Interpolation inequalities and spectral estimates for magnetic operators

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Interpolation inequalities and spectral estimates for magnetic operators

Jean Dolbeault

http://www.ceremade.dauphine.fr/∼dolbeaul Ceremade, Université Paris-Dauphine

May 14, 2018

IML workshop onEigenvlaues and inequalities(14-18/5/2018) Joint work with M.J. Esteban, A. Laptev & M. Loss

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Magnetic interpolation in the Euclidean space

Outline

Magnetic rings

BResults: a one-dimensional magnetic interpolation inequality BConsequences: Keller-Lieb-Thirring estimates,

Bohm-Aharonov magnetic fields and a new Hardy inequality inR²

Interpolation inequalities and spectral estimates for magnetic operators in dimensions 2 and 3

BTheoretical results

BEstimates, numerics and a conjecture

J. Dolbeault Magnetic interpolation inequalities

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Preliminaries: a simple interpolation on the circle

On (−π, π]≈S¹3s, let us consider the uniform probability measure dσ= ds/(2π) and denote bykψk_Lp(S¹)the corresponding L^p norm The inequality

kψ⁰k²_L2(S¹)+αkψk²_L2(S¹)≥µ0,p(α)kψk²_Lp(S¹) (1) holds for some concave functionα7→µ_0,p(α) on (0,+∞)

Lemma

If p >2 and0< α≤1/(p−2), thenµ_0,p(α) =α

If p=−2 andα= 1/(p−2) =−1/4, then µ_0,p(−1/4) =−1/4 In both cases, equality achieved only by constant functions

Casep=−2(Exner, Harrell, Loss, 1998)):

kψ⁰k²_L2(S¹)+1

4kψk²_Lp(S¹)≥1

4kψk²_L2(S¹)

Casep >2: Bakry-Emery method applies toKolmogorov’s inequality

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Carré du champ method

LetF[u] :=ku⁰k²_L2(S¹)+_p−2¹ kuk²_L2(S¹)− kuk²_Lp(S¹)

and consider a positive solution of the parabolic equation

∂u

∂t =u⁰⁰+ (p−1)|u⁰|² u

Ifp=−2 (new application of thecarré du champmethod)

−d

dtF[u(t,·)] = Z π

−π

|u⁰⁰|²− |u⁰|² dσ

| {z }

≥0(Poincaré) +

Z π

−π

|u⁰|⁴ u² dσ

Ifp >1,p6= 2, the method is well known(Bakry, Emery, 85)

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Magnetic interpolation in the Euclidean space Consequences

Magnetic rings

BA magnetic interpolation inequality onS¹: withp >2 kψ⁰+i a ψk²_L2(S¹)+αkψk²_L2(S¹)≥µa,p(α)kψk²_Lp(S¹)

BConsequences

A Keller-Lieb-Thirring inequality

A new Hardy inequality for Bohm-Aharonov magnetic fields inR²

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Magnetic flux, a reduction

Assume thata:R→Ris a 2π-periodic function such that its restriction to (−π, π]≈S¹ is in L¹(S¹) and define the space

Xa:=

ψ∈Cper(R) : ψ⁰+i a ψ∈L²(S¹)

A standard change of gauge (see e.g.(Ilyin, Laptev, Loss, Zelik, 2016))

ψ(s)7→eⁱ Rs

−π(a(s)−¯a) dσ

ψ(s) where ¯a:=Rπ

−πa(s) dσis themagnetic flux, reduces the problem to a is a constant function

For anyk∈Z,ψbys7→e^iksψ(s) shows thatµa,p(α) =µk+a,p(α) a∈[0,1]

µ_a,p(α) =µ_1−a,p(α) because

|ψ⁰+i a ψ|²=|χ⁰+i(1−a)χ|²=

ψ⁰−i a ψ

2 ifχ(s) =e^−isψ(s) a∈[0,1/2]

