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Moser functions and fractional Moser–Trudinger type inequalities
Nonlinear Analysis 146 (2016) 185–210
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Nonlinear Analysis www.elsevier.com/locate/na
Moser functions and fractional Moser–Trudinger type inequalities Ali Hyder Universit¨ at Basel, Switzerland
article
info
abstract
Article history: Received 22 October 2015 Accepted 27 August 2016 Communicated by Enzo Mitidieri
We improve the sharpness of some fractional Moser–Trudinger type inequalities, particularly those studied by Lam–Lu and Martinazzi. As an application, improving upon works of Adimurthi and Lakkis, we prove the existence of weak solutions to
Keywords: Fractional Moser–Trudinger inequalities Nonlocal equations
the problem n
(−∆) 2 u = λuebu
2
in Ω , 0 < λ < λ1 , b > 0,
with Dirichlet boundary condition, forn any domain Ω in Rn with finite measure. Here λ1 is the first eigenvalue of (−∆) 2 on Ω . © 2016 Elsevier Ltd. All rights reserved.
1. Introduction to the problem Let n ≥ 2 and let Ω be a bounded domain in Rn . The Sobolev embedding theorem states that W0k,p (Ω ) np and kp < n. However, it is not true that W0k,p (Ω ) ⊂ L∞ (Ω ) for kp = n. In ⊂ Lq (Ω ) for 1 ≤ q ≤ n−kp the borderline case, as shown by Yudovich [30], Pohozaev [23] and Trudinger [29], W01,n (Ω ) embeds into an Orlicz space and in fact n n−1 sup eα|u| dx < ∞, (1) u∈W01,n (Ω), ∥∇u∥Ln (Ω) ≤1
Ω
for some α > 0. Moser [22] found the best constant α in the inequality (1), obtaining the so called Moser–Trudinger inequality: n 1 1 n−1 sup eαn |u| dx < ∞, αn = n|S n−1 | n−1 . (2) 1,n |Ω | Ω u∈W (Ω), ∥∇u∥Ln (Ω) ≤1 0
The constant αn in (2) is the best constant in the sense that for any α > αn , the supremum in (1) is infinite. A generalized version of Moser–Trudinger inequality is the following theorem of Adams [1]: E-mail address: [email protected]. http://dx.doi.org/10.1016/j.na.2016.08.024 0362-546X/© 2016 Elsevier Ltd. All rights reserved.
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Theorem A ([1]). If k is a positive integer less than n, then there is a constant C = C(k, n) such that n n−k sup eα|u| dx ≤ C|Ω |, u∈Cck (Ω), ∥∇k u∥
n ≤1 L k (Ω)
Ω
where n k k+1 n n−k π22 Γ 2 n−k+1 , k = odd, n Γ 2 α = α(k, n) = n−1 n n k |S | π 2 2 Γ k2 n−k , k = even, Γ n−k 2 and ∇k := ∇∆
k−1 2
k
for k odd and ∇k = ∆ 2 for k even. Moreover the constant α is sharp in the sense that n n−k dx = ∞, (3) sup f (|u|)eα|u| u∈Cck (Ω), ∥∇k u∥
n ≤1 L k (Ω)
Ω
for any f : [0, ∞) → [0, ∞) with limt→∞ f (t) = ∞.1 In a recent work Martinazzi [19] has studied the Adams inequality in a fractional setting. In order to state its result first we recall the space |u(x)| n 1 n Ls (R ) := u ∈ Lloc (R ) : dx < ∞ . n+2s Rn 1 + |x| The operator (−∆)s can be defined on the space Ls (Rn ) via the duality s ⟨(−∆) u, ϕ⟩ := u(−∆)s ϕdx, ϕ ∈ S(Rn ),
(4)
Rn
where (−∆)s ϕ = F −1 |ξ|2s ϕˆ ,
ϕ ∈ S(Rn ),
F is the normalized Fourier transform and S(Rn ) is the Schwartz space. Notice that the integral in (4) is well-defined thanks to [8, Proposition 2.1]. Now for an open set Ω ⊆ Rn (possibly Ω = Rn ), s > 0 and 1 ≤ p ≤ ∞ we define the fractional Sobolev ˜ s,p (Ω ) by space H ˜ s,p (Ω ) := u ∈ Lp (Ω ) : u = 0 on Rn \ Ω , (−∆) 2s u ∈ Lp (Rn ) . H Theorem B ([19]). For any open set Ω ⊂ Rn with finite measure and for any p ∈ (1, ∞) we have p′ sup eαn,p |u| dx ≤ Cn,p |Ω |, n ,p
˜p u∈H
n
(Ω), ∥(−∆) 2p u∥Lp (Ω) ≤1
Ω
where the constant αn,p is given by αn,p =
Γ
n 2p
p′ n n 2p π 2 .
n |S n−1 | Γ np−n 2p
(5)
Moreover, the constant αn,p is sharp in the sense that we cannot replace it with any larger one without making the above supremum infinite. 1 Identity (3) is proven in [1], although not explicitly stated.
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Notice that condition (3) in Theorem A is sharper than only requiring that the constant α in the exponential is sharp, as done in Theorem B. In fact Martinazzi asked whether it is true that p′ sup f (|u|)eαn,p |u| dx = ∞, (6) n ,p
˜p u∈H
n
Ω
(Ω), ∥(−∆) 2p u∥Lp (Ω) ≤1
for any f : [0, ∞) → [0, ∞) with lim f (t) = ∞,
t→∞
f is Borel measurable,
(7)
and αn,p is given by (5). The point here is that Adams constructs smooth and compactly supported test functions similar to the standard Moser functions (constant in a small ball, and decaying logarithmically on an annulus), and then k, n k he estimates their H0 -norms in a very precise way. This becomes much more delicate when k is not integer because instead of computing partial derivatives, one has to estimate the norms of fractional Laplacians (the n term ∥(−∆) 2p u∥Lp (Ω) in (6)). This is indeed done in [19], but the test functions introduced by Martinazzi are not efficient enough to prove (6). As we shall see this has consequences for applications to PDEs. We shall prove that the answer to Martinazzi’s question is positive, indeed in a slightly stronger form, n namely the supremum in (6) is infinite even if we consider the full H p ,p -norm on the whole space. More precisely we have: Theorem 1.1. Let Ω be an open set in Rn with finite measure and let f : [0, ∞) → [0, ∞) satisfy (7). Then p′ sup f (|u|)eαn,p |u| dx = ∞, 1 < p < ∞, n ,p
˜p u∈H
n
≤1 (Ω), ∥u∥p +∥(−∆) 2p u∥p Lp (Rn ) Lp (Ω)
Ω
where the constant αn,p is given by (5). The main difficulty in the proof of Theorem 1.1 is to construct test and cut-off functions in a way that their fractional Laplacians of suitable orders can be estimated precisely. This will be done in Section 2. Here we mention that using a Green’s representation formula, Iula–Maalaoui–Martinazzi [10] proved a ˜ np ,p (Ω ) particular case of Theorem 1.1 in one dimension. Their proof, though, does not extend to spaces H when np > 1 because the function constructed using the Green representation formula do not enjoy enough smoothness at the boundary. Trying to solve this with a smooth cut-off function at the boundary allows to prove (6) only when f grows fast enough at infinity (for instance f (t) ≥ ta for some a > p′ ). Now we move to Moser–Trudinger type inequalities on domains with infinite measure. In this direction we refer to [25,13,21] and the references there in. For our purpose, here we only state the work of Lam–Lu [13]. Theorem C ([13]). Let p ∈ (1, ∞) and τ > 0. Then for every domain Ω ⊂ Rn with finite measure, there exists C = C(n, p, τ ) > 0 such that p′ eαn,p |u| dx ≤ C(|Ω | + 1), sup n ,p
u∈H p
n
(Rn ), ∥(τ I−∆) 2p u∥Lp (Rn ) ≤1
Ω
and sup u∈H
n ,p n p (Rn ), ∥(τ I−∆) 2p
u∥Lp (Rn ) ≤1
′
Φ(αn,p |u|p )dx < ∞,
Rn
where αn,p is given by (5) and Φ(t) := et −
jp −2 j j=0
t , j!
jp := min{j ∈ N : j ≥ p}.
