A Note on Failure of Energy Reversal for Classical Fractional Singular Integrals

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Advance Access Publication December 22, 2014 doi:10.1093/imrn/rnu210

A Note on Failure of Energy Reversal for Classical Fractional Singular Integrals

Eric T. Sawyer

¹

, Chun-Yen Shen

²

, and Ignacio Uriarte-Tuero

³

1

Department of Mathematics & Statistics, McMaster University, 1280 Main Street West, Hamilton, ON, Canada L8S 4K1,

²

Department of Mathematics, National Central University, Chungli 32054, Taiwan, and

3

Department of Mathematics, Michigan State University, East Lansing, MI, USA

Correspondence to be sent to: ignacio@math.msu.edu

For 0≤α <n, we demonstrate the failure of energy reversal for the vector ofα-fractional Riesz transforms, and more generally for the vector of allα-fractional convolution singular integrals having a kernel with vanishing integral on every great circle of the sphere.

1 Introduction

In the recent two-part solution to the Nazarov–Treil–Volberg conjecture for the two weight Hilbert transform inequality (Lacey et al. Part I [2], Lacey Part II [1]), crucial use was made of the following point-wise equivalence for the difference quotients of the Hilbert transform Hμ associated with x,x in an interval J and a positive measureμ

Received April 17, 2014; Revised September 16, 2014; Accepted October 16, 2014 Communicated by Nikolai Makarov

by guest on November 18, 2015http://imrn.oxfordjournals.org/Downloaded from

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supported outside the double 2J: Hμ(x)−Hμ(x)

x−x = 1 x−x

R\2J

1

y−x− 1 y−x

dμ(y)

= 1

|J|

R\2J

|J|

(y−x)(y−x)dμ(y)

≈P(J,1_R\2Jμ)

|J| . (1.1)

Here,Hμ(x)≡

R 1

y−xdμ(y)is the Hilbert transform of the measureμon J, and P(J, ν)≡

R |J|

|J|²+|y−cJ|² dν(y)is the Poisson integral ofνassociated with the interval J, and where c_J is the center of J. The important consequence of this point-wise equivalence is that a difference quotient of the singular integral Hμ is comparable with a positive quan- tity ^P⁽_|^J_J^,μ)_| , that is in turn monotone increasing in the measureμ. A striking difference between the one and higher dimensional settings for singular integrals is that this equivalence fails in both directions and for singular integrals in Rⁿ with n>1.

In fact, the forward direction requires an additional energy term derived by Lacey and Wick [3], and with this addition the forward inequality provides a suitable substi- tute for use in higher dimensions. However, there is a particularly spectacular failure of the reverse inequality, which we refer to as energy reversal, a terminology explained below. In fact, this failure of energy reversal in higher dimensions underlies the restric- tive nature of higher dimensional analogs of the two weight theorem for the Hilbert transform in both [3,5], where in earlier versions of each of these papers, errors were made in assuming some version of energy reversal.

To set direction for this paper, we first recall a special case of Theorem 1 from our paper [5], using notation from that paper, which we now review here, albeit very briefly. Let 0≤α <n. Consider a kernel function K^α(x,y)defined on Rⁿ×Rⁿsatisfying the following fractional size and smoothness conditions of order 1+δ for someδ >0,

|K^α(x,y)| ≤CCZ|x−y|^α−n,

|∇K^α(x,y)| ≤C_CZ|x−y|^α−ⁿ⁻¹,

|∇K^α(x,y)− ∇K^α(x,y)| ≤CCZ

|x−x|

|x−y| _δ

|x−y|^α−n−1, |x−x|

|x−y| ≤1 2,

|∇K^α(x,y)− ∇K^α(x,y)| ≤CCZ

|y−y|

|x−y| _δ

|x−y|^α−ⁿ⁻¹, |y−y|

|x−y| ≤1 2.

(1.2)

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We define a standardα-fractional Calder ´on–Zygmund operator associated with such a kernel as follows.

Definition 1. We say thatT^α is astandardα-fractional integral operator with kernel K^α ifT^α is a bounded linear operator from some L^p(Rⁿ)to some L^q(Rⁿ)for some fixed 1<p≤q<∞, that is

T^αf L^q(Rⁿ)≤C f L^p(Rⁿ), f∈L^p(Rⁿ),

ifK^α(x,y)is defined onRⁿ×Rⁿand satisfies (1.2), and ifT^αandK^αare related by T^αf(x)=

K^α(x,y)f(y)dy, a.e.-x∈/ supp f,

whenever f∈L^p(Rⁿ) has compact support in Rⁿ. We say K^α(x,y) is a standard

α-fractional kernelif it satisfies (1.2).

