|
∈
2
2
f n ( p) ξ ( p) dµ =
||
|
∈
≤
lim f n ( p) 2 ξ ( p) 2 dµ = 0
|
||
|
by the dominated convergence theorem. (b) Suppose µ ν . By Radon-Nikodym, there is a measurable function h such that dν = h dµ, with h nonzero a.e. Define U : L2 (X, ν ) L2 (X, µ) by (U f )(x) = h(x) f (x); then
∼
2
U f
=
|
||
| |
2
|
h(x) f (x) dµ =
X
→
|
f (x) 2 dν = f
X
|
2
34
ARVESON
SPECTRAL THEORY
SOLUTIONS 1 → √ |h(x) | is an inverse for U ,
so that U is an isometry, and since f
it is unitary. Finally, since πµ U = U πν , we see that πµ and πν are unitarily equivalent. Conversely, suppose µ and ν are not mutually absolutely continuous. Then there exists a Borel subset E X such that either µ(E ) = 0 while ν (E ) = 0, or ν (E ) = 0 while µ(E ) = 0; WLOG suppose the former. Then πµ takes 1E to the zero operator, whereas πν takes 1E to a nonzero operator; hence πµ and πν cannot be unitarily equivalent. (c) If (X, µ) is a σ-finite measure space, let X 1 , X 2 , . . . be disjoint measurable subsets of X with each µ(X i ) finite. Define a measure ν by
⊆
ν (E X i ) =
∩
µ(E ) 2i µ(X i )
.
∼
Then ν µ, and ν (X ) = 1. Note that this exercise is very similar to 2.4.1. 2.7. Spectral Measures. Exercise 2.7.1: Let N 1 B(H 1 ) and N 2 B(H 2 ) be two normal operators acting on finite-dimensional Hilbert spaces H 1 , H 2 . Show that there is a unitary operator W : H 1 H 2 such that W N 1 W −1 = N 2 iff N 1 and N 2 have the same spectrum and the same multiplicity function. Solution: Clearly, unitary equivalence preserves both spectrum and multiplicity. For the converse, suppose N 1 and N 2 have the same spectrum and multiplicity. For each λ σ(N 1 ) = σ(N 2 ), let E λ,1 and E λ,2 be the corresponding eigenspaces for N 1 and N 2 . By hypothesis, these have the same dimension, so that there is a unitary W λ : E λ,1 E λ2 . Because the operators are normal, they’re diagonalizable, so that
∈
∈
→
∈
→
H j =
E λ,j
for j = 1, 2.
λ σ(N j )
∈
→
⊕
We can therefore define a unitary W : H 1 H 2 by W = W λ . One has W P λ,1 = P λ,2 W , where P λ,j is the projection onto E λ,j , so that W
λ
λP λ,1 =
λ
λP λ,2 W
⇒
W N 1 = N 2 W.
Exercise 2.7.2: Calculate the spectral measure of the multiplication operator X defined on L2 ([0, 1]) by (Xξ )(t) = tξ (t), 0 t 1. Solution: The spectral measure is given by P (S ) = M (1S ) for each Borel subset S [0, 1], where M (f ) denotes the multiplication operator induced by f . More generally, we will prove the following: Let X be a locally compact Hausdorff space, µ a positive σ-finite Borel measure on X , and φ : X C a bounded Borel function with essential range K C. Then the spectral measure corresponding to the multiplication operator M (φ) on L2 (X, µ) is given by P (S ) = M (1φ 1 (S) ) for Borel subsets S K . The proof amounts to the following observations: (1) In general, if P is the spectral measure for some normal operator T , then P (S ) is the operator 1S (T ) defined via the Borel functional calculus. This observation was made at the end of section 2.7.