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Furthermore, the constant αn,p is sharp in the above inequalities, i.e., if αn,p is replaced by any α > αn,p , then the supremums are infinite. In the spirit of Theorem 1.1 we prove a stronger version of the sharpness of the constant in Theorem C, in the sense that, even without increasing the constant αn,p we can make the two supremums in Theorem C infinite by multiplying the exponential by a function f going to infinity arbitrarily slow. Moreover it is sufficient to consider functions with compact support. Theorem 1.2. Let Ω ⊂ Rn be a domain with finite measure and let f : [0, ∞) → [0, ∞) satisfy (7). Then for any τ > 0 and for any p ∈ (1, ∞) we have (with the notations as in Theorem C) p′ sup f (|u|)eαn,p |u| dx = ∞, n ,p
˜p u∈H
n
(Ω), ∥(τ I−∆) 2p u∥Lp (Rn ) ≤1
Ω
and sup n ,p
˜p u∈H
n
(Ω), ∥(τ I−∆) 2p u∥Lp (Rn ) ≤1
′
f (|u|)Φ(αn,p |u|p )dx = ∞.
Ω
As an application of Theorem 1.1 (in the case p = 2 and f (t) = t2 , compare to (19)) we prove the existence of (weak) solution to a semilinear elliptic equation with exponential nonlinearity. In order to state the theorem first we need the following definition. Definition 1.1. Let Ω be an open set in Rn with finite measure. Let f ∈ Lp (Ω ) for some p ∈ (1, ∞). We say that u is a weak solution of n
(−∆) 2 u = f ˜ if u ∈ H
n 2 ,2
(Ω ) satisfies
n 4
n 4
(−∆) u(−∆) vdx = Rn
in Ω ,
˜ n2 ,2 (Ω ). f vdx for every v ∈ H
Ω
Theorem 1.3. Let Ω be an open set in Rn with finite measure. Let 0 < λ < λ1 and b > 0. Then there exists a nontrivial weak solution to the problem n
(−∆) 2 u = λuebu
2
in Ω .
n
(8)
˜ 2 ,2 (Ω ) ↩→ L2 (Ω ) is compact for any open set Ω with finite measure Due to the fact that the embedding H (see Lemma A.7 in the Appendix), we do not need any regularity assumption or boundedness assumption on the domain Ω . Eq. (8) has been well studied by several authors in even and odd dimensions, with emphasis both on existence and compactness properties see e.g. [3,6,9,11,16–18,20,24,28]. For instance, Lakkis [11], extending a work of Adimurthi [2], proved the existence of solution to (8) in any even dimension. In [12], Lam–Lu have studied Eq. (8) in even dimension with more general right hand side, namely, equation of the form n (−∆) 2 u = f (x, u), where the function f may not satisfy the Ambrosetti–Rabinowitz condition. In a recent work Iannizzotto–Squassina [9] have proven existence of nontrivial weak solution of (8) with Ω = (0, 1) under an assumption, which turns out to be satisfied thanks to our Theorem 1.1, applied with p = 2 (see Lemma 3.5). Finally, we mention that in a recent work Bao–Lam–Lu [4] have studied the existence of positive solutions to a polyharmonic equation on the whole space R2m , more precisely (I − ∆)m u = f (x, u),
in R2m , m ≥ 1,
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where the function f has critical growth at infinity. Moreover, under certain assumptions on f , they also discussed radial symmetry and regularity of solutions. 2. Moser type functions and proof of Theorems 1.1 and 1.2 We construct Moser type functions as follows: First we fix two smooth functions η and ϕ such that 0 ≤ η, ϕ ≤ 1, 3 3 ∞ η ∈ Cc ((−1, 1)), η = 1 on − , , 4 4 and ϕ ∈ Cc∞ ((−2, 2)),
ϕ = 1, on (−1, 1).
For ε > 0, we set ψε (t) =
1 − ϕε (t) η(t)
1 if 0 ≤ t ≤ 2 1 if t ≥ , 2
and − p1 1 1 1 vε (x) = log ϕε (|x|) + log ψε (|x|) log ε ε |x|
x ∈ Rn ,
where ϕε (t) = ϕ
t . ε
Our aim is to show that the supremums (in Theorems 1.1 and 1.2) taken over the functions {vε }ε>0 (up to a proper normalization) are infinite. The following proposition is crucial in the proof of Theorem 1.1. Proposition 2.1. Let uε (x) := |S
1 n−1 − p
|
n p′
n 2
2 π Γ
n 2p′
1 Γ
n 2p
vε (x). γn
Then for 1 < p < ∞ there exists a constant C > 0 such that −1 p1 n 1 ∥(−∆) 2p uε ∥Lp (Rn ) ≤ 1 + C log . ε Proof. Since the proof of above proposition is quite trivial if n is not an integer. the case when 2p
n 2p
is an integer, from now on we only consider
From Lemmas 2.2 and 2.4 we have −1 n 1 ∥(−∆) 2p uε ∥pLp (B3ε ∪B c ) ≤ C log . 2 ε In order to estimate (−∆)σ vε on the domain {x : 3ε < |x| < 2} we consider the function − p1 1 1 Rε (x) = vε (x) − log log =: fε (x) + gε (x) x ∈ Rn , ε |x|
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where
− p1 1 1 vε (x) − log log if |x| < 2ε fε (x) := ε |x| 0 if |x| ≥ 2ε, − p1 1 1 1 log − log ϕε (|x|) = log ε ε |x| and − p1 1 1 vε (x) − log 1 log if |x| > ε |x| 2 gε (x) := 1 0 if |x| ≤ 2 − p1 1 1 = log (η(|x|) − 1) log . ε |x| It is easy to see that for any σ > 0
1 sup |(−∆) gε (x)| ≤ C log ε x∈Rn σ
− p1 .
(9)
With the help of Lemma A.8 and the triangle inequality we bound − p1 n n n 1 1 1 2p 2p 2p (−∆) log |(−∆) uε (x)| = (−∆) Rε (x) + log 1 ε |x| n |S n−1 | p βn, 2p − p1 n 1 1 1 2p ≤ C|(−∆) Rε (x)| + log n . 1 n−1 ε |S | p |x| p Using the elementary inequality (a + b)q ≤ aq + Cq (bq + aq−1 b), we get |(−∆) 3ε<|x|<2
n 2p
1 ≤ q < ∞, a ≥ 0, b ≥ 0,
−1 n 1 1 1 dx + C |(−∆) 2p Rε (x)|p dx uε (x)| dx ≤ log n−1 n ε |S | |x| 3ε<|x|<2 3ε<|x|<2 − 1′ p n 1 1 2p R (x)|dx + C log n |(−∆) ε ′ ε 3ε<|x|<2 |x| p −1 − 1′ p n 1 1 1 2p R (x)|dx, ≤ 1 + C log + C log n |(−∆) ε ′ p ε ε 3ε<|x|<2 |x|
p
where the last inequality follows from Lemma 2.3. Using the pointwise estimate in Lemma 2.3 and (9) one can show that − p1 n 1 1 2p Rε (x)|dx ≤ C log , n |(−∆) ′ ε 3ε<|x|<2 |x| p which completes the proof.
