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Advances in Nonlinear Analysis

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Critical and subcritical fractional Trudinger–Moser-type inequalities on

Futoshi Takahashi
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  • Department of Mathematics, Osaka City University & OCAMI, Sumiyoshi-ku, Osaka, 558-8585, Japan
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Published Online: 2017-11-16 | DOI: https://doi.org/10.1515/anona-2017-0116


In this paper, we are concerned with the critical and subcritical Trudinger–Moser-type inequalities for functions in a fractional Sobolev space H1/2,2 on the whole real line. We prove the relation between two inequalities and discuss the attainability of the suprema.

Keywords: Trudinger–Moser inequality; fractional Sobolev spaces; maximizing problem

MSC 2010: 35A23; 26D10

1 Introduction

Let ΩN, with N2, be a domain with finite volume. Then the Sobolev embedding theorem assures that W01,N(Ω)Lq(Ω) for any q[1,+). However, by a simple example we see that the embedding W01,N(Ω)L(Ω) does not hold. Instead, functions in W01,N(Ω) enjoy the exponential summability:

W01,N(Ω){uLN(Ω):Ωexp(α|u|N/(N-1))𝑑x< for any α>0};

see Yudovich [19], Pohozaev [33] and Trudinger [37]. Later, Moser [26] improved the above embedding, and obtained the following inequality, now known as the Trudinger–Moser inequality:




and ωN-1=|SN-1| denotes the area of the unit sphere in N. On the attainability of TM(Ω,α), Carleson and Chang [5], Struwe [36], Flucher [10] and Lin [24] proved that TM(Ω,α) is attained for any 0<ααN.

On domains with infinite volume, for example on the whole space N, the Trudinger–Moser inequality does not hold as it is. However, several variants are known on the whole space. In the following, let


denote the truncated exponential function.

First, Ogawa [27], Ogawa and Ozawa [28], Cao [4], Ozawa [29] (for small α>0) and finally Adachi and Tanaka [1] proved that the following inequality holds true, which we call Adachi–Tanaka-type Trudinger–Moser inequality:


(see also do Ó [8] and Cassani, Sani and Tarsi [6] for further information). This inequality enjoys the scale invariance under the scaling u(x)uλ(x)=u(λx) for λ>0. Note that the critical exponent α=αN is not allowed for the finiteness of the supremum. Recently, it was proved by Ishiwata, Nakamura and Wadade [16] and Dong and Lu [9] that A(N,α) is attained for any α(0,αN). In this sense, the Adachi–Tanaka-type Trudinger–Moser inequality can be considered as a subcritical inequality.

On the other hand, Ruf [34] and Li and Ruf [22] proved that the following inequality holds true:




is the full Sobolev norm. Note that the scale invariance (uuλ) does not hold for this inequality. Also note that the critical exponent α=αN is permitted to the finiteness. Later, Adimurthi and Yang [2] proved that for all β[0,1) and all τ>0 there holds


by a different method. Clearly, the case β=0 and τ=1 reduces to that of Ruf [34] and Li and Ruf [22].

Concerning the attainability of B(N,α), the following facts have been proved:

  • If N3, then B(N,α) is attained for 0<ααN; see [22].

  • If N=2, then there exists α*>0 such that B(2,α) is attained for α*<αα2 (=4π); see [34] (for α=α2, see [15]).

  • If N=2 and α>0 is sufficiently small, then B(2,α) is not attained; see [15].

The non-attainability of B(2,α) for α sufficiently small is attributed to the non-compactness of “vanishing” maximizing sequences, as described in [15]. Concerning the attainability of AN,β,τ(α), recently Li and Yang [21] proved that AN,β,τ(α) is attained when 0<β<1, τ>0 and ααN. This complements the results by Li and Ruf [22] and Ishiwata [15].

In the following, we focus our attention on the fractional Sobolev spaces.

Let s(0,1), p[1,+) and let ΩN be a bounded Lipschitz domain. For s>0, let us consider the space


For uLs(N), we define the fractional Laplacian (-Δ)s/2u as follows: First, for ϕ𝒮(N), the rapidly decreasing function on N is defined via the normalized Fourier transform as


for xN. Then for uLs(N), the fractional Laplacian (-Δ)s/2u is defined as an element of 𝒮(N), the tempered distributions on N, by the relation


Note that Lp(N)Ls(N) for any p1. Also note that it could happen that supp((-Δ)s/2u)Ω even if supp(u)Ω for some open set Ω in N.