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Optimal interpolation

We want to characterize theoptimal constant in the inequality kψ⁰+i a ψk²_L2(S¹)+αkψk²_L2(S¹)≥µa,p(α)kψk²_Lp(S¹)

written for anyp >2,α∈(−a²,+∞),ψ∈Xa

µa,p(α) := inf

ψ∈Xa\{0}

Rπ

−π |ψ⁰+i a ψ|²+α|ψ|² dσ kψk²_L_p₍

S¹)

p=−2 = 2d/(d−2) withd= 1(Exner, Harrell, Loss, 1998) p= +∞(Galunov, Olienik, 1995) (Ilyin, Laptev, Loss, Zelik, 2016) lim_α→−_a2µ_a,p(α) = 0 (JD, Esteban, Laptev, Loss, 2016)

Using a Fourier seriesψ(s) =P

k∈Zψke^iks, we obtain that kψ⁰+i a ψk²_L2(S¹)=X

k∈Z

(a+k)²|ψ_k|²≥a²kψk²_L2(S¹)

ψ7→ kψ⁰+i a ψk²_L2(S¹)+αkψk²_L2(S¹)is coercive for any α >−a²

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An interpolation result for the magnetic ring

Theorem

For anyp >2,a∈R, andα >−a²,µa,p(α) is achieved and

(i) if a∈[0,1/2]anda²(p+ 2) +α(p−2)≤1, thenµ_a,p(α) =a²+α and equality in (1) is achieved only by the constant functions (ii) if a∈[0,1/2]anda²(p+ 2) +α(p−2)>1, thenµa,p(α)< a²+α

and equality in (1) is not achieved by the constant functions Ifα >−a²,a7→µa,p(α)is monotone increasing on(0,1/2)

-0.2 -0.1 0.1 0.2 0.3 0.4

0.1 0.2 0.3 0.4 0.5

0.2 0.4 0.6 0.8 1.0 1.2 1.4

0.2 0.4 0.6 0.8 1.0 1.2

Figure:α7→µa,p(α) withp= 4 and (left)a= 0.45 or (right)a= 0.2

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Reformulations of the interpolation problem (1/3)

Any minimizerψ∈Xa ofµa,p(α) satisfies the Euler-Lagrange equation

(Ha+α)ψ=|ψ|^p−2ψ , Haψ=− d

ds+i a ²

ψ

up to a multiplication by a constant andv(s) =ψ(s)e^ias satisfies the condition

v(s+ 2π) =e^2iπav(s) ∀s∈R (2) Hence

µa,p(α) = min

v∈Y_a\{0}Qp,α[v]

whereY_a :=

v∈C(R) : v⁰∈L²(S¹), (2) holds and Q_p,α[v] := kv⁰k²_L2(S¹)+αkvk²_L2(S¹)

kvk²_Lp(S¹)

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Reformulations of the interpolation problem (2/3)

Withv=u e^iφ the boundary condition becomes

u(π) =u(−π), φ(π) = 2π(a+k) +φ(−π) (3) for somek∈Z, and kv⁰k²_L2(S¹)=ku⁰k²_L2(S¹)+ku φ⁰k²_L2(S¹)

Hence

µ_a,p(α) = min

(u,φ)∈Za\{0}

ku⁰k²_L2(S¹)+ku φ⁰k²_L2(S¹)+αkuk²_L2(S¹)

kuk²_Lp(S¹)

whereZ_a:=

(u, φ)∈C(R)² : u⁰, u φ⁰∈L²(S¹), (3) holds

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Reformulations of the interpolation problem (3/3)

We use the Euler-Lagrange equations

−u⁰⁰+|φ⁰|²u+α u=|u|^p−2u and (φ⁰u²)⁰= 0

Integrating the second equation, andassuming thatunever vanishes, we find a constantLsuch thatφ⁰ =L/u². Taking (3) into account, we deduce from

