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BY 4.0 license Open Access Published by De Gruyter April 15, 2020

Extremals for Fractional Moser–Trudinger Inequalities in Dimension 1 via Harmonic Extensions and Commutator Estimates

  • Gabriele Mancini ORCID logo EMAIL logo and Luca Martinazzi

Abstract

We prove the existence of extremals for fractional Moser–Trudinger inequalities in an interval and on the whole real line. In both cases we use blow-up analysis for the corresponding Euler–Lagrange equation, which requires new sharp estimates obtained via commutator techniques.

1 Introduction

The celebrated Moser–Trudinger inequality [39] states that for Ωn with finite measure |Ω| we have

(1.1) sup u W 0 1 , n ( Ω ) , u L n ( Ω ) 1 Ω e α n | u | n n - 1 𝑑 x C | Ω | , α n := n ω n - 1 1 n - 1 ,

where ωn-1 is the volume of the unit sphere in n. The constant αn is sharp in the sense that the supremum in (1.1) becomes infinite if αn is replaced by any α>αn. In the case Ω=2, Ruf [46] proved a similar inequality, using the full W1,2-norm instead of the L2-norm of the gradient, which was then generalized to n, n2, by Li and Ruf [31] by

(1.2) sup u W 1 , n ( n ) , u L n ( n ) n + u L n ( n ) n 1 n ( e α n | u | n n - 1 - 1 ) 𝑑 x < .

Higher-order versions of (1.1) were proven by Adams [2] on the space W0k,n/k(Ω) for n>k. The proofs of (1.1) and (1.2) in [39] and [31] rely on symmetrization arguments which cannot be applied when the Pólya-Szegö inequality fails. A rearrangement-free approach was proposed by Lam and Lu to prove Adams-type inequalities for high-order Sobolev spaces on n (see [24]). This approach was also used to obtain inequalities on the Heisenberg group with applications to sub-elliptic PDEs (see, e.g., [23, 25]).

In [21], the authors proved the following 1-dimensional fractional extension of the previous results (for the definition of H1/2,2() and (-Δ)1/4, see (A.4) in Appendix A).

Theorem A.

Set I:=(-1,1)R and H~1/2,2(I):={uH1/2,2(R):u0 on RI}. Then we have

(1.3) sup u H ~ 1 2 , 2 ( I ) , ( - Δ ) 1 4 u L 2 ( I ) 1 I ( e α u 2 - 1 ) 𝑑 x = C α < for  α π ,

and

(1.4) sup u H 1 2 , 2 ( ) , u H 1 2 , 2 ( ) 1 ( e α u 2 - 1 ) 𝑑 x = D α < for  α π ,

where

u H 1 2 , 2 ( ) 2 := ( - Δ ) 1 4 u L 2 ( ) 2 + u L 2 ( ) 2 .

The constant π is sharp in (1.3) and (1.4).

More general results have recently appeared (see, e.g., [1, 16, 24, 38, 47, 50]) in which both the dimension and the (fractional) order of differentiability have been generalized. For instance, (1.3) and (1.4) can be seen as 1-dimensional cases of the more general results of [24, 38, 16] that hold in arbitrary dimension n.

The existence of extremals for this kind of inequalities is a challenging question. Existence of extremals for (1.1) was originally proven by Carleson and Chang [5] in the case of the unit ball, a fundamental result later extended by Struwe [49] and Flucher [15] to the case of general bounded domains in 2, and by Lin [32] to the case of bounded domains in n. In the case of the Li–Ruf inequality (1.2), the existence of extremals appears in [31] when n3, and was proven by Ishiwata [20] when n=2. For the higher-order Adams inequality the existence of extremals has been proven in various cases by, e.g., Li and Ndiaye [30] on a 4-dimensional closed manifold, by Lu and Yang [33] (see also [40]) for a 4-dimensional bounded domain and by DelaTorre and Mancini [9] for a bounded domain in 2m, m1 arbitrary. In recent years, there have been many other papers studying the existence of extremals for similar inequalities on n (see, e.g., [12, 13, 34, 41, 42] and the references therein). Most of the results we mentioned are based on a blow-up analysis approach, but a different method has been recently proposed in [27], where the authors exploit the exact relation between critical and subcritical Moser–Trudinger suprema (see [26, 6]) to prove the existence of extremals.

On the other hand, the existence of extremals for the fractional Moser–Trudinger inequality has remained open until now, with the exception of Takahashi [50] considering a subcritical version of (1.4) of Adachi–Tanaka type [1], and Li and Liu [29] treating the case of a fractional Moser–Trudinger on H1/2,2(M) with M being a compact Riemann surface with boundary. The idea of Li and Liu is that by working on the boundary of a compact manifold one can localize the H1/2,2-norm.

Applying the same method for an interval I creates problems near I, which require additional care in the estimate, and the problem becomes even more challenging when working on the whole . The main purpose of this paper is to handle these two cases and prove that the suprema in (1.3) and (1.4) are attained.

Theorem 1.1.

For any 0<απ, the inequality (1.3) has an extremal, i.e. there exists uαH~1/2,2(I) such that

( - Δ ) 1 4 u α L 2 ( ) 1 𝑎𝑛𝑑 I ( e α u α 2 - 1 ) 𝑑 x = C α .

Theorem 1.1 is rather simple to prove for α(0,π), while the case α=π relies on a delicate blow-up analysis for subcritical extremals.

A similar analysis can be carried out for the Ruf-type inequality (1.4). However, working on the whole real line, we need to face additional difficulties due to the lack of compactness of the embedding of H=H1/2,2() into L2(): vanishing at infinity might occur for maximizing sequences, even in the sub-critical case α(0,π). This issue is not merely technical. Indeed, Takahashi [50] proved that (1.4) has no extremal when α is small enough. Here, in analogy with the results in dimension n2, we prove that the supremum in (1.4) is attained if α is sufficiently close to π.

Theorem 1.2.

There exists α*(0,π) such that for α*απ inequality (1.4) has an extremal, namely there exists u¯αH1/2,2(R) such that

u ¯ α H 1 2 , 2 ( ) 1 𝑎𝑛𝑑 ( e α u ¯ α 2 - 1 ) 𝑑 x = D α .

As for Theorem 1.1, the proof of Theorem 1.2 for α=π is based on blow-up analysis. In fact, we need to study the blow-up of a non-local equation on the whole real line (no boundary conditions), as is done in the following theorem.

Theorem 1.3.

Let (uk)H=H1/2,2(R) be a sequence of non-negative solutions to

(1.5) ( - Δ ) 1 2 u k + u k = λ k u k e α k u k 2 in  ,

where αkπ and λkλ0. Assume uk even and decreasing (uk(-x)=uk(x)uk(y) for xy0) for every k and set μk:=supRuk=uk(0). Assume also that

(1.6) Λ := lim sup k u k H 2 < .

Then, up to extracting a subsequence, we have that one of the following assertions holds:

  1. μ k C , uku in Cloc() for every 0, where uCloc()H solves

    (1.7) ( - Δ ) 1 2 u + u = λ u e π u 2 in  .

  2. μ k , uku weakly in H and strongly in Cloc0({0}) where u is a solution to (1.7). Moreover, setting rk such that

    (1.8) λ k r k μ k 2 e α k μ k 2 = 1 α k

    and

    (1.9) η k ( x ) := 2 α k μ k ( u k ( r k x ) - μ k ) , η ( x ) := - log ( 1 + | x | 2 ) ,

    one has η k η in C loc ( ) for every 0 , supkηkLs()< for any s>0 (cf. (A.2)), and ΛuH2+1.

The proof of Theorem 1.3 is quite delicate because local elliptic estimates of a non-local equation depend on global bounds as we shall prove in Lemma 3.6. This will be based on sharp commutator estimates (Lemma 3.3), as developed in [35] for the case of a bounded domain in n, extending the approach of [37] to the fractional case.

We expect similar existence results to hold for a perturbed version of inequalities (1.3)–(1.4), as in [36, 51] (see also the recent results in [19]), but we will not investigate this issue here.

2 Proof of Theorem 1.1

2.1 Strategy of the Proof

We will focus on the case α=π since the existence of extremals for (1.3) with α(0,π) follows easily by Vitali’s convergence theorem; see, e.g., the argument in [36, Proposition 6].

Let uk be an extremal of (1.3) for α=αk=π-1k. By replacing uk with |uk|, we can assume that uk0. Moreover, (-Δ)1/4ukL2()=1, and uk satisfies the Euler–Lagrange equation

(2.1) ( - Δ ) 1 2 u k = λ k u k e α k u k 2 ,

with bounds on the Lagrange multipliers λk (see (2.4)).

Using the monotone convergence theorem, we also get

(2.2) lim k I ( e α k u k 2 - 1 ) 𝑑 x = lim k C α k = C π ,

where Cαk and Cπ are as in (1.3).

If μk:=maxIuk=O(1) as k, then up to a subsequence uku locally uniformly, where by (2.2), u maximizes (1.3) with α=π. Therefore, we will work by contradiction, assuming

(2.3) lim k μ k = .

By studying the blow-up behavior of uk (see in particular Propositions 2.2 and 2.9), we will show that (2.3) implies Cπ4π (Proposition 2.10), but with suitable test functions we will also prove that Cπ>4π (Proposition 2.11), hence contradicting (2.3) and completing the proof of Theorem 1.1.

2.2 The Blow-Up Analysis

The following proposition is well known in the local case, and its proof in the present setting is similar to the local one. We give it for completeness.

Proposition 2.1.

We have ukC(I)C0,1/2(I¯), uk>0 in I, and uk is symmetric with respect to 0 and decreasing with respect to |x|. Moreover,

(2.4) 0 < λ k < λ 1 ( I ) .

Up to a subsequence, we have λkλ and uku weakly in H~1/2,2(I) and strongly in L2(I), where u solves

(2.5) ( - Δ ) 1 2 u = λ u e π u 2 .

Proof.

For the first claim, see [35, Remark 1.4]. The positivity follows from the maximum principle, and the symmetry and monotonicity follow from the moving point technique; see, e.g., [8, Theorem 11].

Now testing (2.1) with φ1, the first eigenfunction of (-Δ)1/2 in H~1/2,2(I), positive and with eigenvalue λ1(I)>0, we obtain

λ 1 ( I ) I u k φ 1 𝑑 x = λ k I u k e α k u k 2 φ 1 𝑑 x > λ k I u k φ 1 𝑑 x ,

hence proving (2.4). By the theorem of Banach–Alaoglu and the compactness of the Sobolev embedding of H~1/2,2(I)L2(I), we obtain the claimed convergence of uk to u. Finally, to show that u solves (2.5), test with φCc(I):

I u ( - Δ ) 1 2 φ 𝑑 x = lim k I u k ( - Δ ) 1 2 φ 𝑑 x
= lim k I λ k u k e α k u k 2 φ 𝑑 x
= I λ u e π u 2 φ 𝑑 x ,

where the convergence of the last integral is justified by splitting I into

I 1 := { x I : u k ( x ) L } and I 2 := { x I : u k ( x ) > L } ,

applying the dominated convergence on I1 and bounding

I 2 λ k u k e α k u k 2 φ 𝑑 x sup I | φ | L I λ k u k 2 e α k u k 2 𝑑 x
= sup I | φ | L I u k ( - Δ ) 1 2 u k 𝑑 x
= sup I | φ | L ( - Δ ) 1 4 u k L 2 ( ) 2 ,

and letting L. ∎

Let u~k be the harmonic extension of uk to +2 given by the Poisson integral; see (A.5) in Appendix A. Notice that

(2.6) I λ k u k 2 e α k u k 2 𝑑 x = ( - Δ ) 1 4 u k L 2 ( ) 2 = u ~ k L 2 ( + 2 ) 2 = 1 .