By using the above notion, we define the Bessel potential space Hs,p(Ω) for a (possibly unbounded) set ΩN as

Hs,p(N)={uLp(N):(-Δ)s/2uLp(N)},H~s,p(Ω)={uHs,p(N):u0 on NΩ}.

On the other hand, the Sobolev–Slobodeckij space Ws,p(N) is defined as


and for a bounded domain ΩN, we define




It is known that

W~s,p(Ω)={uWs,p(N):u0 on NΩ}

if Ω is a Lipschitz domain, Hs,p(N)=Fp,2s(N) (the Triebel–Lizorkin space), and Ws,p(N)=Bp,ps(N) (the Besov space). Thus Hs,2(N)=Ws,2(N). However in general, Hs,p(N)Ws,p(N) for p2; see [31, 17] and the references therein.

Recently, Martinazzi [25] (see also [18]) proved a fractional Trudinger–Moser-type inequality on H~s,p(Ω) as follows: Let p(1,) and s=Np for N. Then for any open ΩN with |Ω|<, it holds that




We note that, differently from the classical case, the attainability of the supremum is not known even for N=1 and p=2.

On the Sobolev–Slobodeckij spaces W~s,p(Ω) with sp=N, a similar fractional Trudinger–Moser inequality was also proved by Parini and Ruf [31] when N2, and Iula [17] when N=1. They proved the validity of the inequality for sufficiently small values of α>0, and the problem of the sharp exponent is still open.

In the following, we are interested in the simplest one-dimensional case, that is, we put N=1, s=12 and p=2. In this case, the Bessel potential space H1/2,2() coincides with the Sobolev–Slobodeckij space W1/2,2(), and both seminorms are related as


see [7, Proposition 3.6.]. Then the fractional Trudinger–Moser inequality in [25, 18] can be read as in the following proposition.

Proposition 1.1 (A fractional Trudinger–Moser inequality on H~1/2,2(I)).

Let IR be an open bounded interval. Then it holds that


For the fractional Adachi–Tanaka-type Trudinger–Moser inequality on the whole line, put


Then by the precedent results by Ogawa and Ozawa [28] and Ozawa [29] it is known that A(α)< for small exponent α.

On the other hand, a fractional Li–Ruf-type Trudinger–Moser inequality on H1/2,2() is already known as follows.

Proposition 1.2 ([18]).

We have




is the full Sobolev norm on H1/2,2(R).

Concerning A(α) in (1.1), a natural question is to determine the range of the exponent α>0 for which A(α) is finite. As pointed out in [14], this remained an open problem for a while. In this paper, we first prove the finiteness of the supremum in the full range of values of the exponent.

Theorem 1.3 (Full range Adachi–Tanaka-type on H1/2,2(R)).

We have


Ozawa [30] proved that the Adachi–Tanaka-type Trudinger–Moser inequality is equivalent to the Gagliardo–Nirenberg-type inequality, and he obtained an exact relation between the best constants of both inequalities. Actually, he proved the result for general 1<p<, and if p=2, the main result in [30] can be read as follows: Put α0=sup{α>0:A(α)<} and

β0=lim supqsupuH1/2,2(),u0uLq()q1/2(-Δ)1/4uL2()1-2/quL2()2/q.

Then it is shown that 1/α0=2eβ02; see [30, Theorem 1]. Thus, by a direct consequence of Theorem 1.3, we have the next corollary.

Corollary 1.4.

We have β0=(2πe)-1/2.

Furthermore, we obtain the relation between the suprema of both critical and subcritical Trudinger–Moser-type inequalities along the line of [20].

Theorem 1.5 (Relation).

We have


Also we obtain how the Adachi–Tanaka-type supremum A(α) behaves when α tends to π.

Theorem 1.6 (Asymptotic behavior).

There exist C1,C2>0 such that for any α<π which is close enough to π it holds that


Note that the estimate from above follows from Theorem 1.5 and Proposition 1.2. On the other hand, we will see that the estimate from below follows from a computation using the Moser sequence.

Concerning the existence of maximizers of the Adachi–Tanaka-type supremum A(α) in (1.1), we have the following theorem.

Theorem 1.7 (Attainability of A(α)).

A(α) is attained for any α(0,π).

On the other hand, as for B(α) in (1.2), we have the following result.

Theorem 1.8 (Non-attainability of B(α)).

B(α) is not attained for 0<α1.

It is plausible that there exists α*>0 such that B(α) is attained for α*<απ, but we do not have a proof up to now.

Finally, we improve the subcritical Adachi–Tanaka-type inequality along the line of [9].

