Singular Trudinger–Moser inequalities for the Aharonov–Bohm magnetic field

Xumin Wang

doi:10.1515/ans-2023-0187

Article Open Access

Singular Trudinger–Moser inequalities for the Aharonov–Bohm magnetic field

Xumin Wang

Published/Copyright: April 29, 2025

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From the journal Advanced Nonlinear Studies Volume 25 Issue 3

Abstract

The first purpose of this paper is to establish the singular Trudinger–Moser inequality in R 2 for the Aharonov–Bohm magnetic fields. The second purpose is to derive the singular Hardy–Trudinger–Moser inequality in the unit ball B 2 with Aharonov–Bohm magnetic potential. Moreover, we will show the constant 4 π ( 1 − β 2 ) is sharp in these two inequalities. The main skills include providing the asymptotic estimates of the related heat kernel and adapting the level set to derive a global Trudinger–Moser inequality from a local one. These results extend the inequalities established by Lu and Yang [Calc. Var. Partial Differ. Equ., 63 (2024)] into the weighted version.

Keywords: singular Trudinger–Moser inequalities; Hardy–Trudinger–Moser inequality; Aharonov–Bohm magnetic field; sharp constant

2000 Mathematics Subject Classification: Primary 35J20; 46E35

1 Introduction

Let n ≥ 3, the classical Hardy inequality holds for complex-valued. u ∈ C 0 ∞ ( R n ) \ { 0 }

(1.1) ∫ R n | ∇ u | 2 d x ≥ 2 − n 2 2 ∫ R n | u | 2 | x | 2 d x .

The constant 2 − n 2 2 is optimal but cannot be attained by nontrivial functions.

When n = 2, the Hardy inequality is invalid if 2 − n 2 2 is replaced by any positive constant. But if the singular weight 1 | x | 2 is weakened by adding a logarithmic term or includes some additional condition on the function u, then it holds (see [1])

∫ R 2 | ∇ u | 2 d x ≥ C ∫ R 2 | u | 2 | x | 2 1 + | ln ⁡ x | 2 d x if ∫ | x | = 1 u ( x ) d x = 0

∫ R 2 | ∇ u | 2 d x ≥ C ∫ R 2 | u | 2 | x | 2 d x if ∫ | x | = r u ( x ) d x = 0 , ∀ r > 0 .

Nevertheless, in the case of certain magnetic forms, the Hardy inequality (1.1) can be established with a positive constant even in R 2 . Set Z as the set of integers and A as the Aharonov–Bohm magnetic potential

(1.2) A = a | x | 2 ( x 2 , − x 1 ) , a ∈ R .

Laptev and Weidl [2] proved the following Hardy inequality holds

(1.3) ∫ R 2 | ( ∇ + i A ) u | 2 d x ≥ min n ∈ Z ( n − a ) 2 ∫ R 2 | u | 2 | x | 2 d x , u ∈ C 0 ∞ ( R 2 \ { 0 } ) .

Additionally, the constant min n ∈ Z ( n − a ) 2 is sharp.

In this paper, we set re ^iθ = x ₁ + ix ₂ and ∇_A = ∇ + i A. Then

(1.4) | ∇ A u | 2 = ∂ u ∂ r 2 + 1 r 2 ∂ u ∂ θ − i a u 2 .

For p > 2 and λ > − min n ∈ Z ( n − a ) 2 , Bonheure, Dolbeault, Esteban, Laptev and Loss ([3]) demonstrated there is an optimal function λ → μ _p(λ) which is monotonically increasing and concave, leading to the following magnetic Hardy–Sobolev inequality

(1.5) ∫ R 2 | ( ∇ + i A ) u | 2 d x + λ ∫ R 2 | u | 2 | x | 2 d x ≥ μ p ( λ ) ∫ R 2 | u | p | x | 2 d x 2 p , u ∈ H A ( R 2 ) ,

where H A ( R 2 ) is

H A ( R 2 ) = u ∈ L 2 ( R 2 ) : ∇ A u ∈ L 2 ( R 2 ) .

For L ^p-Hardy and L ^p-Rellich inequalities (p ≥ 1) with magnetic potentials, we refer to Evans and Lewis [4], Aermark [5], Lam and Lu [6] and so on. There are also some results about the existence of ground state solutions for equations involving periodic magnetic field, one can see [7].

Recently, Lu and Yang [8] established the Trudinger–Moser inequalities for Aharonov–Bohm magnetic fields in R 2 and Hardy–Trudinger–Moser inequalities for magnetic fields in B 2 . Before describing their theorems, we present a brief history of the Trudinger–Moser inequality which is related to the question under consideration in this paper.

Let Ω be an open and bounded domain in R n and 1 ≤ q ≤ n p n − kp . Classical Sobolev embedding demonstrates when kp < n, W 0 k , p Ω ↪ L q Ω is continuous. However, when kp = n, W 0 k , p ( Ω ) ⊈ L ∞ ( Ω ) . In this critical case, for k = 1, the optimal embedding is the well-known Trudinger–Moser inequality, which was established independently by Trudinger [9], Yudovich [10] and Pohozaev [11]. Later, Moser [12] sharpened the exponent α and proved the following theorem.

