This is the only term that has positive value

Case 2 π/4 ≤ θ ≤ ^{cos−1 1√}₃^{, and} ϕ ∈ D. (further estimate)

In this case π/4 ≤ θ ≤ ^{cos−1 1√}₃ and since we want ω₁ ≥ |ω²|, ^and ω₁ ≥ |ω³|.

Thus we have cotθ ≥ |^sinϕ| ^{and cot}θ ≥ |^cosϕ| in spherical coordinates. As a result, the range of ϕ will be restricted on D which is

On the first quadrant: cos⁻¹(cotθ) ≤ϕ ≤ ^sin−1(cotθ) On the second quadrant: π

2 +cos⁻¹(cotθ) ≤ϕ ≤ π

2 +sin⁻¹(cotθ) On the third quadrant: π +cos⁻¹(cotθ) ≤ϕ ≤ π +^sin−1(cotθ) On the fourth quadrant: 3π

2 +cos⁻¹(cotθ) ≤ϕ ≤ 3π

2 +sin⁻¹(cotθ).

Observe that since f is even so that for all ϕ we have

F^ω1(t)f^ω2(t)f^ω3(t) = F (tcosθ)f (tsinθcosϕ)f (tsinθsinϕ)

= F (tcosθ)f (tsinθcos(ϕ + π/2))f (tsinθsin(ϕ + π/2))

= F (tcosθ)f (tsinθsinϕ)f (tsinθcosϕ)

= F^ω1(t)f^ω3(t)f^ω2(t).

Thus

Z _sin⁻¹_{(cot θ)}

cos⁻¹(cot θ)

(h₁(θ, ϕ) + h₂(θ, ϕ) + h₃(θ, ϕ) + h₄(θ, ϕ))dϕ

=

Z _sin⁻¹(cot θ)+^π₂ cos⁻¹(cot θ)+^π₂

(h₁(θ, ϕ) + h₂(θ, ϕ) + h₃(θ, ϕ) + h₄(θ, ϕ))dϕ

=

Z _sin⁻¹_{(cot θ)+}^2π

2

cos⁻¹(cot θ)+^2π₂

(h₁(θ, ϕ) + h₂(θ, ϕ) + h₃(θ, ϕ) + h₄(θ, ϕ))dϕ

=

Z _sin⁻¹(cot θ)+^3π₂ cos⁻¹(cot θ)+^3π₂

(h₁(θ, ϕ) + h₂(θ, ϕ) + h₃(θ, ϕ) + h₄(θ, ϕ))dϕ.

Hence the integral that we want to estimate is equal to Z _cos^{−1 1}√

3

π/4

cosθ Z

D

(h1(θ, ϕ) + h2(θ, ϕ) + h3(θ, ϕ) + h4(θ, ϕ))dϕsinθdθ (4.10)

=

Z _cos^{−1 1}√ 3

π/4

4cosθ

Z _sin⁻¹(cot θ) cos⁻¹(cot θ)

(h1(θ, ϕ) + h2(θ, ϕ) + h3(θ, ϕ) + h4(θ, ϕ))dϕsinθdθ.

Integrating with respect to ϕ is a closed form as above. However, when integrat-ing with respect to θ, we are unable to find its closed form. The reason is that the range is from cos⁻¹(cotθ) to sin⁻¹(cotθ), and after integrating the variable ϕ these upper and lower limits make the integrand in the variable θ extremely complicated. Therefore this case will be further estimated in the final section.

Second or third part vanishes before the others.

In this case it suffices to consider f^ω2 vanishes before the others. The reasons are the followings. First, we observe that

F^ω1(t)f^ω2(t)f^ω3(t) = F (tcosθ)f (tsinθcosϕ)f (tsinθsinϕ)

= F (tcosθ)f (tsinθcos(ϕ + π/2))f (tsinθsin(ϕ + π/2)), and notice that the if f^ω3 vanishes before the others, the integral range for ϕ in this case is only different from the integral range for ϕ by rotating ^π₂ in the case that f^ω2 vanishes before the others. Now we see that for f^ω2 vanishes before the others, we must have ϕ ∈ [−π/4, π/4] ^and [−3π/4, 5π/4]. Again since f is even, it suffices to only consider the range [−π/4, π/4]^{. Thus}

Z _2π

0

ξ₃(ω)dϕ = Z _2π

0

Z _∞

0

t²F^ω1(t)f^ω2(t)f^ω3(t)dt

= 4 Z π/4

−π/4

Z _∞

0

t²F^ω1(t)f^ω2(t)f^ω3(t)dtdϕ.