Theorem 1.9.

For α>0, set


Then we have


Furthermore, E(α) is attained for all α(0,π).



for t, Theorem 1.9 extends Theorem 1.3. In the classical case, Dong and Lu [9] used a rearrangement technique to reduce the problem to one dimension and obtained a similar inequality by estimating a one-dimensional integral. In the fractional setting H1/2,2, we cannot follow this argument and we need a new idea.

The organization of the paper is as follows: In Section 2, we prove Theorems 1.3, 1.5 and 1.6. In Section 3, we prove Theorems 1.7 and 1.8. In Section 4, we prove Theorem 1.9.


After this work was completed, the author was informed by the anonymous referee that the full range Adachi–Tanaka-type inequality is proven, among other relevant results, in the recent preprints [12, 13] by different methods.

2 Proofs of Theorems 1.3, 1.5 and 1.6

For the proofs of Theorems 1.3, 1.5 and 1.6, we prepare several lemmas.

Lemma 2.1.



Then A~(α)=A(α) for any α>0.


For any uH1/2,2(){0} and λ>0, we put uλ(x)=u(λx) for x. Then we have




Thus for any uH1/2,2(){0} with (-Δ)1/4uL2()1, if we choose λ=uL2()2, then uλH1/2,2() satisfies




which implies A(α)A~(α). The opposite inequality is trivial. ∎

Lemma 2.2.

For any 0<α<π, it holds that



Choose any uH1/2,2() with (-Δ)1/4uL2()1 and uL2()=1. Further, put v(x)=Cu(λx), where C2=απ(0,1) and λ=C2/(1-C2). Then by the scaling rules (2.1) we see


Also we have


Thus, testing B(π) by v, we see


By taking the supremum for uH1/2,2() with (-Δ)1/4uL2()1 and uL2()=1, we have


Finally, Lemma 2.1 implies the result. ∎

Proof of Theorem 1.3.

The assertion that A(α)< for α<π follows from Lemma 2.2 and the fact that B(π)< by Proposition 1.2.

For the proof of A(π)=, we use the Moser sequence

uε={(log(1/ε))1/2if |x|<ε,log(1/|x|)(log(1/ε))1/2if ε<|x|<1,0if 1|x|,(2.2)

and its estimates


as ε0 for some C>0. Note that uεW~1/2,2((-1,1))W1/2,2()=H1/2,2(). For estimate (2.3), we refer to [17, Proposition 2.2]. For estimate (2.4), we refer to [17, (35)]. Actually, after a careful look at the proof of [17, Proposition 2.2], we confirm that


for a positive C>0, which implies (2.4). For (2.5), we compute


as ε0. Thus we obtain (2.5).

By testing A(π) by vε=uε/(-Δ)1/4uεL2(), we have


since et-1(1/2)et for t large and by (2.4). Also since

t1+1/Ct-t=-1/C1+1/Ct-1Cas t,

we see


as t. Put t=log(1/ε). We see


which leads to


for some δ>0 independent of ε0. Therefore, by (2.3), (2.4) and (2.5) we have for δ>0,


as ε0. This proves A(π)=. ∎

Proof of Theorem 1.5.

By Lemma 2.2, we have


Let us prove the opposite inequality. Let


be a maximizing sequence of B(π). We may assume (-Δ)1/4unL2()2<1 for any n. Put


Thus by (2.1) we see


since (-Δ)1/4unL2()2+unL2()21. Thus, setting αn=π(-Δ)1/4unL2()2<π for any n, we may test A(αn) by {vn}, which results in


Here we have used a change of variables y=λnx for the second equality, and vnL2()21 for the first inequality. Letting n, we have the desired result. ∎

Proof of Theorem 1.6.

We need to prove that there exists C1>0 such that for any α<π which is sufficiently close to π it holds that


Again we use the Moser sequence (2.2) and we test A(α) by vε=uε/(-Δ)1/4uεL2(). As in the similar calculations in the proof of Theorem 1.3, we have


where we put


Now, for α<π which is sufficiently close to π, we fix ε>0 small such that


which implies


With this choice of ε>0, we have


Now, we estimate that


where f(t)=2t/(C+1-t) for t[0,1] and we have used (2.6) in the last inequality. We easily see that f(0)=0 and f(t)=2(C+1)/(C+1-t)2>0 for t>0, thus f(t) is strictly increasing in t and maxt[0,1]f(t)=f(1)=2C. Thus we have


which is independent of α. Going back to (2.7) with (2.6), we observe that


which proves the result. ∎

3 Proofs of Theorems 1.7 and 1.8

For uH1/2,2(), we denote by u* its symmetric decreasing rearrangement defined as follows: For a measurable set A, let A* denote an open interval A*=(-|A|/2,|A|/2). We define u* by


where χA denote the indicator function of a measurable set A. Note that u* is nonnegative, even and decreasing on the positive line +=[0,+). It is known that


for any nonnegative measurable function F:++, which is the difference of two monotone increasing functions F1, F2 with F1(0)=F2(0)=0 such that either F1|u| or F2|u| is integrable. Also the inequality of Pólya–Szegő type


holds true for uH1/2,2(); see for example [3, 32, 23].