L Z π

−π

ds u² =

Z π

−π

φ⁰ds= 2π(a+k) that

ku φ⁰k²_L2(S¹)=L² Z π

−π

dσ

u² = (a+k)² ku⁻¹k²_L2(S¹)

Hence

φ(s)−φ(0) = a+k ku⁻¹k²_L2(S¹)

Z s

−π

ds u²

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Let us define

Qa,p,α[u] := ku⁰k²_L2(S¹)+a²ku⁻¹k⁻²_L2(S¹)+αkuk²_L2(S¹)

kuk²_Lp(S¹)

Lemma

For anya∈(0,1/2),p >2,α >−a², µa,p(α) = min

u∈H¹(S¹)\{0}Qa,p,α[u]

is achieved by a functionu >0

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Proofs

The existence proof is done on the original formulation of the problem using the diamagnetic inequality

ψ(s)e^ias=v₁(s) +i v₂(s), solves

−v_j⁰⁰+α v_j= (v₁²+v₂²)^p²⁻¹v_j, j= 1, 2

and the Wronskianw= (v1v₂⁰ − v₁⁰v2) is constant so thatψ(s) = 0 is incompatible with the twisted boundary condition

ifa²(p+ 2) +α(p−2)≤1, thenµa,p(α) =a²+αbecause

ku⁰k²_L2(S¹)+a²ku⁻¹k⁻²_L2(S¹)+αkuk²_L2(S¹)= (1−4a²)ku⁰k²_L2(S¹)+αkuk²_L2(S¹)

+ 4a²

ku⁰k²_L2(S¹)+¹₄ku⁻¹k²_L2(S¹)

ifa²(p+ 2) +α(p−2)>1, the test functionuε:= 1 +ε w1

Qa,p,α[u_ε] =a²+α+ 1−a²(p+ 2)−α(p−2)

ε²+o(ε²) proves the linear instability of the constants andµa,p(α)< a²+α

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Qa,p,α[u] := ^ku

0k²_{L2 (S1 )}+a²ku⁻¹k⁻²_{L2 (S1 )}+αkuk²_{L2 (S1 )} kuk²

Lp(S1 )

, µa,p(α) = min

u∈H¹(S¹)\{0}Qa,p,α[u]

Q_p,α[u] =Qa=0,p,α[u], ν_p(α) := inf

v∈H¹₀(S¹)\{0}

Q_p,α[v]

Proposition

∀p >2,α >−a², we have µa,p(α)< µ_1/2,p(α)≤νp(α) =µ_1/2,p(α)

0.5 1.0 1.5 2.0 2.5 3.0

0.1 0.2 0.3 0.4 0.5 0.6

Figure: p= 4,α= 0,a= 0.40, 0.41,. . . 0.49;u⁰⁰+u^p−1= 0

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A Keller-Lieb-Thirring inequality

Magnetic Schrödinger operatorHa−ϕ=− _ds^d +i a² ψ−ϕ The functionα7→µa,p(α) is monotone increasing, concave, and therefore has an inverse, denoted byαa,p:R⁺→(−a²,+∞), which is monotone increasing, and convex

Corollary

Letp >2,a∈[0,1/2],q=p/(p−2) and assume thatϕ is a non-negative function inL^q(S¹). Then

λ1(Ha−ϕ)≥ −αa,p kϕk_Lq(S¹)

andαa,p(µ) =µ−a² iff4a²+µ(p−2)≤1 (optimalϕis constant) Equality is achieved

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Bohm-Aharonov magnetic fields

On the two-dimensional Euclidean spaceR², let us introduce the polar coordinates (r, ϑ)∈[0,+∞)×S¹ ofx∈R² and consider a magnetic potentialain a transversal (Poincaré) gauge, or Poincaré gauge

(a,er) = 0 and (a,eϑ) =aϑ(r, ϑ) Magnetic Schrödinger energy

Z

R²

|(i∇+a) Ψ|²dx= Z +∞

0

Z π

−π

|∂_rΨ|²+ 1

r²|∂_ϑΨ +i r a_ϑΨ|²

rdϑdr Bohm-Aharonov magnetic fields: aϑ(r, ϑ) =a/rfor some constant a∈R(ais themagnetic flux), with magnetic field b= curla Z

R²

|(i∇+a) Ψ|²dx≥τ Z

R²

ϕ(x/|x|)

|x|² |Ψ|²dx ∀ϕ∈L^q(S¹), q∈(1,+∞)

=⇒τ=τ a,kϕkL^q(S¹)

?