In higher dimensions, there are the twoα-fractional Poisson integrals ofμon a cubeQare given by

P^α(Q, μ)≡

Rⁿ

|Q|¹ⁿ

(|Q|¹ⁿ+ |x−x_Q|)ⁿ⁺¹^−αdμ(x), P^α(Q, μ)≡

Rⁿ

|Q|¹ⁿ (|Q|¹ⁿ+ |x−x_Q|)²

n−α

dμ(x).

(1.3)

We refer to P^α as the standard Poisson integral and to P^α as the reproducing Pois- son integral. We refer the reader to [5] for the more technical definitions of elliptic and strongly elliptic operatorsT^α, and various notions of energy.

Finally, we define theα-energy conditionconstant to be (Eα)²≡ sup

Q= ˙∪Qr

Q,Qr∈Dⁿ

1

|I|σ

∞ r=1

J∈Mr-^deep^(Q^r⁾

P^α(J,1Qσ )

|J|¹ⁿ 2

P^subgood,ω_J x ²_L2(ω)

+sup

≥0

sup

Q

1

|I|_σ

J∈Mr-^deep⁽^Q⁾

P^α(J,1Qσ )

|J|¹ⁿ 2

P^subgood,ω_J x ²_L2(ω),

where the supremum is taken over all dyadic gridsDⁿand all dyadic cubes inDⁿ, and where the goodness parameters r and ε implicit in the definition of Mr-^deep(K) and M_r-^deep(K) are fixed sufficiently large and small, respectively, depending on nandα. Here the collectionMr-^deep(K)consists of themaximalr-deeply embedded dyadic sub- cubes J of a cube K (a subcube J of K is adyadic subcube of K if J∈DwhenD is a

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dyadic grid containingK), and for each≥0, the collectionM_r-^deep(K)consists of those J∈Mr-^deep(πK) (where πK denotes the -fold dyadic parent of K) that are r-deeply embedded inK, that is,

M_r-^deep(K)≡ {J∈Mr-^deep(πK):J⊂Lfor someL∈Mr-^deep(K)}.

Of courseM⁰_r-^deep(K)=Mr-^deep(K), butM_r-^deep(K)is in general a finer subdecomposi- tion of K the largeris, and may in fact be empty. There is a similar definition of the dualα-energy condition constantE_α^∗. We refer the reader to [5], or almost any other paper on the subject, for the detailed definition and properties ofgooddyadic cubes.

Theorem 1. Suppose thatσ andωare locally finite positive Borel measures inRⁿwith no common point masses, and assume the finiteness of theα-energy conditionconstant Eαand its dual constantE_α^∗. LetT^αbe a standard strongly ellipticα-fractional Calder ´on–

Zygmund operator in Euclidean spaceRⁿ. ThenT^αis bounded fromL²(σ )toL²(ω)if and only if theA^α2 condition

A^α₂≡sup

Q∈QⁿP^α(Q, σ) |Q|ω

|Q|¹⁻^αⁿ <∞ (1.4)

and its dual hold, the cube testing conditions

Q

|T^α(1_Qσ)|²ω≤T²_Tα

Q

dσ and

Q

|(T^α)^∗(1_Qω)|²σ≤T²_Tα

Q

dω, (1.5) hold for all cubesQinRⁿ, and the weak boundedness property forT^αholds:

Q

T^α(1Qσ )dω

≤WBPT^α

|Q|ω|Q|σ,

for all cubesQ,Qwith 1

C ≤ |Q|¹ⁿ

|Q|¹ⁿ ≤C,

and eitherQ⊂3Q\QorQ⊂3Q\Q. In [7], we used Theorem1to prove theT1 theorem for the vector of Riesz transforms in Rⁿ in the special case when one of the measures σ, ω is supported on a line inRⁿ. The key to that proof was proving control of the above energy constants E_α and E_α^∗ in terms of the constants in the hypotheses (1.4) and (1.5), and this in turn used an energy reversal that exploited the 1D support of one of the measures. As mentioned above, a number of attempts have been made to prove such control of various different energy conditions by invoking anenergy reversalfor the Riesz transforms and similar operators—see (2.4)—but all of these attempts have been met with failure. The purpose