Remark 3.1.

Note that the radial compactness lemma by Strauss [35] is violated on . More precisely, let


then Hrad1/2,2() cannot be embedded compactly in Lq() for any q>0. To see this, let ψ0 be an even function in Cc() with supp(ψ)(-1,1), and put un(x)=ψ(x-n)+ψ(x+n). Then we see that un is an even, compactly supported smooth function, and un0 weakly in H1/2,2() as n. But {un} does not have any strong convergent subsequence in Lq() because unLq()q=2ψLq()q>0 for any n sufficient large.

However, for a sequence {un}nH1/2,2() with un even, nonnegative and decreasing on +, we have the following compactness result.

Proposition 3.2.

Assume {un}H1/2,2(R) to be a sequence such that un is even, nonnegative and decreasing on R+. Let unu weakly in H1/2,2(R). Then unu strongly in Lq(R) for any q(2,+) for a subsequence.


Since {un}H1/2,2() is a weakly convergent sequence, we have


for some C>0. We also have un(x)u(x) a.e. x for a subsequence, thus u is even, nonnegative and decreasing on +. Now, we use the estimate below, which is referred to a simple radial lemma: If uL2() is even, nonnegative and decreasing on +, then it holds that


Thus un2(x)C2|x| for x0 by


and u2(x)C2|x| by the pointwise convergence. Now, set vn=|un-u|q for q>2. Then we see vn(x)0 a.e. x. Moreover,


as R since q>2. Thus {vn}n is uniformly integrable. Also, by [7, Theorem 6.9] we know that

H1/2,2()Lq0()for any q02, anduLq0()CuH1/2,2().

For any q>2, take q0 such that 2<q<q0<. Since un-u is uniformly bounded in H1/2,2(), we have un-uLq0()C, and


for any bounded measurable set I. Therefore, Ivn𝑑x0 if |I|0, which implies that {vn} is uniformly absolutely continuous. Thus by Vitali’s convergence theorem (see, for example, [11, p. 187]) we obtain vn=|un-u|q0 strongly in L1(), which is the desired conclusion. ∎

Proposition 3.3.

Assume {un}H1/2,2(R) to be a sequence with (-Δ)1/4unL2(R)1. Let unu weakly in H1/2,2(R) for some u and assume un is even, nonnegative and decreasing on R+. Then we have


for any α(0,π).


A similar proposition has already appeared; see [16, Lemma 3.1] and [9, Lemma 5.5]. We prove it here for the reader’s convenience.



Note that Φα(t) is nonnegative, strictly convex and Ψα(t)=2αtΦα(t). Thus by the mean value theorem we have


Thus we have


by Hölder’s inequality, where a,b,c>1 and 1a+1b+1c=1 are chosen later.

First, direct calculation shows that


for all b>1. Thus if we fix 1<b<πα so that bα<π is realized, then we have


Here we used (3.3) for the third inequality and Theorem 1.3 for the last inequality, the use of which is valid since (-Δ)1/4unL2()1 and (-Δ)1/4uL2()1 by the weak lower semicontinuity. Note that {un} satisfies


for some C>0. Thus we have obtained Φα(un)+Φα(u)Lb()=O(1) independent of n.

Next, we estimate the term |un|+|u|La(). Since {un} is a bounded sequence in H1/2,2(), we have by [7, Theorem 6.9] that uLq()CunH1/2,2() for any q2. Thus we see |un|+|u|La()C for some C>0 independent of n for a2. Now, note that if we choose 1<b<πα and a>2 sufficiently large, then we can find c>2 such that 1a+1b+1c=1.

By these choices and Proposition 3.2, we conclude that un-uLc()0 as n. Going back to (3), we conclude that


which is the desired conclusion. ∎

Proof of Theorem 1.7.