≤ ≤
⊆
−
⊆ ⊆
→
ARVESON
SPECTRAL THEORY
SOLUTIONS
35
(2) The Borel functional calculus is applied to multiplication operators by composition: ψ(M (φ)) = M (ψ φ) for a bounded Borel function ψ defined on the essential range of φ. One proves this formula in stages, by “promoting” ψ through the different levels of the functional calculus. If ψ is a polynomial, the formula is immediate. If ψ C (K ), choose polynomials pn ψ in C (K ); then pn φ ψ φ in B∞ (X ), so 2 that M ( pn φ) M (ψ φ) in B(L (X, µ)). Since ψ(M (φ)) is defined by the continuous functional calculus to be the limit of M ( pn φ), this shows that ψ(M (φ)) = M (ψ φ). Finally, to extend to ψ B∞ (K ), note that ψ M (ψ φ) is a σ-representation of B∞ (K ) which extends the continuous functional calculus, and since we know ψ ψ(M (φ)) is the only such, it follows that ψ(M (φ)) = M (ψ φ). (3) Applying our second remark to the function ψ = 1S , we have 1S (M (φ)) = M (1S φ). Applying the first remark, this becomes P (S ) = M (1S φ). Finally, we note that 1S φ = 1φ 1 (S) . Exercise 2.7.3: A resolution of the identity is a function λ : R P λ B(H ) from R to the projections on a Hilbert space with the following properties: λ µ P λ P µ Relative to the strong operator topology,
◦
→ ◦
◦ → →
◦ → ◦
◦
◦
◦
◦
∈
∈ →
◦
◦
◦
−
→
∈
• ≤ ⇒ ≤ •
lim P λ = 0,
→−∞
λ
lim P λ = 1.
→+∞
λ
• Right continuity: For every λ ∈ R, lim P µ = P λ .
µ
→λ+
Early formulations of the spectral theorem made extensive use of resolutions of the identity. It was gradually realized that these ob jects are equivalent to spectral measures, in much the same way that Stieltjes integrals are equivalent to integrals with respect to a measure. This exercise is related to the bijective correspondence that exists between resolutions of the identity and spectral measures on the real line. (a) Consider the Borel space ( R, ) of the real line. Given a spectral measure E : B(H ), show that the function P λ = E (( , λ]), λ R, is a resolution of the identity. (b) Given two spectral measures E, F : B(H ) that give rise to the same resolution of the identity, show that E = F . Solution: (a) (1) If λ µ, then ( , λ] ( , µ], so
B
B→
∈
−∞
B→
≤
−∞ ⊆ −∞ P λ = E ((−∞, λ]) ≤ E ((−∞, µ]) = P µ .
(2) By the previous property, it suffices to prove this with λ replaced by an integer argument m. Now 1H = E (R) = E
n
(n, n + 1]
∈Z
=
n
∈Z
E ((n, n + 1]).
36
ARVESON
SOLUTIONS
SPECTRAL THEORY
We use the fact that the tails of a convergent sum tend to zero. A negative tail of the above sum has the form
−∞, n + 1]) = P m+1,
E ((n, n + 1]) = E ((
n m
≤
from which we see that P m positive tail is of the form E ((m,
→ 0 as m → −∞.
Similarly, a
∞)) = 1 − E ((−∞, m]) = 1 − P m, from which we see that P m → 1 as m → ∞. H
H
H
(3) We write E ((λ, λ + 1]) = E
∞
n=1
∞
E
n=N
1 1 λ+ ,λ+ n+1 n
=
∞
1 1 λ+ ,λ+ n+1 n
E
n=1
and again use the fact that the tails of a convergent sum tend to zero. A tail of this sum has the form 1 1 λ+ ,λ+ n+1 n
= E
→
1 λ, λ + N
= P λ+1/N
− P λ.
→ ∞; appealing to the first property → P λ as µ → λ+.
Thus, P λ+1/N P λ as N again, this implies that P µ
(b) Let
A = {S ∈ B : E (S ) = F (S )}. We will show that A is a σ-algebra which contains the half-open intervals, which implies B ⊆ A and hence that B = A. (1) E (∅) = 0 = F (∅), so that ∅ ∈ A. (2) If S ∈ A, then E (S ) = 1 − E (S ) = 1 − F (S ) = F (S ), so that S ∈ A. Thus A is closed under complementation. (3) If S 1 , S 2 ∈ A, then E (S 1 ∩ S 2 ) = E (S 1 )E (S 2 ) = F (S 1 )F (S 2 ) = F (S 1 ∩ S 2 ), so that S 1 ∩ S 2 ∈ A. Thus A is closed under finite intersections. (4) If T 1 , T 2 , · · · ∈ A, let S 1 = T 1 and j
S j +1 = T j +1
\
j
T i = T j +1
i=1
{}
∩
T i ,
i=1
∈A n
so that S i is a disjoint family with the property that i=1 S i = n for each i, by previi=1 T i for each n. Note also that S i ously established properties of . Then
E
∞
i=1
T i
= E
∞
S i
=
i=1
∞
E (S i ) =
i=1
∞
∞
A
F (S i ) = F
i=1
A
∞
i=1
A
S i
= F
∞
T i
i=1
so that i=1 T i is in . Thus is closed under countable unions, and hence is a σ-algebra. (So far we’ve retraced the standard exercise that a collection of sets is a σ-algebra iff it contains the empty set and is closed under complementation, finite intersections, and countable disjoint unions.)