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Hardy inequalities

(Hoffmann-Ostenhof, Laptev, 2015)proved Hardy’s inequality Z

R^d

|∇Ψ|²dx≥τ Z

R^d

ϕ(x/|x|)

|x|² |Ψ|²dx

where the constantτ depends on the value ofkϕk_Lq(S^d−1)andd≥3 Bohm-Aharonov vector potentialin dimensiond= 2

a(x) =a x2

|x|²,−x1

|x|²

, x= (x1, x2)∈R², a∈R and recall the inequality(Laptev, Weidl, 1999)

Z

R²

|(i∇+a) Ψ|²dx≥min

k∈Z

(a−k)² Z

R²

|Ψ|

|x|²dx

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A new Hardy inequality

Z

R²

|(i∇+a) Ψ|²dx≥τ Z

R²

ϕ(x/|x|)

|x|² |Ψ|²dx ∀ϕ∈L^q(S¹), q∈(1,+∞) Corollary

Letp >2,a∈[0,1/2],q=p/(p−2) and assume thatϕ is a

non-negative function inL^q(S¹). Then the inequality holds withτ >0 given by

αa,p τkϕkL^q(S¹)

= 0

Moreover,τ=a²/kϕkL^q(S¹) if 4a²+kϕkL^q(S¹)(p−2)≤1

For anya∈(0,1/2), by takingϕconstant, small enough in order that 4a²+kϕk_Lq(S¹)(p−2)≤1, we recover the inequality

Z

R²

|(i∇+a) Ψ|²dx≥a² Z

R²

|Ψ|²

|x|² dx

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Proofs (Keller-Lieb-Thirring inequality)

Hölder’s inequality kψ⁰+i a ψk²_L2(S¹)−

Z π

−π

ϕ|ψ|²dσ≥ kψ⁰+i a ψk²_L2(S¹)−µkψk²_Lp(S¹)

whereµ=kϕk_Lq(S¹)and ¹_q +²_p = 1: choose µ_a,p(α) =µ kψ⁰+i a ψk²_L2(S¹)−µkψk²_Lp(S¹)≥ −αkψk²_L2(S¹)

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Proofs (Hardy inequality)

Letτ ≥0,x= (r, ϑ)∈R²be polar coordinates inR² Z

R²

|(i∇+a) Ψ|²−τ ϕ

|x|²|Ψ|²

dx

= Z ∞

0

Z

S¹

r|∂rΨ|²

| {z }

≥0

+1

r|∂ϑΨ +i aΨ|²−τ ϕ r |Ψ|²

dϑdr

≥λ1(Ha−τ ϕ) Z ∞

0

Z

S¹

1

r|Ψ|²dϑdr

≥ −α_a,p(τkϕk_Lq(S¹)) Z ∞

0

Z

S¹

1

r|Ψ|²dϑ Ifτ = 0, thenαa,p(τkϕkL^q(S¹)) =αa,p(0) =−a²

α_a,p(τkϕkL^q(S¹))>0 forτ large

=⇒ ∃!τ >0 such thatα_a,p(τkϕkL^q(S¹)) = 0

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Comments

BThe regiona²(p+ 2) +α(p−2)<1 is exactly the set where the constant functions are linearly stable critical points

BThe proof of therigidity resultis based

- neither on thecarré du champmethod, at least directly

- nor on a Fourier representation of the operator as it was the case in earlier proofs (p= +∞, orp >2 andα= 0)