We now determine the range of θ. Assume f^ω2 vanishes before the others i.e. |ω²| ≥ |ω¹| ^and |ω²| ≥ |ω³|. Notice that ω₁² + ω₂² + ω₃² = 1. If |ω²| =

|^sinθcosϕ| ≥ ^√1₂, then for allϕ,|ω²|is always the largest one. If|ω²| ∈ [^√1₃,^√1

2], then |ω²| ≥ |ω¹| ^{for some} ϕ. If |ω²| ≤ ^√1₃, then by pigeonhole principle one of |ω1|, |ω3| is larger than ^√1

3 which is larger than |ω2|^{. Thus} θ must be in [π/4, π/2]. There are two different situations we need to separately deal with 1 When θ ∈ [^{cos−1 1√}₃, π/2], |ω²| is larger than |ω¹| ^and |ω³| ^{for all} ϕ ∈

[−π/4, π/4] ^.

2 When θ ∈ [π/4,^{cos−1 1√}₃], |ω2| is larger than |ω1| ^and |ω3| ^{for some} ϕ ∈ [−π/4, π/4]^.

Case 3 cos⁻¹(^√1

3) ≤ θ ≤ π/2^.

We also break F^ω1 into 4 pieces as before. And the decompositions in this term are the most complicated one since we need to decide which of the 5 pieces F₁₁^ω1, F₁₁^ω1+ F₁₂^ω1, F₁₁^ω1+ F₁₂^ω1+ F₁₃^ω1,F₁₁^ω1+ F₁₂^ω1+ F₁₃^ω1+ F₁₄^ω1, andf^ω2 vanishes before the others according to θ and ϕ. More precisely, since f^ω2(t) vanishes before the others and F (tω₁) = F₁₁^ω1+ F₁₂^ω1+ F₁₃^ω1+ F₁₄^ω1. Therefore there are 4 possibilities that f^ω2 vanishes before (1) F₁₁^ω1 (2) F₁₁^ω1 + F₁₂^ω1, (3) F₁₁^ω1 + F₁₂^ω1 + F₁₃^ω1 and (4) F₁₁^ω1+ F₁₂^ω1+ F₁₃^ω1+ F₁₄^ω1. We now further decompose these 4 possibilities in terms of the range of ϕ.

1: These 4 possibilities hold for all ϕ ∈ [−π/4, π/4]^.

Part A f^ω2 vanishes before F₁₁^ω1. It gives

Now we define some notations to simplify our expressions. Let g₁(θ, ϕ) : = S₁( 2

Part C ξ3(ω) = h1(θ, ϕ) + h2(θ, ϕ) + g3(θ, ϕ).

Part D ξ₃(ω) = h₁(θ, ϕ) + h₂(θ, ϕ) + h₃(θ, ϕ) + g₄(θ, ϕ).

Since we have further decomposed the integral into these 4 possibilities for all ϕ ∈ [−π/4, π/4], we need to determine the range of θ.

Part A (Closed form)

Since we now have 0≤ _|ω²_2| ≤ _2|ω¹_1| ^or 0≤ _|ω²_3| ≤ _2|ω¹_1| ^{for all} ϕ ∈ [−π/4, π/4]^{. In} order to satisfy the condition max{_|ω²_2|, _|ω²

3|} ≤ _2|ω¹_1|, thus we have max{_|ω²_2|,_|ω²

3|} ≤

2√ 2

sin θ ≤ _{2 cos θ}¹ = _2ω¹

1, which in turn gives that π/2≥ θ ≥ ^cot−1 1

4√ 2. Therefore we get the following integral:

H_A(θ) := 4sinθcosθ Z π/4

−π/4

g₁(θ, ϕ)dϕ = 8sin²θ− 1 sin³θ .

