Integral Equations
and Operator Theory
Vol.
14 (1991)
COMMUTING
0378-620X/91/010001-12$1.50+0.20/0
(c) 1991 Birkhauser
Verlag,
Basel
TOEPLITZOPERATORS
WITH HARMONIC
SYMBOLS
Sheldon Axler* and Zeljko Cuckovic
This paper shows that on the Bergman space, two Toeplitz operators with
harmonic symbols commute only in the obvious cases. The main tool is a
characterization of harmonic functions by a conformally invariant mean value
property.
We begin by recalling some standard notation and definitions.
Let dA denote the usual
area measure on the open unit disk D in the complex plane C. The complex space L2(D, dA) is a
Hilbert space with the inner product
<I, g> = fDI
i
dA.
The Bergman space L~ is the set of those functions in L 2(D, dA) that are analytic on D (the a in
L~ stands for "analytic"). The Bergman space L~ is a closed subspace of L 2(D, dA), and so there
is an orthogonal projection P from L2(D, dA) onto L~. For <p E L OO(D,dA), the Toeplitz
r/
:operator with symbol cp, denoted T <p' is the operator from L~ to L~ defined by T = P( <pl). By a
harmonic function we mean a complex-valued function on D whose Laplacian is identically O.
The following theorem is the main result of this paper.
* The first authorwas partiallysupportedby the NationalScienceFoundation.
3
2
Axler
and
Cuckovic
Axler
Cuckovic
and
THE INV ARrANT MEAN VALUE PROPERTY
THEOREM 1: Suppose that
qJ
and'1f are bounded harmonic functions on D. Then
if and only if
(1.1)
<p
function on the disk D is hannonic if and only if it has the mean value
A continuous
T'{JT1jI= T1jIT'{J
property. In this section we characterize hannonic functions in terms of an invariant mean value
and
'1f
property.
Let Aut(D) denote the set of analytic, one-to-one maps of D onto D ("Aut" stands for
are both analytic on D,
or
(1.2) q; and 'if are both analytic on D,
or
(1.3) there exist constants a, bE
C, not both 0, such that a<p+ b\jl is constant on D.
"Automorphism").
A function h on D is in Aut(D) if and only if there exist a.
aD and ~ E D
E
such that
~
As we will see, the "if' direction of the above theorem is trivial, but the proof of the "only
h(z)
if' direction requires a converse to an invariant form of the mean value property.
The statement of Theorem 1 is similar to the analogous result proved by Arlen Brown and
basis for the Hardy space H2(aD»
On the Bergman space L~, Toeplitz operators
techniques
used by Brown and Halmos do not seem to work in this context.
do not have nice matrices, and the
E
Aut(D)
and every r
E
»
dfl = u(h(O»
2n;
[0, I). Here "invariant" refers to conformal invariance,
meaning invariance under composition with elements of Aut(D). If u is harmonic on D, then so is
I was proved by Sheldon Axler and Pamela Gorkin ([2],
converse is also true (see [4], Corollary 2 to Theorem 4.2.4): if a function u E C(D) has the
uoh for every hE Aut(D); thus hannonic functions have the invariant mean value property. The
Theorem 7), using function theory techniques quite different from those that we use here. Dechao
Zheng ([5], Theorem 5) also proved a special case of Theorem I; our proof makes use of some of
his ideas.
Functions in L~(aD) correspond, via the Poisson integral, to bounded hannonic functions
on D, so perhaps the restriction
for every h
'S
Thus function
theory, rather than matrix manipulations, plays a large role in our proof.
A special case of Theorem
u(h(re1
f2Jt
o
of products of Hardy space Toeplitz
operators.
I _~z
A function u E C(D) is said to have the invariant mean value property if
Brown and Halmos proved their result by examining the matrix (with respect to
the usual orthonormal
a.
for all zED.
Paul Halmos ([3], Theorem 9) for Toeplitz operators with symbol in L~(aD) acting on the Hardy
space H2(aD).
=
in Theorem 1 to consideration
only of Toeplitz operators with
harmonic symbols (as opposed to Toeplitz operators with symbol in L~(D, dA» is natural.
importantly, Theorem 1 does not hold if "harmonic" is replaced by "measurable".
Paul Bourdon has pointed out to us that if <pand
\jI
invariant mean value property, then u is harmonic on D.
The invariant mean value property concerns averages over circles with respect to arc length
measure.