BMagnetic rings: see(Bonnaillie-Noël, Hérau, Raymond, 2017) BDeducingHardy’s inequalityapplied withBohm-Aharonovmagnetic fields from aKeller-Lieb-Thirring inequalityis an extension of

(Hoffmann-Ostenhof, Laptev, 2015)to the magnetic case BOur results are not limited to the semi-classical regime

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Magnetic interpolation in the Euclidean space Estimates in dimensiond= 2 for constant magnetic fields

Magnetic interpolation inequalities in the Euclidean space

BThree interpolation inequalities and their dual forms BEstimates in dimensiond= 2 for constant magnetic fields

Lower estimates

Upper estimates and numerical results

A linear stability result (numerical) and an open question Warning: assumptions are not repeated

Estimates are given only in the casep >2 but similar estimates hold in the other cases

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Magnetic Laplacian and spectral gap

In dimensionsd= 2 andd= 3: themagnetic Laplacianis

−∆Aψ=−∆ψ−2iA· ∇ψ+|A|²ψ− i(divA)ψ

where the magnetic potential (resp. field) isA(resp.B= curlA) and H_A¹(R^d) :=

ψ∈L²(R^d) : ∇Aψ∈L²(R^d) , ∇A:=∇+iA Spectral gap inequality

k∇Aψk²₂≥Λ[B]kψk²₂ ∀ψ∈H_A¹(R^d) (4) Λ depends only onB= curlA

Assumption: equality in (4)holds for someψ∈H_A¹(R^d) IfBis a constant magnetic field, Λ[B] =|B|

Ifd= 2, spec(−∆A) ={(2j+ 1)|B| : j∈N}is generated by the Landau levels. TheLowest Landau Levelcorresponds to j= 0

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Magnetic interpolation inequalities

k∇Aψk²₂+αkψk²₂≥µB(α)kψk²_p ∀ψ∈H_A¹(R^d) (5) for anyα∈(−Λ[B],+∞) and anyp∈(2,2^∗),

k∇Aψk²₂+βkψk²_p≥ν_B(β)kψk²₂ ∀ψ∈H_A¹(R^d) (6) for anyβ∈(0,+∞) and anyp∈(1,2)

k∇Aψk²₂≥γ Z

R^d

|ψ|² log |ψ|²

kψk²₂

dx+ξB(γ)kψk²₂ ∀ψ∈H_A¹(R^d) (7) (limit case corresponding top= 2) for anyγ∈(0,+∞)

C_p:=







min_u∈H1(R^d)\{0}

k∇uk²₂+kuk²₂

kuk²_p if p∈(2,2^∗) min_u∈H1(R^d)\{0}

k∇uk²₂+kuk²_p

kuk²₂ if p∈(1,2) µ0(1) =Cpifp∈(2,2^∗),ν0(1) =Cp ifp∈(1,2)

ξ0(γ) =γlog π e²/γ

ifp= 2

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Technical assumptions

A∈L^α_loc(R^d),α >2 ifd= 2 orα= 3 ifd= 3 and

σ→+∞lim σ^d−2 Z

R^d

|A(x)|²e^−σ^|x|dx= 0 if p∈(2,2^∗)

σ→+∞lim σ^d²⁻¹

logσ Z

R^d

|A(x)|²e^−σ^|x|²dx= 0 if p= 2

σ→+∞lim σ^d−2 Z

|x|<1/σ

|A(x)|²dx if p∈(1,2)