Let

r k = 1 α k λ k μ k 2 e α k μ k 2 and η k ( x ) := 2 α k μ k ( u k ( r k x ) - μ k )

be as in (1.8) and (1.9), and set

η ~ k ( x , y ) := 2 α k μ k ( u ~ k ( r k x , r k y ) - μ k ) .

Note that η~k is the Poisson integral of ηk.

Proposition 2.2.

We have rk0 and η~kη~ in Cloc(R+2¯) for every 0, where

η ~ ( x , y ) = - log ( ( 1 + y ) 2 + x 2 )

is the Poisson integral (compare to (A.5)) of η:=-log(1+x2), and

(2.7) ( - Δ ) 1 2 η = 2 e η , e η 𝑑 x = π .

Proof.

According to [35, Lemma 2.2, Theorem 1.5 and Proposition 2.7], we have rk0 and ηkη in Cloc() for every 0 and (ηk) is uniformly bounded in L1/2() (see (A.2)).

To obtain the local convergence of η~k, fix R>0 and split the integral in the Poisson integral (A.5) of η~k into an integral over (-R,R) and an integral over (-R,R), for R large. The former is bounded by the convergence of ηk locally, the latter by the boundedness of ηk in L1/2(), provided

( x , y ) B R 2 + 2 .

As a consequence, we get that η~k is locally uniformly bounded in +2¯. Since η~k is harmonic, we conclude by elliptic estimates. ∎

Remark 2.3.

As L, we have

(2.8) + 2 B L | η ~ | 2 𝑑 x 𝑑 y = 4 π log ( L 2 ) + O ( log L L ) .

Moreover, the same estimate holds if BL is replaced by BL(0,-1).

Proof.

As L, we have

η ~ ( x , y ) = - 2 log L + O ( L - 1 ) and η ( x , y ) ( x , y ) | ( x , y ) | = - 2 L + O ( L - 2 )

for (x,y)+2BL. Then, integrating by parts and using (2.7), we get that

+ 2 B L | η ~ | 2 𝑑 x 𝑑 y = + 2 B L η ~ η ~ ν 𝑑 σ + 2 - L L η e η 𝑑 x
= 4 π log L + 2 η e η 𝑑 x + O ( log L L ) .

The definition of the Poisson integral (see (A.5)) gives

2 η e η 𝑑 x = 2 η 1 + x 2 𝑑 x = 2 π η ~ ( 0 , 1 ) = - 4 π log 2 .

This proves (2.8). Finally, observe that BLBL(0,-1) and BL(0,-1)BL are contained in AL:=BL+1BL-1. Since |η~|2=O(L-2) in +2AL, we get

| + 2 B L | η ~ | 2 𝑑 x 𝑑 y - + 2 B L ( 0 , - 1 ) | η ~ | 2 𝑑 x 𝑑 y | + 2 A L | η ~ | 2 𝑑 x 𝑑 y = O ( L - 1 ) .

Corollary 2.4.

For R>0 and i=0,1,2, we have

(2.9) lim k - R r k R r k λ k μ k i u k 2 - i e α k u k 2 𝑑 x = 1 π - R R e η 𝑑 x .

Moreover, u0, i.e. up to a subsequence uk0 in L2(I), weakly in H~1/2,2(I), and a.e. in I.

Proof.

With the change of variables ξ=xrk, writing uk(rk)=μk+ηk2αkμk and using (1.8) and Proposition 2.2, we see that

- R r k R r k λ k μ k i u k 2 - i e α k u k 2 𝑑 x = r k λ k μ k 2 e α k μ k 2 = 1 α k - R R ( 1 + η k 2 α k μ k 2 ) 2 - i e η k + η k 2 4 α k μ k 2 𝑑 ξ 1 π - R R e η 𝑑 ξ

as k, as claimed in (2.9).

In order to prove the last statement, recalling that (-Δ)1/4ukL2=1, we write

1 = - R r k R r k λ k u k 2 e α k u k 2 d x + I ( - R r k , R r k ) λ k u k 2 e α k u k 2 d x = : (I) k + (II) k .

By (2.7) and (2.9), we get

lim k ( I ) k = 1 π - R R e η 𝑑 x = 1 + o ( 1 ) ,

with o(1)0 as R. This in turn implies that

lim R lim k (II) k = 0 ,

which is possible only if u0 or λ=0 (by Fatou’s lemma). But on account of (2.5), also in the latter case we have u0. ∎

Lemma 2.5.

For A>1, set ukA:=min{uk,μkA}. Then we have

lim sup k ( - Δ ) 1 4 u k A L 2 ( ) 2 1 A .

Proof.

We set u¯kA:=min{u~k,μkA}. Since u¯kA is an extension (in general not harmonic) of ukA, we have

(2.10) ( - Δ ) 1 4 u k A L 2 ( ) 2 + 2 | u ¯ k A | 2 𝑑 x 𝑑 y .

Using integration by parts and the harmonicity of u~k, we get

+ 2 | u ¯ k A | 2 𝑑 x 𝑑 y = + 2 u ¯ k A u ~ k d x d y
= - u k A ( x ) u ~ k ( x , 0 ) y 𝑑 x
(2.11) = ( - Δ ) 1 2 u k u k A 𝑑 x .

Note that, even if +2 is unbounded, the integration by parts above holds since |u~(x,y)|=O(|(x,y)|-1) and |u~(x,y)|=O(|(x,y)|-2) for |(x,y)| large (see Lemma A.4). Proposition 2.2 implies that ukA(rkx)=μkA for |x|R and kk0(R). Then, with (2.7) and (2.9), we obtain

( - Δ ) 1 2 u k u k A 𝑑 x - R r k R r k λ k u k e α k u k 2 u k A 𝑑 x
k 1 π A - R R e η 𝑑 ξ
R 1 A .

Set now vkA:=(uk-μkA)+=uk-ukA. With similar computations, we get

( - Δ ) 1 2 u k v k A 𝑑 x - R r k R r k λ k u k v k A e α k u k 2 𝑑 x
k 1 π ( 1 - 1 A ) - R R e η 𝑑 ξ
R A - 1 A .

Since

( - Δ ) 1 2 u k u k A 𝑑 x + ( - Δ ) 1 2 u k v k A 𝑑 x = ( - Δ ) 1 2 u k u k 𝑑 x = 1 ,

we get that

lim k ( - Δ ) 1 2 u k u k A 𝑑 x = 1 A .

Then we conclude using (2.10) and (2.11). ∎

Proposition 2.6.

We have

(2.12) C π = lim k 1 λ k μ k 2 .

Moreover,

(2.13) lim k μ k λ k = 0 .

Proof.

Fix A>1 and let ukA be defined as in Lemma 2.5. We split

I ( e α k u k 2 - 1 ) d x = I { u k μ k A } ( e α k ( u k A ) 2 - 1 ) d x + I { u k > μ k A } ( e α k u k 2 - 1 ) d x = : (I) + (II) .

Using Corollary 2.4 and Vitali’s theorem, we see that

(I) I ( e α k ( u k A ) 2 - 1 ) 𝑑 x 0 as  k

since eαk(ukA)2 is uniformly bounded in LA(I) by Lemma 2.5 together with Theorem A.

By (2.6) and Corollary 2.4, we now estimate

(II) A 2 λ k μ k 2 I { u k > μ k A } λ k u k 2 ( e α k u k 2 - 1 ) 𝑑 x A 2 λ k μ k 2 ( 1 + o ( 1 ) ) ,

with o(1)0 as k. Together with (2.2), and by letting A1, this gives

C π lim k 1 λ k μ k 2 .

The converse inequality follows from (2.9) as follows:

I ( e α k u k 2 - 1 ) 𝑑 x - R r k R r k e α k u k 2 𝑑 x + o ( 1 ) = 1 λ k μ k 2 ( 1 π - R R e η 𝑑 x + o ( 1 ) ) + o ( 1 ) ,

with o(1)0 as k. Letting R and recalling (2.7), we obtain (2.12).

Finally, (2.13) follows at once from (2.12) because otherwise we would have Cπ=0, which is clearly impossible. ∎

Proposition 2.7.

Let us set fk:=λkμkukeαkuk2. Then we have

I f k φ 𝑑 x φ ( 0 )

as k for any φC(I¯). In particular, fkδ0 in the sense of Radon measures in I.

Proof.

Take φC(I¯). For given R>0 and A>1, we split

I φ f k d x = - R r k R r k φ f k d x + { u k > μ k A } ( - R r k , R r k ) φ f k d x + { u k μ k A } φ f k d x = : I 1 + I 2 + I 3 .

On {ukμkA} we have uk=ukA, and Lemma 2.5 and Theorem A imply that ukeαkuk2 is uniformly bounded in L1 (depending on A). Thus using (2.13), we get I30.

With (2.6) and (2.9) we also get

I 2 A φ L ( I ) { u k > μ k A } ( - R r k , R r k ) λ k u k 2 e α k u k 2 𝑑 x
A φ L ( I ) ( 1 - - R r k R r k λ k u k 2 e α k u k 2 𝑑 x )
= A φ L ( I ) ( 1 - 1 π - R R e η 𝑑 x + o ( 1 ) ) ,

with o(1)0 as k. Thanks to (2.7), we conclude that I20 as k and R.

As for I1, again with (2.9) we compute

I 1 = ( φ ( 0 ) + o ( 1 ) ) ( 1 π - R R e η 𝑑 x + o ( 1 ) ) ,

so that I1φ(0) as k and R. ∎

Given xI, let Gx:{0} be the Green’s function of (-Δ)1/2 on I with singularity at x. We recall that we have the explicit formula (see, e.g., [3])

(2.14) G x ( y ) := { 1 π log ( 1 - x y + ( 1 - x 2 ) ( 1 - y 2 ) | x - y | ) , y I , 0 , y I .

In the following, we further denote

(2.15) S ( x , y ) := G x ( y ) - 1 π log 1 | x - y | .

Lemma 2.8.

We have μkukG:=G0 in Lloc(I¯{0})L1(I) as k+.

Proof.

Let us set

v k := μ k u k - G and f k = μ k λ k u k e α k u k 2 .

Arguing as in Proposition 2.7, we show that fkL1()1 as k. Moreover, since uk is decreasing with respect to |x|, we get that uk0 and fk0 locally uniformly in I¯{0} as k. By Green’s representation formula, we have

| v k ( x ) | = | I G x ( y ) f k ( y ) 𝑑 y - G ( x ) |
(2.16) I | G x ( y ) - G ( x ) | f k ( y ) 𝑑 y + | f k L 1 ( I ) - 1 | | G ( x ) | , x I .

Fix σ(0,1). If we assume |x|σ and |y|σ2, then we have

| G x ( y ) - G ( x ) | 1 π | log | x | | x - y | | + | S ( x , y ) - S ( x , 0 ) |
1 π | log | x | x | - y | x | | | + sup | x | σ , | y | σ 2 | y S ( x , y ) | | y |
C | y | ,

where C is a constant depending only on σ. Then, for any ε(0,σ2), we can write

| v k ( x ) | I | G x ( y ) - G ( x ) | f k ( y ) 𝑑 y + o ( 1 )
= - ε ε | G x ( y ) - G ( x ) | f k ( y ) 𝑑 y + I ( - ε , ε ) | G x ( y ) - G ( x ) | f k ( y ) 𝑑 y + o ( 1 )
C ε f k L 1 ( - ε , ε ) + ( sup z I G z L 1 ( I ) + | G ( x ) | ) f k L ( I ( - ε , ε ) ) + o ( 1 )
(2.17) C ε + o ( 1 ) ,

where o(1)0 uniformly in I(-σ,σ) as k. Clearly, (2.17) implies

lim sup k v k L ( I ( - σ , σ ) ) C ε .