We will show that A(α) in (1.1) is attained for any 0<α<π. Since A(α)=A~(α) by Lemma 2.1, we choose a maximizing sequence for A~(α):


Here {un}nH1/2,2() satisfies (-Δ)1/4unL2()1 and unL2()=1. Appealing to the use of rearrangement, moreover we may assume that un is nonnegative, even and decreasing on +. Since {un}nH1/2,2() is a bounded sequence, we have uH1/2,2() such that unu in H1/2,2(). By Proposition 3.3, we see


as n. Therefore, since unL2()2=1, we have, letting n,


Next, we claim that A(α)>α for any 0<α<π. Indeed, take any u0H1/2,2() such that u00, (-Δ)1/4u0L2()1 and u0L2()=1. Then we have


Now, since eαt2-1-αt2>0 for any t>0, we have


for u00, which results in A(α)>α, the claim.

By the claim and (3.4), we conclude that the weak limit u satisfies u0. By the weak lower semi-continuity, we have that u0 satisfies (-Δ)1/4uL2()1 and uL2()1. Thus, by (3.4) again, we see


Thus we have shown that uH1/2,2() maximizes A(α). ∎

Next, we prove Theorem 1.8. We follow Ishiwata’s argument in [15]. Let


Actually, we will show a stronger claim that Jα has no critical point on M for sufficiently small α>0. Assume to the contrary that there exists a critical point vM of Jα for small α>0. Then we define an orbit on M through v as


Note that w1=v, thus it must hold that


By the scaling rules (2.1), we see for any p2,


Now, we see




with a=(-Δ)1/4v22, b=v22 and c=v2j2j. Since


and (-Δ)1/4v22+v22=1, we calculate


Here, we need the following lemma.

Lemma 3.4 (Ogawa and Ozawa [28]).

There exists C>0 such that for any uH1/2,2(R) and p2, it holds that



For p=2j, Lemma 3.4 implies


Thus, for 0<α1 sufficiently small (it would be enough that α<12e), Stirling’s formula j!jje-j2πj implies that


for some C>0 independent of α. Therefore we have


for small α, which is the desired contradiction. ∎

4 Proof of Theorem 1.9

In order to prove Theorem 1.9, first we set


for β>0. Then we obtain the following result.

Proposition 4.1.

We have F(β)< for β<π.


We follow the proof of [18, Theorem 1.5]. Take any uH1/2,2() with uH1/2,2()1 in the admissible sets for F(β) in (4.1). By appealing to the rearrangement, we may assume that u is even, nonnegative and decreasing on +. We divide the integral


where I=(-12,12).

First, we estimate (I). By the radial lemma (3.1), we see for any k, k2,

u2k(x)(uL2()22|x|)k=uL2()2k2k1|x|kfor x0.



Therefore, we have


Now by the constraint uH1/2,2()1, we have uL2()1. Also if we put


then k=2ak converges since ak+1/ak=β(k-1)/k20 as k. Thus we obtain


where C>0 is independent of uH1/2,2() with uH1/2,2()1.

Next, we estimate (II). Set


Then by the argument of [18] we know that




for xI. Put


Then we have wH~1/2,2(I) since v0 on I, and


Thus we may use the fractional Trudinger–Moser inequality (Proposition 1.1) to w to obtain


for some C>0 independent of u. By u2w2+2 on I, we conclude that


Now, since β<π, there is an absolute constant C0 such that


for any t. Finally, we obtain


Proposition 4.1 follows from the estimates (I) and (II). ∎

By using Proposition 4.1 and arguing as in the proof of Theorem 1.3 (after establishing claims similar to those in Lemma 2.1 and Lemma 2.2), it is easy to obtain the following proposition.

Proposition 4.2.

For any 0<α<β<π, we have


Since F(β)< for any β<π, this proves the first part of Theorem 1.9. For the attainability of E(α) for α(0,π) it is enough to argue as in the proof of Theorem 1.7. We omit the details.


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About the article

Received: 2017-05-22

Revised: 2017-08-10

Accepted: 2017-09-21

Published Online: 2017-11-16

Funding Source: Japan Society for the Promotion of Science

Award identifier / Grant number: 15H03631

Award identifier / Grant number: 26610030

Part of this work was supported by JSPS Grant-in-Aid for Scientific Research (B), no. 15H03631, and JSPS Grant-in-Aid for Challenging Exploratory Research, no. 26610030.

Citation Information: Advances in Nonlinear Analysis, Volume 8, Issue 1, Pages 868–884, ISSN (Online) 2191-950X, ISSN (Print) 2191-9496, DOI: https://doi.org/10.1515/anona-2017-0116.

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