Theorem A.

([12]) For any α ≤ α n = n ω n − 1 1 n − 1 , there exists a constant C > 0 such that for any u ∈ W 0 1 , n ( Ω ) , the following inequality holds

(1.6) sup ∇ u L n Ω ≤ 1 1 | Ω | ∫ Ω ⁡ exp α u n n − 1 d x ≤ C ,

where ω _n−1 denotes the area of the unit sphere in R n .

Employing a rearrangement argument and a change of variables, Adimurthi and Sandeep generalized the Trudinger–Moser inequality (1.6) to a singular version in [13]. They proved for α > 0 and 0 ≤ β < n satisfying α α n + β n ≤ 1 , then the following inequality holds

sup u ∈ W 0 1 , n ( Ω ) , ‖ ∇ u ‖ L n ( Ω ) ≤ 1 ∫ Ω exp α | u | n n − 1 | x | β d x ≤ C n .

Another interesting but nontrivial promotion is to establish Trudinger–Moser inequality in the entire space. Adachi and Tanaka [14] and Do Ó [15] established the subcritical Trudinger–Moser inequality in R n . Later, Ruf [16] and Li and Ruf [17] proved the critical Trudinger–Moser inequality in R n . Adimurthi and Yang [18] established the Trudinger–Moser inequality with singular term. Proofs of Trudinger–Moser inequalities often require rearrangement argument and the Polyá–Szegö inequality. In the work [19], [20], Lam and Lu developed a symmetrization-free approach to give a simple proof for both critical and subcritical sharp Trudinger–Moser inequalities in W 1 , n ( R n ) . This method can be improved and applied to other settings where symmetrization argument does not work, such as on the high-order Sobolev spaces, Heisenberg group and Riemannian manifolds, we refer to [21], [22], [23], [24], [25], [26], [27]. In recent years, there are some results about Trudinger–Moser inequality in one dimension, one can see [28], [29], [30].

In 2012, Wang and Ye [31] established the Hardy–Trudinger–Moser inequality on the unit disc B 2 , which improves the classical Moser–Trudinger inequality and the classical Hardy inequality at the same time. In 2016, Lu and Yang [32] showed the Hardy–Trudinger–Moser inequality holds on any bounded and convex domain in R 2 via the Riemann mapping theorem. For the Hardy–Trudinger–Moser inequalities with dimension n ≥ 3, one can see [33], [34]. For higher order Hardy–Adams inequalities, we refer to [35], [36], [37], [38], [39] and so on.

Now we are ready to state the Trudinger–Moser inequality and the Hardy–Trudinger–Moser inequality for Aharonov–Bohm magnetic fields, which have been established by Lu and Yang in [8]. They also proved the constant 4π is sharp.

Theorem B.

([8]) Let A be the Aharonov–Bohm magnetic potential and 0 ≤ a ≤ 1 2 and λ > − a ². For any complex-valued function u ∈ C 0 ∞ ( R 2 \ { 0 } ) with

∫ R 2 | ∇ A u | 2 d x + λ ∫ R 2 | u | 2 | x | 2 d x ≤ 1 ,

then there exists a constant C > 0 such that

∫ R 2 e 4 π | u | 2 − 1 | x | 2 d x ≤ C .

Theorem C.

([8]) Let 0 ≤ a ≤ 1 2 and λ > − a ². For any complex-valued function u ∈ C 0 ∞ ( B 2 \ { 0 } ) with

∫ B 2 | ∇ A u | 2 d x + λ ∫ B 2 | u | 2 | x | 2 d x − 1 4 ∫ B 2 | u | 2 | x | 2 ⁡ ln 2 | x | d x ≤ 1 ,

then there exists a constant C > 0 such that

∫ B 2 e 4 π | u | 2 − 1 | x | 2 d x ≤ C .

The motivation of this paper is to extend the results in [8] into singular version. That is, we will establish the sharp weighted Trudinger–Moser inequality in R 2 and sharp weighted Hardy–Trudinger–Moser inequality in B 2 with Aharonov–Bohm magnetic potential. We state the results as the following two theorems.

Theorem 1.1.

Let A be the Aharonov–Bohm magnetic potential and 0 ≤ a ≤ 1 2 , 0 < β < 2 and λ > − a ². Set x = re ^iθ, then there exists C > 0 such that

(1.7) ∫ R 2 e 4 π 1 − β 2 | u | 2 − 1 | x | 2 ( | ln | x ‖ 2 + θ 2 ) β d x ≤ C

for any complex-valued function u ∈ C 0 ∞ ( R 2 \ { 0 } ) with

∫ R 2 | ∇ A u | 2 d x + λ ∫ R 2 | u | 2 | x | 2 d x ≤ 1 .

Moreover, the constant 4 π ( 1 − β 2 ) is sharp.

Theorem 1.2.