Moreover the integral of H_A(θ) with respect to θ is a closed form. Finally we can explicitly compute the vaule

Z π/2 cot^{−1 1}₄√

2

H_A(θ)dθ ≈ −0.0146.

Part B (Closed form) Since we have _2ω¹

1 ≤ _|ω²_2| ≤ _2ω²₁ ^{for all} ϕ. In other words we have ²_{sin θ}^√2 ≤ _{2 cos θ}^{2 ,} and _{2 cos θ}¹ ≤ _{sin θ}² . Hence we get range of θ

cot⁻¹ 1

4 ≥ θ ≥ ^cot−1 2 4√

2.

Therefore

H_B(θ) : = 4sinθcosθ Z π/4

−π/4

h₁(θ, ϕ) + g₂(θ, ϕ)dϕ

= −(24^cosθsin⁵θ−^sin6θ− 1024^cos6θ + 2048cos⁵θsinθ − 40π^sin4θ + 40πsin⁶θ + 5120log(√

2 + 1)cos⁵θsinθ + 1024√

2cos⁵θsinθ)/(1280cos⁴θsin³θ).

Similarly, the integral of H_B(θ) is a closed form and finally we have Z _cot^{−1 1}

4

cot^{−1 2}₄√ 2

H_B(θ)dθ ≈ −0.0655.

Part C (Closed form) Since we have: _2ω²

1 ≤ _|ω²_2| ≤ _2ω³₁ ^{for all} ϕ ∈ [−π/4, π/4]. As above, we will have that the range of θ is

cot⁻¹ 2

4 ≥ θ ≥ ^cot−1 3 4√

2. Therefore

H_C(θ) : = 4sinθcosθ Z π/4

π/4

h₁(θ, ϕ) + h₂(θ, ϕ) + g₃(θ, ϕ)dϕ

= −[61^sinθ− 696^cosθ + 2088cos³θ +cos⁵θ(5120log(√

2 + 1) + 1024√

2− 40)

−^cos7θ(5120log(√

2 + 1) + 1024√

2 + 1352) + (520π− 183)(^sinθ−^sin3θ)

−^cos4θsinθ(1040π− 183) + ^cos6θsinθ(520π− 10301)/(3840^cos4θsin⁴θ)], and finally we have

Z _cot^{−1 2}₄

cot^{−1 3}₄√ 2

H_C(θ)dθ ≈ −0.0139.

Part D (Closed form)

Since we have: _2ω³ and finally we have

Z _cot^{−1 3}

However there are still some ranges of θ that we have not dealt with in parts A, B, C, D above and the ranges are For these 3 ranges, we need to further estimate the integrals.

Remark 4. Those 3 parts are more complicated than above 4 parts since in each cases, sup{^suppf^ω2} = _|ω²_2| is not contained in one of [0,_2ω¹ three parts are not part A,B,C and D. Fortunately, it is contained in one of [0, _2ω² to split the integrals into two subcases according to ϕ.

Part E θ ∈ [^{cot−1 1}₄,cot^{−1 1}

4√

2]. (negative value) When θ ∈ [^{cot−1 1}₄,cot^{−1 1}

4√

2] the range [−π/4, π/4] ^of ϕ will be split into two cases. One of the ranges will give ξ₃(ω) = g₁(θ, ϕ), and the other range will give ξ₃(ϕ) = g₂(θ, ϕ). More precisely, when θ ∈ [^{cot−1 1}₄,cot^{−1 1}

4√

2], the variable ϕ will have two possibilities. One possibility is that f^ω2 vanishes before F₁₁^ω1 i.e.

we will have 0 ≤ _|ω²_2| ≤ _|2ω¹_1|. The other possibility is that f^ω2 vanishes before F₁₂^ω1 and after F₁₁^ω1, which gives _2ω¹

1 ≤ _|ω²_2| ≤ _2ω²₁. As a result, we split the range of ϕ according to which above possibility occurs.

1 0 ≤ _|ω²_2| ≤ _2ω¹₁ ⁽ξ3(ω) = g1(θ, ϕ)).