Because we are dealing with the Bergman space L~, we need an invariant condition
stated in terms of an area average over D. Thus we say that a function u E C(D) n L I (D, dA)
has the area version of the invariant mean value property if
More
f D uoh
For example,
are any two radial functions in L~(D, dA), then
TipT1jI == T 1jIT (a function is called radial if its value at z depends only on Izl). Thus the following
open problem may be hard: Find conditions on functions <pand \jI in L~(D, dA) that are necessary
for every h
E
Aut(D).
If u is in C(D)
dA.
n; = u(h(O»
n LI(D,
dA), then so is uoh for every h
E
Aut(D), so the
left-hand side of the above equation makes sense. Note that the area version of the invariant mean
'{J
and sufficient for T'{J to commute with T 0/'
The next section of this paper discusses an invariant mean value property that will be used
value property deals with integrals over all of D, as opposed to integrals over rD for r E (0, 1).
If u is harmonic on D and'in L I(D, dA), then so is uoh for every h E Aut(D). Thus, by
the mean value property, harmonic functions have the area version of the invariant mean value
in the proof of Theorem 1. The final section of the paper contains the proof of Theorem I, along
property.
with a corollary that describes the normal Toeplitz operators with hannonic symbol.
u
Whether
or not the converse
is true is an open question.
In other words, if
n L I (D ,dA) has the area version of the invariant mean value property, must u be
harmonic? This question has an affirmative answer if we replace the hypothesis that u is in
E C (D)
C(D)nLI(D,
[1], Proposition
dA) with the stronger hypothesis thatu is in CCD); see [4], Proposition
10.2.
13.4.4 or
",
4
/
Axler and Cuckovic
We need to consider functions that are not necessarily continuous on the closed disk, so
the result mentioned in the last sentence will not suffice. However, our functions do have the
property that their radializations
5
Axler and Cuckovic
The inverse (undercomposition)fe-1
a
-y
I-t'z
ie
.-I(z)=a
prove that this property, along with the area version of the invariant mean value property, is
of D, so there exist
for all zED,
Thus
enough to imply harmonicity; see Lemma 2.
denoted ':R(u), is the function on D defined by
ofu,
=
9«u)(w)
is also an analytic automorphism
~ - z
(defined below) are continuous on the closed disk, and we will
If u E C(D), then the radialization
off a
E aD and ~ E D such that
u(we1
12lt
o
)
lifa-1)'(z)1 = 1 -1~122
11 -
dfr '
'a 2it
~zl
1 + I~I
In the following lemma, which will be a key tool in our proof of Theorem
:J<.(u'h) E C(15) means ':R(u'h) can be extended to a continuous complex-valued
for all zED.
$; --
1, the statement
1 - I~I
function on D.
Note that ~ = fa (0) = h(g(O)eie);
we are thinking of h and g as fixed, so the above inequality
shows there is a constant K (depending only on h and g) such that
n
LEMMA 2: Suppose that u E C(D)
Ll(D,
dA).
Then u
is
harmonic
on D if and
lifa -1)'(z)1
$;
K
for all zED
and all
e
E [0, 2it],
only if
Now
f D u'h
(3)
dA.
= u(h(O»
it
r2lt J D lu(h(g(w)eia»I.dA.U0 it
Jo
and
':R(u'h)
E C(D)
for
every hE
= Jo
plt
dSt
2it
$; K2 JD
r
(4)
Aut(D).
r
JD
<
lu(z)llifa-lnz)12
d.A!..iJ
it dfr
2it
lu(z)1 dA.W
it
00
PROOF: We first prove the easy direction (the other direction will be the one that we need
in the proof of Theorem
1).
Suppose
that u is harmonic on D.
Let h E Aut(D).
As we
discussed earlier, U'h is harmonic and so (3) holds. The mean value property implies that ':R(u'h)
Thus we can apply Fubini's Theorem to (5), getting
plt J
JrD v'g dA
it = J 0
D
u(h(g(w)ei9»
dA(w)
it
dSt
2it
is a constant function on D, with value u(h(O», so (4) also holds.
To prove the other direction, suppose that (3) and (4) hold. Let h E Aut(D), and let
v
By (4), v
= J2lt
o J D (u'fe)(w)
= ':R(U'h).
=
E C(D).
We want to show that v has the area version of the invariant mean value property.
this, fix g E Aut(D).
= J2lt
o
dA
it
=
f D ':R(u'h)(g(w»
Thus v is a continuous function on
To check that interchanging
(by (3»
dfr
2it
u(h(g(O)eia)
dJi
2it
=
9«u'h)(g(O»
. = v(g(O».
dA(w)
it
= J D J 02lt u(h(g(w)eie»
u(ja(O»
To do
Then
f D v'g
eE
J2lt
o
dA.LJ1Udfr
it
2it
de
2it dA(w).
it
(5)
the order of integration in the last integral is valid, for each
15
that has the area version of the invariant mean value
property.