Lemma 2.2. Let p ∈ (1, ∞). Then there exists a constant C = C(n, p, σ) > 0 such that − p1 1 n |(−∆)σ vε (x)| ≤ C log ε−2σ for |x| ≤ 3ε, 0 < σ < . ε 2
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Moreover, ∥(−∆)
n 2p
vε ∥pLp (B3ε )
1 ≤ C log ε
−1 .
Proof. We claim that for every nonzero multiindex α ∈ Nn there exists C = C(n, α) > 0 such that − p1 1 α |D vε (x)| ≤ C log ε−|α| , x ∈ Rn . ε
(10)
In order to prove (10), first notice that 1 1 p′ if |x| ≤ ε log ε vε (x) = − p1 1 1 1 log η(|x|) if |x| ≥ , log ε |x| 2 and that η ∈ Cc∞ ((−1, 1)). Therefore, the estimate in (10) holds for x ∈ Bε ∪ B c1 . For x ∈ B 21 \ Bε and for 2 a nonzero multiindex α ∈ Nn we compute 1 1 p α 1 1 1 log D vε (x) = Dα ϕε (|x|) log + Dα ψε (|x|) log + Cα,β Dβ log Dα−β ψε (|x|) ε ε |x| |x| 0<β≤α 1 1 1 + Cα,β Dβ log Dα−β ψε (|x|), = Dα ϕε (|x|) log − log ε |x| |x| 0<β≤α
where Cα,β ’s are some constants, and in the second equality we have used that Dα ψε (|x|) = −Dα ϕε (|x|), which follows from the fact that ϕε + ψε = 1 on B 12 . Moreover, from the definition of ϕ and ϕε we have that Cε−|α| if x ∈ B2ε \ Bε α |D ϕε (|x|)| ≤ c 0 if x ∈ B2ε ∪ Bε . Therefore, 1 1 |D ϕε (|x|)| log − log ≤ Cε−|α| . ε |x| α
It is easy to see that for 0 < β ≤ α β D log(|x|) Dα−β ψε (|x|) ≤ C(α, β)|x|−|β| ε−|α−β| ≤ C(α, β)ε−|α| ,
for x ∈ B 12 \ Bε .
This completes the proof of (10). In the case when σ is not an integer then we write σ = m + s where 0 < s < 1 and m is a nonnegative integer. Then for |x| ≤ 3ε we have (the following equivalent definition of fractional Laplacian can be found in [26,5]) (−∆)m vε (x + y) + (−∆)m vε (x − y) − 2(−∆)m vε (x) σ (−∆) vε (x) = Cn,s dy. |y|n+2s Rn Splitting Rn into A1 = {x : |x| ≤ 2ε} ,
A2 =
x : 2ε < |x| ≤
1 4
and A3 =
we have σ
(−∆) vε (x) = Cn,s
3 i=1
Ii ,
x : |x| >
1 4
,
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where Ii := Ai
(−∆)m vε (x + y) + (−∆)m vε (x − y) − 2(−∆)m vε (x) dy. |y|n+2s
For y ∈ A1 , using (10) we have |(−∆)m vε (x + y) + (−∆)m vε (x − y) − 2(−∆)m vε (x)| ≤ |y|2 ∥D2 (−∆)m vε ∥L∞ − p1 1 ≤ C|y|2 ε−2m−2 log , ε and hence − p1 − p1 1 1 dy |I1 | ≤ Cε−2m−2 log ≤ C log ε−2σ . n+2s−2 ε |y| ε A1 For m ≥ 1, again by (10) − p1 1 |(−∆)m vε (x + y) − (−∆)m vε (x)| ≤ C log ε−2m . ε Therefore, − p1 − p1 1 1 dy −2m |I2 + I3 | ≤ C log ≤ C log ε ε−2σ . n+2s ε ε |y|>ε |y| Since on A2 |x + y| ≤ 3ε + 41 < 12 , one has 1 1 1 p |vε (x + y) − vε (x)| = log log (ϕε (|x + y|) + ψε (|x + y|) − ϕε (|x|) − ψε (|x|)) ε ε ε ε + log ψε (|x + y|) − log ψε (|x|) |x + y| |x| ε ε = log ψε (|x + y|) − log ψε (|x|) |x + y| |x| ε ≤ C + log ψε (|x + y|) . |x + y| Hence, for m = 0, changing the variable y → εz − p1 1 |I2 | ≤ C log ε − p1 1 ≤ C log ε − p1 1 ≤ C log ε − p1 1 ≤ C log ε
ε − p1 ψε (|x + y|) log |x+y| 1 −2s ε + C log dy ε |y|n+2s ε<|y|< 41 − p1 log | x + z| ψε (ε| x + z|) 1 −2s −2s ε ε ε + C log ε dz n+2s ε |z| |z|>1 − p1 1 log (3 + |z|) −2s −2s ε + C log ε dz ε |z|n+2s |z|>1 ε−2s .
− 1 Finally, for m = 0, using that |vε | ≤ C log 1ε p on B c1 , we bound 8
− p1 − p1 1 dy 1 |I3 | ≤ C log ≤ C log . n+2s ε ε |y|≥ 14 |y| The lemma follows immediately.
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Lemma 2.3. For |x| ≥ 3ε we have
|(−∆)σ fε (x)| ≤ C
1 |x|2σ
log
1 ε
1 −p
n ε |x| n−2m ε |x|
if 0 < σ < 1 if 1 < σ = m + s <
n , 2
where m is a positive integer and 0 < s < 1. In particular − p1 n 1 ∥(−∆) 2p Rε ∥Lp (B2 \B3ε ) ≤ C log . ε Proof. Notice that for every nonzero multiindex α ∈ Nn we have 1 if |x| < ε − p1 |x||α| 1 α 1 |D fε (x)| ≤ C log if ε < |x| ≤ 2ε ε ε|α| 0 if |x| ≥ 2ε. First we consider 0 < σ < 1. Using that |ϕε | ≤ 1, changing the variable y → εy and by H¨older inequality we obtain fε (x) − fε (y) σ dy |(−∆) fε (x)| = C n+2σ Rn |x − y| 1 1 − p1 log − log ϕ (|y|) ε ε |y| 1 = C log dy n+2σ ε |x − y| |y|<2ε p1 1′ − p1 p dy 1 p′ |log |y|| dy ≤ Cε log np+2pσ ε |y|<2 |x − εy| |y|<2 p1 − p1 n |x| 1 dy 1 ≤ Cεn log , x np+2pσ 2ε | ε εn |x|np+2pσ |y|< |x| |x| − y| n − p1 1 ε 1 , ≤ C 2σ log |x| |x| ε n
where in the second last inequality we have used a change of variable y → |x| ε y and the last inequality follows from the uniform bound 1 2ε ≤ C for every |x| ≥ 3ε, |y| ≤ . (11) x np+2pσ | |x| − y| |x| For σ > 1, changing the variable y → |x|y and by (11) we have (−∆)m fε (x) − (−∆)m fε (y) |(−∆)σ fε (x)| = C dy |x − y|n+2s n R (−∆)m fε (y) =C dy |y|<2ε |x − y|n+2s − p1 1 1 1 ≤ C log dy 2m |x − y|n+2s ε |y| |y|<2ε n−2m − p1 1 ε 1 ≤ C 2σ log . |x| |x| ε We conclude the lemma by (9).