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of this paper is to show first that energy reversal is false, not only for the vector of α-fractional Riesz transforms in the plane when 0≤α <2, but also for the vectors of classicalα-fractional singular integrals in the plane,

T^α_M≡ {T_Ω^α:Ω∈PM}, PM≡ {cosnθ,sinnθ}^Mn=1, where T_Ω^α has convolution kernel ^Ω(

x

|x|)

|x|^2−α =_|x|^Ω(θ)2−α and 0≤α <2. The linear space LM of trigonometric polynomials with vanishing mean and degree at mostMis spanned by the monomialsPM, and so we also obtain the failure of energy reversal for the infinite vec- torT^α_M≡ {T_Ω:Ω∈LM}. A standard limiting argument applied to the proof below extends this failure to all sufficiently smoothΩ(θ)with vanishing mean on the circle. Finally, we embed an analog of the planar measure constructed below into Euclidean spaceRⁿto obtain the failure of energy reversal forany vector of classical convolution Calder ´on–

Zygmund operators with odd kernel inRⁿ—and more generally for kernels _|^Ω(_x_|_n−α^x⁾ where Ω has vanishing integral on every great circle in the sphereSⁿ⁻¹. A key to our proof is the positivity of the determinants det[_{Γ (z−|i−}^{Γ (z)}j|)Γ (z+|i−² j|)]ⁿ_i,_j=1for alln≥1, whereΓ (z)is the gamma function. See also [3] for related results regarding fractional Riesz transforms in higher dimensions.

2 Failure of Reversal of Energy

Recall the energyE(J, ω)of a measureωon a cube J,

E(J, ω)²≡ 1

|J|ω

1

|J|ω

J

x−z

|J|¹ⁿ

²dω(x)dω(z)=2 1

|J|ω

J

x−E^ω_Jx

|J|¹ⁿ

²dω(x).

Define its associatedcoordinateenergiesE^j(J, ω)by

E^j(J, ω)²≡ 1

|J|_ω 1

|J|_ω

J

x^j−z^j

|J|¹ⁿ

²dω(x)dω(z), j=1,2, . . . ,n,

and the rotationsE_R^j (J, ω)of the coordinate energies by a rotationR∈SO(n), which we refer to aspartialenergies,

E_R^j (J, ω)²≡ 1

|J|ω

1

|J|ω

J

x_R^j −z_R^j

|J|¹ⁿ

²dω(x)dω(z), j=1,2, . . . ,n,

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where for R∈SO(n), x_R=(x_R^j)ⁿ_j₌₁=R(x^j)ⁿ_j₌₁=Rx. Set E_R(J, ω)²≡E¹_R(J, ω)²+ · · · + Eⁿ_R(J, ω)². We have the following elementary computations.

Lemma 1. ForR∈SO(n)we have

E_R(J, ω)²=E¹_R(J, ω)²+ · · · +Eⁿ_R(J, ω)²=E(J, ω)². (2.1) More generally, ifR= {Rj}ⁿ_j=1⊂SO(n)is a collection of rotations such that the matrix M_R=

_R

1e¹ Rn...e¹

with rowsRe¹is nonsingular, then

E(J, ω)²≤ 1 R

n

=1

E¹_R(J, ω)², (2.2)

where_Ris the least eigenvalue ofM_R^∗M_R.

Proof. We have

|x_R¹ −z¹_R|²+ · · · + |xⁿ_R−zⁿ_R|²= |R(x−z)|²

= |x−z|²= |x¹−z¹|²+ · · · + |xⁿ−zⁿ|², so that

E_R(J, ω)²≡E¹_R(J, ω)²+ · · · +Eⁿ_R(J, ω)²

=E¹(J, ω)²+ · · · +Eⁿ(J, ω)²=E(J, ω)². More generally, ifM_R denotes theth row of the matrixM_R, we have

R|x−z|²≤(x−z)^trM_R^∗ M_R(x−z)

= n

=1

|Re¹·(x−z)|²,

so that

RE(J, ω)²=

1

|J|ω|J|¹ⁿ 2

J

R|x−z|²dω(x)dω(z)

≤

1

|J|_ω|J|¹ⁿ 2

J

_n

=1

|Re¹·(x−z)|²

dω(x)dω(z)

= n

=1

E¹_R(J, ω)².