Hence (see [4], Proposition 13.4.4 and Remark 4.1.4 or [1], Proposition 10.2,
combined with a change of variables) v is harmonic on D. Because v is also a radial function, the
mean value property implies that v is a constant function on D, with value v(O).
Recall that v = ':R(u'h), so
[0,21t] define fa E Aut(D) by
fe(w)
=
h(g(w)eie).
7
6
Axler
121t
o
(uoh)(rel
'S
for every r E [0, 1) and for each h E Aut(D).
)
and
Cuckovi6
Cuckovic
and
To complete the proof, now suppose that u is a function in C(D)
= u(h(O»
d.fr
21t
area version of the invariant mean value property.
In other words, u has the invariant mean value
property (the usual version, not the area version).
Axler
Thus (see [4], Corollary 2 to Theorem 4.2.4
and Remark 4.1.4) u is harmonic on D .•
Let h E Aut(D).
n L leD,
dA) having the
Clearly 'J<.(u·h)is a radial
function on D, and, as shown in the proof of Lemma 2, 'J<.(u·h) has the area version of the
invariant mean value property.
By the above paragraph, 'J<.(u·h)is constant on D. In particular,
'J<.(uoh)E C(D), and so by Lemma 2, u is harmonic .•
As mentioner earlier, it is unknown whether Lemma 2 remains true if (4) is deleted.
We
believe that the following proposition, which reduces this question to a tempting integral equation,
is the best way to attack this problem.
Patrick Ahern and Walter Rudin also independently proved
TOEPLITZ OPERATORS
Lemma 2 and Proposition 6 (with similar proofs) at about the same time we did.
For h E Aut(D), define an operator U h on L~ by
PROPOSITION
V E C([O,
1»
nL1[0,
6: Suppose
I]
functions
in C(D)
2111+ts
o (1 - ts)
n L lCD, dA)
<
yes) ds
UII= (f·h)h'.
are the only functions
such that
Vet) = (I - t)
Then every function
that the constant
for every t
E
[0, 1).
(7)
A simple computation shows that Uh is a unitary operator from L~ onto L~, with inverse Uh-1•
We will need the following lemma in the proof of Theorem 1.
having the area version of the invariant mean value
LEMMA
property is harmonic.
PROOF: First suppose that v is a radial function in C(D)
version of the invariant
u. E
For
a - z
v(a) =
f D (v'ha)(z)
d.t1i1.)
1t
=(I-a) 22Jl o
PROOF: Define Vh:L2(D, dA)
operatorfrom
E L ~(D, dA).
Then
-t
L2(D,dA)
by VJ=
(f·h)h'.
Then Vh is a unitary
L2(D, dA) ontO L2(D, dA). Obviously VhIL~ = Uh· Thus Vh maps L~ onto L~, so
=--.
1 - az
Note that ha is its own inverse under composition.
<p
UhTi!fUh* = Ti!f'h .
be defined by
ha(z)
Aut(D) and let
dA) having the area
We will show that v is constant on D.
mean value property.
[0, 1), let ha E Aut(D)
n L 1(D,
8: Let hE
(9)
PVh = VhP.
For each
= f D v(w)
aE
If
[0, 1) we have
Iha'(w)12
f
E L~, then
Ti!f"hU
II = T i!f.h((f·h)h ,) = P((<p'h)(f'h)h
dACw)
1t
') = p(V h(<Pj)
= Vh(P(<pj)
v(r)rS4 ]21t
0
11 -
1
are!
I
(by
(9»
dJldr
21t
= UhT {.
a22Jl
)
= (1 -
2 2 < vCr)dr
o r (1(I - +a a2r2)
r )
Thus T i!f'hUh = UhTi!f' and because Uh is unitary, this implies the desired result. •
Let HP(D) denote the usual Hardy space on the disk. It is well known that HI (D)
= (1 - a 22]11+a2s
)
2 3 v('1
s) ds.
_r:
o (1 - a s)
In the above equation, replace
transforming
claimed.
a
with
(t
and define a function V on [0, 1) by Vet) =
the above equation into (7). Hence V is constant on [0,1),
v(I/t),
and thus so is v, as
c L~
(the proof follows easily from Hardy's Inequality). In the proof of Theorem 1 we will use,
without comment, the following consequence: If f, g E H2(D), then fg E L~, and thus
jgE
L2(D,dA).
We have now assembled all the ingredients needed to prove Theorem 1.
