Let , with , be a domain with finite volume. Then the Sobolev embedding theorem assures that for any . However, by a simple example we see that the embedding does not hold. Instead, functions in enjoy the exponential summability:
On domains with infinite volume, for example on the whole space , 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 , Ogawa and Ozawa , Cao , Ozawa  (for small ) and finally Adachi and Tanaka  proved that the following inequality holds true, which we call Adachi–Tanaka-type Trudinger–Moser inequality:
(see also do Ó  and Cassani, Sani and Tarsi  for further information). This inequality enjoys the scale invariance under the scaling for . Note that the critical exponent is not allowed for the finiteness of the supremum. Recently, it was proved by Ishiwata, Nakamura and Wadade  and Dong and Lu  that is attained for any . In this sense, the Adachi–Tanaka-type Trudinger–Moser inequality can be considered as a subcritical inequality.
is the full Sobolev norm. Note that the scale invariance () does not hold for this inequality. Also note that the critical exponent is permitted to the finiteness. Later, Adimurthi and Yang  proved that for all and all there holds
Concerning the attainability of , the following facts have been proved:
If , then is attained for ; see .
If and is sufficiently small, then is not attained; see .
The non-attainability of for α sufficiently small is attributed to the non-compactness of “vanishing” maximizing sequences, as described in . Concerning the attainability of , recently Li and Yang  proved that is attained when , and . This complements the results by Li and Ruf  and Ishiwata .
In the following, we focus our attention on the fractional Sobolev spaces.
Let , and let be a bounded Lipschitz domain. For , let us consider the space
For , we define the fractional Laplacian as follows: First, for , the rapidly decreasing function on is defined via the normalized Fourier transform as
for . Then for , the fractional Laplacian is defined as an element of , the tempered distributions on , by the relation
Note that for any . Also note that it could happen that even if for some open set Ω in .
By using the above notion, we define the Bessel potential space for a (possibly unbounded) set as
On the other hand, the Sobolev–Slobodeckij space is defined as
and for a bounded domain , we define
It is known that
We note that, differently from the classical case, the attainability of the supremum is not known even for and .
On the Sobolev–Slobodeckij spaces with , a similar fractional Trudinger–Moser inequality was also proved by Parini and Ruf  when , and Iula  when . They proved the validity of the inequality for sufficiently small values of , 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 , and . In this case, the Bessel potential space coincides with the Sobolev–Slobodeckij space , and both seminorms are related as
Proposition 1.1 (A fractional Trudinger–Moser inequality on ).
Let be an open bounded interval. Then it holds that
For the fractional Adachi–Tanaka-type Trudinger–Moser inequality on the whole line, put
On the other hand, a fractional Li–Ruf-type Trudinger–Moser inequality on is already known as follows.
Proposition 1.2 ().
is the full Sobolev norm on .
Concerning in (1.1), a natural question is to determine the range of the exponent for which is finite. As pointed out in , 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 ).
Ozawa  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 , and if , the main result in  can be read as follows: Put and
We have .
Furthermore, we obtain the relation between the suprema of both critical and subcritical Trudinger–Moser-type inequalities along the line of .
Theorem 1.5 (Relation).
Also we obtain how the Adachi–Tanaka-type supremum behaves when α tends to π.
Theorem 1.6 (Asymptotic behavior).
There exist such that for any which is close enough to π it holds that
Concerning the existence of maximizers of the Adachi–Tanaka-type supremum in (1.1), we have the following theorem.
Theorem 1.7 (Attainability of ).
is attained for any .
On the other hand, as for in (1.2), we have the following result.
Theorem 1.8 (Non-attainability of ).
is not attained for .
It is plausible that there exists such that is attained for , but we do not have a proof up to now.
Finally, we improve the subcritical Adachi–Tanaka-type inequality along the line of .
For , set
Then we have
Furthermore, is attained for all .
for , Theorem 1.9 extends Theorem 1.3. In the classical case, Dong and Lu  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 , we cannot follow this argument and we need a new idea.
Then for any .
For any and , we put for . Then we have
Thus for any with , if we choose , then satisfies
which implies . The opposite inequality is trivial. ∎
For any , it holds that
Choose any with and . Further, put , where and . Then by the scaling rules (2.1) we see
Also we have
Thus, testing by v, we see
By taking the supremum for with and , we have
Finally, Lemma 2.1 implies the result. ∎
Proof of Theorem 1.3.
For the proof of , we use the Moser sequence
and its estimates
as for some . Note that . 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
as . Thus we obtain (2.5).
By testing by , we have
since for t large and by (2.4). Also since
as . Put . We see
which leads to
as . This proves . ∎
Proof of Theorem 1.5.