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Lemma 2.4. For 0 < σ <
n 2
there exists a constant C = C(n, σ) such that
− p1 1 1 |(−∆)σ vε (x)| ≤ C log ε |x|n+2σ
for every x ∈ B2c .
Moreover, −1 n 1 ∥(−∆) 2p vε ∥pLp (B c ) ≤ C log . 2 ε Proof. If 0 < σ < 1 then
vε (y) dy, |x| > 2 |x − y|n+2σ |y|<1 1 ≤ C n+2σ vε (y)dy |x| |y|<1 − p1 1 1 (log |y| + log 2) dy ≤ C log ε |x|n+2σ |y|<1 − p1 1 1 ≤ C log . n+2σ ε |x|
σ
|(−∆) vε (x)| = C
(12)
Since the integral in the right hand side of (12) is a proper integral, differentiating under the integral sign one can prove the lemma in a similar way. Proof of Theorem 1.1. Without loss of generality we can assume that B1 ⊆ Ω . Let uε be defined as in Proposition 2.1. We set u ¯ε (x) =
uε (x) ∥uε ∥pLp (Ω)
+ ∥(−∆)
n 2p
uε ∥pLp (Rn )
x ∈ Rn .
p1 ,
n ˜ np ,p (Ω ) and ∥¯ Then u ¯ε ∈ H u∥pLp (Ω) + ∥(−∆) 2p u ¯∥pLp (Rn ) = 1. We claim that there exists a constant δ > 0 such that ′ lim sup exp αn,p |¯ uε |p dx =: lim sup Iε ≥ δ. (13)
ε→0
ε→0
Bε
Noticing that (restriction of u ¯ε on Bε is a constant function) lim inf u ¯ε (x) = ∞,
ε→0 x∈Bε
and by (7), we obtain f (|u|)e
sup ˜ u∈H
n n ,p p (Ω), ∥u∥p +∥(−∆) 2p Lp (Ω)
u∥p ≤1 Lp (Rn )
Ω
′
αn,p |u|p
p′
f (|¯ uε |)eαn,p |¯uε | dx
dx ≥ lim sup ε→0
Ω
p′
f (|¯ uε |)eαn,p |¯uε | dx ε→0 Bε ≥ lim sup Iε inf f (|¯ uε (x)|) ≥ lim sup
ε→0
x∈Bε
≥ δ lim sup inf f (|¯ uε (x)|) ε→0
= ∞.
x∈Bε
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To prove (13) we choose ε = e−k . Noticing that − pp′ p′ C = −C , lim −k + k 1 + k→∞ k p −1 1 , ∥uε ∥pLp (Rn ) ≤ C log ε and using Proposition 2.1 we have n n log
Iε ≥ |B1 |ε e for some δ > 0.
1 ε
1+C (log
1 ε
)
−1
− pp′
C
= |B1 |e−kn+kn(1+ k )
−
p′ p
≥ δ,
In order to prove Theorem 1.2, first we prove the following proposition which gives a similar type of estimate as in Proposition 2.1. Proposition 2.5. Let τ > 0 and 1 < p < ∞. Then there exists a constant C > 0 such that −1 p1 n 1 ∥(τ I − ∆) 2p uε ∥Lp (Rn ) ≤ 1 + C log . ε Proof. We set n
n
wε (x) = (τ I − ∆) 2p uε (x) − (−∆) 2p uε (x). We observe that there exists C = C(p) > 0 such that 1
h(t) = (1 + t)p − 1 − C(tp + tp−1 + t 2 ) < 0,
for every t > 0, 1 ≤ p < ∞,
which follows from the fact that h(0) = 0 and h′ (t) < 0 for every t > 0. Therefore, there holds 1
1
(a + b)p ≤ ap + Cp (bp + abp−1 + b 2 ap− 2 ),
a ≥ 0, b ≥ 0, 1 ≤ p < ∞,
for some constant Cp > 0 and using this inequality we bound n n |(τ I − ∆) 2p uε (x)|p dx = |wε (x) + (−∆) 2p uε (x)|p dx n n R R n n ≤ |(−∆) 2p uε (x)|p + C |wε (x)|p dx + C |(−∆) 2p uε (x)||wε (x)|p−1 dx Rn Rn Rn n 1 1 +C |(−∆) 2p uε (x)|p− 2 |wε (x)| 2 dx Rn
=: I1 + I2 + I3 + I4 . From Proposition 2.1 we have
1 I1 ≤ 1 + C log ε
−1 .
To estimate I2 , I3 and I4 we will use pointwise estimates on (−∆)σ uε , (−∆)σ wε and Lp estimates on (−∆)σ wε . For 0 < σ < n2 combining Lemmas 2.2–2.4, A.8, and (9) we get −2σ − p1 if |x| < 3ε ε 1 σ |(−∆) uε (x)| ≤ C log (14) |x|−2σ if 3ε < |x| < 2 ε |x|−n−2σ if |x| > 2.
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With the help of (14) one can verify that − p1
σ
∥(−∆) uε ∥Lp (Rn )
1 ≤ C(n, p, σ) log ε
n , 2p
(15)
p2 . p+1
(16)
1 ≤ p < ∞, 0 ≤ σ <
,
and together with Lemma A.2 −1
1 I2 ≤ C log ε
.
We conclude the proposition by showing that −1 n 1 p−q q 2p |wε | |(−∆) vε | dx ≤ C(n, p, q) log , ε Rn
0
It follows from Lemma A.1 that
1 |wε (x)| ≤ C log ε and for
n 2p
− p1
x ∈ Rn ,
,
n < 1, 2p
>1 n − +2 − p1 ε p n 1 |wε (x)| ≤ C log |x|− p +2 ε 1
if |x| < 3ε if 3ε < |x| < 2 if |x| > 2,
thanks to (14) and (15). Splitting Rn into A1 = {x : |x| ≤ 2}
and A2 = {x : |x| > 2} ,
we have
n
|wε |q |(−∆) 2p vε |p−q dx =
Rn
2
Ji ,
n
|wε |q |(−∆) 2p vε |p−q dx,
Ji :=
i = 1, 2.
Ai
i=1
−1 Using (14) one can show that J1 ≤ C log 1ε and together with q < gives (16).
p2 p+1
−1 one has J2 ≤ C log 1ε , which
Proof of Theorem 1.2. Here also we can assume that B1 ⊆ Ω . We choose M > 0 large enough such that p′ ′ 1 Φ(αn,p tp ) ≥ eαn,p t , t ≥ M. 2 We set uε u ¯ε = . n ∥(τ I − ∆) 2p uε ∥Lp (Rn ) Then we have
′ f (|¯ uε |)Φ αn,p |¯ uε |p dx ≥
Rn
′ f (|¯ uε |)Φ αn,p |¯ uε |p dx
uε ≥M
1 ≥ 2
p′
f (|¯ uε |)eαn,p |¯uε | dx,
Bε
for ε > 0 small enough. Now the proof follows as in Theorem 1.1, thanks to Proposition 2.5.