Since ε and σ can be arbitrarily small, this shows that vk0 in Lloc(I¯{0}). With a similar argument, we prove the L1 convergence. Indeed, integrating (2.16), for ε(0,1) we get

v k L 1 ( I ) I I | G x ( y ) - G ( x ) | f k ( y ) 𝑑 y 𝑑 x + | f k L 1 ( I ) - 1 | G L 1 ( I )
I f k ( y ) I | G x ( y ) - G ( x ) | 𝑑 x 𝑑 y + o ( 1 )
- ε ε f k ( y ) I | G x ( y ) - G ( x ) | 𝑑 x 𝑑 y + 2 sup z I G z L 1 ( I ) f k L ( I ( - ε , ε ) ) + o ( 1 )
(2.18) = - ε ε f k ( y ) I | G x ( y ) - G ( x ) | 𝑑 x 𝑑 y + o ( 1 ) .

Since

sup y ( - ε , ε ) sup x I | S ( x , y ) - S ( x , 0 ) | = O ( ε ) ,

we get

- ε ε f k ( y ) I | G y ( x ) - G ( x ) | 𝑑 x 𝑑 y = 1 π - ε ε f k ( y ) I | log | x | | x - y | | 𝑑 x 𝑑 y + O ( ε ) .

Moreover, using the change of variables x=yz, we obtain

I | log | x | | x - y | | 𝑑 x = | y | - 1 | y | 1 | y | | log | z | | z - 1 | | 𝑑 z = O ( | y | log 1 | y | ) .

Then we have

(2.19) - ε ε f k ( y ) I | G y ( x ) - G 0 ( x ) | 𝑑 x 𝑑 y = - ε ε f k ( y ) O ( | y | log 1 | y | ) 𝑑 y + O ( ε ) = O ( ε log 1 ε ) .

Clearly, (2.18) and (2.19) yield lim supk+vk-GL1(I)=O(εlog1ε). Since ε can be arbitrarily small, we get the conclusion. ∎

Proposition 2.9.

We have μku~kG~ in

C loc 0 ( + 2 ¯ { ( 0 , 0 ) } ) C loc 1 ( + 2 ) ,

where G~ is the Poisson extension of G.

Proof.

As in the proof of Lemma 2.8, we denote vk:=μkuk-G. Let us consider the Poisson extension v~k=μku~k-G~. For any fixed ε>0, we can split

v ~ k ( x , y ) = 1 π - ε ε y v k ( ξ ) ( x - ξ ) 2 + y 2 𝑑 ξ + 1 π I ( - ε , ε ) y v k ( ξ ) ( x - ξ ) 2 + y 2 𝑑 ξ .

By Lemma 2.8, we have

| 1 π I ( - ε , ε ) y v k ( ξ ) ( x - ξ ) 2 + y 2 𝑑 ξ | 1 π v k L ( I ( - ε , ε ) ) y ( x - ξ ) 2 + y 2 𝑑 ξ
= v k L ( I ( - ε , ε ) ) 0

as k. Moreover, assuming (x,y)+2B2ε(0,0), we get

| 1 π - ε ε y v k ( ξ ) ( x - ξ ) 2 + y 2 𝑑 ξ | 1 π - ε ε y | v k ( ξ ) | | ( x , y ) - ( ξ , 0 ) | 2 𝑑 ξ y π ε 2 v k L 1 ( I ) 0 .

Hence v~k0 in Cloc0(+2B2ε(0,0)). Finally, since can ε be arbitrarily small and v~k is harmonic in +2, we get v~k0 in

C loc 0 ( + 2 ¯ { ( 0 , 0 ) } ) C loc 1 ( + 2 ) .

2.3 The Two Main Estimates and Completion of the Proof

We shall now conclude our contradiction argument by showing the incompatibility of (2.3) with (2.2) and the definition of Cπ. In this final part of the proof, we will use the precise asymptotic of G~ near (0,0). Since log|(x,y)| is the Poisson integral of log|x| (see Proposition A.3), and since S(0,)C(), equation (2.15) guarantees the existence of the limit

S 0 := lim ( x , y ) ( 0 , 0 ) G ~ ( x , y ) + 1 π log | ( x , y ) | = lim x 0 G ( x ) + 1 π log | x | .

In fact, using (2.14), we get S0=log2π. More precisely, noting that S(0,)C(I), we can write

(2.20) G ~ ( x , y ) = 1 π log 1 | ( x , y ) | + S 0 + h ( x , y ) ,

with

h C ( + 2 ¯ B 1 ( 0 , 0 ) ) C ( + 2 ¯ ) and h ( 0 , 0 ) = 0 .

Proposition 2.10.

If (2.3) holds, then Cπ2πeπS0=4π.

Proof.

For a fixed large L>0 and a fixed and small δ>0, set

a k := inf B L r k + 2 u ~ k , b k := sup B δ + 2 u ~ k , v ~ k := ( u ~ k a k ) b k .

Recalling that u~kL22=1, we have

( B δ B L r k ) + 2 | v ~ k | 2 𝑑 x 𝑑 y 1 - + 2 B δ | u ~ k | 2 𝑑 x 𝑑 y - + 2 B L r k | u ~ k | 2 𝑑 x 𝑑 y .

Clearly, the left-hand side bounds

inf u ~ | + 2 B L r k = a k u ~ | + 2 B δ = b k ( B δ B L r k ) + 2 | u ~ | 2 𝑑 x 𝑑 y = ( B δ B L r k ) + 2 | Φ ~ k | 2 𝑑 x 𝑑 y = π ( a k - b k ) 2 log δ - log ( L r k ) ,

where the function Φ~k is the unique solution to

{ Δ Φ ~ k = 0 in  + 2 ( B δ B L r k ) , Φ ~ k = a k on  + 2 B L r k , Φ ~ k = b k on  + 2 B δ , Φ ~ k y = 0 on  + 2 ( B δ B L r k ) ,

given explicitly by

Φ ~ k = b k - a k log δ - log ( L r k ) log | ( x , y ) | + a k log δ - b k log L r k log δ - log ( L r k ) .

Using Proposition 2.2, we obtain

a k = μ k + - 1 π log L + O ( L - 1 ) + o ( 1 ) μ k ,

where for fixed L>0 we have o(1)0 as k, and |O(L-1)|CL uniformly for L and k large. Moreover, using Proposition 2.9 and (2.20), we obtain

b k = - 1 π log δ + S 0 + O ( δ ) + o ( 1 ) μ k ,

where for fixed δ>0 we have o(1)0 as k, and |O(δ)|Cδ uniformly for δ small and k large.

Still with Proposition 2.2, we get

lim k μ k 2 + 2 B L r k | u ~ k | 2 𝑑 x 𝑑 y = 1 4 π 2 + 2 B L | η ~ | 2 𝑑 x 𝑑 y = 1 π log L 2 + O ( log L L ) .

Similarly with Proposition 2.9 we get

lim inf k μ k 2 + 2 B δ | u ~ k | 2 𝑑 x 𝑑 y + 2 B δ | G ~ | 2 𝑑 x 𝑑 y
= + 2 B δ - G ~ r G ~ d σ + ( - δ , δ ) - G ~ ( x , 0 ) y G ( x ) d x
= + 2 B δ ( 1 π δ + O ( 1 ) ) ( - 1 π log δ + S 0 + O ( δ ) ) 𝑑 σ
= - 1 π log δ + S 0 + O ( δ log δ ) ,

where we used Lemma A.4, the expansion in (2.20) and the boundary conditions

{ G ~ ( x , 0 ) = G ( x ) = 0 for  x I , - G ~ ( x , 0 ) y = ( - Δ ) 1 2 G ( x ) = 0 for  x I { 0 } .

We then get

π ( a k - b k ) 2 log δ - log ( L r k ) 1 - - 1 π log δ + S 0 + O ( δ log δ ) + 1 π log L 2 + O ( log L L ) μ k 2

or

π ( a k - b k ) 2 = π μ k 2 - 2 log L + O ( L - 1 ) + 2 log δ - 2 π S 0 + O ( δ ) + o ( 1 ) + O ( log 2 L + log 2 δ ) μ k 2
( log δ - log L + log ( λ k μ k 2 ) + α k μ k 2 + log α k )
× ( 1 - - 1 π log δ + S 0 + O ( δ log δ ) + 1 π log L 2 + O ( log L L ) μ k 2 )
= log δ L + log ( λ k μ k 2 ) + α k μ k 2 + log α k + α k ( 1 π log 2 δ L - S 0 )
+ O ( δ log δ ) + O ( log L L ) + O ( log 2 δ ) + O ( log 2 L ) + O ( 1 ) μ k 2 .

Rearranging gives

log 1 λ k μ k 2 ( 1 - α k π ) log L δ + ( α k - π ) μ k 2 + ( 2 π - α k ) S 0 + α k π log 2 + log α k + O ( δ log δ ) + O ( log L L ) + o ( 1 ) ,

with o(1)0 as k. Then, recalling that αkπ and letting first k and then L and δ0, we obtain

lim sup k log 1 λ k μ k 2 π S 0 + log ( 2 π ) = log ( 4 π ) .

Using Proposition 2.6, we conclude. ∎

Proposition 2.11.

There exists a function uH~1/2,2(I) with (-Δ)1/4uL2(R)1 such that

I ( e π u 2 - 1 ) 𝑑 x > 2 π e π S 0 = 4 π .

Proof.

For ε>0 choose L=L(ε)>0 such that, as ε0, we have L and Lε0. Fix

Γ L ε := { ( x , y ) + 2 : G ~ ( x , y ) = γ L ε := min + 2 B L ε G ~ }

and

Ω L ε := { ( x , y ) + 2 : G ~ ( x , y ) > γ L ε } .

By the maximum principle, we have

+ 2 B L ε Ω L ε .

Indeed, G~ is harmonic in +2, G~γLε on (+2BLε){(0,0)}, and G~+ as (x,y)(0,0). Notice also that (2.20) gives

(2.21) γ L ε = - 1 π log ( L ε ) + S 0 + O ( L ε ) .

For some constants B and c to be fixed, we set

U ε ( x , y ) := { c - log ( x 2 ε 2 + ( 1 + y ε ) 2 ) + 2 B 2 π c for  ( x , y ) + 2 B L ε ( 0 , - ε ) , γ L ε c for  ( x , y ) Ω L ε B L ε ( 0 , - ε ) , G ~ ( x , y ) c for  ( x , y ) + 2 Ω L ε .

Observe that +2BLε(0,-ε)+2BLεΩLε. To have continuity on +2BLε(0,-ε) we impose

- log L 2 - 2 B 2 π c + c = γ L ε c ,

which, together with (2.21), gives the relation

(2.22) B = π c 2 + log ε - π S 0 + O ( L ε ) .

Moreover,

+ 2 B L ε ( 0 , - ε ) | U ε | 2 𝑑 x 𝑑 y = 1 4 π 2 c 2 + 2 B L ( 0 , - 1 ) | log ( x 2 + ( 1 + y ) 2 ) | 2 𝑑 x 𝑑 y = 1 π log ( L 2 ) + O ( log L L ) c 2

and

+ 2 Ω L ε | U ε | 2 𝑑 x 𝑑 y = 1 c 2 + 2 Ω L ε | G ~ | 2 𝑑 x 𝑑 y
= 1 c 2 + 2 Ω L ε G ~ ν G ~ 𝑑 σ - 1 c 2 ( × { 0 } ) Ω ¯ L ε G ~ y G ~ 𝑑 x = 0
= 1 π log ( 1 L ε ) + S 0 + O ( L ε log ( L ε ) ) c 2 ,

where the last equality follows from (2.20). We now impose UεL2(+2)=1, obtaining

(2.23) - log ε - log 2 + π S 0 + O ( L ε log ( L ε ) ) + O ( log L L ) = π c 2 ,

which, together with (2.22), implies

(2.24) B = - log 2 + O ( L ε log ( L ε ) ) + O ( log L L ) .