ARVESON
SPECTRAL THEORY
SOLUTIONS
37
(5) Finally, given any half-open interval (a, b]), we have E ((a, b]) = P b so that (a, b]
− P a = F ((a, b])
∈ A.
2.8. Compact Operators. Exercise 2.8.1: Let A be a compact operator on a Hilbert space H . Show that for every sequence of mutually orthogonal unit vectors ξ 1 , ξ 2 , H we have
·· · ∈
→∞
lim Aξ n = 0.
n
Solution: If not, we may assume, by taking a subsequence and rescaling A if needed, that Aξ n 1 for all n. Taking another subsequence if needed and using the compactness of A, we have Aξ n η for some η 1 (which automatically satisfies η 1). This implies Re Aξ n , η 2 for sufficiently large n; WLOG, by taking a tail if needed, for all n. Now let xN = ξ 1 + ξ 2 + + ξ N . We have xN = N , but
≥
→
≥ √
···
≥
≥ N 2 ⇒ |AxN , η| ≥ N 2 ⇒ AxN η ≥ N 2 √ N N ⇒ AxN η ≥ 2 ⇒ A ≥ 2η Re AxN , η
for all N , contradicting the boundedness of A. Exercise 2.8.2: Let e1 , e2 , . . . be an orthonormal basis for a Hilbert space H and let A B(H ). Show that A is compact iff
∈
lim
n
→∞
(1 − E n)A(1 − E n) = 0,
where E n denotes the projection onto spane1 , . . . , en . Solution: If A has this property, it is the norm limit of finite-rank operators, hence is compact. Conversely, if A is compact, the preceding exercise implies that for any > 0 there exists N such that Aen < for n N ; this implies that A(1 E n ) < , so that (1 E n )A(1 E n ) < . Exercise 2.8.3: Verify the polarization formula for bounded operators on a Hilbert space H :
−
4B ∗ A =
−
−
≥
3
ik (A + ik B)∗ (A + ik B).
k=0
Solution: It’s tempting to cite the fact that ( A, B) B ∗ A is an inner product, but since this is B(H )-valued rather than C-valued, we’d have to develop the theory of Hilbert C ∗ -modules to substantiate this. So we can
→
38
ARVESON
SOLUTIONS
SPECTRAL THEORY
make the direct calculation: 3
3
k
k
i (A + i B)∗ (A + ik B) =
k=0
− − − ik (A∗
ik B ∗ )(A + ik B)
k=0 3
=
ik (A∗ A
ik B ∗ A + ik A∗ B + B ∗ B)
k=0 3
=
3
k
i
A∗ A +
3
k=0
k=0
3
k
1 B ∗A +
( 1)
k=0
k=0
= 4B ∗ A.
1/2
Exercise 2.8.4: Let A 2 = A, A 2 for every Hilbert-Schmidt operator A. (a) Let A1 , A2 , . . . be a sequence in 2 that satisfies
L lim Am − An 2 = 0. m,n→∞ ∈
− →
Show that there is an operator A B(H ) such that An A 0 as n . (b) Show that 2 is a Hilbert space relative to the inner product A, B 2 = trace B ∗ A. Solution: (a) This will follow from the more general fact that A A 2 . For this, let x H ; then
→∞ ∈
A∗ x2 =
L
≤
|
A∗ x, en
n
|2 =
| n
x,Aen
|2 ≤
Aen
n
2x2
by Parseval’s identity and the Cauchy-Schwarz inequality. It follows that A 2 = A∗ 2 A 22 . (b) Let An be Cauchy in 2 . Let A be the norm limit from part (a). Let > 0, and choose N N such that Aj Ak 2 < for j, k N . Let k N . We seek to prove that A Ak 2 . It is tempting to use the computation
≤ { } L ∈ − ≥ − ≤
≥
A − Ak 2 = lim Aj − Ak 2 = lim Aj − Ak 2 ≤ . j j However, the notation limj here denotes the norm limit, which cannot be moved through the 2 symbol because the Hilbert-Schmidt norm is not continuous in the operator-norm topology if H is infinitedimensional. (For instance, consider a projection of rank n; this has operator norm 1 but Hilbert-Schmidt norm n.) Thus, another proof approach must be found. We will use the fact that an infinite sum of nonnegative terms is the supremum over finite sums. For any fixed m N,
·
√
∈
(A
− Ak )em = (lim Aj − Ak )em = lim j j
(Aj
− Ak )em
,
ik
A∗ B +
B ∗B
ARVESON
SPECTRAL THEORY
so that for any M M
m=1
SOLUTIONS
∈ N,
M
(A
− Ak )em2 =
m=1
= sup
≥
39
M
lim (Aj j
− Ak )em2 ≤
m=1
M
j N m=1
(Aj
j ≥N
sup (Aj
− Ak )em2
− Ak )em2 ≤ jsup Aj − Ak 22 ≤ . ≥N
Thus, M
Ak 22
A −
= sup M
(A
∈N m=1
− Ak )em2 ≤ .