Remark. From the point of view of conductor capacity estimate (see e.g. [1, p. 393], [12, p. 2193]), it would be interesting to know whether n
∥(−∆) 2p uε ∥Lp (Rn ) ≥ 1,
n
and ∥(τ I − ∆) 2p uε ∥Lp (Rn ) ≥ 1,
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or not. Here we recall that the “original” Moser functions (see [22]), that is 0 for x ∈ R2 \ B1 1 u(ε, x) := √ 1/ log(1/ε) log(1/|x|) for x ∈ B1 \ Bε 2π log(1/ε) for x ∈ Bε , satisfy ∥∇u(ε, ·)∥L2 (B1 ) = 1. Lemma 2.6. Let uε be as in Proposition 2.1. Then there exists C > 0 (depends on n, p, τ, ϕ, η) such that −1 −1 n n 1 1 p p 2p 2p ∥(−∆) uε ∥Lp (B1 ) ≥ 1 − C log , ∥(τ I − ∆) uε ∥Lp (B1 ) ≥ 1 − C log . ε ε Moreover, if
n 2p
n
is an integer, then ∥(−∆) 2p uε ∥Lp (B1 ) > 1.
Proof. Using the inequality ||a| − |b||p ≥ |a|p − p|a|p−1 |b|,
p ≥ 1, a, b ∈ R,
we obtain
n
|(−∆) 2p uε |p dx
3ε<|x|<1
−1 − 1′ p n 1 1 1 1 1 2p R |dx, ≥ log dx − C log n |(−∆) ε n−1 n ′ p ε |S | |x| ε 3ε<|x|<1 3ε<|x|<1 |x|
and
n
|(τ I − ∆) 2p uε |p dx ≥
3ε<|x|<1
n
|(−∆) 2p uε |p dx − C
3ε<|x|<1
n
|(−∆) 2p uε |p−1 |wε |dx,
3ε<|x|<1
where Rε and wε are defined in the proof of Propositions 2.1 and 2.5 respectively. First part of the lemma follows as in Propositions 2.1 and 2.5. n
n We choose r < 1 so that uε ∈ Cc∞ (Br ). If 2p is an integer, then the support of (−∆) 2p uε is a subset of Br \ Bε . Therefore, by H¨ older inequality n 1 uε (0) = Kn,p (−∆) 2p uε (x) n dx |x| p′ ε<|x|
where the first identity follows from the fact that the function n Γ 2p′ 1 1 −n −n p 2 Kn,p π n := 2 n , ′ ′ n p |x| Γ 2p |x| p n
is a fundamental solution of the operator (−∆) 2p . Since 1′ p n 1 n n 1 1 n−1 − p ′ uε (0) = |S | 2p π 2 Γ log ′ 2p Γ n γ ε 2p
1
= Kn,p |S n−1 | p′ n
we have ∥(−∆) 2p uε ∥Lp (B1 ) > 1.
log
1 ε
1′ p
,
n
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3. Proof of Theorem 1.3 n ˜ n2 ,2 (Ω ) and α0 = αn,2 . Throughout this section we use the notation ∥u∥ = ∥(−∆) 4 u∥L2 (Rn ) , H = H To prove Theorem 1.3 we follow the approach in [2,11]. First we prove that λ1 > 0, which makes the statement of Theorem 1.3 meaningful.
Lemma 3.1. Let Ω be an open set in Rn with finite measure. Then λ1 > 0 and there exists a function u ∈ H such that ∥u∥L2 (Ω) = 1,
and
∥u∥2 = λ1 .
Proof. We recall that λ1 = inf ∥u∥2 : u ∈ H, ∥u∥L2 (Ω) = 1 . Let {uk }∞ k=1 ⊂ H be a sequence such that lim ∥uk ∥2 = λ1 ,
∥uk ∥L2 (Ω) = 1
uk ⇀ u0
u k → u0
k→∞
for every k.
Then up to a subsequence in H,
in L2 (Ω ),
where the latter one follows from the compact embedding H ↩→ L2 (Ω ) (see Lemma A.7). Therefore, λ1 ≤ ∥u0 ∥2 ≤ lim inf ∥uk ∥2 = λ1 ,
∥u0 ∥L2 (Ω) = 1.
k→∞
Let us now define the functional 1 J(u) = ∥u∥2 − 2
u ∈ H,
G(u)dx, Ω
where G(t) =
t
2
g(r) := λrebr ,
g(r)dr,
0 < λ < λ1 , b > 0.
0
Then J is C 2 and the Fr´echet derivative of J is given by n n 4 4 g(u)vdx, DJ(u)(v) = (−∆) u(−∆) vdx − Rn
v ∈ H.
Ω
We also define 2
F (u) = DJ(u)(u) = ∥u∥ −
g(u)udx, Ω
1 I(u) = J(u) − F (u), 2
S = {u ∈ H : u ̸= 0, F (u) = 0} . Observe that if u ∈ H is a nontrivial weak solution of (8) then u ∈ S. With the above notations we have: Lemma 3.2. The set S is closed in the norm topology and α0 0 < s2 < , s := 2 inf J(u). u∈S b Proof. Since F is continuous (actually F is C 1 as J is C 2 ), it is enough to show that 0 is an isolated point of S. If not, then there exists a sequence {uk } ⊂ S such that ∥uk ∥ → 0 as k → ∞. We set vk = ∥uukk ∥ . From the compactness of the embedding H ↩→ Lq (Ω ) for any 1 ≤ q < ∞, we can assume that (up to a subsequence)
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vk ⇀ v in H and vk → v almost everywhere in Ω . By Lemma 3.4 we get 1 k→∞ bu2k 2 1=λ e vk dx −−−−→ λ v 2 dx ≤ λ ∥v∥2 < 1, λ1 Ω Ω which is a contradiction. Hence S is closed. Since, 1 bt2 1 2 e + > 0, f (t) := t − b b
for t > 0, b > 0,
which follows from f (0) = 0 and f ′ (t) > 0 for t > 0, we have λ 1 bu2 1 I(u) = u2 − e + dx > 0, 2 Ω b b
if u ∈ H \ {0},
(17)
and in particular J(u) = I(u) > 0 for u ∈ S. If possible, we assume that s = 0. Then there exists a sequence {uk } ⊂ S such that J(uk ) → 0 as k → ∞. Moreover, 2 2 2 u2k ebuk dx + λ u2k ebuk dx ∥uk ∥2 = λ u2k ebuk dx = λ u2k ≤ 2b
u2k > 2b
Ω
λ ≤4 2
u2k > 2b
u2k
1 − b
e
bu2k
1 + b
dx + λ u2k ≤ 2b
2
u2k ebuk dx,
≤ 4J(uk ) + λ u2k ≤ 2b
2
u2k ebuk dx (18)
and hence uk is bounded in H. Then up to a subsequence uk → u, a.e. in Ω and uk ⇀ u. Using Fatou lemma and (ii) in Lemma 3.4 we obtain λ 1 bu2 1 I(u) = u2 − e + dx ≤ lim inf I(uk ) = lim inf J(uk ) = 0, k→∞ k→∞ 2 Ω b b and hence u = 0, thanks to (17). It follows from (18) that uk → 0 in H which is a contradiction as S is closed. We prove now s2 < α0 b−1 . First we fix u ∈ H with ∥u∥ = 1. We consider the function 2 2 2 Fu (t) := F (tu) = ∥tu∥ − λ t2 u2 ebt u dx, t ≥ 0. Ω
Then 2 2 Fu (t) ≥ t2 λ1 u2 dx − λ u2 ebt u dx > 0, Ω
Ω
for t > 0 sufficiently small and limt→∞ Fu (t) = −∞. Hence, the continuity of Fu implies that there exists tu > 0 such that Fu (tu ) = 0, i.e., tu u ∈ S. Thus 1 1 s2 ≤ J(tu u) ≤ ∥tu u∥2 = t2u . 2 2 2 Again using that tu u ∈ S we have 2 1 1 1 2 bs2 u2 u e dx ≤ 2 λ (tu u)2 eb(tu u) dx = 2 ∥tu u∥2 = , λt λt λ Ω Ω u u which implies that sup ∥u∥≤1, u∈H
and by Theorem 1.1 we deduce that s2 < α0 b−1 .