Let now

I L , ε 1 = ( - ε L 2 - 1 , ε L 2 - 1 )

and ILε2 be the disjoint sub-intervals of I obtained by intersecting I×{0} respectively with

B L ε ( 0 , - ε ) and + 2 Ω L ε ¯ .

Then, for uε(x):=Uε(x,0), using a change of variables and (2.23)–(2.24), we get

I L , ε 1 e π u ε 2 𝑑 x = ε - L 2 - 1 L 2 - 1 exp ( π ( c - log ( 1 + x 2 ) + 2 B 2 π c ) 2 ) 𝑑 x
> ε e π c 2 - 2 B - L 2 - 1 L 2 - 1 1 1 + x 2 𝑑 x
= 2 e π S 0 + O ( L ε log ( L ε ) ) + O ( log L L ) π ( 1 + O ( 1 L ) )
= 2 π e π S 0 + O ( L ε log ( L ε ) ) + O ( log L L ) .

Moreover,

I L ε 2 ( e π u ε 2 - 1 ) d x I L ε 2 π u ε 2 d x = 1 c 2 I L ε 2 π G 2 d x = : ν L ε c 2 ,

with

ν L ε > ν 1 2 > 0 for  L ε < 1 2 .

Now observe that c2=-logεπ+O(1) by (2.23), and choose L=log2ε to obtain

O ( L ε log ( L ε ) ) + O ( log L L ) = O ( log log ε log 2 ε ) = o ( 1 c 2 ) ,

so that

I ( e π u ε 2 - 1 ) 𝑑 x 2 π e π S 0 + ν 1 2 c 2 + o ( 1 c 2 ) > 2 π e π S 0

for ε small enough.

Finally, notice that

( - Δ ) 1 4 u ε L 2 ( ) 2 = + 2 | u ~ ε | 2 𝑑 x 𝑑 y + 2 | U ε | 2 𝑑 x 𝑑 y 1

since the Poisson extension u~ε minimizes the Dirichlet energy among extensions with finite energy. ∎

3 Proof of Theorem 1.3

Let ukHC() be a sequence of positive even and decreasing solutions to (1.5) satisfying the energy bound (1.6) and with λkλ0 as k.

First we show that case (i) holds when μkC.

Lemma 3.1.

If μkC, then (i) holds.

Proof.

By assumption, we know that uk and

f k := ( - Δ ) 1 2 u k = λ k u k e α k u k 2 - u k

are uniformly bounded in L(). Then, by elliptic estimates and a bootstrap argument (see [11, Theorem 1.5] and [22, Corollary 25]), we can find uC() such that, up to a subsequence, uku in Cloc() for every 0. To prove that u satisfies (1.7), note that fkf:=λueπu2-u locally uniformly on and set M=supk(fkL()+μk). For any φ𝒮() (the Schwarz space of rapidly decreasing functions) and any R>0, we have that

| f k - f | | φ | 𝑑 x f k - f L ( ( - R , R ) ) - R R | φ | + 2 M φ L 1 ( ( - R , R ) c )
k + M φ L 1 ( ( - R , R ) c )
R + 0 .

Similarly, recalling that (-Δ)12φ has quadratic decay at infinity (see, e.g., [18, Proposition 2.1]), we get

| u k - u | | ( - Δ ) 1 2 φ | 𝑑 x ( - Δ ) 1 2 φ L ( ( - R , R ) ) u k - u L 1 ( ( - R , R ) ) + C ( - R , R ) c | u k ( x ) - u ( x ) | | x | 2 𝑑 x
( - Δ ) 1 2 φ L ( ( - R , R ) ) u k - u L 1 ( ( - R , R ) ) + 2 C M ( - R , R ) c d x x 2 𝑑 x
k , R + 0 .

Hence u is a weak solution of (1.7). ∎

From now on we will assume that μk+ and prove that Theorem 1.3 (ii) holds.

Lemma 3.2.

Let ηk be defined as in Theorem 1.3. Then ηk is bounded in Cloc0,α(R) for α(0,1).

Proof.

Note that

r k μ k 2 = 1 α k λ k e α k μ k 2 = 1 α k u k H 2 e α k μ k 2 u k 2 e α k u k 2 𝑑 x
C 1 α k u k H 2 e α k 2 μ k 2 u k 2 e α k 2 u k 2 𝑑 x
C u k L 4 2 D α k α k u k H 2 e α k 2 μ k 2
C D π α k e α k 2 μ k 2 0 .

Moreover, we have that

( - Δ ) 1 2 η k = 2 u k ( r k ) μ k e α k u k 2 ( r k ) - α k μ k 2 - 2 α k r k μ k 2 u k ( r k ) μ k

is bounded in L. Since ηk0 and ηk(0)=0, this implies that ηk is bounded in Lloc() and then in Clocα() for any α(0,1). ∎

The bound of Lemma 3.2 implies that, up to a subsequence, ηkη in Cloc0,α() for some function η. However, it does not provide a limit equation for η. In order to prove that η solves

( - Δ ) 1 2 η = 2 e η ,

we will prove that ηk is bounded in Ls() for any s>0. This bound can be obtained thanks to the commutator estimates proved in [35]. Part of the argument must be modified since the uk’s are not compactly supported. We start by recalling the following technical lemma, which is a consequence of the estimates in [35].

Lemma 3.3.

For any s(0,1), there exists a constant C=C(s) such that, for any φ,ψCc(Rn) and ρR+, we have

φ ( - Δ ) s 2 ψ L ( 1 s , ) ( ( - ρ , ρ ) ) C ( E 1 ( φ , ψ ) + E 2 , 2 ρ ( φ , ψ ) ) ,

where

E 1 ( φ , ψ ) = ( - Δ ) 1 4 φ L 2 ( ) ( - Δ ) 1 4 ψ L 2 ( )
E 2 , ρ ( φ , ψ ) = ( - Δ ) 1 4 φ L 2 ( ) ( - Δ ) 1 2 ψ L log 1 2 L ( - ρ , ρ ) .

Proof.

Let θCc((-2,2)) be a cut-off function such that θ1 on (-1,1) and 0θ1. Let us denote θρ=θ(ρ). Let us also introduce the Riesz operators

I 1 - s u := κ s | | - s * u for  s ( 0 , 1 ) ,

where the constant κs is defined by the identity

κ s | | - s ^ = | | s - 1 .

With this definition, I1-s is the inverse of (-Δ)1-s2. Then we can split

φ ( - Δ ) s 2 ψ = φ I 1 - s ( - Δ ) 1 2 ψ
= φ I 1 - s ( θ 2 ρ ( - Δ ) 1 2 ψ ) + φ I 1 - s ( ( 1 - θ 2 ρ ) ( - Δ ) 1 2 ψ )
= φ I 1 - s ( θ 2 ρ ( - Δ ) 1 2 ψ ) + [ φ , I 1 - s ] ( ( 1 - θ 2 ρ ) ( - Δ ) 1 2 ψ ) + I 1 - s ( ( 1 - θ 2 ρ ) φ ( - Δ ) 1 2 ψ )
= : J 1 + J 2 + J 3 ,

where we use the commutator notation [u,I1-s](v)=uI1-sv-I1-s(uv) for any u,vCc(). Applying respectively [35, Proposition 3.2, Proposition 3.4 and Proposition A.3], we get that

J 1 L ( 1 s , ) ( - ρ , ρ ) = I 1 2 ( ( - Δ ) 1 4 φ ) I 1 - s ( θ 2 ρ ( - Δ ) 1 2 ψ ) L ( 1 s , ) ( - ρ , ρ )
C ( - Δ ) 1 4 φ L 2 ( ) ( - Δ ) 1 2 ψ L log 1 2 L ( - 2 ρ , 2 ρ )
= C E 2 , 2 ρ ( φ , ψ ) ,

that

J 2 L ( 1 s , ) ( - ρ , ρ ) = [ φ , I 1 - s ] ( ( 1 - θ 2 ρ ) ( - Δ ) 1 4 ( - Δ ) 1 4 ψ ) L ( 1 s , ) ( - ρ , ρ )
C ( - Δ ) 1 4 φ L 2 ( ) ( - Δ ) 1 s ψ L 2 ( )
= C E 1 ( φ , ψ ) ,

and that

J 3 L ( 1 s , ) ( - ρ , ρ ) I 1 - s ( φ ( - Δ ) 1 2 ψ ) L ( 1 s , ) ( )
C φ ( - Δ ) 1 2 ψ L 1 ( )
= C ( - Δ ) 1 4 φ ( - Δ ) 1 4 ψ L 1 ( )
C E 1 ( φ , ψ ) ,

as desired. ∎

As a consequence of Lemma 3.3, we obtain the following crucial estimate.

Lemma 3.4.

For any s(0,1) there exists a constant C=C(s) such that

( - ρ , ρ ) | u ( - Δ ) s 2 u | 𝑑 x C ρ 1 - s ( E 1 ( u , u ) + E 2 , 2 ρ ( u , u ) )

for any ρ>0 and uHC(R). Here E1 and E2,2ρ are defined as in Lemma 3.3.

Proof.

By the Hölder inequality for Lorentz spaces (see, e.g., [43, Theorem 3.5]), we have

u ( - Δ ) s 2 u L 1 ( - ρ , ρ ) χ ( - ρ , ρ ) L ( 1 1 - s , 1 ) ( ) u ( - Δ ) s 2 u L ( 1 s , ) ( - ρ , ρ )
(3.1) C ρ 1 - s u ( - Δ ) s 2 u L ( 1 s , ) ( - ρ , ρ ) .

We shall bound the right-hand side of (3.1) by approximating u with compactly supported functions and applying Lemma 3.3. To this purpose, we take a sequence of cut-off functions (τj)jCc() such that τj(x)=1 for |x|j, τj(x)=0 for |x|j+1, 0τj1, and |τj|2. We define uj:=τju. We claim that

(3.2) u j u in  H 1 2 , 2 ( ) L q ( ) , q ( 2 , ) ,

and

(3.3) ( - Δ ) s 2 u j ( - Δ ) s 2 u in  L loc ( ) ,

The first claim is proved in [14, Lemma 12]. We shall prove the second claim. Set vj=uj-u. Then, for any fixed R0>0 and x(-R0,R0), if j>2R0, we have

| ( - Δ ) s 2 v j | K s ( - j , j ) | v j ( y ) | | x - y | 1 + s 𝑑 y 2 1 + s K s ( - j , j ) | u ( y ) | | y | 1 + s 𝑑 y C u L 2 ( ) j 1 + 2 s .

with C depending only on s. As j, we get (3.3).

Now, By Lemma 3.3, we know that, for any j,

(3.4) u j ( - Δ ) s 2 u j L ( 1 s , ) ( - ρ , ρ ) C ( E 1 ( u j , u j ) + E 2 , 2 ρ ( u j , u j ) ) ,

where C depends only on s. Clearly, (3.2) yields

E 1 ( u j , u j ) E 1 ( u , u ) .

Moreover,

E 2 , 2 ρ ( u j , u j ) = E 2 , 2 ρ ( u , u ) for  j 2 ρ .

Finally, (3.2) and (3.3) imply that uj(-Δ)s/2uju(-Δ)s/2u in Llocq() for every q[1,), and therefore in L(1/s,)(-ρ,ρ). Then, passing to the limit in (3.4), we get

u ( - Δ ) s 2 u L ( 1 s , ) ( - ρ , ρ ) C ( E 1 ( u , u ) + E 2 , 2 ρ ( u , u ) ) ,

and together with (3.1) we conclude. ∎

We can now apply Lemma 3.4 to uk. After scaling, we get the following bound on ηk.

Lemma 3.5.

