Proof of Theorem 3.2.4: Necessity part

Finally, we upgrade Proposition 3.3.11 to the strong type estimate. Observe that it is sufficient to show the desired estimate for all (¹_q,¹_p) belonging to intOGA. Indeed, once this is established, we may employ complex interpolation once again with the estimates in Theorem 3.1.1 on the line segment [B, A) to obtain desired estimates for all (¹_q,¹_p) belonging to intOAB.

As we shall soon see, the advantage of first considering the regionOGAis that

α^∗(p, q)≤p≤q (3.3.36)

whenever (¹_q,¹_p) belongs to OGA. Indeed, one easily sees that α^∗(p, q) ≤ pis equivalent to

1

p ≤ _(d−1)q^d which means below the line [O, A] (andp≤qobviously holds inOGAsinceGlies on the diagonal ¹_q =¹_p).

Since we wish to use real interpolation, we fixs ∈(0,^d₂) and take any two points (_q¹

i,_p¹

i) from intOGAsuch that _p²

i +_q^d

i =d−2sfor i= 0,1. From Proposition 3.3.11 we know that O^s(pi, qi; (α^∗(pi, qi),1)):

X

j

λ_j|e^it∆D^−sf_j|² _Lpi

t L^qi_x

.kλk_`α∗(pi,qi),1

holds fori= 0,1. This means that if we fix an orthonormal system (fj)j in the common space H˙^s, then real interpolation, (3.3.4) and (3.3.5) yield

X

j

λj|e^it∆fj|² _Lp

tL^q,p_x

.kλk_`α∗(p,q),p (3.3.37)

with ¹_p = ^1−θ_p

0 + _p^θ

1, ¹_q = ^1−θ_q

0 + _q^θ

1 and any θ ∈ (0,1); that is, (3.3.37) holds for all (¹_q,¹_p) belonging to intOGA. Thanks to (3.3.36), we may deduce from the nesting of Lorentz spaces

that

X

j

λj|e^it∆fj|² _Lp

tL^qx

.kλk_`α∗(p,q)

and therefore desired estimates holds for all (¹_q,_p¹) belonging to intOGAwith 2s= ²_p+^d_q, as claimed.

equals 1 onB(0,¹₃). Remark that|v| ∼1. For eachv definefv by fbv=χv(ξ) :=R^d2χ(R(ξ−v)).

A simple computation shows that

|e^it∆f_v(x)|&R^−d2χ_Tv(x, t), (3.3.39) with the implicit constant independent ofv, R, where

T_v={(x, t) :|x+tv| ≤cR, |t| ≤cR²}

andcis a sufficiently small number. In fact, by change of variables we write

|e^it∆fv(x)|=R^−d2 Z

R^d

e^{iR−1ξ·(x+2tv)}e^it|R−1ξ|2χ(ξ)dξ .

Sincet|R⁻¹ξ|²= O(c) if|t| ≤cR², we see that|e^it∆f_v(x)| ∼R^−d if|x+tv| ≤cR, |t| ≤cR². Hence, (3.3.39) follows.

Then, since thefv’s have disjoint the Fourier support andkfvkH˙^s∼1 ³, (3.3.38) implies R^−d

X

v

χT_v

_Lp

t(−cR²,cR²);L^q_x(B(0,cR)). X

v

1_α¹

∼R^αd.

Note that allT_v contain (−˜cR, cR)×B(0,˜cR) with a small enough ˜c. Hence, the above gives R^1p+dq .R^dα. LettingR→ ∞gives

1 p+d

q ≤ d α. This gives the boundα≤α^∗(p, q).

Let us show that α≤pis also necessary. Take any nonnegative (λ_j)_j ∈ `^a and φ∈ S so that supp( ˆφ)⊂B(0,1) and kφk_L2 = 1. Further, we choose{v_j}_j ⊂R^d so thatB(v_j,100)∩ B(v_i,100) =∅for anyi6=j. Then we define

fj(x) :=e^−i2j∆[φ·e^i·vj](x).

We first check the orthonormality of the system{fj}j. Ifj6=k, then by Plancherel’s Theorem Z

R^d

fj(x)fk(x)dx= Z

R^d

e^−i2j|ξ|2φ(ξˆ −vj)e^−i2k|ξ|2φ(ξˆ −vk)dξ= 0

since the support of the function ˆφ(·−vj) is contained inB(v_j,1) andB(v_j,100)∩B(vk,100) =∅ as long asj6=k. Meanwhile, we clearly havekf_jk_L2 =kφk_L2 = 1. So,{f_j}_j is an orthonormal system, and hence we may apply (3.3.38) to obtain that

X

j

λj|e^it∆fj| _Lp

tL^q_x

.k(λj)jk`^a. (3.3.40) Let us now evaluate the left-hand side of (3.3.40) from below. We claim here that

|e^it∆fj(x)|=|e^i(t−2j)∆φ(x+ 2(t−2^j)vj)|. (3.3.41)

3Since|v| ∼1, we havekfvkH˙^s∼ kfvk_L2= 1.