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The point of the estimate (2.2) is that it could hopefully be used to help obtain a reversal of energy for a vector transformTⁿ^,α= {Tⁿ^,α}ⁿ₌₁, where the convolution kernel Kⁿ^,α(w)of the operatorTⁿ^,αhas the form

K^n,α(w)=Ωⁿ

|w|w

|w|ⁿ^−α , (2.3)

and where Ωⁿ is smooth on the sphere Sⁿ⁻¹. We refer to the operator Tⁿ^,α as an α-fractionalconvolutionCalder ´on–Zygmund operator. If in addition we require thatΩⁿ has vanishing integral on the sphere Sⁿ⁻¹, we refer to Tⁿ^,α as a classical α-fractional Calder ´on–Zygmund operator.

However, we now dash this hope, at least for the most familiar singular operators in the plane, in a spectacular way. Here is the key definition we work with.

Definition 2. A vectorT^α= {T^α}₌^N₁ofα-fractional transforms in Euclidean spaceRⁿsat- isfies astrongreversal ofω-energy on a cube J if there is a positive constantC₀ such that for allγ≥2 sufficiently large and for all positive measuresμsupported outsideγJ, we have the inequality

E(J, ω)²P^α(J, μ)²≤C0Edω(x) J Edω(z)

J |T^αμ(x)−T^αμ(z)|². (2.4) Note that in dimensionn=1, (2.4) is an immediate consequence of (1.1)—simply squareHμ(x)−Hμ(x)≈^x_|J|⁻^xP(J, μ)and takeω-expectations over J in bothxandx. We show that (2.4) is false in higher dimensions by stating and proving a variant of Lemma 9 in an earlier paper [6] that has since been withdrawn.

Lemma 2(Failure of Reverse Energy). Suppose thatJis a square in the planeR², 0≤α <

2,γ >2 and thatR^α= {R^α}²₌₁is the vector ofα-fractional Riesz transforms in the plane R²with kernelsK^α(w)=^Ω_|w|⁽2^|w|^w−α⁾ andΩ(_|w|^w)=_|w|^w. Finally, suppose thatC0>0 is given. For γ sufficiently large, there exists a positive measureμ onR² supported outsideγJ and depending only onαandγ, such that the strong reversal ofω-energy inequality (2.4)fails for a certain class of measuresω. Moreover, we can chooseμas above so that in addition, for anyM≥1, the strong reversal ofω-energy inequality (2.4) fails for the vectorT^α_M. As a corollary of the proof of this lemma we easily obtain an extension to higher dimensions by simply embedding an appropriate planar measure into Euclidean spaceRⁿ.

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Corollary 1(of the proof of Lemma 2). Suppose that J is a cube inRⁿ, 0≤α <n,γ >2 and suppose thatC₀>0 is given. Forγ sufficiently large, there exists a positive mea- sureμon Rⁿsupported outsideγJ and depending only onn,α, andγ, such that for a certain class of measuresωthe strong reversal ofω-energy inequality (2.4)failsfor any vectorT^α= {T^α}₌₁^N ofα-fractional smooth Calder ´on–Zygmund operators inRⁿwith ker- nelsK^α(w)=^Ω_|w|⁽n−α^|w|^w⁾, whereΩ has vanishing integral on every great circle in the sphere

Sⁿ⁻¹—in particular this holds if eachK^αis odd.

2.1 Failure of weak reversal of energy

The right-hand side of (2.4) is clearly dominated by C₀Edω

J |T^αμ|², and so we say that T^α= {T^α}²₌₁satisfies aweakreversal ofω-energy on a cube Jif forγ andμas above, we have the weaker inequality

E(J, ω)²P^α(J, μ)²≤C0Edω

J |T^αμ|². (2.5)

Now we show that even this weaker form of energy reversal fails, although not as spec- tacularly as the strong energy reversal. We content ourselves with the following special case for the Riesz transform vectorR^α.

Lemma 3(Failure of Weak Reverse Energy). Suppose thatJ is a square in the planeR², 0≤α <2,γ >2 and thatR^α= {R^α}²₌₁is the vector ofα-fractional Riesz transforms with kernels K^α(w)=^Ω_|w|⁽2−α^|w|^w⁾. Finally suppose thatC₀>0 is given. Forγ >2 sufficiently large, there is a positive measureμ=μα,γ,T^α on R² supported outsideγJ and depending only onα,γ, andR^α, such that for a certain class of measuresωtheweakreversal ofω-energy

inequality (2.5)fails.