9
8
Axler
PROOF OF 11IEOREM 1: We begin with the easy direction.
so that
cp
and
Cuckovic
Axler
Cuckovic
and
*
*
"
*
UhT9Uh UhT'l'Uh = UhT'l'Uh UhT9Uh .
First suppose that (1.1) holds,
and 'V are analytic on D, which means that T9 and T'l' are, respectively, the operators on
Lemma 8 now shows that
(13)
L~ of multiplication by cp and 'V. Thus T9T'l' = T'l'T9'
Now suppose that (1.2) holds, so that q; and \if are analytic on D. By the paragraph above,
T
,l Iii = T liiT~. Take the adjoint
of both sides of this equation, and use the identity T ~* = T 9 to
conclude that T9T 'l' = T'l'T9'
Finally (for the easy direction)
bEe,
not both 0, such that
Y E C such that
cp
acp
+ b\jl is constant on D. If
a
0, then there exist constants
#-
= ~'V + yon D, which means that T = ~T'l' + yJ (here
of Theorem
Because
= /1 +
h
and
\jI = gl + gz
cp
=fIg
I + gz(O)ft
cp
and 'V are bounded
fD
(j2gl-/Igz)'hdA
on D.
rc
(10)
on D.
= h(h(O»)gl(h(O»)-fj(h(O»)gz(h(O»).
the above equation becomes
f D u'h
+gz(O»)
4.t1.
rc
= u(h(O»).
In other words, u has the area version of the invariant mean value property.
We want to show that u is harmonic on D. By the above equation and Lemma 2, we need
only show that 'R(u-h) E C(D).
To do this, represent the analytic functions!z'h
~
(jz'h)(z)
=
1>
+ gz(O)/1 + hgl
+ h(O) gz(O) dA
= rc[h (O)g I(0) + ft (O)gz(O) + h(O) gz(O)] + ID hgl dA.
A sirnilarformula
HZ(D),
L anzn
n=O
and
(gl'h)(z)
so
=
L ~nzn
n=O
for all zED.
n=O
Now for zED
Because T <pT'l'= T'l'T <p'we can set the right-hand side of (11) equal to the corresponding
and
L I~nlz <
n=O
hgj-flgzd.d
7t
= fz(O)gj(O)-ftCO)gz(O).
(12)
Let hE Aut(D). Multiplying both sides of the equation T <pT'l'= T'l'T <pby Uh on the left
and byUh* on the right, and recalling that Uhis unitary (so Uh*Uh = I), we get
(15)
00
.
we have
'R(chgl).h)(z)
= fZ1t
(gl'h)(zei6)
o (f2'h)(zei6)
.,
formula for <T'l'T<p1,1>, getting
JD
and gl'h are in
00
I. Janlz < 00
(11)
(interchangiIlg thef's and th.e g's) can be obtained for <I' 'l'T<pI,1>.
and gt"h as
~
Because cp'h and 'V'h are bounded harmonic functions on D, (14) implies that!z'h
= 5/lg1
(14)
u =hgl-ftfi,
Taylor series:
I +!z(O) gz(O)'
'V'h = gl'h + gz'h
Letting
+ Pchg I) +h(O) gz(O) .
+ hg
and
(14), says that (12) is still valid when we replace each function in it by its composition with h. In
other words,
+ gz(O)J)
I + gz(O)ft
+ h"h
Equation (12) was derived under the assumption that T <pT'l' = T'l'T <p;thus (13), combined with
J
Thus
<T<pT'l'l, 1> = <jig
cp'h = kh
and \jI are harmonic on D, there
T<pT'l'l = T<p(P1jf)= TGl(P(gl +gz»)=T<p(gl
+ h][gl
cp'h and 'V'h as the sum of an analytic function and a conjugate analytic
function:
~,
cp and 'V are bounded on D, the functions/I,h,
gl' and gz must be in HZ(D).
Let 1 denote the constant function 1 on D. Then
= P(lft
both sides of the equations in (10) with h expresses each of the bounded
functions
denotes the identity
1, suppose now that
harmonic functions on D such that T GlT'l' = T 'l'T Gl' Because
existfunctionsfl,h,
gl' and gz analytic on D such that
cp
#-
Composing
harmonic
0, we conclude in a similar
Gl
operator on L~), which clearly implies that T GlT'l' = T'l'T 9' If b
fashion that T9 and T'l' commute.
To prove the other direction
a,
suppose that (1.3) holds, so there exist constants
T Gl.hT'l'.h = T'l"hT <p·h·
=
I. ~Pn
n=O
Izl2n.
de
Zrc
10
Axler and Cuckovic
The inequalities in (15) imply that 2:;;'=0 lan~nl < 00, so the above formula for :R«(hgl)'h)
that :R(J;gl)'h)
E C(D); similarly,
we get
:R«(/li2H)
E
qD).