2
u2 ebs
Ω
u2
dx < ∞,
(19)
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Lemma 3.3. Let u ∈ S be a minimizer of J on S. Then DJ(u) = 0. Proof. We fix a function v ∈ H \ {0} and consider the function γ > 0, t ∈ R.
Fu,v (γ, t) := F (γu + tv),
Differentiating Fu,v with respect to γ and using that F (u) = 0, we get 2 ∂Fu,v (1, 0) = −2bλ u4 ebu dx < 0. ∂γ n R Hence, by implicit function theorem, there exists δ > 0 such that we can write γ = γ(t) as a C 1 function of t on the interval (−δ, δ) which satisfies γ(0) = 1,
Fu,v (γ(t), t) = 0,
for every t ∈ (−δ, δ).
Moreover, choosing δ > 0 smaller if necessary, we have γ(t)u + tv ∈ S for every t ∈ (−δ, δ). We write J(u + tv) − J(u) t J(γ(t)u + tv) − J(u) J(γ(t)u + tv) − J(u + tv) − = lim . t→0 t t
DJ(u)(v) = lim
t→0
Since J is C 1 , a first order expansion of J yields J(γ(t)u + tv) − J(u + tv) = J((u + tv) + (γ(t) − 1)u) − J(u + tv) = DJ(u + tv)((γ(t) − 1)u) + o ((γ(t) − 1)∥u∥) = (γ(t) − 1)DJ(u + tv)(u) + (γ(t) − 1)∥u∥o(1). Therefore, using that F (u) = 0, J(γ(t)u + tv) − J(u + tv) = γ ′ (0)DJ(u)(u) = 0. t On the other hand, since u is a minimizer of J on S and γ(t)u + tv ∈ S, J(γ(t)u + tv) − J(u) ≥ 0 if t ≥ 0 = t ≤ 0 if t ≤ 0, lim
t→0
implies that (since it exists) J(γ(t)u + tv) − J(u) = 0. t This shows that DJ(u)(v) = 0 for every v ∈ H, i.e., DJ(u) = 0. lim
t→0
Proof of Theorem 1.3. Let {uk } be a sequence in S such that limk→∞ J(uk ) → bounded sequence in H and consequently, up to a subsequence uk → u, a.e.
uk ⇀ u,
s2 2 .
ℓ := lim ∥uk ∥,
in Ω ,
k→∞
for some u ∈ H. First we claim that u ̸= 0. Assuming u = 0, by (ii) in Lemma 3.4 we get λ α0 2 bu2k (e − 1)dx = s2 < , lim ∥uk ∥ = lim 2 J(uk ) + k→∞ k→∞ 2b Ω b and hence by (i) in Lemma 3.4 lim ∥uk ∥2 = lim λ
k→∞
a contradiction as S is closed.
k→∞
Ω
2
u2k ebuk dx = 0,
Then by (18) uk is a
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We claim that ℓ = ∥u∥. Then uk → u in H and applying Lemmas 3.2 and 3.3 we have Theorem 1.3. If the claim is false then necessarily we shall have ℓ > ∥u∥. One has
λ bu2k lim ∥uk ∥ = lim 2 J(uk ) + (e − 1)dx k→∞ k→∞ 2b Ω 2 2 s λ =2 + (ebu − 1)dx, 2 2b Ω 2
= s2 − 2J(u) + ∥u∥2 . We divide the proof in two cases, namely J(u) ≤ 0 and J(u) > 0. Case 1. We consider that J(u) ≤ 0. Since u ̸= 0, 2 2 λ (ebu − 1)dx < λ u2 ebu dx, ∥u∥2 ≤ b Ω Ω where the second inequality follows from (17). It is easy to see that we can choose 0 < t0 < 1 such that 2 ∥t0 u∥2 = λ (t0 u)2 eb(t0 u) dx, Ω
that means t0 u ∈ S. Using that I(tu) is strictly monotone increasing in t, which follows from the expression in (17), we obtain s2 s2 ≤ J(t0 u) = I(t0 u) < I(u) ≤ lim inf J(uk ) = , k→∞ 2 2 a contradiction. Case 2. Here we assume that J(u) > 0. Then ℓ2 = lim ∥uk ∥2 = s2 − 2J(u) + ∥u∥2 < s2 + ∥u∥2 < k→∞
Taking vk =
uk ∥uk ∥
α0 + ∥u∥2 . b
(20)
we see that (up to a subsequence) vk ⇀ v :=
u , ℓ
vk → v, a.e.
in Ω ,
and by Lemma A.5, for every p < (1 − ∥v∥2 )−1 2 sup epα0 vk dx < ∞. k∈N
Ω
Taking (20) into account we have 0 < ℓ2 − ∥u∥2 = s2 − 2J(u) <
α0 , b
and therefore, we can choose ε0 > 0 such that 1 + ε0 =
α0 1 , 2 b ℓ − ∥u∥2
i.e., ℓ2 (1 + ε0 ) =
α0 b
1−
∥u∥2 ℓ2
−1 .
For k large enough such that ∥uk ∥2 ≤ ℓ2 (1 + ε20 ) holds, we observe that b∥uk ∥2 ≤ p0 α0 for some 1 < p0 < (1 − ∥v∥2 )−1 . Thus, for some p1 > 1, p2 > 1 with p1 p2 p0 < (1 − ∥v∥2 )−1 we obtain 2 2 p1 p1 p0 α0 vk 1 ′ u2k ebuk dx ≤ sup ∥u2p ∥Lp2 (Ω) < ∞, sup k ∥Lp2 (Ω) ∥e k∈N
k∈N
Ω
and together with Lemma A.9 lim
k→∞
Ω
2
u2k ebuk dx =
Ω
2
u2 ebu dx.
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Indeed, 2
2
2
∥u∥ < ℓ = lim ∥uk ∥ = λ lim k→∞
k→∞
2 u2k ebuk dx
=λ
Ω
2
u2 ebu dx,
Ω
and we can now proceed as in Case 1.
Lemma 3.4. Let uk , vk ∈ H such that uk ⇀ u in H, uk → u, a.e. in Ω , vk ⇀ v in H and vk → v, a.e. in Ω . Then (i) If lim sup ∥uk ∥2 < k→∞
α0 , b
then for every integer ℓ ≥ 1
2
ebuk vkℓ dx =
lim
k→∞
Ω
2
ebu v ℓ dx.
Ω
(ii) If
2
u2k ebuk dx < ∞,
lim sup k→∞
Ω
then
2
ebuk dx =
lim
k→∞
Ω
2
ebu dx.
Ω
Proof. We prove the lemma with the help of Lemma A.9 (in the Appendix). We choose p > 1 such that for k large enough p∥uk ∥2 < αb0 holds and together with Theorem C we have 2 sup epbuk dx < ∞. k∈N
˜ Since the embedding H
n 2 ,2
Ω
(Ω ) ↩→ Lq (Ω ) is compact (see Lemma A.7) for every 1 ≤ q < ∞, we have vkq → v q
in L1 (Ω ).
Indeed, 2
2
sup ∥ebuk vkℓ ∥Lp (Ω) ≤ ∥vkℓ ∥Lp′ (Ω) ∥ebuk ∥Lp (Ω) < ∞, k∈N
and we conclude (i). Now (ii) follows from e
bu2k
u2k >M
1 dx ≤ M
2
u2k >M
2
u2k ebuk dx ≤
C , M
which implies that the function fk := ebuk satisfies the condition in Lemma A.9.