It follows that A Ak 2 0, so that 2 is complete. Exercise 2.8.5: Show that a multiplication operator M f is self-adjoint and has nonnegative spectrum iff M f ξ, ξ 0 for every ξ L2 (X, µ). Solution: Suppose M f is self-adjoint and has nonnegative spectrum. Since the spectrum is the essential range, it follows that f is nonnegative a.e. Then for any ξ L2 (X, µ),
− → L ≥
∈
M f ξ, ξ =
(M f ξ )(x)ξ (x) dµ(x) =
X
∈
f (x)ξ (x)ξ (x) dµ(x)
X
≥0
because the integrand is nonnegative a.e. Conversely, suppose f does not have nonnegative spectrum. Let λ σ(M f ) [0, ), and let N be a neighborhood of λ such that one of the conditions (1) Re z < 0 (2) Im z < 0 (3) Im z > 0 holds throughout N ; for convenience, we assume the first. Let V = f −1 (N ), which has positive measure because λ is in the essential range of f . Then
∈
\ ∞
M f 1
V
, 1V =
f (x) dµ(x)
V
which has strictly negative real part, because the integrand does a.e. Hence there exists ξ L2 such that M f ξ, ξ is not nonnegative.
∈
2.9. Adjoining a Unit to a C ∗ -Algebra. Exercise 2.9.1: Let A be a nonunital C ∗ -algebra and let π : A Ae be the natural map of A into its unitalization. Considering Ae as a C ∗ -algebra, suppose that there is an isometric *-homomorphism σ : A B of A into ∗ another unital C -algebra B such that σ(A) is an ideal of codimension 1 in B. Show that there is a unique isometric *-isomorphism θ : Ae B such that θ π = σ. Solution: The map θ : Ae B defined by θ(a + λ1) = σ(a) + λ1 is a *homomorphism with the property θ π = σ. It is surjective because its range includes σ(A) and 1. It is injective because if θ(a + λ1) = 0, then σ(a) = λ1, and since σ(A) C1 = 0 it follows that σ(a) = λ = 0 and hence, by the injectivity of σ, that a = λ = 0.
→ →
◦
−
→
→
∩
◦ {}
40
ARVESON
SPECTRAL THEORY
SOLUTIONS
Exercise 2.9.2: Let be the C ∗ -algebra of compact operators on a Hilbert space H . Show that the space of operators λ1 + K : λ C, K is a ∗ e C -algebra *-isomorphic to K . Solution: It is easy to see by direct calculation that the space of operators in question is a unital C ∗ -algebra in which K is an ideal of codimension one. The previous exercise implies that it is isomorphic to K e . Exercise 2.9.3: Let X be a compact Hausdorff space and let F be a proper closed subset of X . Let A be the ideal of all functions f C (X ) that vanish throughout F . Note that A is a C ∗ -algebra in its own right. (a) Show that A has a unit if and only if F is both closed and open. (b) Assuming that F is not open, identify the unitalization of A in concrete terms by exhibiting a compact Hausdorff space Y such that Ae C (Y ), describing the precise relationship of Y to X and F . Solution: (a) If F is clopen, the function 1 1F is continuous, and is a unit for A. Conversely, suppose A has a unit φ. Given any point x X F , there exists f C (X ) with f (x) = 1 and f = 0 on F ; since f A, φf = f , so that φ(x) = 1. Hence φ = 1 1F . Since this is continuous, F is clopen. (b) If F is not open, X F is locally compact but not compact. The algebra A is isomorphic to C 0 (X F ), so its unitization is isomorphic
K
{
∈
∈ K}
∈
−
∈
−
\
to C (X F ), where
\
∈ \ ∈
\
∼ denotes the one-point compactification.