Thus :R(u'h)
shows
E C(D),
as
desired. Thus at this stage of the proof we know that u is harmonic.
I
•••••'J
11
~
Axler and Cuckovic
Recall that an operator is called normal if it commutes with its adjoint.
I
Theorem 1 to prove the following corollary, which states that for
cp
We can use
a bounded harmonic function
on D, the Toeplitz operator Trp is normal only in the obvious case.
~,
Let d/dZ and d/dZ denote the usual operators defined (on smooth functions on D) by
'#,
W:
dZ = ~(iL
2 dX
iL
dY
_ iiL)
Ifl is analytic, then the Cauchy-Riemann
fl'
2 ax + iiL).
ay
= ~(iL
dZ
d_
and
t~
equations show that
17: Suppose that
COROLLARY
normal operator il and only if
cp
is a bounded harmonic junction on D. Then T cp is a
lies on some line in C.
cp(D)
r·,·
~
dZ =
I,
~
,~~
d-Z = 0,
dZ = 0,
f'
~
It is easy to check that d/dZ and d/dZ obey the usual addition and multiplication
derivatives and that
aZ
a
dZ
formulas for
d
U
az
dZ =4
a (au)
=4 (/z'gl'-/1'gZ')
,r.
operator, and hence T<P' which equals a-I(T arp + ~ - ~l), is a normal operator.
To prove the other direction, suppose now that T cp is a normal operator.
Thus T cpTip =
Cji
The latter condition implies that
az
d (aJ;gl az- 11i2») =4
lies on some line in C. Then there exist constants a,
0, such that a<p + ~ is real valued on D. Thus T acp + ~ is a self-adjoint
T
is harmonic, we have
0=4
~ E C, with a
cp(D)
cp T CP' and so Theorem
1 implies that cp and are both analytic on D (in which case <pis constant,
so we are done) or there are constants a, b E C, not both 0, such that acp + b~ is constant on D.
axz + dYZ = 4 az dZ .
Thus, because
PROOF: First suppose that
az = 1 .
and
az
a (-hg1'-fr'gZ
_)
cp(D)
We conclude with a question.
lies on a line .•
Does Theorem 1 remain true if we replace the disk by an
arbitrary connected region in the plane? (So the Toeplitz operators now act on the Bergman space
onD.
of this new region.)
The statement of Theorem 1 certainly makes sense in this context and is
Hence
plausible.
11'gZ' = /z'gl'
onD.
(16)
We finish the proof by showing that the above equation implies that (1.1), (1.2), or (l.3)
holds.
If gl' is identically 0 on D, then (l6) shows that either gz' is identically 0 on D (so
would be constant on D and (1.3) would hold) or fr' is identically 0 on D (so both
Cji
J-
1.
2.
hand side is the complex conjugate of an analytic function on the same domain, and so both sides
C. Thusfr'=cg1'and/z'=cgz'onD.
holds and the proof of Theorem 1 is complete .•
cp -
Hence/1-cg1
use of the set of analytic
REFERENCES
3.
the above equation is an analytic function (on D with the zeroes of gl 'gz' deleted), and the right-
are constant on D, and so their sum, which equals
1 made extensive
will not work in general, because nonsimp1y-
\jf
at all points of D except the countable set consisting of the zeroes of gl 'gz'. The left-hand side of
mustequalaconstantcE
This approach
and \ji would
that either (1.3) or (1.1) would hold. Thus we may assume that neither g l' nor gz' is identically 0
on D, and so (16) shows that
~z',
the proof of Theorem
of the disk.
connected regions have too few analytic automorphisms to provide any useful information.
be analytic on D and (l.2) would hold). Similarly, if gz' is identically 0 on D, then (l6) shows
~11" =
However,
automorphisms
andfz-cfi
c\jf, is constant on D; in other words, (1.3)
4.
J. ARAZY, S. D. FISHER, AND J. PEETRE, Hankel operators on weighted Bergman
spaces, American J. Math. 110 (1988), 989-1053.
SHELDON AXLER AND PAMELA GORKIN, Algebras on the disk and doubly commuting
multiplication operators, Trans. Amer. Math. Soc. 309 (1988),711-723.
ARLEN BROWN AND P. R. HALMOS, Algebraic properties of Toeplitz operators, J. Reine
Angew. Math. 213 (1964), 89-102.
W ALTER RUDIN, Function Theory on the Unit Bail ofCn, Springer-Verlag, New York,
1980 ..