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A statement

Theorem

p∈(2,2^∗): µBis monotone increasing on (−Λ[B],+∞), concave and lim

α→(−Λ[B])+

µB(α) = 0 and lim

α→+∞µB(α)α^d−2² ⁻^d^p =Cp

p∈(1,2): νB is monotone increasing on(0,+∞), concave and

β→0lim₊νB(β) = Λ[B] and lim

β→+∞νB(β)β⁻²^p+d(2−p)²^p =Cp

ξB is continuous on(0,+∞), concave,ξB(0) = Λ[B]and ξB(γ) =^d₂γlog ^{π e}_γ²

(1 +o(1)) as γ→+∞

Constant magnetic fields: equality is achieved

Nonconstant magnetic fields: only partial answers are known

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2 4 6 8 10

2 4 6 8

Figure: Cased= 2,p= 3,B= 1: plot ofα7→(2π)²^p⁻¹µB(α)

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0 1 2 3 4 5 6

0 2 4 6 8

Figure: Cased= 2,p= 1.4,B= 1: plot ofβ7→νB(β). The horizontal axis is measured in units of (2π)¹⁻²^pβ

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Magnetic Keller-Lieb-Thirring inequalities

λ_A,V is the principal eigenvalue of−∆A+V

α_B: (0,+∞)→(−Λ,+∞) is the inverse function ofα7→µ_B(α) Corollary

(i) For any q=p/(p−2)∈(d/2,+∞)and any potential 0≥V ∈L^q(R^d)

λ_A,V ≥ −α_B(kVk_q)

lim_µ→0₊αB(µ) = Λand lim_µ→+∞αB(µ)µ^d−2−2^{2 (q+1)}^q =−C

2 (q+1) d−2−2q

p

(ii) For any q=p/(2−p)∈(1,+∞)and any 0< W⁻¹∈L^q(R^d) λA,W ≥νB kW⁻¹k⁻¹_q

(iii) For any γ >0 and any W ≥0 s.t.e^−W/γ∈L¹(R^d) λA,W ≥ξB(γ)−γ log R

R^de^−W/γdx

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A more general duality result

Proposition

Letd= 2 or3. Letφ∈L¹_loc(R^d)be an arbitrary potential (i) If q > d/2,p=_q−1²^q , we have

λA,φ≥ −(αB(k(λ−φ)kq,+)−λ) (ii) If q∈(1,+∞),p= _q+1²^q , we have

λA,φ≥λ+νB k(φ−λ)⁻¹k⁻¹_q

These estimates hold for anyλ∈Rsuch that all above norms are well defined, with the additional condition thatφ≥λa.e. in Case (ii)

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Preliminaries: interpolation without magnetic field

Assume thatp >2 and letCp denote the best constant in k∇uk²₂+kuk²₂≥Cpkuk²_p ∀u∈H¹(R^d) By scaling, if we test the inequality byu ·/λ

, we find that k∇uk²₂+λ²kuk²₂≥Cpλ²⁻^d⁽¹⁻^p²⁾kuk²_p ∀u∈H¹(R^d) ∀λ >0 An optimization onλ >0 shows that the best constant in the scale-invariant inequality

k∇uk^d⁽¹⁻

2 p)

2 kuk^2−d⁽¹⁻

2 p)

2 ≥Spkuk²_p ∀u∈H¹(R^d) is given by

Sp= ₂¹_p (2p−d(p−2))^1−d

p−2

2p (d(p−2))

d(p−2) 2p Cp

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... and with magnetic field

Proposition

Letd= 2 or3. For anyp∈(2,+∞), anyα >−Λ =−Λ[B]<0

µB(α)≥µinterp(α) :=







Sp(α+ Λ) Λ^−d^p−2²^p ifα∈h

−Λ,^{Λ (2}_d^p−d_(p−2)^(p−2))i Cpα^1−d^p−2²^p ifα≥ ^{Λ (2}_d^p−d_(p−2)^(p−2)) Lett∈[0,1]. From the diamagnetic inequalityk∇|ψ|k2≤ k∇Aψk2

and from the inequality withλ=^α+Λ_1−t^t, we deduce that k∇_Aψk²₂+αkψk²₂≥t k∇_Aψk²₂−Λkψk²₂