2.10. Quotients of C ∗ -Algebras. Exercise 2.10.1: Let e1 , e2 , . . . be an orthonormal basis for a separable Hilbert space H , and let E n be the projection on the span of e1 , . . . , en . Show that an operator T B(H ) is compact iff
{
}
{
∈
}
lim
n
→∞ T − T E n = 0,
and deduce that E n : n N is an approximate unit for the C ∗ -algebra . Solution: The “if” direction is true because a norm-limit of compact operators is compact; this is true even on Banach spaces (Exercise 1.4.2). The “only if” direction is a consequence of the spectral theorem for compact normal operators (Theorem 2.8.2). Exercise 2.10.2: Let U be a unitary operator on a Hilbert space H . Then T, and hence there is a unique representation ρ σ(U ) rep(C (T), H ) satisfying ρ(f ) = f (U ) for f C (T). Identify ker ρ as an ideal in C (T), identify the quotient C (T)/ ker ρ in concrete terms as a commutative C ∗ algebra, and similarly describe the natural factorization ρ = ρ˙ π, where
{
K
∈ }
⊆
∈
π : C (T)
→ C (T)/ ker ρ
∈
◦
is the natural map onto the quotient C ∗ -algebra. Solution: The existence and uniqueness of ρ deserves a little explanation, since ρ is not just the continuous functional calculus for U if σ(U ) is a proper subset of T. It is Moreover, even the symbol f (U ) needs defining ˆ in this case. We define it to mean f (U ) where f ˆ is the restriction of f to σ(U ). Then ρ is unique because it is determined on the set of polynomials, which are dense in C (T); for existence, note that the composition F r
◦
ARVESON
SPECTRAL THEORY
SOLUTIONS
41
→
has the desired properties, where F : C (σ(U )) B(H ) is the continuous functional calculus and r : C (T) C (σ(U )) is restriction. Now on to the exercise. Because F is injective,
→
{ ∈ C (T) : f = 0 on σ(U )}.
ker ρ = ker r = f
◦
Now the restriction map r is surjective (by Urysohn’s lemma), so r = r˙ π where π : C (T) C (T)/ ker r is the quotient map (the same as the π mentioned in the problem, since ker ρ = ker r) and r˙ is an isomorphism. Hence C (T)/ ker ρ = C (T)/ ker r C (σ(U )). Finally, ρ = F r = F r˙ π, so that ρ˙ = F r˙ is the composition of the continuous functional calculus with the identification between C (T)/ ker ρ and C (σ(U )). Exercise 2.10.3: Show that there is a compact Hausdorff space β R and an isometric *-isomorphic of C b (R) onto C (β R). Solution: Since C b (R) is a unital C ∗ -algebra, the Gelfand spectrum Σ(C b (R)) fits the bill. Exercise 2.10.4: For every t R, show that there is a ( naturally defined) point tˆ β R, and that the map t tˆ is a homeomorphism of R onto a dense subspace of β R. Solution: We define tˆ to be evaluation at t, that is, tˆ(f ) = f (t) for f C b (R). t tˆ is continuous because if tα t in R, then f C b (R) : f (tα ) f (t); that is, f C b (R) : tˆα (f ) tˆ(f ) in C. By the definition of the Gelfand topology, this is the same as tˆα tˆ in β R. t tˆ is injective because if t1 = t2 , we can use Urysohn’s lemma to find a function g C b (R) with g(t1 ) = 0 and g(t2 ) = 1; then tˆ1 (g) = tˆ2 (g), so tˆ1 = tˆ2 . Note that in a metric space, you don’t need Urysohn; you can use the explicit formula
→
◦
◦ ◦
◦
∈
∈
• →
→
→ →
∀ ∈
• →
g(x) = max 0, 1
→
→
∈
∈
∀ ∈
−
2d(x, t2 ) d(t2 , t1 )
.
• t → tˆ is open because if tα → t in R, there exists (by taking a subnet if necessary) a neighborhood N of t such that tα ∈ / N for all α. Then by Urysohn’s lemma, there exists g ∈ C b (R) with g(t) = 1 and with g supported in N . Then
0 = tˆα (g)
→ tˆ(g) = 1.