Since 0≤ _|ω²_2| ≤ _2ω¹₁ ^{which is} sin θ cos ϕ² ≤ _{2 cos θ}^{1 . Thus}

−^cos−14cotθ ≤ ϕ ≤ ^cos−14cotθ.

Therefore we can estimate the integral below H_E1(θ) := 4sinθcosθ

Z _cos⁻¹_{(4 cot θ)}

− cos⁻¹(4 cot θ)

g₁(θ, ϕ)dϕ,

which is a closed form in variable ϕ. However after plugging in the upper and lower limits, we are unable to show whether the integral R

HE1(θ)dθ has a closed form. But it is easy to show that its value is negative. Notice that g₁(θ, ϕ) = ^R4/2 sin θ cos ϕ

0 t²F₁₁^ω1(t)f^ω2(t)f^ω3(t)dt, and for allt,t², f^ω2(t), and f^ω2(t) are positive but F₁₁^ω1(t) is negative so that g₁(θ, ϕ) ≤ 0 ^{for all} θ and ϕ. Hence

Z cot^{−1 1}

4√ 2

cot^{−1 1}₄

H_E1(θ)dθ ≤ 0.

2 _2ω¹ ≤ _|ω²_| ≤ _2ω^{2 (}ξ₃(ω) = h₁ + g₂).

The integral range of ϕ is just the complement of the range in case 1 above.

Hence we have

H_E2(θ) := 4sinθcosθ

"Z _π/4

cos⁻¹(4 cot θ)

h₁(θ, ϕ) + g₂(θ, ϕ)dϕ +

Z _{− cos}⁻¹(4 cot θ)

−π/4

h₁(θ, ϕ) + g₂(θ, ϕ)dϕ

# . Again we can explicitly compute H_E2(θ) because the above integrals are closed

forms in variable ϕ. However after plugging in the upper and lower limits, the integral R

H_E2(θ)dθ is difficult to see if it has a closed form. But it is easy to show that its value is negative. Notice that h₁(θ, ϕ) + g₂(θ, ϕ) = R4/2 sin θ cos ϕ

0 t²[F₁₁^ω1(t) + F₁₂^ω1(t)] f^ω2(t)f^ω3(t)dt, and for allt,t²,f^ω2(t), andf^ω2(t) are positive but F₁₁^ω1(t), and F₁₂^ω1(t) are negative so that h₁(θ, ϕ) + g₂(θ, ϕ) ≤ 0 for all θ and ϕ. Hence

Z _cot^{−1 1}

4√ 2

cot^{−1 1}₄

HE2(θ)dθ ≤ 0.

Part F θ ∈ [^{cot−1 2}₄,cot^{−1 2}

4√

2](further estimate)

As in part E above, there will be two cases when θ ∈ [^{cot−1 2}₄,cot^{−1 2}

4√

2]. One case is that _2ω¹

1 ≤ _|ω²_2| ≤ _2ω²₁, and the other case is _2ω²

1 ≤ _|ω²_2| ≤ _2ω³₁^.

1 _2ω¹

1 ≤ _|ω²_2| ≤ _2ω²₁ ⁽ξ₃(ω) = h₁ + g₂).

Since _2ω¹

1 ≤ _|ω²_2| ≤ _2ω²₁, it gives that 0≤ ϕ ≤ ^cos−1(2cotθ). Hence H_{F 1}(θ) := 4sinθcosθ

Z _cos⁻¹(2 cot θ)

− cos⁻¹(2 cot θ)

h₁(θ, ϕ) + g₂(θ, ϕ)dϕ.

H_{F 1}(θ) can be explicitly computed, but integral R

H_{F 1}(θ)dθ is difficult to show

if it has a closed form. Therefore this case will be further estimated in the final section.

2 _2ω²

1 ≤ _|ω²_2| ≤ _2ω³₁ ⁽ξ₃(ω) = h₁ + h₂ + g₃) Since _2ω²

1 ≤ _|ω²_2| ≤ _2ω³₁. It gives us that H_{F 2}(θ) := 4sinθcosθ(

Z π/4

cos⁻¹(2 cot θ)

h₁(θ, ϕ) + h₂(θ, ϕ) + g₃(θ, ϕ)dϕ +

Z _{− cos}⁻¹_{(2 cot θ)}

−π/4 h₁(θ, ϕ) + h₂(θ, ϕ) + g₃(θ, ϕ)dϕ).