+ (1−t)

k∇|ψ|k₂+α+ Λt 1−t kψk²₂

≥Cp(1−t)^d(p−2)²^p (α+tΛ)^1−d^p−2²^p kψk²_p and optimize ont∈[max{0,−α/Λ},1]

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Lower estimates: d = 2, constant magnetic field

Assume thatB= (0, B) is constant, d= 2 and choose A1=^B₂x2, A2=−^B₂x1 ∀x= (x1, x2)∈R² Proposition

(Loss, Thaller, 1997)Consider a constant magnetic field with field strengthB in two dimensions. For everyc∈[0,1], we have

Z

R²

|∇Aψ|²dx≥ 1−c² Z

R²

|∇ψ|²dx+c B Z

R²

ψ²dx

and equality holds withψ=u e^iS andu >0 if and only if

−∂2u², ∂1u²

= 2u²

c (A+∇S)

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... a computation (d = 2, constant magnetic field)

Z

R²

|∇Aψ|²dx= Z

R²

|∇u|²dx+ Z

R²

|A+∇S|²u²dx

= 1−c² Z

R²

|∇u|²dx+ Z

R²

c²|∇u|²+|A+∇S|²u² dx

| {z }

≥R

R22c|∇u| |A+∇S|u dx

with equality only ifc|∇u|=|A+∇S|u

2|∇u| |A+∇S|u=|∇u²| |A+∇S| ≥ ∇u²^⊥

·(A+∇S) where ∇u²⊥

:= −∂2u², ∂1u² Equality case: −∂2u², ∂1u²

=γ(A+∇S) forγ= 2u²/c Integration by parts yields

Z

R²

c²|∇u|²+|A+∇S|²u²

dx≥B c Z

R²

u²dx

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Lower estimate (d = 2, constant magnetic field):

a result

Proposition

Consider a constant magnetic field with field strengthB in two dimensions. Given anyp∈(2,+∞), and any α >−B, we have

µB(α)≥Cp 1−c²¹⁻²p (α+c B)²^p =:µLT(α) with

c=c(p, η) =

pη²+p−1−η

p−1 = 1

η+p

η²+p−1 ∈(0,1) andη=α(p−2)/(2B)

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Upper estimate (1): d = 2, constant magnetic field

For every integerk∈Nwe introduce the special symmetry class ψ(x) =

x2+i x1

|x|

^k

v(|x|) ∀x= (x1, x2)∈R² (Ck) (Esteban, Lions, 1989): ifψ∈Ck, then

1 2π

Z

R²

|∇Aψ|²dx= Z +∞

0

|v⁰|²r dr+ Z +∞

0 k

r−^{B r}₂ ²

|v|²r dr and optimality is achieved inCk

Test functionv_σ(r) =e⁻^r²^/(2^σ): an optimization on σ >0 provides an explicit expression ofµ_Gauss(α) such that

Proposition Ifp >2, then

µ_B(α)≤µ_Gauss(α) ∀α >−Λ[B]

This estimate is not optimal becausev_σ does not solve the Euler-Lagrange equations

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Upper estimate (2): d = 2, constant magnetic field

A more numerical point of view. The Euler-Lagrange equation inC0 is

−v⁰⁰−v⁰ r +

B²

4 r²+α

v=µ_EL(α) Z +∞

0

|v|^pr dr ²p−1

|v|^p−2v

We can restrict the problem to positive solutions such that µEL(α) =

Z +∞

0

|v|^pr dr ¹⁻p²

and then we have to solve the reduced problem

−v⁰⁰− v⁰ r +

B²

4 r²+α

v=|v|^p−2v

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2 4 6 8 10

-1 1 2 3

0.004 0.003 0.002 0.001 0.001 0.002 0.003

-0.02

-0.04

-0.06

-0.08

2 4

-1

6

Figure: Cased= 2,p= 3,B= 1: comparison of the upper estimates α7→µGauss(α) andα7→µEL(α) with the lower estimatesα7→µinterp(α) andα7→µLT(α)

Plots represent the curves log₁₀(µGauss/µEL), log₁₀(µLT/µEL) and log₁₀(µinterp/µEL) so thatα7→µEL(α) corresponds to a straight line at level 0. The exact value associated withµBlies in the grey area

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Asymptotics (1): Lowest Landau Level

Proposition

Letd= 2 and consider a constant magnetic field with field strengthB.