Let 0 ≤ a ≤ 1 2 , 0 < β < 2 and λ > − a ². Then there exists C > 0 such that

(1.8) ∫ B 2 e 4 π 1 − β 2 | u | 2 − 1 | x | 2 ( | ln | x ‖ 2 + θ 2 ) β d x ≤ C

for any complex-valued function u ∈ C 0 ∞ ( B 2 \ { 0 } ) with

∫ B 2 | ∇ A u | 2 d x + λ ∫ B 2 | u | 2 | x | 2 d x − 1 4 ∫ B 2 | u | 2 | x | 2 ⁡ ln 2 | x | d x ≤ 1 .

Moreover, the constant 4 π ( 1 − β 2 ) is sharp.

Remark 1.3.

If β = 0, the inequalities in Theorems 1.1 and 1.2 are the results in Lu and Yang [8].

We organize the paper as follows: Section 2 contains some notations and preliminaries. Sections 3 and 4 separately provides the proof of Theorem 1.1 for cases a = 0 and 0 < a ≤ 1 2 . In Section 5, we prove Theorem 1.2. In the Appendix section, we give the proofs of the sharpness of the constant in above inequalities.

2 Notations and preliminaries

2.1 Heat kernel on S 1

Denote S 1 = e i θ : − π ≤ θ < π as the unit circle in R 2 . Set Δ S 1 = ∂ θ 2 the Laplace–Beltrami operator on S 1 . Recall the explicit formula of the heat kernel on S 1 (see e.g. [40], Section 8.5):

e t Δ S 1 = 1 2 π ∑ n ∈ Z e − n 2 ⁡ t e i n θ − θ ′ .

Applying the elementary fact

∫ − ∞ ∞ e i θ ξ e − ξ 2 t d ξ = π t e − θ 2 4 t , t > 0

and the Poisson summation formula, we have (see also [41], Page 116, (4.25))

(2.1) e t Δ S 1 = 1 4 π t ∑ n ∈ Z e − θ − θ ′ − 2 n π 2 4 t .

2.2 O’Neil’s lemma

Set f be a measurable function on a measure space (X, μ). Set E _s = {x: |f(x)| > s}, denote by m(f, s) = μ(E _s). Define f* as the non-increasing rearrangement of f:

f * ( t ) = inf { s ∈ [ 0 , ∞ ) : m ( f , s ) ≤ t } .

For t > 0, set

f * * ( t ) = 1 t ∫ 0 t f * ( s ) d s .

f** is also monotone non-increasing. Moreover, if f = f ₁ + f ₂, there holds [42]

f * * ( t ) ≤ f 1 * * ( t ) + f 2 * * ( t ) , t > 0 .

Let (X ₁, μ ₁), (X ₂, μ ₂) and (X ₃, μ ₃) be three measure spaces. Set T: (X ₁, μ ₁) × (X ₂, μ ₂) → (X ₃, μ ₃) be a convolution operator. Denote h = T(f, g), the following inequality holds (see O’Neil [42])

(2.2) h * * ( t ) ≤ t g * * ( t ) f * * ( t ) + ∫ t ∞ f * ( s ) g * ( s ) d s .

3 Proof of Theorem 1.1: case a = 0

In what follows, we write a ≲ b to express a ≤ Cb and a ∼ b to stand for C ⁻¹ b ≤ a ≤ Cb with a positive constant C, where C may be different from line to line.

From now on, we will set x = re ^iθ and ω = ln⁡r. Then

(3.1) ∫ R 2 e 4 π 1 − β 2 | u | 2 − 1 | x | 2 ( | ln | x ‖ 2 + θ 2 ) β d x = ∫ 0 ∞ ∫ 0 2 π e 4 π 1 − β 2 | u | 2 − 1 r 2 ( | ln ⁡ r | 2 + θ 2 ) β r d r d θ = ∫ R × S 1 e 4 π 1 − β 2 | u | 2 − 1 ( ω 2 + θ 2 ) β d ω d θ

and

∫ R 2 | ∇ A u | 2 d x + λ ∫ R 2 | u | 2 | x | 2 d x = ∫ − ∞ ∞ ∫ 0 2 π ∂ u ∂ ω 2 + ∂ ∂ θ − i a u 2 d ω d θ + λ ∫ − ∞ ∞ ∫ 0 2 π u 2 d ω d θ .

In the case a = 0, it’s obvious to get

(3.2) ∫ R 2 | ∇ A u | 2 d x + λ ∫ R 2 | u | 2 | x | 2 d x = ∫ R × S 1 | ∂ ω u | 2 + | ∂ θ u | 2 + λ | u | 2 d ω d θ .

This section is devoted to proving Theorem 1.1 in the case a = 0. We firstly recall the pointwise estimates for the kernel of the fractional powers in [8]. By (2.1), the heat kernel for − ∂ w 2 − Δ S 1 + λ is

e t ∂ w 2 + Δ S 1 − λ = e − λ t 1 4 π t e − w − w ′ 2 4 t ∑ n ∈ Z e − θ − θ ′ − 2 n π 2 4 t .