Again H_{F 2}(θ) can be explicitly computed, but integral R

H_{F 2}(θ)dθ is difficult to show if it has a closed form. Therefore this case will be further estimated in the final section.

Part G θ ∈ [^{cot−1 3}₄,cot^{−1 3}

4√

2] (further estimate)

1 _2ω²

1 ≤ _|ω²_2| ≤ _2ω³₁ ⁽ξ₃(ω) = h₁ + h₂ + g₃).

Since _2ω²

1 ≤ _|ω²_2| ≤ _2ω³₁, it gives us that H_G1(θ) := 4sinθcosθ

Z _cos⁻¹(³₄cot θ)

− cos⁻¹(³₄cot θ)

h₁(θ, ϕ) + h₂(θ, ϕ) + g₃(θ, ϕ)dϕ.

H_G1(θ) can be explicitly computed, but integral R

H_G1(θ) is difficult to show if it has a closed form. Therefore this case will be further estimated in the final section.

2 _2ω³

HG2(θ) can be explicitly computed, but integral R

HG2(θ) is difficult to show if it has a closed form. Therefore this case will be further estimated in the final section.

Finally the remaining case is below. Case 4: ^π₄ ≤ θ ≤ ^cos−1(^√1

3) (further estimate)

For this case ^π₄ ≤ θ ≤ ^cos−1(^√1

3), the range of ϕ is actually the complement of the range in case 2 above. In other words, the integral range of ϕ is D^c (see page 8 for the definition of D). Just as what we observe in case 2 the integral that we want to estimate can be reduced to

Z _cos^{−1 1}√

Now here is the key observations that since ϕ ∈ [−^cos−1cotθ,cos⁻¹cotθ] so that

ω₂ = sinθcosϕ ∈ [^cosθ,sinθ], Hence we have

2

|^sinθ| ≤ 2

|sinθ^cosϕ| ≤ 2

|^cosθ| = 2

|ω¹|.

Therefore we will only have f^ω2 vanishes before F₁₁^ω1 + F₁₂^ω1 + F₁₃^ω1 + F₁₄^ω1 and after F₁₁^ω1 + F₁₂^ω1 + F₁₃^ω1 for all ϕ ∈ [−^cos−1cotθ,cos⁻¹cotθ]. This gives us that

| sin θ|2 ≥ _2|ω³_1| and hence we have π

4 ≤ θ ≤ ^cot−1 3 4. Therefore the other part is

cot⁻¹ 3

4 ≤ θ ≤ ^cos−1 1

√3.

Part A’ : ^π₄ ≤ θ ≤ ^{cot−1 3}₄ In this case we have

KA(θ) := 4cosθsinθ

Z _cos⁻¹_{cot θ}

− cos⁻¹cot θ

h1(θ, ϕ) + h2(θ, ϕ) + h3(θ, ϕ) + g4(θ, ϕ)dϕ, which shows we need to estimate the integral below for this case.

Z _cot^{−1 3}₄

π/4

K_A(θ)dθ.

Still, we are not able to find its closed form and hence this case will be further estimated in the final section.

Part B’ : cot^{−1 3}₄ ≤ θ ≤ ^{cos−1 1√}₃

When cot^{−1 3}₄ ≤ θ ≤ ^{cos−1 1√}₃, there will be two cases (1) f^ω2(t) vanishes after F₁₁^ω1 + F₁₂^ω1 + F₁₃^ω1, which gives _ω²

2 ≥ _2ω³₁^;

(2)f^ω2(t)vanishes beforeF₁₁^ω1+F₁₂^ω1+F₁₃^ω1 afterF₁₁^ω1+F₁₂^ω1, which gives _|ω²

2| ≤ _2ω³₁^.

1 _|ω²

2| ≥ _2ω³₁ ⁽ξ₃(ω) = h₁ + h₂ + h₃ + g₄).