3 The Proofs

Proof of Lemma2for the Riesz transform vector. Let ε >0. We let ^Ω⁽

|w|w)

|w|²^−α be an arbitrary standard kernel for the moment with smoothness indexδ in (1.2). With K^α(x,y)= K^α(x−y)andx,z∈J, there is a pointξx,zon the line joiningxandzsuch that

T^αμ(x)−T^αμ(z)=

{K^α(x−y)−K^α(z−y)}dμ(y)

=

{(x−z)· ∇K^α(ξx,z−y)}dμ(y)

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=

{(x−z)· ∇K^α(c_J−y)}dμ(y) +

{(x−z)·[∇K^α(ξx,z−y)− ∇K^α(c_J−y)]}dμ(y)

≡Λ^α +E_,^α_x_,_z.

Ifγ >2 is sufficiently large, (1.2) gives

|E^α_,_x_,_z| ≤

|x−z||∇K^α(ξx,z−y)− ∇K^α(cJ−y)|dμ(y)

≤

(γJ)^c|x−z|CCZ

|ξx,z−c_J|

|c_J−y| _δ

|cJ−y|^α−3dμ(y)

≤C_CZ|x−z|

|ξx,z−cJ|

|cJ−y| _δ

|c_J−y|^α−³dμ(y),

and so from (1.3) withn=2, we obtain

|E_,x,z^α | ≤C 1 γ^δ

P^α(J, μ)

|J|¹² |x−z| ≤εP^α(J, μ)

|J|¹² |x−z|. (3.1) The point of this inequality (3.1) is that it permits the replacement of the difference T^αμ(x)−T^αμ(z) in (2.4) by the linear part Λ^α of the Taylor expansion of the kernelK^α.

Now we make the choice

Ω(w)=Ω(θ(w));

θ(w)≡tan⁻¹(−1)w

w , 1≤≤2,

wherewdenotes the coordinate variable other thanw, that is,+=3. Thus,θ1is the usual angular coordinate on the circle andθ2=θ1+^π₂. We now use

∇|w|^α−²= ∂

∂w¹((w¹)²+(w²)²)^α−2² , ∂

∂w²((w¹)²+(w²)²)^α−2²

=α−2

2 ((w¹)²+(w²)²)^α−2² ⁻¹2w

=(α−2)|w|^α−⁴w.

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and

∂

∂wtan⁻¹w

w = 1

1+(^w_w)²

−w

(w)²=−w

|w|²,

∂

∂w tan⁻¹w

w = 1

1+(^w_w)² 1 w= w

|w|². to calculate that the gradient of the convolution kernel

K^α(w)=Ω(w)

|w|^2−α =Ω(θ(w))

|w|^2−α =Ω(tan⁻¹^w_w)

|w|^2−α , is given by,

∇K^α(w)= ∇

Ω(w)

|w|^2−α

=Ω(θ(w))∇|w|^α−²+ |w|^α−²Ω(θ(w))∇θ

=(α−2)Ω(θ(w))w+Ω(θ(w))w^⊥

|w|⁴^−α .

Thus, the linear partΛ^α in the Taylor expansion ofT^αμis given by Λ^α=(x−z)·

∇K^α(c_J−y)dμ(y)≡(x−z)·Z^α_Ω

(c_J;μ), where

Z^α_Ω(cJ;μ)=

R²

(α−2)Ω(θ(c_J−y))(c_J−y)+Ω(θ(c_J−y))(c_J−y)^⊥

|cJ−y|^4−α dμ(y)

=

w∈S¹{(α−2)Ω(θ(w))w¹−Ω(θ(w))w²}e¹dΨ_μ(w) +

w∈S¹{(α−2)Ω(θ(w))w²+Ω(θ(w))w¹}e²dΨ_μ(w),

andeis the coordinate vector with a 1 in theth position. Here the measureΨ_μ is an arbitrarypositive finite measure on the circleS¹given formally by

dΨμ(w)=dΨ_μ^J(w)= _∞

0

r^α−³dμ_w^J(r)= _∞

0

r^α−³dμ(c_J+rw), w∈S¹.

Note that conversely, if we are given a positive finite measure Ψ on the circle, we can simply translate and dilateΨ, preserving its mass, to be supported in a circle centered atc_Jwith radiusR. Then, the resulting measureμwill satisfy dΨ_μ^J=dΨ. Below we will typically apply this converse observation withRchosen to beγ|J|¹² to obtain our counterexamples.