Ifψ_α is a minimizer forµB(α)such thatkψ_αk_p= 1, then there exists a non trivialϕ_α∈LLL such that

lim

α→(−B)+

kψα−ϕαk_H1

A(R²)= 0

Letψα∈H_A¹(R²) be an optimal function for (5) such thatkψαkp= 1 and let us decompose it asψα=ϕα+χα, where ϕα∈LLL andχαis in the orthogonal of LLL

µB(α)≥(α+B)kϕ_αk²₂+(α+3B)kχ_αk²₂≥(α+3B)kχ_αk²₂∼2Bkχ_αk²₂ asα→(−B)+ becausek∇χαk²₂≥3Bkχαk²₂

Since lim_α→(−B)₊µB(α) = 0, lim_α→(−B)₊kχαk2= 0 and

µB(α) = (α+B)kϕαk²₂+ k∇Aχαk²₂+αkχαk²₂≥²₃k∇Aχαk²₂ concludes the proof

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Asymptotics (2): semi-classical regime

Let us consider the small magnetic field regime. We assume that the magnetic potential is given by

A1=^B₂x₂, A2=−^B₂x₁ ∀x= (x₁, x₂)∈R² ifd= 2. In dimension d= 3, we chooseA= ^B₂(−x2, x₁,0) and observe that the constant magnetic field isB= (0,0, B), while the spectral gap in (4) is Λ[B] =B.

Proposition

Letd= 2 or3 and consider a constant magnetic fieldB of intensityB with magnetic potential A

For anyp∈(2,2^∗)and any fixedαandµ >0, we have

ε→0lim+

µεB(α) =Cpα^d^p⁻^d−2² Consider any functionψ∈H_A¹(R^d) and letψ(x) =χ(√

ε x),

√εA x/√ ε

=A(x) with our conventions onA

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Numerical stability of radial optimal functions

Let us denote byψ₀ an optimal function in (C0) such that

−ψ⁰⁰₀− ψ₀⁰ r +

B²

4 r²+α

ψ₀=|ψ0|^p−2ψ₀ and consider the test function

ψε=ψ0+ε e^{i θ}v wherev=v(r) ande^{i θ}= (x₁+i x₂)/r Asε→0₊, the leading order term is

2π Z

R²

|v⁰|²dx+ Z

R²

1 r−^{B r}₂ ²

+α

|v|²dx−^p₂ Z +∞

0

|ψ₀|^p−2v²r dr

ε²

and we have to solve the eigenvalue problem

−v⁰⁰− v⁰ r +

1

r−^{B r}₂ ² +α

v−^p₂|ψ₀|^p−2v=µ v

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2 4 6 8

1 2 3 4

Figure: Casep= 3 andB= 1: plot of the eigenvalueµas a function ofα A careful investigation shows thatµis always positive, including in the limiting case asα→(−B)+, thus proving the numerical stability of the optimal function inC0with respect to perturbations inC1

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An open question of symmetry

(Bonheure, Nys, Van Schaftingen, 2016)for a fixed α >0 and forB small enough, the optimal functions are radially symmetric functions,i.e., belong toC0

This regime is equivalent to the regime asα→+∞for a givenB, at least if the magnetic field is constant

Numerically our upper and lower bounds are (in dimensiond= 2, for a constant magnetic field) numerically extremely close

The optimal function inC0with respect to perturbations inC1

Prove that the optimality case is achieved among radial function if d= 2 andB is a constant magnetic field