From this condition, we get the integral range of ϕ is cos⁻¹ 4

3^cotθ ≤ ϕ ≤ ^cos−1cotθ.

Therefore we have K_B1(θ) := 4cosθsinθ(

Z _cos⁻¹_{cot θ}

cos^{−1 4}₃cot θ

h₁(θ, ϕ) + h₂(θ, ϕ) + h₃(θ, ϕ) + g₄(θ, ϕ)dϕ +

Z _{− cos}^{−1 4}_{3cot θ}

− cos⁻¹cot θ

h1(θ, ϕ) + h2(θ, ϕ) + h3(θ, ϕ) + g4(θ, ϕ)dϕ.

K_B1 can be explicitly computed, but integral R

K_B1(θ)is difficult to see if it has a closed form. Therefore this case will be further estimated in the final section.

2 _2ω²

1 ≤ _|ω²_2| ≤ _2ω³₁ ⁽ξ₃(ω) = h₁ + h₂ + g₃).

From this condition, we get the integral range of ϕ is 0≤ ϕ ≤ ^cos−1 4

3^cotθ.

K_B2(θ) := 4cosθsinθ

Z _cos^{−1 4}_{3cot θ}

− cos^{−1 4}₃cot θ

h₁(θ, ϕ) + h₂(θ, ϕ) + g₃(θ, ϕ)dϕ.

K_B2(θ)can be explicitly computed, but integral R

K_B2 is difficult to see if it has a closed form. Therefore this case will be further estimated in the final section.

Some Remarks (1) We use the program of Matlab [14] to find the closed forms for some of above integrals. It can be also directly checked that all the indefinite integrals are correct. (2) The variable-precision floating-point arithmetic (VPA) that we use in the program of Matlab is 32 digits, thus the precision of the values for closed forms is accurate up to 10⁻³² error which would not effect our final value.

Upper bounds for all the further estimate cases Recall that for all the closed forms above, their values add up to be negative. In addition, part E is proved to be negative. Therefore our goal in the section is to give upper bounds for all the further estimate cases above and show that the values of the upper bounds are all negative which after all shows the integral (4.2) is negative.

Note that ξ₃(ω) =^R₀^∞t²F (tω₁)f (tω₂)f (tω₃)dt and F (tω₁) = F₁₁^ω1(t) + F₁₂^ω1(t) + F₁₃^ω1(t) + F₁₄^ω1(t), also recall that for all t, ω, F₁₁^ω1(t), F₁₂^ω1(t) are negative, and F₁₃^ω1(t), F₁₄^ω1(t), f (tω₂), f (tω₃) are positive. In previous sections, we have shown that for those further estimate cases, it is difficult to see if they have closed forms. Therefore the ideas for these remaining cases are to combine the integrals and split the combined integrals into positive and negative integrals. Finally, we are able to find upper bounds for these negative and positive integrals and show that these upper bounds have closed forms. Case 2 and case 4 Recall that in case 2, the integral (4.10) is

Z _cos^{−1 1}√ 3

π/4

cosθsinθ Z

ϕ∈Dξ₃(ω)dϕdθ.

In case 4, the integral (4.11) is

We now separate ξ₃(ω) into negative part and positive part i.e.

(Negative) that the negative part becomes

Z _cos^{−1 1}√

Also this integral has closed form Z

Plugging in the exact integral range, we thus obtain Z _cos^{−1 1}√

Thus the negative part of case 2 +case 4 is about -0.0607.

Positive part For positive part, we are unable to get the exact value. Instead, we will find an upper bound for the positive part and show that the upper bound has a closed form. Recall that in case 2, and part A’ and part B’1 of case 4, the function t²F (tω₁)f (tω₂)f (tω₃) vanishes after F₁₃^ω1 and they are

Case 2

Z _cos^{−1 1}√ 3

π/4

cosθsinθ Z

D

h₃(θ, ϕ) + h₄(θ, ϕ)dϕdθ, Case 4,A’

Z _cot^{−1 3}

4

π/4

cosθsinθ Z

D^c

h3(θ, ϕ) + g4(θ, ϕ)dϕdθ, Case 4,B’1

Z _cos^{−1 1}√ 3

cot^{−1 3}₄

cosθsinθ Z

R

h₃(θ, ϕ) + g₄(θ, ϕ)dϕdθ, where R is the integral range of ϕ in B’1 of case 4.