Employing the Mellin type expression (see e.g. [43], Section 4.2), we obtain the kernel of the fractional powers

− ∂ w 2 − Δ S 1 + λ − 1 / 2 = 1 Γ ( 1 / 2 ) ∫ 0 ∞ t − 1 / 2 e t ∂ w 2 + Δ S 1 − λ d t = 1 4 π 3 / 2 ∑ n ∈ Z ∫ 0 ∞ t − 3 / 2 e − λ t − w − w ′ 2 + θ − θ ′ − 2 n π 2 4 t d t .

Denote K _ν(z) the modified Bessel function of the second kind, one can see [44] and for more properties. Then

(3.3) ∫ 0 ∞ x ν − 1 e − β x − γ x d x = 2 β γ ν / 2 K ν ( 2 β γ ) , Re β > 0 , Re γ > 0 .

Changing the variable t as 1 t and using modified Bessel function (3.3), we derive

(3.4) − ∂ w 2 − Δ S 1 + λ − 1 / 2 = 1 4 π 3 / 2 ∑ n ∈ Z ∫ 0 ∞ t − 1 / 2 e − λ t − w − w ′ 2 + θ − θ ′ − 2 n π 2 4 t d t = 1 4 π 3 / 2 ∑ n ∈ Z 2 4 λ | w − w ′ | 2 + θ − θ ′ − 2 n π 2 1 4 × K 1 2 λ w − w ′ 2 + θ − θ ′ − 2 n π 2 = 1 2 π ∑ n ∈ Z 1 w − w ′ 2 + θ − θ ′ − 2 n π 2 e − λ w − w ′ 2 + θ − θ ′ − 2 n π 2 .

To obtain the last equation, we apply the fact K 1 / 2 ( z ) = π 2 z − 1 / 2 e − z .

For the sake of brevity, we denote ϕ 1 w − w ′ , θ − θ ′ = − ∂ w 2 − Δ S 1 + λ − 1 / 2 . Applying (3.4), it’s easy to get there exist A ₁ > 0 and δ ₁ > 0 such that

(3.5) ϕ 1 w − w ′ , θ − θ ′ ≤ 1 2 π 1 w − w ′ 2 + θ − θ ′ 2 + A 1 , w − w ′ ≤ 1 , θ − θ ′ ≤ π ; ϕ 1 w − w ′ , θ − θ ′ ≲ e − δ 1 w − w ′ , w − w ′ ≥ 1 , θ − θ ′ ≤ π .

Consequently, there exist A ₂ > 0 and δ ₂ > 0 such that the non-increasing rearrangement of ϕ ₁ satisfies (see [8])

ϕ 1 * ( t ) ≤ 1 4 π t + A 2 , 0 < t ≤ 1 , ϕ 1 * ( t ) ≲ e − δ 2 t , t > 1 .

Now we are ready to prove Theorem 1.1 for the case a = 0. The main method is to adapt the level set developed by Lam and Lu to derive a global Trudinger–Moser inequality from a local one (see [20]). In virtue of equalities (3.1) and (3.2), Theorem 1.1 is equivalent to the following theorem in the case a = 0.

Theorem 3.1.

Let λ > 0 and 0 < β < 2. Then there exists C > 0 such that

∫ R × S 1 e 4 π 1 − β 2 | u | 2 − 1 ( ω 2 + θ 2 ) β d ω d θ ≤ C

for any complex-valued function u ∈ C 0 ∞ ( R × S 1 ) with

(3.6) ∫ R × S 1 | ∂ ω u | 2 + | ∂ θ u | 2 + λ | u | 2 d ω d θ ≤ 1 ,

where x = re ^iθ and ω = ln r.

Proof.

Let u satisfy the condition (3.6), then

∫ R × S 1 | − ∂ ω − △ S 1 + λ 1 2 u | 2 d ω d θ = ∫ R × S 1 | ∂ ω u | 2 + | ∂ θ u | 2 + λ | u | 2 d ω d θ ≤ 1 .

Set Ω ( u ) = { ( ω , θ ) ∈ R × S 1 : | u ( ω , θ ) ≥ 1 | } , then we can get

(3.7) | Ω ( u ) | = ∫ Ω ( u ) d ω d θ ≤ ∫ Ω ( u ) | u | 2 d ω d θ ≤ ∫ R × S 1 | u | 2 d ω d θ ≤ λ − 1 .

We divide the integral into two parts

(3.8) ∫ R × S 1 e 4 π 1 − β 2 | u | 2 − 1 ( ω 2 + θ 2 ) β d ω d θ = ∫ R × S 1 \ Ω ( u ) e 4 π 1 − β 2 | u | 2 − 1 ( ω 2 + θ 2 ) β d ω d θ + ∫ Ω ( u ) e 4 π 1 − β 2 | u | 2 − 1 ( ω 2 + θ 2 ) β d ω d θ = : I 1 + I 2 .

For I ₁, we write it as follows

(3.9) I 1 = ∫ R × S 1 \ Ω ( u ) e 4 π 1 − β 2 | u | 2 − 1 ( ω 2 + θ 2 ) β d ω d θ = ∫ R × S 1 \ Ω ( u ) ∑ n = 1 ∞ 4 π 1 − β 2 | u | 2 n n ! 1 ( ω 2 + θ 2 ) β d ω d θ = ∫ R × S 1 \ Ω ( u ) ⋂ ω 2 + θ 2 < ‖ u ‖ L 2 ( R × S 1 ) + ∫ R × S 1 \ Ω ( u ) ⋂ { ω 2 + θ 2 ≥ ‖ u ‖ L 2 ( R × S 1 ) } = : I 3 + I 4 .