• Suppose the range Rˆ is not dense.
Let φ β R be a point and U a ˜ . Using the basis for the weak-* neighborhood of φ disjoint from R topology, there exist an > 0, an n N, and function f 1 , . . . , fn C b (R) such that φ(f k ) ω(f k ) < for all ω U and all k = 1, . . . , n. Let
|
−
|
∈
∈
∈
∈
n
f (x) =
|
k=1
f k (x)
− φ(f k )|2.
Since no x ˆ is in U , we must have f (x) 2 for every x R. Thus, f ˆ is invertible, which implies φ(f ) is nonzero, a contradiction. Hence R is dense in β R. This density argument is taken from chapter 6.5 of ( [Zhu93]).
≥
∈
42
ARVESON
SPECTRAL THEORY
SOLUTIONS
Exercise 2.10.5: Identifying R with its image in β R, the subspace β R R is called the corona of R. Show that the corona is closed (and hence, R is an open subset of β R). Solution: As suggested in the hint, we will show that the corona is precisely the closed subset
\
{ω ∈ β R : ∀f ∈ C 0(R) : ω(f ) = 0}.
R then there Clearly this subset is contained in the corona, since if t exists f C 0 (R) (in fact, f can be compactly supported) with f (t) = 0 and hence with tˆ(f ) = 0. On the other hand, suppose ω β R R. Let f C 0 (R). Let > 0 and choose compact K R such that f < on ˆ is dense in β R, there is a net tα R K . Because R R such that tˆα ω in β R. Then every subnet converges to ω as well; this implies that only finitely many points of the net lie in K , as otherwise a subnet of tα would converge to a point t K , and therefore the corresponding subnet of tˆα would converge to tˆ rather than ω. Then tˆα (f ) < for sufficiently large α, so that ω(f ) = lim tˆα (f ) .
∈
∈ \
⊆ { }⊆
\ ||
{ }
∈
|
∈ ∈
| |
| |≤
α
|
→
{ }
This is true for arbitrary > 0, so ω(f ) = 0. Exercise 2.10.6: Deduce that the quotient C ∗ -algebra C b (R)/C 0 (R) is isometrically isomorphic to C (β R R). Solution: Define ρ : C b (R) C (β R R) by ρ = r Γ where Γ : C b (R) R ( ) is the Gelfand transform and r : C (β R) C β C (β R R) is restriction. By Urysohn’s lemma, r is surjective, and since Γ is an isomorphism, ρ is surjective as well. Moreover, because Γ is an isomorphism, ker ρ = Γ−1 (ker r), which equals C 0 (R) by the previous exercise. Hence C b (R)/C 0 (R) C (β R R) by the First Isomorphism Theorem. Exercise 2.10.7: A compactification of R is a pair (φ, Y ) where Y is a compact Hausdorff space and φ : R Y is a continuous map such that φ(R) is dense in Y . Show that (t tˆ, β R) is a universal compactification of R in the following sense: If (φ, Y ) is any compactification of R, then there is a unique extension of φ : R Y to a continuous surjection φˆ : β R Y . Hint: ∗ The map φ induces a *-isomorphism of C (Y ) onto a unital C -subalgebra of C b (R). Solution: Let Ψ : R β R denote the map t tˆ. Define Φ : C (Y ) C b (R) by Φ = φ; that is, for any f C (Y ), Φ(f ) = f φ. To see that Φ is injective, suppose Φ(f ) = 0, so f φ = 0. Thus, f is zero on the range of φ; but this is dense in Y , so that f must be zero. ˆ : C (Y ) ˆ = Γ Φ where Γ : C b (R) Let Φ C (β R) be the map Φ ˆ is C (β R) is the Gelfand transform; since the latter is an isomorphism, Φ also injective. By Exercise 2.2.6, there exists a unique continuous function ˆ = ˆ implies the u : β R Y with Φ u; moreover, the injectivity of Φ ˆ and taking the corresponding surjectivity of u. Finally, we have Φ = Γ−1 Φ maps on topological spaces as in Exercise 2.2.6, we get u Ψ = φ.
→
\
\
◦ →
\
→
→
→
·◦
→
→
→
∈ ◦
→
→
3.
→
\
→
◦
◦
·◦
◦
→
◦
Asymptotics: Compact Perturbations and Fredholm Theory
3.1. The Calkin Algebra.
View more...
Comments