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We use,

tanθ(w)=(−1)w w ,

cscθ(w)=(−1)

1+cot²θ(w)=(−1)

1+ w

w 2

= |w|

(−1)w, sinθ(w)=(−1)w

|w| and cosθ(w)= w

|w|, forw=0, to obtain

Z^α_Ω₁(c_J;μ)=

S¹{(α−2)Ω(θ1(w))cosθ1(w)−Ω(θ1(w))sinθ1(w)}e¹dΨμ

+

S¹{(α−2)Ω(θ1(w))sinθ1(w)+Ω(θ1(w))cosθ1(w)}e²dΨμ

≡

S¹{A¹_α(θ1(w))e¹+B_α¹(θ1(w))e²}dΨ_μ, and

Z^α_Ω₂(c_J;μ)=

S¹{−(α−2)Ω(θ2(w))sinθ2(w)−Ω(θ2(w))cosθ2(w)}e¹dΨμ

+

S¹{(α−2)Ω(θ2(w))cosθ2(w)−Ω(θ2(w))sinθ2(w)}e²dΨ_μ

≡

S¹{A²_α(θ2(w))e¹+B_α²(θ2(w))e²}dΨ_μ, with

A¹_α(t)=(α−2)Ω(t)cost−Ω(t)sint=B_α²(t), B_α¹(t)=(α−2)Ω(t)sint+Ω(t)cost= −A²_α(t).

(3.2)

Now we show below in (3.7) that a necessary condition for reversal of energy on J is that the span of the pair of vectors{Z^α_Ω

(c_J;μ)}²₌₁is all ofR²:

Span{Z^α_Ω(c_J;μ)}²₌1=R². (3.3) So it suffices to show the failure of (3.3), that is, thatZ^α_Ω₁(c_J;μ)andZ^α_Ω₂(c_J;μ)are parallel, or at least one of them is the zero vector.

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At this point, we take=1 and setθ=θ1(w)so that we obtain A_α(θ)≡A¹_α(θ1(w))=(α−2)Ω(θ)cosθ−Ω(θ)sinθ, B_α(θ)≡B_α¹(θ1(w))=(α−2)Ω(θ)sinθ+Ω(θ)cosθ.

(3.4)

In the caseα=1, these coefficients are perfect derivatives:

A1(θ)= −Ω(θ)cosθ−Ω(θ)sinθ= −[Ω(θ)sinθ], B1(θ)= −Ω(θ)sinθ+Ω(θ)cosθ= −[Ω(θ)cosθ],

and so have vanishing integral on the circle. Thus with the choice dΨ_μ(θ)=dθwe have Z_Ω(c_J;μ)=

S¹{A₁(θ)e¹+B₁(θ)e²}dθ=0 the zero vector, foreverychoice of differentiableΩ on the circle.

In the case 0≤α <2 withα=1, it is no longer possible to find a nontrivial mea- sureμso that Z^α_Ω(cJ;μ)vanishes for all differentiable Ω, but we will see that we can always find a positive measureμsuch that the vectorsZ^α_Ω₁(cJ;μ)andZ^α_Ω₂(cJ;μ)are parallel for the choiceΩ(θ)=cosθthat corresponds to the vector of Riesz transforms.

Indeed, in the special case thatΩ(t)=cost, and recalling thatθ2(w)=θ1(w)+^π₂= θ+^π₂, we have

A¹_α(θ1(w))=A¹_α(θ)=(α−2)cos²θ+sin²θ;

B_α¹(θ1(w))=B_α¹(θ)=(α−3)cosθsinθ;

A²_α(θ2(w))= −B_α¹ θ+π

2

= −(α−3)cos θ+π

2

sin θ+π

2

=(α−3)cosθsinθ;

B_α²(θ2(w))=A¹_α θ+π

2

=(α−2)cos² θ+π

2

+sin² θ+π

2

=(α−2)sin²θ+cos²θ.

Thus we also have Z^α_Ω₁(c_J;μ)=

S¹{A¹_α(θ1(w))e¹+B_α¹(θ1(w))e²}dΨμ

=

S¹{[(α−2)cos²θ+sin²θ]e¹+[(α−3)cosθsinθ]e²}dΨμ

=

S¹[(α−2)cos²θ+sin²θ] dΨ_μ

e¹+

S¹[(α−3)cosθsinθ] dΨ_μ

e²

(13)

and

Z^α_Ω₂(cJ;μ)=

S¹{A²_α(θ2(w))e¹+B_α²(θ2(w))e²}dΨμ

=

S¹{[(α−3)cosθsinθ]e¹+[(α−2)sin²θ+cos²θ]e²}dΨμ

=

S¹[(α−3)cosθsinθ] dΨ_μ

e¹+

S¹[(α−2)sin²θ+cos²θ] dΨ_μ

e².