We claim I ₃ ≤ C. To prove this claim, recall the magnetic Hardy–Sobolev inequality (1.5)

∫ R 2 | ( ∇ + i A ) u | 2 d x + λ ∫ R 2 | u | 2 | x | 2 d x ≥ μ p ( λ ) ∫ R 2 | u | p | x | 2 d x 2 p , u ∈ H A ( R 2 ) .

Set x = re ^iθ and ω = ln r, then it will become the following form in the case a = 0

(3.10) ∫ R × S 1 | ∂ ω u | 2 + | ∂ θ u | 2 + λ | u | 2 d ω d θ ≥ μ p ( λ ) ∫ R × S 1 | u | p d x 2 p , p > 2 .

Noting β < 2, we can choose some q > 2 such that q q − 2 β < 2 . According to 2 q + q − 2 q = 1 and employing Hölder inequality, we derive

(3.11) I 3 = ∫ R × S 1 \ Ω ( u ) ⋂ ω 2 + θ 2 < ‖ u ‖ L 2 ( R × S 1 ) ∑ n = 1 ∞ 4 π 1 − β 2 | u | 2 n n ! 1 ( ω 2 + θ 2 ) β d ω d θ = ∑ n = 1 ∞ 4 π 1 − β 2 n n ! ∫ R × S 1 \ Ω ( u ) ⋂ ω 2 + θ 2 < ‖ u ‖ L 2 ( R × S 1 ) | u | 2 ( ω 2 + θ 2 ) β d ω d θ ≲ ∫ R × S 1 | u | q d ω d θ 2 q ∫ ω 2 + θ 2 < ‖ u ‖ L 2 ( R × S 1 ) 1 ( ω 2 + θ 2 ) β q q − 2 d ω d θ q − 2 q ≲ 1 μ q ( λ ) ∫ ω 2 + θ 2 < 1 λ 1 ( ω 2 + θ 2 ) β q q − 2 d ω d θ q − 2 q < ∞ .

For I ₄, recall (3.7), it is easy to get

(3.12) I 4 = ∫ R × S 1 \ Ω ( u ) ⋂ ω 2 + θ 2 ≥ ‖ u ‖ L 2 ( R × S 1 ) ∑ n = 1 ∞ 4 π 1 − β 2 | u | 2 n n ! 1 ( ω 2 + θ 2 ) β d ω d θ ≤ ∑ n = 1 ∞ 4 π 1 − β 2 n n ! ∫ R × S 1 \ Ω ( u ) ⋂ ω 2 + θ 2 ≥ ‖ u ‖ L 2 ( R × S 1 ) | u | 2 ( ω 2 + θ 2 ) β d ω d θ ≤ ∑ n = 1 ∞ 4 π 1 − β 2 n n ! ‖ u ‖ L 2 ( R × S 1 ) β ∫ R × S 1 | u | 2 d ω d θ ≲ ‖ u ‖ L 2 ( R × S 1 ) 2 − β ≲ λ β 2 − 1 .

Combining inequalities (3.9), (3.11) with (3.12), we obtain I ₁ ≤ C.

For I ₂, exploiting the same approach as Li, Lu and Yang [36], we set

v = − ∂ w 2 − Δ S 1 + λ u ,

then

∫ R × S 1 | v | 2 d ω d θ = ∫ R × S 1 | − ∂ w 2 − Δ S 1 + λ u | 2 d ω d θ ≤ 1 .

We can write u as the potential form

u = v ∗ − ∂ w 2 − Δ S 1 + λ − 1 / 2 .

Let ϕ 1 = − ∂ w 2 − Δ S 1 + λ − 1 / 2 , then u = v∗ϕ ₁. Set g = 1 ( ω 2 + θ 2 ) β , then the rearrangement of g satisfies g * ( t ) ≤ ( π t ) β 2 for 0 < t ≤ 1 (refer to [8]). Recall (3.7), we may assume |Ω(u)| ≤ Ω₀. Applying O’Neil’s inequality (2.2) and changing of variables, a straightforward calculation leads to the following

I 2 ≤ ∫ 0 | Ω 0 | ⁡ exp 4 π 1 − β 2 u * ( t ) 2 g * ( t ) d t ≤ ∫ 0 | Ω 0 | ⁡ exp 4 π 1 − β 2 1 t ∫ 0 t v * ( s ) d s ∫ 0 t ϕ 1 * ( s ) d s + ∫ t ∞ v * ( s ) ϕ 1 * ( s ) d s 2 g * ( t ) d t = Ω 0 ∫ 0 ∞ ⁡ exp 4 π 1 − β 2 1 Ω 0 e − t ∫ 0 Ω 0 e − t v * ( s ) d s ∫ 0 Ω 0 e − t ϕ 1 * ( s ) d s + ∫ Ω 0 e − t ∞ v * ( s ) ϕ 1 * ( s ) d s 2 g * Ω 0 e − t e − t d t = Ω 0 ∫ 0 ∞ e − F ( t ) d t ,

where

F ( t ) = t − 4 π 1 − β 2 1 Ω 0 e − t ∫ 0 Ω 0 e − t v * ( s ) d s ∫ 0 Ω 0 e − t ϕ 1 * ( s ) d s + ∫ Ω 0 e − t ∞ v * ( s ) ϕ 1 * ( s ) d s 2 − ln g * Ω 0 e − t .