For the time being, we admit (3.3.41). We chooseδ=δφ>0 depending only onφso that for any|t−s| ≤δ,

ke^i(t−s)∆φ−φk_L2q x ≤ 1

2kφk_L2q. (3.3.42)

This is true since we knowe^ih∆φ→φinL^2q_x ash→0. Then we obtain from (3.3.41) that ⁴

X

j

λ_j|e^it∆f_j|² _Lp

tL^q_x

≥





 X

j

Z 2^j+δ 2^j

"

Z

R^d

X

k

λ_k|e^it∆f_k(x)|²

!^q dx

#

p q

dt







1 p

≥





 X

j

Z 2^j+δ 2^j

Z

R^d

λ_j|e^it∆f_j(x)|^2q dx

^p_q dt







1 p

=





 X

j

λ^p_j Z 2^j+δ

2^j

Z

R^d

|e^i(t−2j)∆φ(x+ 2(t−2^j)vj)|^2qdx ^p_q

dt







1 p

=





 X

j

λ^p_j Z 2^j+δ

2^j

ke^i(t−2j)∆φk^2p

L^2q_x dt







1 p

where we changed the variable: x+ 2(t−2^j)vj 7→y in the last equality. Now, we notice from (3.3.42) that

ke^i(t−2j)∆φk_L2q

x =ke^i(t−2j)∆φ−φ+φk_L2q x

≥ kφk_L2q− ke^i(t−2j)∆φ−φk_L2q x

≥ kφk_L2q−1

2kφk_L2q = 1 2kφk_L2q

as long as|t−2^j|< δ. So, we arrive at

X

j

λj|e^it∆fj|² _Lp

tL^q_x

≥





 X

j

λ^p_j Z 2^j+δ

2^j

1 2kφk_L2q

2p

dt







1 p

=Cφk(λj)jk`^p,

where the constantCφ:= 2⁻¹kφk_L2q

2

δ^1p is depending only onφ∈ S, that is,Cφ is a global constant. Hence, it follows from (3.3.40) that

k(λj)_jk`<sup>p</sup> .k(λj)<sub>j</sub>k`^a

which meansa≤p. So, our task to finish the proof is to ensure (3.3.41). To this end, we first show the following formula:

|e^is∆[φ·e^i·vj](x)|=|e^is∆φ(x+ 2sv_j)| (3.3.43) for generals∈R. Based on the definitions, we calculate

|e^is∆[φ·e^i·vj](x)|= F⁻¹h

e^is|·|2Fφ(· −v_j)i (x)

= F⁻¹h

e^is|·−vj|2e^{2isvj(·−vj)}e^is|vj|2Fφ(· −v_j)i (x)

= F⁻¹hn

e^is|·−vj|2Fφ(· −vj)o

e^{2isvj(·−vj)}i (x)

.

4In below, we use the disjointness of {(2^j,2^j+δ)}j. Of course, this may not be true for small j, but considering sufficiently largej≥Jδ, it can be ensured.

We notice⁵

e^is|ξ−vj|2Fφ(ξ−vj) =F[e^is∆φ](ξ−vj).

So, by changing variables: ξ−v_j=η,

|e^is∆[φ·e^i·vj](x)|= Z

R^d

F[e^is∆φ](ξ−vj)e^{2isvj·(ξ−vj)}e^ix·ξdξ

= Z

R^d

F[e^is∆φ](η)e^2isvj·ηe^ix·ηe^ix·vjdη

= Z

R^d

F[e^is∆φ](η)e^{i(x+2svj)·η}dη

=

e^is∆φ(x+ 2svj) which means (3.3.43). Using this formula, we obtain (3.3.41):

|e^it∆fj(x)|=|e^i(t−2j)∆[φ·e^i·vj](x)|=|e^i(t−2j)∆φ(x+ 2(t−2^j)vj)|

and it completes the proof.

Semi-classical limiting argument

In order to prove the negative results on [O, A], we show that certain induced estimates for the velocity average of the kinetic transport equation fail on [O, A]. This requires us to first make the following observation based on a semi-classical limiting argument.