Notice that in part B’2 of case 4, the function t²F (tω₁)f (tω₂)f (tω₃) vanishes before F₁₄^ω1 and the integral is

Case 4,B’2

Z _cos^{−1 1}√ 3

cot^{−1 3}₄

cosθsinθ Z

R^c

g₃(θ, ϕ)dϕdθ.

To obtain an upper bound, we introduce a function which is a linear extension of F₁₃^ω1.

F˜₁₃^ω1(t) = 1

2(tω₁ − 1).

Where the dotted line is F˜₁₃^ω1(t) = ¹₂(tω₁ − 1).

Now for case 2

h₃(θ, ϕ) + h₄(θ, ϕ) =

Z 3/2 cos θ 2/2 cos θ

t²F₁₃^ω1(t)f^ω2(t)f^ω3(t)dt +

Z 4/2 cos θ 3/2 cos θ

t²F₁₄^ω1(t)f^ω2(t)f^ω3(t)dt

≤

Z _{3/2 cos θ}

2/2 cos θ

t²F˜₁₃^ω1(t)f^ω2(t)f^ω3(t)dt +

Z _{4/2 cos θ}

3/2 cos θ

t²F˜₁₃^ω1(t)f^ω2(t)f^ω3(t)dt

≤

Z 4/2 sin θ cos ϕ 2/2 cos θ

t²F˜₁₃^ω1(t)f^ω2(t)f^ω3(t)dt.

The last inequality comes from that fact that F vanishes before f^ω2 in case 2, i.e.

4

2cosθ ≤ 4

2sinθcosϕ.

For case 4A and 4B’1

For the case 4B’2, we note that g3(θ, ϕ) =

Therefore combining all the inequalities above to get an upper bound which is Z

Plugging in the exact integral range, we thus obtain Z _cos^{−1 1}√

Thus the positive part of case 2 +case 4 is bounded by 0.08717. Part F in

where h₁, h₂, g₂ represent the negative part and g₃ represents the positive part.

Negative part of F We first notice that the negative part of F is bounded by Z

Plugging in the exact integral range, we thus obtain Z cot⁻¹( ¹

Thus the negative part of F is bounded by -0.026. Positive part of F The positive part of F is

Z _cot⁻¹( ¹

≤ But we observe that

Z _cot⁻¹( ¹

Hence (4.12) is bounded by (4.13)+(4.14) which is Z _cot⁻¹( ¹

It has closed form Z

Plugging in the exact integral range, we thus obtain Z _cot⁻¹( ¹

Thus the positive part of F is bounded by 0.0064. part G in case 3 Part G is negative part of G is

Z _cot⁻¹₍ ³

Thus the negative part of G is bounded by -0.0694. Positive part of G The positive part of G can be split into two terms.

(G1)

Now Hence the positive part of G is bounded by (4.15)+(4.16), which has closed form

Plugging in the exact integral range, we thus obtain Z _cot⁻¹( ³

4√ 2) cot⁻¹(³₄)

4cosθsinθ Z π/4

−π/4

Z 4/2 sin θ cos ϕ 2/2 cos θ

t²F˜₁₃^ω1(t)f^ω2(t)f^ω3(t)dtdϕdθ

≈0.0139.

Therefore the positive part of G is bounded by 0.0139. Altogether The sum of all further estimate cases is

(Negative part of case 2+4)+ (Positive part of case 2+4)+(Negative part of F) +(Positive part of F)+(Negative part of G)+(Positive part of G)

≤ − 0.0607 + 0.08718 + (−0.026) + 0.0064 + (−0.0694) + 0.0139 < 0.

Remark 5. It is very likely to extend the ideas to all dimensions n ≥ 4 ^and show the correspondent integral is negative.

5 General Calderon-Zygmund operators and sharp A2

在文檔中 奇異積分的二次平均以及最加加權上界 (頁 37-59)