For any s(0,1), there exists a constant C=C(s)>0 such that

- R R | ( - Δ ) s 2 η k | 𝑑 x C R 1 - s for any  R > 0 and  k k 0 ( R ) .

Proof.

First we observe that

f k := ( - Δ ) 1 2 u k = λ k u k e α k u k 2 - u k

is bounded in Llog12Lloc(). Indeed, we have

log 1 2 ( 2 + | f k | ) C ( 1 + u k ) ,

so that

| f k | log 1 2 ( 2 + | f k | ) C | f k | ( 1 + u k ) = O ( | f k | u k + 1 ) .

Since |fk|uk is bounded in L1() by (1.5) and (1.6), we get that fk is bounded in Llog12Lloc().

Then Lemma 3.4 and (1.6) imply the existence of C=C(s) such that

- ρ ρ | u k ( - Δ ) s 2 u k | 𝑑 x C ρ 1 - s , ρ ( 0 , 1 ) .

For any R>0, we can apply this with ρ=Rrk and rewrite it in terms of ηk. Then we obtain

- R R ( 1 + η k μ k 2 ) | ( - Δ ) s 2 η k | 𝑑 x C R 1 - s .

Since, by Lemma 3.2, ηk is locally bounded, if k is sufficiently large, we get

1 + η k μ k 2 1 2

and the proof is complete. ∎

Lemma 3.6.

The sequence (ηk) is bounded in Ls(R) for any s>0.

Proof.

It is sufficient to prove the statement for s(0,12). Since ηk0, Lemma 3.5 gives

C 1 K s - 1 1 | ( - Δ ) s η k | 𝑑 x
| - 1 1 η k ( x ) - η k ( y ) | x - y | 1 + 2 s 𝑑 y 𝑑 x |
- 1 1 - 2 2 η k ( x ) - η k ( y ) | x - y | 1 + 2 s 𝑑 y 𝑑 x = : I 1 + - 1 1 ( - 2 , 2 ) c η k ( x ) d y d x | x - y | 1 + 2 s = : I 2 + - 1 1 ( - 2 , 2 ) c - η k ( y ) d y d x | x - y | 1 + 2 s = : I 3 .

Take 2s<α<1. Since ηk is bounded in Clocα() by Lemma 3.2, we have that

| I 1 | C - 1 1 - 2 2 d y d x | x - y | 1 + 2 s - α C - 3 3 d z | z | 1 + 2 s - α = C .

Similarly,

| I 2 | - 1 1 | η k ( x ) | ( x - 1 , x + 1 ) c 1 | x - y | 1 + 2 s 𝑑 y 𝑑 x C .

Therefore, we obtain that

I 3 = - 1 1 ( - 2 , 2 ) c | η k ( y ) | | x - y | 1 + 2 s 𝑑 y 𝑑 x C .

But for x(-1,1) and y(-2,2) we have

| x - y | | y | + | x | 2 | y | 2 ( 1 + | y | 1 + 2 s ) 1 1 + 2 s .

Hence

I 3 = - 1 1 ( - 2 , 2 ) c | η k ( y ) | | x - y | 1 + 2 s 𝑑 y 𝑑 x 1 2 2 s ( - 2 , 2 ) c | η k ( y ) | 1 + | y | 1 + 2 s 𝑑 y .

This and Lemma 3.2 imply that ηk is bounded in Ls(). ∎

Proof of Theorem 1.3 (completed).

By Lemma 3.2, up to a subsequence, we can assume that ηkη in Clocα() for any α(0,1), with ηClocα(). Let us set

f k := ( - Δ ) 1 2 η k = 2 ( 1 + η k 2 α k μ k 2 ) e η k + η k 2 4 α k μ k 2 - 2 r k α k μ k 2 ( 1 + η k 2 α k μ k 2 ) .

As observed in the proof of Lemma 3.2, we have rkμk20 as k, and thus fk2eη locally uniformly on . Moreover, fk is bounded in L(). Then, for any Schwarz function φ𝒮(), we have

| f k - 2 e η | | φ | 𝑑 x o ( 1 ) ( - R , R ) | φ | 𝑑 x + ( f k L ( ) + 2 e η L ( ) ) ( - R , R ) c | φ | 𝑑 x 0

as k,R+. On the other hand, we know by Lemma 3.6 that ηk is bounded in Ls() and, consequently, ηLs(), s>0. In particular, for s(0,12), letting first k and then R, we get

| η k - η | | ( - Δ ) 1 2 φ | 𝑑 x ( - Δ ) 1 2 φ L ( - R , R ) η k - η L 1 ( - R , R ) + C ( - R , R ) c | η k ( x ) - η ( x ) | | x | 2 𝑑 x
C η k - η L 1 ( - R , R ) + C R 2 s - 1 ( η k L s ( ) + η L s ( ) ) 0 .

Then η is a weak solution (-Δ)12η=2eη and ηLs() for any s. Moreover, repeating the argument of Corollary 2.4 and using (1.6), we get

1 π - R R e η 𝑑 ξ = lim k - R r k R r k λ k u k 2 e α k u k 2 𝑑 x lim sup k u k H 2 = Λ ,

which implies eηL1(). Then η(x)=-log(1+x2); see, e.g., [7, Theorem 1.8].

To complete the proof, we shall study the properties of the weak limit u of uk in H. First, we show that u is a weak solution of (1.7). Let us denote

g k := λ k u k e α k u k 2 , g := λ u e π u 2 .

Take any function φ𝒮(). On the one hand, since (-Δ)12φL2() and uku weakly in L2(), we have

( u k - u ) ( - Δ ) 1 2 φ 𝑑 x + ( u k - u ) φ 𝑑 x 0

as k. On the other hand, for any large t>0 we get

| g k - g | | φ | 𝑑 x { u k t } | g k - g | | φ | 𝑑 x + φ L ( ) t u k ( g k + g ) 𝑑 x = o ( 1 ) + O ( t - 1 ) 0

as k,t, where we used that gL2() by Theorem A (see, e.g., [21, Lemma 2.3]) together with the dominated convergence theorem and the bounds ukgkL1()Λ and ukL2()Λ. Then u is a weak solution of (1.7).

Now, observe that

u k H 2 = g k u k 𝑑 x = - R r k R r k g k u k 𝑑 x + ( - R r k , R r k ) g k u k 𝑑 x

with

lim k - R r k R r k u k g k 𝑑 x = 1 π - R R e η 𝑑 x 1

as R, and

lim inf k ( - R r k , R r k ) g k u k 𝑑 x = g u 𝑑 x = u H 2

for any R>1, by Fatou’s lemma. Thus we conclude that

u k H 2 u H 2 + 1 .

Finally, to prove that uku in Cloc({0}) for every 0, we use the monotonicity of uk, which implies that uk is locally bounded away from 0. Hence we can conclude by elliptic estimates, as in Lemma 3.1. ∎

4 Proof of Theorem 1.2

Let us set

E α ( u ) = ( e α u 2 - 1 ) 𝑑 x , D α := sup u H : u H 1 E α ( u ) .

The proof of Theorem 1.2 is organized as follows. First, we prove that Dα is attained for α(0,π) sufficiently close to π. Then we fix a sequence (αk)kN such that αkπ as k+, and for any large k we take a positive extremal ukH for Dαk. With a contradiction argument similar to the one of Section 2, we show that μk:=supukC. Finally, we show that uku in Lloc()L2(), where u is a maximizer for Dπ.

4.1 Subcritical Extremals: Ruling out Vanishing

The following lemma describes the effect of the lack of compactness of the embedding HL2() on Eα, and holds uniformly for α[0,π].

Lemma 4.1.

Let (αk)[0,π] and (uk)H be two sequences such that the following conditions hold:

  1. α k α [ 0 , π ] as k .

  2. u k H 1 , uku weakly in H, uku a.e. in , and eαkuk2eαu2 in Lloc1() as k.

  3. The u k ’s are even and monotone decreasing, i.e. u k ( - x ) = u k ( x ) u k ( y ) for 0 x y .

Then we have

E α k ( u k ) = E α ( u ) + α ( u k L 2 ( ) 2 - u L 2 ( ) 2 ) + o ( 1 )

as k.

Proof.

Since uk is even and decreasing, we know that

(4.1) u k ( x ) 2 u k L 2 ( ) 2 2 | x | 1 2 | x |

for any x{0}. In particular, there exists a constant C>0, such that

e α k u k 2 ( x ) - 1 - α k u k 2 ( x ) C | x | - 4

for |x|1. Applying the dominated convergence theorem for |x|1, using the assumption that

e α k u k 2 e α u 2 in  L loc 1 ( ) ,

and recalling that (uk) is precompact in Lloc1(), we find that

( e α k u k 2 - 1 - α k u k 2 ) 𝑑 x ( e α u 2 - 1 - α u 2 ) 𝑑 x ,

and the lemma follows. ∎

Lemma 4.2.

Take α(0,π). If Dα>α, then Dα is attained by an even and decreasing function, i.e. there exists uαH even and decreasing such that uαH=1 and Eα(uα)=Dα.

Proof.

Let (uk)H be a maximizing sequence for Eα. Without loss of generality, we can assume ukuH weakly in H and a.e. on . Moreover, up to replacing uk with its symmetric decreasing rearrangement, we can assume that uk is even and decreasing (see [44]). Since α(0,π), the sequence eαuk2-1 is bounded in Lπα(), with πα>1. Then, by Vitali’s theorem, we get eαuk2eαu2 in Lloc1(), and Lemma 4.1 yields

(4.2) E α ( u k ) = E α ( u ) + α ( u k L 2 ( ) 2 - u L 2 ( ) 2 ) + o ( 1 ) .

This implies that u0 since otherwise we have Eα(uk)=αukL2()2+o(1)α+o(1), which contradicts the assumption Dα>α. Let us denote

L := lim sup k u k L 2 ( ) 2 , τ := u L 2 ( ) 2 L .

Observe that L,τ(0,1]. Let us consider the sequence vk(x)=uk(τx). Clearly, we have vkv weakly in H, where v(x):=u(τx). Moreover, since vL22=L and

( - Δ ) 1 4 v L 2 2 lim inf k ( - Δ ) 1 4 v k L 2 2 = lim inf k ( - Δ ) 1 4 u k L 2 2 1 - L ,

we get vH1. By (4.2), we have

(4.3) D α E α ( u ) + α L ( 1 - τ ) = τ E α ( v ) + α L ( 1 - τ ) τ D α + α L ( 1 - τ ) .

If τ<1, this implies DααLα, contradicting the assumptions. Hence τ=1 and (4.3) gives Dα=Eα(u). Finally, we have uH=1 since otherwise

E α ( u u H ) > E α ( u ) = D α .

Lemma 4.3.

There exists α*(0,π) such that Dα>α for any α(α*,π]. In particular, Dα is attained by an even and decreasing function uα for any α(α*,π) by Lemma 4.2.

Proof.

This follows from Proposition 4.14 by continuity. Indeed Proposition 4.14 gives Dπ>2πe-γ>π. ∎

4.2 The Critical Case

Next, we take a sequence αk such that αkπ as k. For any large k, Lemma 4.3 yields the existence of ukH even and decreasing such that Dαk=Eαk(uk). Each uk satisfies

( - Δ ) 1 2 u k + u k = λ k u k e α k u k 2

and ukH=1. Note that ukC() by elliptic estimates (see [28, Theorem 13], [11, Theorem 1.5] and [22, Corollary 25]). Multiplying the equation by uk and using the basic inequality tetet-1 for t0, we infer

1 λ k = u k 2 e α k u k 2 𝑑 x 1 α k E α k ( u k ) = 1 α k D α k .

Since DαkDπ>0, we get that λk is uniformly bounded.

Then the sequence uk satisfies the alternative of Theorem 1.3. If case (i) holds, then we can argue as in Lemma 4.2 and Lemma 4.3 and prove that Dπ is attained. Therefore, we shall assume by contradiction that case (ii) occurs.