Proposition 3.3.13. Let p, q ∈ [1,∞], r,er, β ∈ [1,∞) and s ∈ [0,^d₂) be such that 2s = d−(_p²+^d_q). Ifα=α^∗(p, q)and

X

j

λ_j|e^it∆f_j|² _Lp,r

t L^q,e_x^r

.kλk_`α,β (3.3.44)

holds for all orthonormal systems(f_j)_j inH˙^s(R^d)and all sequencesλ= (λ_j)_j in`^α,β(C), then

Z

R^d

f(x−tv, v) dv

|v|^2s _Lp,r

t L^q,ex^r

.kfk_Lα,β (3.3.45)

wheneverf ∈L^α,β.

Before giving a proof of Proposition 3.3.13, we would like to make some remarks. The functionF(x, v, t) =f(x−tv, v) satisfies the kinetic transport equation

(∂t+v· ∇x)F(x, v, t) = 0, F(x, v,0) =f(x, v) for (x, v, t)∈R^d×R^d×R, and

ρf(x, t) = Z

R^d

f(x−tv, v) dv (3.3.46)

is the velocity average of the solution. Estimates of the form (3.3.45) are typically referred to as the Strichartz estimates for the kinetic transport equation; the classical case iss = 0 and

5Indeed, recall the definition:

F[e^is∆φ](η) =e^is|η|2Fφ(η).

r=α^∗(p, q), in which case it is known that (3.3.45) holds if and only if (¹_q,¹_p) belongs to [B, A).

The positive results were obtained in [22] and [64], and the failure at the endpointAwas shown in [3] for alld≥1. The argument establishing failure at the endpointAprofitably used duality and it was shown that

Z

R

g(x+tv, t) dt

_Ld+1.kgk

L^d+1_t L

d+1 x2

(3.3.47) fails by multiplying out the norm on the left-hand side and ultimately testing on smooth and rapidly decaying g whose Fourier transform is non-zero at the origin. A simplification was given in [6] by showing the left-hand side of (3.3.47) is infinite on the centred gaussian g(x, t) = e^{−π(t2+|x|2)}, and such an explicit argument did not rely on the fortuitous fact that Lebesgue exponent coincides with an integer.

We also note that the failure of (3.3.45) at A when s = 0 and r = α^∗(p, q) was shown whend= 1 prior to [3] by Guo–Peng [48] and Ovcharov [82]. The argument of Ovcharov used characteristic functions of Besicovitch (or Kakeya) sets; these are sets containing a unit line segment in all possible directions and a famous argument of Besicovitch generates such sets with arbitrarily small measure; this argument is particularly relevant to the present discussion and will be used to disprove Conjecture 3.2.3 whend= 1.

The connection between the solutions of the free Schr¨odinger equation and the kinetic transport equation is well documented and proceeds by a semi-classical limiting argument.

Thus, we are not viewing Proposition 3.3.13 as particularly novel and we present its statement and proof below for completeness and since we were not able to find elsewhere in the literature the statement in the form that we need. Sabin presented the special caser=α^∗in Lemma 9 of [85] and we use a similar argument to extend his observations to the setting of Lorentz spaces.

Proof of Proposition 3.3.13. First we note that (3.3.44) implies kργ(t)(x)k_Lp,r

t L^q,e_x^r .kγ0k_Cα,β(L²), (3.3.48) whereγ(t) =|D|^−se^it∆γ0e^−it∆|D|^−s. Next, we fix anyf in the Schwartz classS(R^d×R^d) and test (3.3.48) on the semi-classical Weyl quantisation γ0 =γ0(f;h) of f, whose kernel is given by

γ0(x, x⁰) = Z

R^d

f(^x+x₂ ⁰, v)e^i(x−x

0)·v

h dv.

The parameterhwill later be sent to zero. The Fourier transform ofγ₀onR^d×R^dis given by γb₀(v, v⁰) = (2πh)^dF_xf(·,^h₂(v−v⁰))(v+v⁰), (3.3.49) whereF_xdenotes the Fourier transform in the xvariable.

A direct computation, making use of (3.3.49), reveals that if eγ(t) =h^−2sγ(ht),

then ρ

eγ(t)(x) = h^−2s (2π)^2d

Z

R^2d

e^{−ith2|v−v0|2}e^ith2|v0|2|v−v⁰|^−s|v⁰|^−sγb₀(v−v⁰, v⁰) dv⁰e^ix·vdv

=h^d−2s (2π)^d

Z

R^2d

e^{−ith2(v·(v−2v0))}|v−v⁰|^−s|v⁰|^−sFxf(·,^h₂(v−2v⁰))(v) dv⁰e^ix·vdv

and therefore, by a change of variables, ρeγ(t)(x) = 1

(2π)^d Z

R^2d

e^−itv·v00|v⁰⁰+^h₂v|^−s|v⁰⁰−^h₂v|^−sFxf(·, v⁰⁰)(v) dv⁰⁰e^ix·vdv.