In the following lemma we prove that the assumption H ′ (v) in [9] is true under certain conditions. 2
Lemma 3.5. Let α0 > 0. Let f (t) = eα0 t h(t) satisfies H(i)–(iii) in [9]. Let h ≥ 0 on [0, ∞) and h(−t) = −h(t). Let s f (st) be a monotone increasing function with respect to t on (0, ∞), s ̸= 0. If t
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√ ˜ 21 ,2 ((0, 1)) such that 2π∥(−∆) 14 u∥L2 (R) = 1 and limt→∞ h(t)t = ∞ then there exists u ∈ H 2 1 ω t F (tu)dx < sup Φ(tu) := sup − , 4π 2α0 t>0 t>0 0 where
t
f (s)ds,
F (t) = 0
and ω is as in [9]. ˜ 21 ,2 ((0, 1)) such that Proof. For a given M > 0 we can choose u ∈ H 1 2 √ 1 2π f u udx > M, 2π∥(−∆) 4 u∥L2 (R) = 1, α0 0 thanks to Theorem 1.1. Differentiating with respect to t one has 1 f (tu) 1 − udx . Φ ′ (tu) = t 2π t 0 2 Hence, for t ≥ 2π α0 =: t0 and 2πM > t0 1 f (t0 u) 1 ′ − udx < 0. Φ (tu) ≤ t 2π t0 0 Thus Φ ′ (tu) ≤ 0 on (t0 − ε, ∞) for some ε > 0 and therefore, sup Φ(tu) = t>0
Φ(tu) ≤
sup t∈(0, t0 −ε)
t2 π < . 4π 2α 0 t∈(0, t0 −ε) sup
Since ω = π, thanks to Theorem B, we conclude the lemma.
Acknowledgments I would like to thank my advisor Prof. Luca Martinazzi for all the useful discussion and encouragement. I would also like to thank the reviewers for carefully reading the manuscript and many useful comments and suggestions. The author is supported by the Swiss National Science Foundation, project no. PP00P2-144669. Appendix Lemma A.1 (Pointwise Estimate). Let s > 0 and not an integer. Let m be the smallest integer greater than s. Then for any τ > 0 |(τ I − ∆)s u(x) − (−∆)s u(x)| ≤ C
m−1
|(−∆)s−j u(x)| + C∥(−∆)σ u∥L1 (Rn ) ,
u ∈ S(Rn ),
j=1
where σ ∈ max{ n2 − m + s, 0}, can be interpreted as zero.
n 2
, the constant C depends only on n, s, σ, τ and for m = 1 the above sum
Proof. We set f (t) = ts on R+ . By Taylor’s expansion we have f (t + τ ) = f (t) + τ f ′ (t) + · · · +
τ m−1 m−1 τm m f (t) + f (ξt ), (m − 1)! m!
for some t < ξt < t + τ.
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In particular (τ + t2 )s = t2s + c1 t2s−2 + c2 t2s−4 + · · · + cm−1 t2s−2m+2 + E(t), where the function E satisfies the estimate |E(t)| ≤ C(1 + t)2s−2m ,
t > 0.
Therefore, for u ∈ S(Rn ) F((τ I − ∆)s u)(ξ) = (τ + |ξ|2 )s u ˆ 2s = |ξ| + c1 |ξ|2s−2 + · · · + cm−1 |ξ|2s−2m+2 + E(|ξ|) u ˆ =
m−1
cj |ξ|2s−2j u ˆ + E(|ξ|)ˆ u(ξ)
j=0
=
m−1
cj F((−∆)s−j u) + E(|ξ|)ˆ u(ξ),
j=0
and hence (τ I − ∆)s u(x) =
m−1
cj (−∆)s−j u(x) + F −1 (E u ˆ)(x).
j=0
To estimate the term F observe that
−1
(E u ˆ) (uniformly in x) in terms of L1 (Rn ) norm of (fractional) derivative of u, we 1 σ |E(|ξ|)ˆ u(ξ)| = E(|ξ|) 2σ (−∆) u(ξ) |ξ| C σ u(ξ) ≤ 2σ (−∆) |ξ| (1 + |ξ|2 )m−s C ≤ 2σ ∥(−∆)σ u∥L1 (Rn ) . |ξ| (1 + |ξ|2 )m−s
Thus −1 F (E u ˆ)(x) ≤ C∥E u ˆ∥L1 (Rn ) ≤ C∥(−∆)σ u∥L1 (Rn ) , and we complete the proof.
Lemma A.2 (Lp Estimate). Let s > 0 be a noninteger. Let τ > 0 be any fixed number. Then for p ∈ (1, ∞) there exists C = C(n, s, p, τ ) > 0 such that ∥u∥Lp (Rn ) if s < 1 ∥(τ I − ∆)s u − (−∆)s u∥Lp (Rn ) ≤ C ∥u + (−∆)s−1 u∥Lp (Rn ) if s > 1. Proof. We have F((τ I − ∆)s u)(ξ) − F((−∆)s u)(ξ) = (τ + |ξ|2 )s − |ξ|2s u ˆ(ξ) ˆ(ξ) if s < 1 (τ + |ξ|2 )s − |ξ|2s u = (τ + |ξ|2 )s − |ξ|2s (1 + |ξ|2s−2 )ˆ u(ξ) if s > 1 1 + |ξ|2s−2 m(ξ)ˆ u(ξ) if s < 1 =: m(ξ)F u + (−∆)s−1 u (ξ) if s > 1. Now the proof follows from the Hormander multiplier theorem (see [27, p. 96]).
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The following lemma appears already in [7, p. 46], but for the reader’s convenience we give a more detailed proof. Lemma A.3 (Equivalence of Norms). Let σ > 0. Then for p ∈ (1, ∞) there exists a constant C > 0 such that for every u ∈ S(Rn ) 1 ∥u∥Lp (Rn ) + ∥(−∆)σ u∥Lp (Rn ) ≤ ∥(I − ∆)σ u∥Lp (Rn ) C ≤ C ∥u∥Lp (Rn ) + ∥(−∆)σ u∥Lp (Rn ) . Proof. We set Gσ (x) =
1 1 σ σ 2 (4π) Γ 2
∞
e−π
|x|2 t
t
e− 4π t
0
−n+σ 2
dt , t
which is the Bessel potential of order σ (see [27, p. 130]). Then 1 ˆ σ (x) = 1 n Gσ (x)dx = 1, G σ . 2 (1 + |x|2 ) 2 n (2π) R Setting f = (I − ∆)σ u we can write u = G2σ ∗ f and by Young’s inequality one has ∥u∥Lp (Rn ) ≤ ∥f ∥Lp (Rn ) . Again writing u = G2σ ∗ f and taking Fourier transform we obtain 1 F((−∆)σ u) = |ξ|2σ u ˆ = |ξ|2σ fˆ =: m(ξ)fˆ, (1 + |ξ|2 )σ and by Hormander multiplier theorem we get ∥(−∆)σ u∥Lp (Rn ) ≤ C∥f ∥Lp (Rn ) . Thus, ∥u∥Lp (Rn ) + ∥(−∆)σ u∥Lp (Rn ) ≤ C∥(I − ∆)σ u∥Lp (Rn ) . To conclude the lemma, it is sufficient to show that ∥(−∆)s u∥Lp (Rn ) ≤ C(n, s, σ, p)(∥u∥Lp (Rn ) + ∥(−∆)σ u∥Lp (Rn ) ),
0 < s < σ,
(21)
thanks to Lemma A.2. In order to prove (21) we fix a function ϕ ∈ Cc∞ (B2 ) such that ϕ = 1 on B1 . Then F((−∆)s u) = |ξ|2s u ˆ = |ξ|2s ϕˆ u + |ξ|2s (1 − ϕ)ˆ u = m1 (ξ)ˆ u + m2 (ξ)F((−∆)σ u), where m1 (ξ) = |ξ|2s ϕ(ξ), m2 (ξ) = |ξ|2s−2σ (1 − ϕ(ξ)) are multipliers and we conclude (21) by Hormander multiplier theorem. Lemma A.4 (Embedding to An Orlicz Space). Let Ω be an open set with finite measure. Then for every ˜ n2 ,2 (Ω ) u∈H 2 eu dx < ∞. Ω n
Proof. We set f = (−∆) 4 u. By [19, Proposition 8] we have u(x) = G(x, y)f (y)dy, 0 ≤ G(x, y) ≤ Ω
Cn n , |x − y| 2
where G is a Greens function. We choose M > 0 large enough such that ∥f˜∥L2 Cn < α0 , where f˜ = f − f χ{|f |≤M } . Then |f˜(y)| ˜ ˜ n n |u(x)| ≤ C(M ) + Cn I 2 f (x), I 2 f (x) := n dy, Ω |x − y| 2 and by [1, Theorem 2] we conclude the proof.