Denote

ψ ( t ) = Ω 0 e − t v * Ω 0 e − t , ϕ ( t ) = 4 π Ω 0 e − t ϕ 1 * Ω 0 e − t .

Then F(t) can be illustrated as

F ( t ) = t − 1 − β 2 ∫ − ∞ + ∞ a ( s , t ) ψ ( s ) d s 2 − ln g * Ω 0 e − t ,

where a(s, t) is follows

a ( s , t ) = ϕ ( s ) , s < t , e t ∫ t ∞ e − r 2 ϕ ( r ) d r e − s 2 , s > t .

Next we need to show ∫ 0 + ∞ e − F ( t ) d t < C , where C is a constant independent of ψ. To prove this we only need to show

There exists a constant C ≥ 0 independent of ψ satisfying inf t ≥ 0 F ( t ) ≥ − C ;
Let E _λ = {t ≥ 0: F(t) ≤ λ}, then there exist constants C ₁ and C ₂ both independent of ψ satisfying E λ ≤ C 1 | λ | + C 2 .

Utilizing analogous method to [36], [45], we know

∫ − ∞ + ∞ a ( s , t ) ψ ( s ) d s 2 ≤ t + C ,

where C ≥ 0 is a constant and independent of ψ. Recall g * ( t ) ≤ ( π t ) β 2 , then

− ln g * Ω 0 e − t ≥ − β 2 t + β 2 t ⁡ ln Ω 0 π .

Hence, we have

F ( t ) = t − 1 − β 2 ∫ − ∞ + ∞ a ( s , t ) ψ ( s ) d s 2 − ln g * Ω 0 e − t ≥ t − 1 − β 2 ( t + C ) − β 2 t + β 2 ln Ω 0 π ≳ − C .

Thus we get (i).

Next we prove (ii). Let R > 0, without loss of generality, we suppose E _λ ∩ [R, ∞) ≠ ∅. Take t ₁, t ₂ ∈ E _λ ∩ [R, ∞) ≠ ∅ and t ₁ < t ₂. Then by careful calculations, we derive

(3.13) t 2 − λ ≤ 1 − β 2 ∫ − ∞ + ∞ a ( s , t 2 ) ψ ( s ) d s 2 + ln g * Ω 0 e − t 2 ≤ 1 − β 2 ∫ − ∞ + ∞ a ( s , t 2 ) ψ ( s ) d s 2 + β 2 t 2 − β 2 ln Ω 0 π .

Then we obtain

1 − β 2 t 2 − λ ≤ 1 − β 2 ∫ − ∞ + ∞ a ( s , t 2 ) ψ ( s ) d s 2 − β 2 ln Ω 0 π .

Since the later calculation is similar to [46], [36], [45], here we omit the detailed calculations. Then we can get I ₂ ≤ C. Therefore we complete the proof.□

4 Proof of Theorem 1.1 for case 0 < a ≤ 1 2

Next we consider the case 0 < a ≤ 1 2 . Define

T a = − ∑ n ∈ Z n 2 − 2 a n 2 n 2 + ϵ P n ,

where ϵ > 0 and P n : L 2 ( S 1 ) → C e i n θ ( n ∈ Z ) is the orthogonal projection. According to [8], we have the following lemma.

Lemma 4.1.

([8]) For complex-valued function u ∈ C 0 ∞ ( R × S 1 ) , there exists λ′ > 0 for ϵ small enough such that

(4.1) − ∫ R × S 1 u ̄ T a u d ω d θ + ∫ R × S 1 | ∂ ω u | 2 + λ ′ | u | 2 d ω d θ ≤ ∫ R 2 | ∇ A u | 2 d x + λ ∫ R 2 | u | 2 | x | 2 d x .

Consequently, if we want to prove Theorem 1.1 for case 0 < a ≤ 1 2 , it’s sufficient to prove the following theorem.

Theorem 4.2.

Let λ > 0, 0 < a ≤ 1 2 and 0 < β < 2. Then there exists C > 0 such that

(4.2) ∫ R × S 1 e 4 π 1 − β 2 | u | 2 − 1 ( ω 2 + θ 2 ) β d ω d θ ≤ C

for any complex-valued function u ∈ C 0 ∞ ( R × S 1 ) with

− ∫ R × S 1 u ̄ T a u d ω d θ + ∫ R × S 1 | ∂ ω u | 2 + λ | u | 2 d ω d θ ≤ 1 .

Proof.

To prove this theorem, we need some asymptotic estimates of the kernel

ϕ 2 = − ∂ w 2 − T a + λ − 1 2 .