It follows that kρ

eγ(t)(x)k_Lp,r t L^q,e_x^r →

Z

R^d

f(x−tv⁰⁰, v⁰⁰)|v⁰⁰|^−2sdv⁰⁰ _Lp,r

t L^q,e_x^r

ash→0, or equivalently, h^−(p1+2s)kρ_γ(t)(x)k_Lp,r

t L^q,ex^r → Z

R^d

f(x−tv⁰⁰, v⁰⁰)|v⁰⁰|^−2sdv⁰⁰ _Lp,r

t L^q,e_x^r

(3.3.50) ash→0.

For the right-hand side, we observe that kγ0k_Cα,β .h^d−αd X

0≤|n|+|m|≤2d+2

h^|n|+|m|2 k∂_xⁿ∂_v^mfk_Lα,β

x,v (3.3.51)

forα, β ∈ [1,∞), which follows, for example, by using Propositions 2.1 and 2.2 in [1] for the endpoint casesα=β = 1 andα=β=∞, along with real interpolation in the classical Sobolev spaces (see [34]).

From (3.3.48), (3.3.50) and (3.3.51), and using the assumption thatα=α^∗(p, q) along with the scaling condition 2s=d−(²_p +^d_q), we obtain (3.3.45) in the limith→0.

By putting together Theorem 3.2.4 and Proposition 3.3.13 we immediately obtain the fol-lowing weighted estimates for the solution of the kinetic transport equation.

Theorem 3.3.14. Let d≥1. Suppose that(¹_p,¹_q)belongs tointOAB∪[B, A). Ifα=α^∗(p, q) and2s=d−(²_p+^d_q), then

Z

R^d

f(x−tv, v) dv

|v|^2s _Lp

tL^qx

.kfkL^α

for allf ∈L^α.

The endpoint failure

Recall that when the cases= 0, there is an endpoint Aand (3.1.4) withα= _q+1^2q fails at the point, see Theorem 3.1.1. This is a cause of trouble for the orthonormal Strichartz estimates on beyond region (A, C). Corresponding to this, there is a critical line (O, A) whereα^∗(p, q) =p.

As before, if the orthonormal Srichartz estimate (3.1.4) holds withα=α^∗(p, q) at the critical line (¹_q,¹_p) ∈ (O, A), then one would find a sharp estimate on whole region of intOAB and intOACD. However, we observed that this is not the case.

Proposition 3.3.15. Let d, p, q≥1 be(¹_q,¹_p)∈(O, A]and2s=d−(²_p+^d_q). Assume for any λ= (λj)j ⊂Cand any orthonormal system (fj)j in H˙^s,

X

j

λ_j|e^it∆f_j|² _Lp

tL^q_x

.kλk`^α. (3.3.52)

Thenα < p.

Before the proof, we remark that it is easy to see the failure of (3.3.52) for the case (p, q) = (∞,∞), which yieldss= ^d₂, α^∗(∞,∞) =∞. In fact, such a case is prohibited even when the one function case;ke^it∆fkL^∞_t L^∞_x .kfk_˙

H^d2. We can easily disprove this estimate. Indeed, if we assume (3.3.52) holds for (p, q) = (∞,∞), then we have

kfkL^∞_x ≤lim inf

t→0 ke^it∆fkL^∞_x .kfk_˙

H^d2

which implies the embedding ˙H^d2 ⊂L^∞. But this is clearly a contradiction. ⁶ So, we hereafter consider the case s < ^d₂. Once we restrict the case to s < ^d₂, then Proposition 3.3.15 is a consequence of the combination of Proposition 3.3.13 and the following lemma. The strategy is the combination of the duality argument and putting the Gaussian which is motivated by [3, 6].

Lemma 3.3.16. Let d, p, q≥1 be (¹_q,¹_p)∈(O, A]and2s=d−(_p²+^d_q). Then

Z

R^d

f(x−tv, v)|v|^−2sdv _Lp

tL^q_x

.kfk_L^p_x,v (3.3.53) fails.