Let rk and ηk be as in Theorem 1.3. Let η~k denote the Poisson integral of ηk.

Proposition 4.4.

We have η~kη~ in Cloc(R+2¯) for every 0, where

η ~ ( x , y ) = - log ( ( 1 + y ) 2 + x 2 )

is the Poisson integral (compare to (A.5)) of η:=-log(1+x2).

Proof.

By Theorem 1.3, we know that ηkη in Cloc() and that ηk is bounded in L1/2. Then we can repeat the argument of the proof of Proposition 2.2. ∎

Remark 4.5.

As in (2.9), the convergence ηkη in Lloc() implies

lim k - r k R r k R λ k μ k i u k 2 - i e α k u k 2 = 1 π - π π e η 𝑑 x

for i=0,1,2 and for any R>0.

Lemma 4.6.

We have uk0 in L2(R).

Proof.

Indeed, otherwise up to a subsequence we would have

( - Δ ) 1 4 u k L 2 ( ) 1 A

for some A>1. Consider the function vk=(uk-uk(1))+. Then

v k H ~ 1 2 , 2 ( I ) and ( - Δ ) 1 4 v k L 2 ( ) 1 A .

The Moser–Trudinger inequality (1.3) gives that eαkvk2 is bounded in LA(). Since

u k 2 ( 1 + ε ) v k 2 + 1 ε ( u k - v k ) 2

and |vk-uk|uk(1)0 as k, we get that eαkuk2 is uniformly bounded in Lp() for every 1<p<A. Therefore, we have

( - 1 , 1 ) ( e α k u k 2 - 1 ) 𝑑 x 0

as k. But then, by Lemma 4.1, we find Dππ, which contradicts Lemma 4.3. ∎

Lemma 4.7.

For A>1, set ukA:=min{uk,μkA}. Then we have

lim sup k ( - Δ ) 1 4 u k A L 2 ( ) 2 1 A .

Proof.

The proof is similar to the one of Lemma 2.5. We set

u ¯ k A := min { u ~ k , μ k A } .

Note that u¯kA=u~k for |(x,y)| by (4.1) and Lemma A.6. Then, since (-Δ)1/2ukL2(), we get (see (A.9))

lim R + B R + 2 u ¯ k A u ~ k ν 𝑑 σ = lim R + B R + 2 u ~ k u ~ k ν = 0 .

Since u¯kA is an extension of ukA, using integration by parts and the harmonicity of u~k, we get

( - Δ ) 1 4 u k A L 2 ( ) 2 + 2 | u ¯ k A | 2 𝑑 x 𝑑 y = + 2 u ¯ k A u ~ k d x d y
= - u k A ( x ) u ~ k ( x , 0 ) y 𝑑 x
(4.4) = ( - Δ ) 1 2 u k u k A 𝑑 x .

Proposition 4.4 implies that ukA(rkx)=μkA for |x|R and kk0(R). Noting that ukAuk and using Lemma 4.6 and Remark 4.5, we get

( - Δ ) 1 2 u k u k A 𝑑 x - R r k R r k λ k u k e α k u k 2 u k A 𝑑 x - u k u k A 𝑑 x
= 1 A - R r k R r k λ k μ k u k e α k u k 2 u k A 𝑑 x + O ( u k L 2 ( ) 2 )
k 1 π A - R R e η 𝑑 ξ
R 1 A .

Set now vkA:=(uk-μkA)+. With similar computations, we get

( - Δ ) 1 2 u k v k A 𝑑 x - R r k R r k λ k u k v k A e α k u k 2 𝑑 x + O ( u k L 2 ( ) 2 )
k 1 π ( 1 - 1 A ) - R R e η 𝑑 ξ
R A - 1 A .

Since

( - Δ ) 1 2 u k u k A 𝑑 x + ( - Δ ) 1 2 u k v k A 𝑑 x = ( - Δ ) 1 2 u k u k 𝑑 x = 1 - u k L 2 ( ) 2 1

as k, we get that

lim n ( - Δ ) 1 2 u k u k A 𝑑 x = 1 A .

Then we conclude using (4.4). ∎

Proposition 4.8.

We have

(4.5) D π = lim k 1 λ k μ k 2 .

Moreover,

(4.6) lim k μ k λ k = 0 .

Proof.

Fix A>1 and write

( e α k u k 2 - 1 ) 𝑑 x = (I) + (II) + (III) ,

where (I), (II) and (III) denote respectively the integrals over the regions

{ u k μ k A } ( - 1 , 1 ) , { u k μ k A } ( - 1 , 1 ) c , { u k > μ k A } .

Using Lemmas 4.11 and 4.7 together with Theorem A, we see that

(I) - 1 1 ( e α k ( u k A ) 2 - 1 ) 0 as  k

since eαk(ukA)2-1 is uniformly bounded in Lp for any 1p<A. By (4.1) and Lemma 4.6, we find

(II) ( - 1 , 1 ) c ( e α k u k 2 - 1 ) 𝑑 x C u k 2 𝑑 x 0 as  k .

We now estimate

(III) A 2 λ k μ k 2 { u k > μ k A } λ k u k 2 e α k u k 2 𝑑 x A 2 λ k μ k 2 ( 1 + o ( 1 ) ) ,

with o(1)0 as k, where we used that

I { u k > μ k A } λ k u k 2 e α k u k 2 𝑑 x u k H 2 = 1 .

Letting A1 gives

sup H E π lim k 1 λ k μ k 2 .

The converse inequality follows from Remark 4.5:

( e α k u k 2 - 1 ) 𝑑 x - R r k R r k e α k u k 2 𝑑 x + o ( 1 ) = 1 λ k μ k 2 ( - R R e η 𝑑 x + o ( 1 ) ) + o ( 1 ) .

with o(1)0 as k. Letting R, we obtain (4.5).

Finally, (4.6) follows at once from (4.5), because otherwise we would have Dπ=0, which is clearly impossible. ∎

Lemma 4.9.

We have

f k := λ k μ k u k e α k u k 2 δ 0

as k, in the sense of Radon measures in R.

Proof.

The proof follows step by step the proof of Proposition 2.7, with (1.4), Proposition 4.4, Remark 4.5 Lemma 4.6 and Lemma 4.7 used in place of (1.3), Proposition 2.2, equality (2.9), Lemma 2.4, and Lemma 2.5. We omit the details. ∎

For x, let Gx be the Green function of (-Δ)12+Id on with singularity at x. In the following, we set G:=G0. By translation invariance, we get Gx(y)=G(y-x) for any x,y, xy. Moreover, the inversion formula for the Fourier transform implies that

(4.7) G ( x ) = 1 2 sin | x | - 1 π sin ( | x | ) Si ( | x | ) - 1 π cos ( | x | ) Ci ( | x | ) ,

where

Si ( x ) = 0 x sin t t 𝑑 t and Ci ( x ) = - x + cos t t 𝑑 t .

We recall that the identity

(4.8) Ci ( x ) = log x + γ + 0 x cos t - 1 t 𝑑 t

holds for any x{0}, where γ denotes the Euler–Mascheroni constant; see, e.g., [17, Chapter 12.2].

Proposition 4.10.

The function G satisfies the following properties:

  1. We have G C ( { 0 } ) and

    G ( x ) = - 1 π log | x | - γ π + O ( | x | ) , G ( x ) = - 1 π x + O ( 1 )    as  x 0 .

  2. We have G ( x ) = O ( | x | - 2 ) , G(x)=O(|x|-3) and G′′(x)=O(|x|-4) as |x|.

  3. Let G ~ be the Poisson extension of G . There exists a function

    f C 1 ( + 2 ¯ )

    such that f ( 0 , 0 ) = 0 and

    G ~ ( x , y ) = - 1 π ln | ( x , y ) | - γ π + x π arctan x y - y 2 π log ( x 2 + y 2 ) + f ( x , y ) in  + 2 .

Proof.

Property (i) follows directly by formula (4.7) and the identity in (4.8). Similarly, since

S i ( t ) = π 2 - cos t t - sin t t 2 + O ( t - 3 ) , C i ( t ) = sin t t - cos t t 2 + O ( t - 3 )

as t+, we get (ii).

Given R>0, let ψCc() be a cut-off function with ψ1 on (-R,R). Let us set g0:=-1πlog||-γπ, g1:=12||ψ and g2:=G-g0-g1. By Proposition A.3, we have

g ~ 0 ( x , y ) = - 1 π log | ( x , y ) | - γ π , ( x , y ) + 2 .

Let us set θ(x,y):=arctanxy, the angle between the y-axis and the segment connecting the origin to (x,y). A direct computation shows that

Δ ( x θ ( x , y ) ) = 2 y x 2 + y 2 = 1 2 Δ ( y log ( x 2 + y 2 ) ) .

Then the function

h ( x , y ) := g ~ 1 ( x , y ) - 1 π x θ ( x , y ) + 1 2 π y log ( x 2 + y 2 )

is harmonic in +2, continuous on ¯+2, and identically 0 on (-R,R)×{0}. By [45, Theorem C], we get that

h C ( + 2 ¯ B R ( 0 , 0 ) ) .

Finally, note that formula (4.7) implies g2(0)=0 and g2C1,α() for any α(0,1). Hence, standard elliptic regularity yields

g ~ 2 C 1 , α ( + 2 ¯ B R ( 0 , 0 ) )

for any α(0,1). In particular, g~2(0,0)=g2(0)=0. ∎

Lemma 4.11.

We have μkukG in L2(R)L(R(-ε,ε)) for any ε>0.

Proof.

Let us set vk:=μkuk-G and fk=μkλkukeαkuk2. By Lemma 4.9, we have fkL1(I)1 as k+, I=(-1,1). Then, arguing as in Lemma 2.9, we get

| v k ( x ) | = | G ( y - x ) f k ( y ) 𝑑 y - G ( x ) |
(4.9) I | G ( x - y ) - G ( x ) | f k ( y ) 𝑑 y + | f k L 1 ( I ) - 1 | = o ( 1 ) | G ( x ) | + I G ( x - y ) f k ( y ) d y . = : w k ( x )

Using (4.1), Lemma 4.6 and (4.6), we get that fk0 in L2(I). In particular,

| w k ( x ) | f k L 2 ( I ) G L 2 ( ) 0 .

Fix σ(0,1) and assume |x|σ. If we further take |y|σ2, then Proposition 4.10 implies

| G ( x - y ) - G ( x ) | C | y | ,

where C is a constant depending only on σ. Thus, for any ε(0,σ2), setting Iε=(-ε,ε), we can write

| v k ( x ) | I | G ( x - y ) - G ( x ) | f k ( y ) 𝑑 y + o ( 1 ) G L ( ( - σ , σ ) ) + o ( 1 )
C - ε ε | y | f k ( y ) 𝑑 y + I ε | G ( x - y ) | f k ( y ) 𝑑 y + | G ( x ) | I ε f k ( y ) 𝑑 y + o ( 1 )
C ε f k L 1 ( I ) + f k L ( I ε ) ( G L 1 ( ) + G L ( ( - σ , σ ) ) ) + o ( 1 )
(4.10) C ε + o ( 1 ) ,

where o(1)0 as k (depending on ε and σ). Here, we used that fk0 in L((-ε,ε)) by (4.1) and (4.6). Since ε is arbitrarily small, (4.10) shows that vk0 in L((-σ,σ)).