Recall Lemma 4.4 in [8], there exist some δ, δ ₃, δ ₄ such that

(4.3) | ϕ 2 | ≤ 1 2 π 1 w − w ′ 2 + θ − θ ′ 2 + O 1 | ω − ω ′ | δ 3 , | ω − ω ′ | ≤ 1 , | θ − θ ′ | ≤ π , | ϕ 2 | ≲ e − δ 4 | ω − ω ′ | , | ω − ω ′ | ≥ 1 , | θ − θ ′ | ≤ π

and

(4.4) ϕ 2 * * ( t ) ≤ 1 π t + O t − δ 3 , 0 < t ≤ 1 , ϕ 2 * * ( t ) ≲ e − δ 4 t , t > 1 .

Set Ω ( u ) = { ( ω , θ ) ∈ R × S 1 : | u ( ω , θ ) ≥ 1 | } , then

(4.5) ∫ R × S 1 e 4 π 1 − β 2 | u | 2 − 1 ( ω 2 + θ 2 ) β d ω d θ = ∫ R × S 1 \ Ω ( u ) e 4 π 1 − β 2 | u | 2 − 1 ( ω 2 + θ 2 ) β d ω d θ + ∫ Ω ( u ) e 4 π 1 − β 2 | u | 2 − 1 ( ω 2 + θ 2 ) β d ω d θ = : I 1 + I 2 .

Similar to the previous proof of Theorem 1.1 in the case a = 0, it’s not difficulty to derive I ₁ < ∞. For I ₂, choose two constants A ₃ > 0 and A ₄ > 0 such that

φ ( t ) = 1 4 π t + A 3 t − δ 3 , 0 < t ≤ 1 , A 4 e − δ 4 t , t > 1

is continuous on (0, ∞) and satisfies

ϕ 2 * * ( t ) ≤ 1 t ∫ 0 t φ ( s ) d s , 0 < t ≤ 1 , ϕ 2 * ( t ) ≤ φ ( t ) , t > 1 .

Utilizing calculations similar to [8], we get

ϕ 2 * * ( t ) ≤ 1 t ∫ 0 t φ ( s ) d s , t > 0 .

Set v = − ∂ w 2 − T a + λ 1 2 u . Then we have

| u ( ω , θ ) | = | v ∗ − ∂ w 2 − T a + λ − 1 2 | ≤ ∫ R × S 1 | v ( ω ′ , θ ′ ) ϕ 2 ( ω − ω ′ , θ − θ ′ ) | d ω ′ d θ ′

Using O’Neil’s lemma, one can get

| u | * ( t ) ≤ | u | * * ( t ) | ≤ | v | * * ( t ) ∫ 0 t φ ( s ) d s + ∫ t ∞ | v | * ( s ) φ ( s ) d s , t > 0 .

Then, using the similar method to [36] and Section 3, we obtain

I 2 = ∫ Ω ( u ) e 4 π 1 − β 2 | u | 2 − 1 ( ω 2 + θ 2 ) β d ω d θ ≤ ∫ 0 | Ω | e 4 π 1 − β 2 u * ( t ) 2 π t β 2 d t ≤ C .

The later method is similar to Theorem 3.1, we omit the details.□

Theorem 4.2 implies Theorem 1.1. In this way, we get Theorem 1.1 in the case 0 < a ≤ 1 2 .

5 Proof of Theorem 1.2

It’s easy to see

(5.1) ∫ B 2 | ∇ A u | 2 d x + λ ∫ B 2 | u | 2 | x | 2 d x − 1 4 ∫ B 2 | u | 2 | x | 2 ⁡ ln 2 | x | d x = ∫ 0 ∞ ∫ S 1 | ∂ ω u | 2 + | ( ∂ θ − i a ) u | 2 + λ | u | 2 − 1 4 ω 2 | u | 2 d ω d θ .

The main idea to prove Theorem 1.2 is similar to Theorem 1.1. We only need to prove the singular Trudinger–Moser inequality for − ∂ ω 2 − T a − 1 4 ω 2 + λ . Recall Lemma 5.2 in paper [8].

Lemma 5.1.

([8]) Set ϕ 3 = − ∂ ω 2 − T a − 1 4 ω 2 + λ − 1 2 . For some δ ₅ > 0 and δ ₆ > 0, there holds

| ϕ 3 | ≤ 1 2 π 1 w − w ′ 2 + θ − θ ′ 2 + O 1 | w − w ′ | δ 6 , w − w ′ ≤ 1 , θ − θ ′ ≤ π | ϕ 3 | ≲ e − δ 5 w − w ′ , w − w ′ ≥ 1 , θ − θ ′ ≤ π .

Using Lemma 5.1, similar to Section 3, we can get the following theorem.

Theorem 5.2.