By Lemma 2.2, we have
Let us prove the opposite inequality. Let
be a maximizing sequence of . We may assume for any . Put
Thus by (2.1) we see
since . Thus, setting for any , we may test by , which results in
Here we have used a change of variables for the second equality, and for the first inequality. Letting , we have the desired result. ∎
Proof of Theorem 1.6.
We need to prove that there exists such that for any which is sufficiently close to π it holds that
where we put
Now, for which is sufficiently close to π, we fix small such that
With this choice of , we have
Now, we estimate that
where for and we have used (2.6) in the last inequality. We easily see that and for , thus is strictly increasing in t and . Thus we have
which proves the result. ∎
For , we denote by its symmetric decreasing rearrangement defined as follows: For a measurable set , let denote an open interval . We define by
where denote the indicator function of a measurable set . Note that is nonnegative, even and decreasing on the positive line . It is known that
for any nonnegative measurable function , which is the difference of two monotone increasing functions , with such that either or is integrable. Also the inequality of Pólya–Szegő type
Note that the radial compactness lemma by Strauss  is violated on . More precisely, let
then cannot be embedded compactly in for any . To see this, let be an even function in with , and put . Then we see that is an even, compactly supported smooth function, and weakly in as . But does not have any strong convergent subsequence in because for any n sufficient large.
However, for a sequence with even, nonnegative and decreasing on , we have the following compactness result.
Assume to be a sequence such that is even, nonnegative and decreasing on . Let weakly in . Then strongly in for any for a subsequence.
Since is a weakly convergent sequence, we have
for some . We also have a.e. 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 is even, nonnegative and decreasing on , then it holds that
Thus for by
and by the pointwise convergence. Now, set for . Then we see a.e. . Moreover,
as since . Thus is uniformly integrable. Also, by [7, Theorem 6.9] we know that
For any , take such that . Since is uniformly bounded in , we have , and
for any bounded measurable set . Therefore, if , which implies that is uniformly absolutely continuous. Thus by Vitali’s convergence theorem (see, for example, [11, p. 187]) we obtain strongly in , which is the desired conclusion. ∎
Assume to be a sequence with . Let weakly in for some u and assume is even, nonnegative and decreasing on . Then we have
for any .
Note that is nonnegative, strictly convex and . Thus by the mean value theorem we have
Thus we have
by Hölder’s inequality, where and are chosen later.
First, direct calculation shows that
for all . Thus if we fix so that is realized, then we have
for some . Thus we have obtained independent of n.
Next, we estimate the term . Since is a bounded sequence in , we have by [7, Theorem 6.9] that for any . Thus we see for some independent of n for . Now, note that if we choose and sufficiently large, then we can find such that .
which is the desired conclusion. ∎
Proof of Theorem 1.7.
Here satisfies and . Appealing to the use of rearrangement, moreover we may assume that is nonnegative, even and decreasing on . Since is a bounded sequence, we have such that in . By Proposition 3.3, we see
as . Therefore, since , we have, letting ,
Next, we claim that for any . Indeed, take any such that , and . Then we have
Now, since for any , we have
for , which results in , the claim.
Thus we have shown that maximizes . ∎
Actually, we will show a stronger claim that has no critical point on M for sufficiently small . Assume to the contrary that there exists a critical point of for small . Then we define an orbit on M through v as
Note that , thus it must hold that
By the scaling rules (2.1), we see for any ,
Now, we see
with , and . Since
and , we calculate
Here, we need the following lemma.
Lemma 3.4 (Ogawa and Ozawa ).
There exists such that for any and , it holds that
For , Lemma 3.4 implies
Thus, for sufficiently small (it would be enough that ), Stirling’s formula implies that
for some 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 . Then we obtain the following result.
We have for .
We follow the proof of [18, Theorem 1.5]. Take any with in the admissible sets for in (4.1). By appealing to the rearrangement, we may assume that u is even, nonnegative and decreasing on . We divide the integral
First, we estimate (I). By the radial lemma (3.1), we see for any , ,
Therefore, we have
Now by the constraint , we have . Also if we put
then converges since as . Thus we obtain
where is independent of with .
Next, we estimate (II). Set
Then by the argument of  we know that
for . Put
Then we have since on , and
Thus we may use the fractional Trudinger–Moser inequality (Proposition 1.1) to w to obtain
for some independent of u. By on I, we conclude that
Now, since , there is an absolute constant such that
for any . Finally, we obtain
Proposition 4.1 follows from the estimates (I) and (II). ∎
For any , we have
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About the article
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.
© 2019 Walter de Gruyter GmbH, Berlin/Boston. This work is licensed under the Creative Commons Attribution 4.0 Public License. BY 4.0