Next, we prove the L2 convergence. First, Hölder’s inequality and Fubini’s theorem give

w k L 2 ( ) 2 = ( I G ( x - y ) f k ( y ) 𝑑 y ) 2 𝑑 x G L 1 ( ) 2 f k L 2 ( I ) 2 0

as k. With a similar argument, after integrating (4.9) and using the triangular inequality in L2, we find

v k L 2 ( ) ( ( I | G ( x - y ) - G ( x ) | f k ( y ) 𝑑 y ) 2 𝑑 x ) 1 2 + | f k L 1 ( I ) - 1 | G L 2 ( ) + w k L 2 ( )
( I f k ( y ) 𝑑 y ) 1 2 = 1 + o ( 1 ) ( I f k ( y ) | G ( x - y ) - G ( x ) | 2 𝑑 x 𝑑 y ) 1 2 + o ( 1 ) .

Since GL2(), the function ψ(y):=|G(x-y)-G(x)|2𝑑x is continuous on and ψ(0)=0. Let φC() be a compactly supported function such that φψ on I. Then Lemma 4.9 implies

I f k ( y ) | G ( x - y ) - G ( x ) | 2 𝑑 x 𝑑 y = f k ( y ) φ ( y ) 𝑑 y + o ( 1 ) 0

as k, and the conclusion follows. ∎

Repeating the argument of Proposition 2.9, we get the following lemma.

Lemma 4.12.

We have μku~kG~ in

C loc 0 ( + 2 ¯ { ( 0 , 0 ) } ) C loc 1 ( + 2 ) ,

where G~ is the Poisson extension of G.

With Proposition 4.4 and Lemma 4.12, we can give an upper bound on Dπ.

Proposition 4.13.

Under the assumption that μk as k, we have Dπ2πe-γ.

Proof.

For a fixed and small δ>0, set

a k := inf B L r k + 2 u ~ k , b k := sup B δ + 2 u ~ k , v ~ k := ( u ~ k a k ) b k .

Recalling that

u ~ k L 2 ( + 2 ) 2 = ( - Δ ) 1 4 u k L 2 ( ) 2 = 1 - u k L 2 ( ) 2 ,

we have

( B δ B L r k ) + 2 | v ~ k | 2 𝑑 x 𝑑 y 1 - u k L 2 2 - + 2 B δ | u ~ k | 2 𝑑 x - + 2 B L r k | u ~ k | 2 𝑑 x .

Clearly, the left-hand side bounds

inf u ~ | + 2 B L r k = a k u ~ | + 2 B δ = b k ( B δ B L r k ) + 2 | u ~ | 2 𝑑 x 𝑑 y = π ( a k - b k ) 2 log δ - log ( L r k ) .

Using Proposition 4.4, Proposition 4.10 and Lemma 4.12, we obtain

(4.11) { a k = μ k + - 1 π log L + O ( L - 1 ) + o ( 1 ) μ k , b k = - 1 π log δ - γ π + O ( δ | log δ | ) + o ( 1 ) μ k ,

where o(1)0 as k for fixed L>0, δ>0, and |O(L-1)|CL-1, |O(δ|logδ|)|Cδ|logδ|, uniformly for δ small and L, k large. Still by Proposition 4.4, we get

lim k μ k 2 B L r k + | u ~ k | 2 𝑑 x 𝑑 y = 1 4 π 2 B L + | η ~ | 2 𝑑 x 𝑑 y = 1 π log L 2 + O ( log L L ) .

Similarly Lemma 4.12 and Proposition 4.10 yield

lim inf k μ k 2 + 2 B δ | u ~ k | 2 𝑑 x 𝑑 y + 2 B δ | G ~ | 2 𝑑 x 𝑑 y

with

+ 2 B δ | G ~ | 2 𝑑 x 𝑑 y = + 2 B δ - G ~ r G ~ d σ + ( - δ , δ ) - G ~ ( x , 0 ) y G ( x ) d x
= + 2 B δ ( 1 π δ + O ( | log δ | ) ) ( - 1 π log δ - γ π + O ( δ | log δ | ) ) 𝑑 σ - ( - δ , δ ) G ( x ) 2 𝑑 x
= - 1 π log δ - γ π - G L 2 ( ) 2 + O ( δ log 2 δ ) ,

where we used Lemma A.5 and

- G ~ ( x , 0 ) y = ( - Δ ) 1 2 G ( x ) = - G ( x ) for  x { 0 } .

From Lemma 4.11 we get that μkukG in L2(), and hence

u k L 2 ( ) 2 = G L 2 ( ) 2 + o ( 1 ) μ k 2

as k+. We then get

π ( a k - b k ) 2 log δ - log ( L r k ) 1 - 1 π log L 2 δ - γ π + O ( δ log 2 δ ) + O ( log L L ) + o ( 1 ) μ k 2 .

Using (4.11) and rearranging as in the proof of Proposition 2.10, we find

log 1 λ k μ k 2 ( 1 - α k π ) log L δ + ( α k - π ) μ k 2 + ( α k π - 2 ) γ + α k π log 2 + log α k
+ O ( δ log 2 δ ) + O ( log L L ) + o ( 1 ) ,

with o(1)0 as k. Then, recalling that αkπ and letting first k and then L and δ0, we obtain

lim sup k log 1 λ k μ k 2 - γ + log ( 2 π ) .

Using Proposition 4.8 we conclude. ∎

Proposition 4.14.

There exists a function uH1/2,2(R) such that uH1 and Eπ(u)>2πe-γ.

Proof.

For ε>0 choose L=L(ε)>0 such that, as ε0, we have L and Lε0. Fix

Γ L ε := { ( x , y ) + 2 : G ~ ( x , y ) = γ L ε := min + 2 B L ε G ~ }

and

Ω L ε := { ( x , y ) + 2 : G ~ ( x , y ) > γ L ε } .

By the maximum principle, we have +2BLεΩLε. Notice also that Proposition 4.10 gives

γ L ε = - 1 π log ( L ε ) - γ π + O ( L ε | log ( L ε ) | )

and ΩLε+2B2Lε. For suitable constants B,c to be fixed, we set

U ε ( x , y ) := { c - log ( x 2 ε 2 + ( 1 + y ε ) 2 ) + 2 B 2 π c for  ( x , y ) B L ε ( 0 , - ε ) + 2 , γ L ε c for  ( x , y ) Ω L ε B L ε ( 0 , - ε ) , G ~ ( x , y ) c for  ( x , y ) + 2 Ω L ε .

Observe that +2BLε(0,-ε)2BLεΩLε. We choose B in order to have continuity on +2BLε(0,-ε), i.e. we impose

- log L 2 - 2 B 2 π c + c = γ L ε c ,

which gives the relation

(4.12) B = π c 2 + log ε + γ + O ( L ε | log ( L ε ) | ) .

This choice of B also implies that the function cUε does not depend on the value of c. Then we can choose c by imposing

(4.13) U ε L 2 ( + 2 ) 2 + u ε L 2 ( ) 2 = 1 ,

where we set uε(x)=Uε(x,0). Since the harmonic extension u~ε minimizes the Dirichlet energy among extensions with finite energy, we have

( - Δ ) 1 4 u ε L 2 ( ) 2 = + 2 | u ~ ε | 2 𝑑 x 𝑑 y + 2 | U ε | 2 𝑑 x 𝑑 y ,

and (4.13) implies

u ε H 1 2 , 2 ( ) 2 1 .

In order to obtain a more precise expansion of B and c, we compute

B L ε ( 0 , - ε ) + 2 | U ε | 2 𝑑 x 𝑑 y = 1 4 π 2 c 2 B L ( 0 , - 1 ) + 2 | log ( x 2 + ( 1 + y ) 2 ) | 2 𝑑 x 𝑑 y
(4.14) = 1 π log ( L 2 ) + O ( log L L ) c 2

and

+ 2 Ω L ε | U ε | 2 𝑑 x 𝑑 y = 1 c 2 + 2 Ω L ε | G ~ | 2 𝑑 x 𝑑 y
= - 1 c 2 + 2 Ω L ε G ~ ν G ~ 𝑑 σ - 1 c 2 ( × { 0 } ) Ω ¯ L ε G ~ y G ~ 𝑑 σ
= (I) + (II) .

By the divergence theorem, for τ<Lε and letting τ0, we have

(I) = - γ L ε c 2 ( × { 0 } ) ( Ω ¯ L ε B τ ) G ~ ν 𝑑 σ - γ L ε c 2 + 2 B τ G ~ ν 𝑑 σ
= γ L ε c 2 ( ( × { 0 } ) Ω ¯ L ε G 𝑑 σ + 1 )
= γ L ε c 2 ( 1 + O ( L ε log ( L ε ) ) )
(4.15) = 1 π log ( 1 L ε ) - γ π + O ( L ε log 2 ( L ε ) ) c 2 ,

where in the third identity we used that ΩLεB2Lε for Lε small enough. Observe also that

u ε L 2 ( ) 2 = 1 c 2 ( × { 0 } ) Ω ¯ L ε G 2 𝑑 x + O ( L ε log 2 ( L ε ) ) c 2 = - (II) + O ( L ε log 2 ( L ε ) ) c 2 .

Together with (4.13)–(4.15) this gives

- log ε - log 2 - γ + O ( L ε log 2 ( L ε ) ) + O ( log L L ) = π c 2 ,

which, together with (4.12), implies

B = - log 2 + O ( L ε log 2 ( L ε ) ) + O ( log L L ) .

Now, observe that

B L ε ( 0 , - ε ) ( × { 0 } ) = ( - ε L 2 - 1 , ε L 2 - 1 )

and that

- ε L 2 - 1 ε L 2 - 1 e π u ε 2 𝑑 x = ε - L 2 - 1 L 2 - 1 exp ( π ( c - log ( 1 + x 2 ) + 2 B 2 π c ) 2 ) 𝑑 x
> ε e π c 2 - 2 B - L 2 - 1 L 2 - 1 1 1 + x 2 𝑑 x
= 2 e - γ + O ( L ε log 2 ( L ε ) ) + O ( log L L ) π ( 1 + O ( 1 L ) )
= 2 π e - γ + O ( L ε log 2 ( L ε ) ) + O ( log L L ) .

Moreover,

( × { 0 } ) Ω ¯ L ε ( e π u ε 2 - 1 ) d x ( × { 0 } ) Ω ¯ L ε π u ε 2 d x = 1 c 2 ( × { 0 } ) Ω ¯ L ε π G 2 d x = : ν L ε c 2 ,

with

ν L ε > ν 1 2 > 0 for  L ε < 1 2 .

Now choose L=log2ε to obtain

O ( L ε log 2 ( L ε ) ) + O ( log L L ) = O ( log log ε log 2 ε ) = o ( 1 c 2 ) ,

so that

E π ( u ε ) = ( e π u ε 2 - 1 ) 𝑑 x 2 π e - γ + ν 1 2 c 2 + o ( 1 c 2 ) > 2 π e - γ

for ε small enough. ∎

Proof of Theorem 1.2 (completed).

By Propositions 2.11 and 4.14, we know that μkC. By the dominated convergence theorem, we have eαkuk2eπu2 in Lloc1(). Then, by Lemma 4.2, we infer

(4.16) E α k ( u k ) = E π ( u ) + π ( u k L 2 ( ) 2 - u L 2 ( ) 2 ) + o ( 1 ) .

This implies that u0; otherwise we would have

E α k ( u k ) π u k L 2 ( ) 2 + o ( 1 ) π + o ( 1 ) ,

which contradicts the strict inequality Dπ>2πe-γ>π since Eαk(uk)=DαkDπ as k.

Let us set

L := lim sup k u k 2 2 and τ = u L 2 ( ) 2 L

and observe that L,τ(0,1]. Let us consider the sequence vk(x)=uk(τx). Clearly, we have vkv in H, where v(x):=u(τx). Since vL22=L and

( - Δ ) 1 4 v L 2 2 lim inf k ( - Δ ) 1 4 v k L 2 2 = lim inf k ( - Δ ) 1 4 u k L 2 2 1 - L ,

we get vH1/2,21. By (4.16), we have

D π = E π ( u ) + π L ( 1 - τ ) = τ E π ( v ) + π L ( 1 - τ ) τ D π + π L ( 1 - τ ) .