Using

(α−2)cos²θ+sin²θ=(α−3)cos²θ+1, (α−2)sin²θ+cos²θ=(α−3)sin²θ+1,

sinθcosθ=1

2sin 2θ, cos²θ=1+cos 2θ

2 , sin²θ=1−cos 2θ

2 ,

(3.5)

we see that

(α−2)cos²θ+sin²θ=(α−3)1+cos 2θ

2 +1=α−3

2 cos 2θ+α−1 2 , (α−2)sin²θ+cos²θ=(α−3)1−cos 2θ

2 +1= −α−3

2 cos 2θ+α−1 2 , (α−3)cosθsinθ=α−3

2 sin 2θ.

Plugging these formulas into those forZ^α_Ω₁(c_J;μ)andZ^α_Ω₂(c_J;μ),we obtain

det

Z^α_Ω₁(cJ;μ) Z^α_Ω₂(cJ;μ)

=det

⎡

⎢⎢

⎢⎣

S¹

α−3

2 cos 2θ+α−1 2

dΨμ

S¹

α−3 2 sin 2θ

dΨμ

S¹

α−3 2 sin 2θ

dΨμ

S¹

−α−3

2 cos 2θ+α−1 2

dΨμ

⎤

⎥⎥

⎥⎦

= α−3

2

S¹

cos 2θdΨμ+α−1

2 Ψμ −α−3 2

S¹

cos 2θdΨμ+α−1 2 Ψμ

−

α−3 2

S¹

sin 2θdΨ_μ 2

= α−1

2 Ψμ

2

−

α−3 2

S¹

cos 2θdΨμ

2

+ α−3

2

S¹

sin 2θdΨμ

2 .

(14)

Thus det_Zα Ω1(cJ;μ) Z^α_Ω2(cJ;μ)

=0 if and only if the length of the vector

α−3 2

⎛

⎜⎜

⎝

S¹cos 2θdΨμ

S¹sin 2θdΨ_μ

⎞

⎟⎟

⎠

equals ^|α−₂¹^| Ψμ , that is,

""

⎛

⎜⎜

⎝

S¹cos 2θdΨμ

S¹sin 2θdΨ_μ

⎞

⎟⎟

⎠

""

""=|α−1|

|α−3| Ψ_μ . (3.6)

To construct a positive probability measure dΨμon the circle that satisfies (3.6), we first observe that if dΨμ=δ(1,0)is the unit point mass at(1,0), then

""

⎛

⎜⎜

⎝

S¹cos 2θdΨμ

S¹sin 2θdΨ_μ

⎞

⎟⎟

⎠

""

""=

""

⎛

⎝

S¹

dΨ_μ 0

⎞

⎠""

""

""= Ψ_μ , and since|α−1|<|α−3|for all 0≤α <2, we have

""

⎛

⎜⎜

⎝

S¹cos 2θdΨ_μ

S¹

sin 2θdΨ_μ

⎞

⎟⎟

⎠

""

"">|α−1|

|α−3| Ψμ ,

in this case. On the other hand, if dΨμ(θ)=₂¹_πdθis normalized Lebesgue measure on the circle, we have

""

⎛

⎜⎜

⎝

S¹

cos 2θdΨ_μ

S¹

sin 2θdΨμ

⎞

⎟⎟

⎠

""

""=""

""

"

0 0

""

"=0<|α−1|

|α−3| Ψμ .

It is now easy to see that there is a convex combination dΨ_μ=(1−λ)δ₍1,0)+λ₂¹_πdθ such that (3.6) holds. Thus (3.3) fails, and we now show that energy reversal fails.

In fact, we may assume that both Z^α_Ω₁(cJ;μ) and Z^α_Ω₂(cJ;μ) are parallel to the coordinate vectore₂, and in this case we will see that we can reverse at most the coordinate energyE²(J, ω), defined above by

E²(J, ω)²≡ 1

|J|ω

1

|J|ω

J

x²−z²

|J|¹ⁿ

² dω(x)dω(z),