Proof. It follows from the scaling that 2s=d−(²_p+^d_q). Clearly (3.3.53) fails ifs≥ ^d₂. So, we fix any s∈ [0,^d₂). To show Lemma 3.3.16, it is enough to show the failure of (3.3.53) under

1

p =_d−1^{d 1}_q. In other words, from 2s=d−(²_p+^d_q) we show the failure of (3.3.53) for (p, q) such that ¹_p = _d+1^d (1−^2s_d) and ¹_q = ^d−1_d+1(1−^2s_d). Hereafter, we fix suchp, q andα=α^∗(p, q) =p.

For that purpose, we further dualize (3.3.53):

Z

R

g(t, x+tv)dt|v|^−2s _Lα0

x,v

.kgk_Lp0

t L^q_x⁰, g∈L^p_t⁰L^q_x⁰,

where _α¹0 = _p¹0 = ^2s+1_d+1 and _q¹0 = _d+1¹ (2 +^d−1_d 2s).Now we putg(t, x) =e^−(t2+|x|2), the gaussian input. It follows from the direct computation that

Z

R

g(t, x+tv)dt∼ 1

(1 +|v|²)¹²e^−|x|

2+^(x·v)2

1+|v|2.

From this calculation and polar coordinate we see

Z

R

g(t, x+tv)dt|v|^−2s

α⁰

L^α_x,v⁰

∼ Z ∞

0

Z ∞ 0

1 (t²+r²)^sα0

1 (1 +t²+r²)^{α0 −d2}

r^d−2 (1 +r²)^d2drdt

&

Z ∞ 1

Z ∞ 1

1

(t²+r²)¹²dtr⁻²dr=∞ sincesα⁰+^α0₂^−d =¹₂. This gives the failure of (3.3.53) when ¹_p = _d−1^{d 1}_q.

6On the other hand, we have the embedding ˙H^d2 ⊂BMO. From this point of view, it seem natural to expect ke^it∆fk_L^∞

t BMOx.kfk

H˙^d2.

Failure of Frank-Lewin-Lieb-Seiringer conjecture ond= 1

Recall that on Conjecture 3.2.3, the exponents are p =α = ^d+1_d and q = ^d+1_d−1 which satisfy

1

p+^d_q = _α^d. So, thanks to Proposition 3.3.13, to show the failure of Conjecture 3.2.3, we have only to show the failure of the corresponding Strichartz inequality for the velocity average:

Z

R^d

f(x−tv, v)dv _L^d+1_d

t L

d+1 d−1 x

≤Ckfk

L

d+1 d ,1 x,v

, f ∈L

d+1 d ,1 x,v .

We notice that_d−1^d+1 =∞ifd= 1. From this speciality, we may prove the following proposition.

Theorem 3.3.17. Let d= 1. Then the weak type Strichartz estimate for the velocity average:

Z

R

f(x−tv, v)dv _L2

tL^∞_x

≤Ckfk_L2,1

x,v, f ∈L^2,1_x,v (3.3.54) fails.

Proof. Assume (3.3.54) is true and let us get a contradiction. For the sake of simplicity, we write

ρf(x, t) = Z

R

f(x−tv, v)dv

in this proof. Let us denote the Kakeya set on R² by E which means |E|_R2 = 0 and for any direction θ ∈ S¹, there exists a line segment`θ ⊂E whose length is one such that θ and `θ

are parallel. Take any smallδ > 0 and fatE to Eδ. Applyingf =χE_δ to the inequality we supposed, it follows that

kρχE_δk_L2

tL^∞_x ≤CkχE_δk_L2,1

x,v =|Eδ|¹². (3.3.55) For the right-hand side, we claim the estimate

sup

x∈R

ρχ_Eδ(x, t)≥ 1

(t²+ 1)¹² (t∈R) (3.3.56) which implies

kρχE_δk_L²

tL^∞_x =

sup

x∈R

ρχE_δ(x, t) _L2

t

≥√ π=

Z

R

dt t²+ 1

¹₂ .

So, once we have (3.3.56), then we obtain from (3.3.55) a contradiction:

√π≤C|Eδ|¹² →0 (δ→0).

To establish (3.3.56), we fix anyt∈Rand choose`<sub>t</sub>⊂E<sub>δ</sub> so that `_tand (−t,1)^t are parallel.

Further, we choosex_t∈Rso that x_t is an intersection point between the line extension of`_t and thex-axis. Then it follows that

sup

x∈R

ρχE_δ(x, t)≥ρχE_δ(xt, t)≥ Z v₂

v₁

χE_δ(xt−vt, v)dv=v2−v1,

where we chosev1, v2 ∈Rso that the line segment combining points (xt−v1t, v1)^tand (xt− v2t, v2)^t corresponds to the line segment `t which implies v2−v1 = (t²+ 1)⁻¹². This yields (3.3.56).

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