If τ<1, this implies DππL, which is not possible. Hence, we must have τ=1 and Eπ(u)=Dπ. ∎


Communicated by Guozhen Lu


Award Identifier / Grant number: PP00P2-144669

Award Identifier / Grant number: PP00P2-170588/1

Award Identifier / Grant number: P2BSP2-172064

Funding statement: This work was supported by the Swiss National Science Foundation project nos. PP00P2-144669, PP00P2-170588/1 and P2BSP2-172064. The first author has received support by the INdAM group Gruppo Nazionale per l’Analisi Matematica, la Probabilità e le loro Applicazioni (GNAMPA).

A Appendix: The Half-Laplacian on

For u𝒮 (the Schwarz space of rapidly decaying functions), we set

( - Δ ) s u ^ ( ξ ) = | ξ | 2 s u ^ ( ξ ) , f ^ ( ξ ) := f ( x ) e - i x ξ 𝑑 x .

One can prove that it holds (see, e.g., [10, Section 3])

(A.1) ( - Δ ) s u ( x ) = K s P . V . u ( x ) - u ( y ) | x - y | 1 + 2 s 𝑑 y := K s lim ε 0 [ - ε , ε ] u ( x ) - u ( y ) | x - y | 1 + 2 s 𝑑 y ,

from which it follows that

sup x | ( 1 + x 1 + 2 s ) ( - Δ ) s φ ( x ) | < for every  φ 𝒮 .

Then one can set

(A.2) L s ( ) := { u L loc 1 ( ) : u L s := | u ( x ) | 1 + | x | 1 + 2 s 𝑑 x < } ,

and for every uLs() one defines the tempered distribution (-Δ)su by

(A.3) ( - Δ ) s u , φ := u ( - Δ ) s φ 𝑑 x = u - 1 ( | ξ | φ ^ ( ξ ) ) 𝑑 x for every  φ 𝒮 .

Moreover, for p1 and s(0,1) we will define

(A.4) H s , p ( ) := { u L p ( ) : ( - Δ ) s 2 u L p ( ) } .

In the case s=12, we have K1/2=1π in (A.1) and a simple alternative definition of (-Δ)12 can be given via the Poisson integral. For uL1/2(), define the Poisson integral

(A.5) u ~ ( x , y ) := 1 π y u ( ξ ) ( y 2 + ( x - ξ ) 2 ) 𝑑 ξ , y > 0 ,

which is harmonic in +2=×(0,) and satisfies the boundary condition u~|×{0}=u in the following sense.

Proposition A.1.

If uL1/2(R), then u~(,y)Lloc1(R) for y(0,) and u~(,y)u in the sense of distributions as y0+. If uL1/2(R)C((a,b)) for some interval (a,b)R, then u~ extends continuously to (a,b)×{0} and u~(x,0)=u(x) for any x(a,b). If uH1/2(R), then u~H1(R+2), the identity

u ~ L 2 ( + 2 ) = ( - Δ ) 1 4 u L 2 ( )

holds, and u~|R×{0}=u in the sense of traces.

Then we have (see, e.g., [4])

(A.6) ( - Δ ) 1 2 u = - u ~ y | y = 0 ,

where the identity is pointwise if u is regular enough (for instance Cloc1,α()), and has to be read in the sense of tempered distributions in general, with

(A.7) - u ~ y | y = 0 , φ := u , - φ ~ y | y = 0 , φ 𝒮 , φ ~  as in (A.5) .

More precisely, we obtain the following proposition.

Proposition A.2.

If uL1/2(R)Cloc1,α((a,b)) for some interval (a,b)R and some α(0,1), then the tempered distribution (-Δ)1/2u defined in (A.3) coincides on the interval (a,b) with the functions given by (A.1) and (A.6). For general uL1/2(R), the definitions (A.3) and (A.6) are equivalent, where the right-hand side of (A.6) is defined by (A.7).

It is known that the Poisson integral of a function uL1/2() is the unique harmonic extension of u under some growth constraints at infinity. In fact, combining [48, Theorem 2.1 and Corollary 3.1] and [45, Theorem C], we get the following proposition.

Proposition A.3.

For any uL1/2(R), the Poisson extension u~ satisfies

u ~ ( x , y ) = o ( y - 1 ( x 2 + y 2 ) ) as  | ( x , y ) | .

Moreover, if U is a harmonic function in R+2 which satisfies U(x,y)=o(y-1(x2+y2)) as |(x,y)| and U(,y)u as y0+ in the sense of distributions, then U=u~ in R+2.

Assume that uH1/2,2() solves (-Δ)1/2u=f in with fL2(). The identity

(A.8) u ~ L 2 ( + 2 ) = ( - Δ ) 1 4 u L 2 ( ) = u f 𝑑 x

of Proposition A.1 can be seen as an integration-by-parts formula on +2 for u~, since

u ~ y | y = 0 = - f

in the sense of distributions. On the other hand, the standard integration-by-parts formula for u~ on +2BR implies

u ~ L 2 ( + 2 B R ) = - R R u f 𝑑 x + B R + 2 u ~ u ~ ν 𝑑 σ ,

where

u ~ ν ( x , y ) = u ~ ( x , y ) ( x , y ) | ( x , y ) |

is the normal derivative of u~ on BR. Then the validity of the integration-by-parts formula (A.8) is equivalent to

(A.9) lim + B R + 2 u ~ u ~ ν 𝑑 σ = 0 .

We shall prove that (A.9) holds even when uH1/2,2(), provided u has a good decay at infinity. This provides integration by parts formulas for the Poisson extension of Green’s functions, which are not in H1/2,2(). In the following, we set

P ( x , y ) := 1 π y x 2 + y 2 .

Lemma A.4.

Assume that uL1(R) and u0 in RI. Then R0>0 such that

(A.10) | u ~ ( x , y ) | C x 2 + y 2 𝑎𝑛𝑑 | u ~ ( x , y ) | C x 2 + y 2

for any (x,y)R+2 with |(x,y)|R0. In particular, (A.9) holds.

Proof.

Note that we have P(x-ξ,y)1πy and |P(x-ξ,y)|1yP(x-ξ,y) for any (x,y)+2. In particular, we have

| u ~ ( x , y ) | 1 π y | u ( ξ ) | 𝑑 ξ and | u ~ ( x , y ) | 1 π y 2 | u ( ξ ) | 𝑑 ξ .

Moreover, for |x|2 we have |x-ξ||x|2 for any ξI, and hence

| u ~ ( x , y ) | y x 2 4 + y 2 u L 1 ( ) 4 u L 1 ( ) x 2 + y 2 .

Similarly, using again that |P(x-ξ)|1yP(x,y), we get

| u ~ ( x , y ) | 4 u L 1 ( ) x 2 + y 2 .

Lemma A.5.

Let uL1(R)C2(R(-R,R)) for some R>0. If |u′′(x)|=O(|x|-4) as |x|, then (A.10) and (A.9) hold.

Proof.

It is sufficient to prove the lemma for uC2(). Indeed, we can write u=u1+u2, with u1=uφ and u2=u(1-φ), where φCc() is a cut-off function which equals 1 in (-R,R). Note that u~2C2() and that it coincides with u when |x| is large. By lemma A.4, inequalities (A.10) hold for u~1. Hence, (A.10) holds for u~=u~1+u~2 if it holds for u~2.

Thus, we assume that uC2(). Let us consider the complex map f(z)=iz+1z+i, z, which defines a bi-holomorphic map between the half plane and the complex disk. In Euclidean coordinates, f and its inverse correspond respectively to the conformal diffeomorphisms Φ:+2B12, Ψ:B1+2 given by

Φ ( x , y ) := ( 2 x x 2 + ( y + 1 ) 2 , x 2 + y 2 - 1 x 2 + ( y + 1 ) 2 ) ,

Ψ ( x , y ) := ( 2 x x 2 + ( y - 1 ) 2 , 1 - x 2 - y 2 x 2 + ( y - 1 ) 2 ) .

Note that Φ and Ψ extend continuously to diffeomorphisms

Φ : + 2 ¯ B 1 ¯ { ( 0 , 1 ) } , Ψ : B 1 ¯ { ( 0 , 1 ) } + 2 ¯ .

In fact, we have Ψ|B1(z)=πN(z) and Φ(x,0)=πN-1(x), where πN:B1, π(z1,z2)=z11-z2, is the stereographic projection from the north pole. Let us consider the function

v ( x ) := { u ( π N ( z ) ) , z B 1 ( 0 , 1 ) , 0 , z = ( 0 , 1 ) .

Near the point (0,1), in local coordinates given by the stereographic projection

π S - 1 ( t ) = ( 2 t 1 + t 2 , 1 - t 2 1 + t 2 ) , t ,

we get that v(πS-1(t))=u(t-1) for any t0. The decay assumption on u′′ implies that |dkdtku(t-t)|=O(t2+k) for k=0,1,2 as t0. In particular, we get

lim t 0 u ( t - 1 ) = 0 , lim t 0 d d t u ( t - 1 ) = 0 , u ′′ ( t - 1 ) = O ( 1 )  as  t 0 .

This implies that v is of class C1,1(B1). Let then V be the unique harmonic extension of v to B1, which can be defined by the Poisson formula for B1. The regularity of v implies that VC1,α(B1¯) for any α(0,1). In particular, V and the elements of its Jacobian matrix DV are bounded on B1¯. Since Φ is a conformal map and V is bounded, Proposition A.3 implies that u~(x,y)=V(Φ(x,y)). Moreover, we have |V(z)|C|z-(0,1)| for any zB1. Then we get

| u ~ ( x , y ) | C | Φ ( x , y ) - ( 0 , 1 ) | = 2 x 2 + ( 1 + y ) 2 = O ( 1 x 2 + y 2 ) .

Similarly, since |DV|=O(1), we find

| u ~ ( x , y ) | C | Φ y ( x , y ) | = 2 C x 2 + ( 1 + y ) 2 = O ( 1 x 2 + y 2 ) ,

as desired. ∎

With no regularity assumptions, one can still prove that u~ decays at infinity if u does.

Lemma A.6.

Assume that uLp(R) for some p1. If lim|x|u(x)=0, then lim|(x,y)|0u~(x,y)=0.

Proof.

By Hölder’s inequality, we have

| u ~ ( x , y ) | 1 π ( P ( x - ξ , y ) p p - 1 𝑑 ξ ) p - 1 p u L p ( ) C y 1 / p 0

as y+. Then it is sufficient to prove that lim|x|+|u~(x,y)|=0, uniformly with respect to y(0,). To see this, we write the integral in the definition of u~ as the sum of integrals on I(x):=(-|x|2,|2|2) and on I(x). For ξI(x), we have |x-ξ||x|2 and

I ( x ) P ( x - ξ , y ) u ( ξ ) 𝑑 ξ 4 y x 2 + y 2 - | x | 2 | x | 2 | u ( ξ ) | 𝑑 ξ 4 y | x | 1 - 1 p u L p ( ) x 2 + y 2 2 u L p ( ) | x | 1 p .

Instead, for |ξ||x|2, we have

- | x | 2 | x | 2 P ( x - ξ , y ) u ( ξ ) 𝑑 ξ sup | ξ | | x | 2 | u ( ξ ) | P ( x - ξ , y ) 𝑑 ξ = sup | ξ | | x | 2 | u ( ξ ) | 0

as |x|+. This gives the conclusion. ∎

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Received: 2019-06-22
Revised: 2020-03-17
Accepted: 2020-03-18
Published Online: 2020-04-15
Published in Print: 2020-08-01

© 2020 Walter de Gruyter GmbH, Berlin/Boston

This work is licensed under the Creative Commons Attribution 4.0 International License.

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