Let λ > 0, 0 ≤ a ≤ 1 2 and 0 < β < 2. Then there exists C > 0 such that

(5.2) ∫ 0 ∞ ∫ S 1 e 4 π 1 − β 2 | u | 2 − 1 ( ω 2 + θ 2 ) β d ω d θ ≤ C

for any complex-valued function u ∈ C 0 ∞ ( ( 0 , ∞ ) × S 1 ) with

− ∫ 0 ∞ ∫ S 1 u ̄ T a u d ω d θ + ∫ 0 ∞ ∫ S 1 | ∂ ω u | 2 − 1 4 ω 2 | u | 2 + λ | u | 2 d ω d θ ≤ 1 .

Now, we give the Proof of Theorem 1.2.

Proof of Theorem 1.2

Let λ′ > 0 be such that for ϵ small enough,

− ∫ 0 ∞ ∫ S 1 u ̄ T a u d ω d θ + ∫ 0 ∞ ∫ S 1 | ∂ ω u | 2 − 1 4 ω 2 | u | 2 + λ ′ | u | 2 d ω d θ ≤ ∫ 0 ∞ ∫ S 1 | ∂ ω u | 2 − 1 4 ω 2 | u | 2 + | ( ∂ θ − i a ) u | 2 + λ | u | 2 d ω d θ = ∫ B 2 | ∇ A u | 2 d x + λ ∫ B 2 | u | 2 | x | 2 d x − 1 4 ∫ B 2 | u | 2 | x | 2 ⁡ ln 2 | x | d x ≤ 1 .

Applying Theorem 5.2, we have

∫ 0 ∞ ∫ S 1 e 4 π 1 − β 2 | u | 2 − 1 ( ω 2 + θ 2 ) β d ω d θ ≤ C .

Therefore, we finish the proof.□

Corresponding author: Xumin Wang, College of Science, Beijing Forestry University, Beijing 100083, People’s Republic of China, E-mail: xmwang@bjfu.edu.cn

Funding source: National Natural Science Foundation of China

Award Identifier / Grant number: (No.12101050)

Acknowledgment

The authors wish to sincerely thank the referees for their very careful reading of the article and for many helpful comments that have improved the exposition of this article.

Research ethics: Not applicable.
Informed consent: Not applicable.
Author contributions: The author has accepted responsibility for the entire content of this manuscript and approved its submission.
Use of Large Language Models, AI and Machine Learning Tools: None declared.
Conflict of interest: The author states no conflict of interest.
Research funding: The author’s research was supported by the National Natural Science Foundation of China (No.12101050).
Data availability: Not applicable.

Appendix A: Proofs of the sharpness of the constant

In this section, we show the constant 4 π ( 1 − β 2 ) is sharp in Theorems 1.1 and 1.2. Since the method is analogous, here we only prove the sharpness of the constant in Theorem 1.1.

A.1 Proof of the sharpness of the constant

For 0 < δ < 1, define

(6.1) u δ ( ω , θ ) = − ln ⁡ δ if ω 2 + θ 2 < δ 2 , − 1 2 ln ( ω 2 + θ 2 ) if δ 2 ≤ ω 2 + θ 2 ≤ 1 , 0 if ω 2 + θ 2 > 1 .

By calculations, we have

(6.2) ∫ R 2 | ∇ A u δ | 2 d x + λ ∫ R 2 | u δ | 2 | x | 2 d x = ∫ R × S 1 ( | ∂ ω u δ | 2 + | ∂ θ u δ | 2 + ( λ + a 2 ) | u δ | 2 ) d ω d θ = − 2 π ⁡ ln ⁡ δ + ( λ + a 2 ) π 1 2 − 1 2 δ 2 + δ 2 ⁡ ln ⁡ δ .

Set

u ̃ δ ( ω , θ ) = 1 − 2 π ⁡ ln ⁡ δ + ( λ + a 2 ) π 1 2 − 1 2 δ 2 + δ 2 ⁡ ln ⁡ δ u δ ( ω , θ ) .

Then using equation (6.2), it’s obvious

∫ R 2 | ∇ A u δ | 2 d x + λ ∫ R 2 | u δ | 2 | x | 2 d x = 1 .

Suppose

∫ R × S 1 e α | u ̃ δ | 2 − 1 ( ω 2 + θ 2 ) β d ω d θ ≤ C .

Then

∫ ω 2 + θ 2 < δ 2 e α | u ̃ δ | 2 − 1 ( ω 2 + θ 2 ) β d ω d θ = exp α ⁡ ln 2 ⁡ δ − 2 π ⁡ ln ⁡ δ + ( λ + a 2 ) π 1 2 − 1 2 δ 2 + δ 2 ⁡ ln ⁡ δ 2 π δ 2 − β 2 − β ≤ C .

Consequently,

α ≤ ln C 2 π δ 2 − β 2 − β + 1 × − 2 π ⁡ ln ⁡ δ + ( λ + a 2 ) π 1 2 − 1 2 δ 2 + δ 2 ⁡ ln ⁡ δ ln 2 ⁡ δ → 4 π 1 − β 2 , δ → 0 .

Therefore, we show the best constant is 4 π ( 1 − β 2 ) .

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Received: 2024-10-04

Accepted: 2025-04-10

Published Online: 2025-04-29

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Keywords for this article

singular Trudinger–Moser inequalities; Hardy–Trudinger–Moser inequality; Aharonov–Bohm magnetic field; sharp constant

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