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As a consequence of the above lemma one can prove a higher dimensional generalization of Lions lemma [15] (for a simple proof see e.g. [9, Lemma 2.6]), namely ˜ n2 ,2 (Ω ) such that Lemma A.5 (Lions). Let uk be a sequence in H n
n
n
˜ 2 ,2 (Ω ), uk ⇀ u in H ∥(−∆) 4 uk ∥L2 (Rn ) = 1. 0 < ∥(−∆) 4 u∥L2 (Rn ) < 1, −1 n ∞ Then for every 0 < p < 1 − ∥(−∆) 4 u∥2L2 (Rn ) , the sequence {eα0 puk }1 is bounded in L1 (Ω ). Lemma A.6 (Poincar´e Inequality). Let Ω be an open set with finite measure. Then there exists a constant C > 0 such that s
∥u∥L2 (Ω) ≤ C∥(−∆) 2 u∥L2 (Rn ) ,
˜ s,2 (Ω ). for every u ∈ H
Proof. We have |ˆ u(ξ)| ≤
1 1 1 2 n ∥u∥L1 (Ω) ≤ n |Ω | ∥u∥L2 (Ω) , (2π) 2 (2π) 2
and hence ∥u∥2L2 (Ω) =
|ˆ u|2 dξ =
Rn
|ˆ u|2 dξ +
|ξ|<δ
|ˆ u|2 dξ
|ξ|≥δ
1 2 n −2s |Ω |∥u∥ |B |δ + δ |ξ|2s |ˆ u|2 dξ 2 1 L (Ω) (2π)n |ξ|≥δ s 1 n 2 −2s ≤ |Ω ||B |δ ∥u∥ + δ |F((−∆) 2 u)(ξ)|2 dξ. 2 1 L (Ω) n (2π) Rn
≤
Choosing δ > 0 so that
1 n (2π)n |Ω ||B1 |δ
=
1 2
we complete the proof.
Lemma A.7 (Compact Embedding). Let Ω be an open set in Rn with finite measure. Then the embedding ˜ s,2 (Ω ) ↩→ H ˜ r,2 (Ω ) is compact for any 0 ≤ r < s (with the notation H ˜ 0,2 (Ω ) = L2 (Ω )). Moreover, H n ˜ 2 ,2 (Ω ) ↩→ Lp (Ω ) is compact for any p ∈ [1, ∞). H Proof. We prove the lemma in few steps. ˜ s,2 (Ω ) ↩→ H ˜ r,2 (Ω ) is continuous for any 0 ≤ r < s. Step 1 The embedding H With the notation (−∆)0 u = u we see that r 2 2r 2 2r 2 2 ∥(−∆) u∥L2 (Rn ) = |ξ| |ˆ u| dξ = |ξ| |ˆ u| dξ + |ξ|2r |ˆ u|2 dξ Rn |ξ|≤1 |ξ|>1 s 2 ≤ |ˆ u| dξ + |ξ|2s |ˆ u|2 dξ ≤ ∥u∥2L2 (Ω) + ∥(−∆) 2 u∥2L2 (Rn ) , |ξ|≤1
|ξ|>1
which is Step 1, thanks to Lemma A.6 Step 2 For a given s > 0 and a given ε > 0 there exists R > 0 such that ∥u∥L2 (Ω∩BRc ) ≤ ε∥u∥H˜ s,2 (Ω) ,
˜ s,2 (Ω ). for every u ∈ H
To prove Step 2 it is sufficient to consider 0 < s < 1, thanks to Step 1.
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We fix ϕ ∈ Cc∞ (B2 ) such that ϕ = 1 on B1 and 0 ≤ ϕ ≤ 1. Setting ϕr (x) = ϕ( xr ) we get ∥(1 − ϕr )u∥2L2 (Rn ) = ∥F((1 − ϕr )u)∥2L2 (Rn ) 2 = |F((1 − ϕr )u)| dξ + |F((1 − ϕr )u)|2 dξ |ξ|
1 |BR1 | ≤ (2π)n
|ξ|≥R1 2
|(1 − ϕr )u|dx
+
Rn
R1−2s
|ξ|2s |F((1 − ϕr )u)|2 dξ
|ξ|≥R1
=: I1 + I2 . Using that supp (1 − ϕr )u ⊂ Ω ∩ Brc and by H¨ older inequality we bound 1 1 I1 ≤ |BR1 ||Ω ∩ Brc | |BR1 ||Ω ∩ Brc |∥u∥2L2 (Ω) . |(1 − ϕr )u|2 dx ≤ n (2π)n (2π) c Ω∩Br From [5, Proposition 3.4] we have |ξ|2s |ˆ u|2 dξ = C(n, s) Rn
Rn
Rn
|u(x) − u(y)|2 dxdy, |x − y|n+2s
and hence I2 ≤ =
R1−2s
|ξ|2s |F((1 − ϕr )u)|2 dξ
Rn
C0 R1−2s
Rn ×Rn
((1 − ϕr (x))u(x) − (1 − ϕr (y))u(y))2 dxdy |x − y|n+2s 2
((1 − ϕr (x))(u(x) − u(y)) − u(y)(ϕr (x) − ϕr (y))) dxdy |x − y|n+2s Rn ×Rn u2 (y)(ϕr (x) − ϕr (y))2 (1 − ϕr (x))2 (u(x) − u(y))2 + dxdy ≤ 2C0 R1−2s |x − y|n+2s |x − y|n+2s Rn ×Rn (ϕr (x) − ϕr (y))2 (u(x) − u(y))2 −2s −2s 2 dxdy + 2C R u (y) dxdy ≤ 2C0 R1 0 1 |x − y|n+2s |x − y|n+2s Rn Rn Rn ×Rn = C0 R1−2s
≤ C1 R1−2s (∥(−∆)s u|∥2L2 (Rn ) + ∥u∥2L2 (Ω) ), where in the last inequality we have used that (ϕr (x) − ϕr (y))2 dx ≤ C, |x − y|n+2s Rn c Thus we have Step 2 by choosing R so that |BR1 ||Ω ∩ BR |<
y ∈ Rn , r ≥ 1. ε 2
where C1 R1−2s = 2ε .
˜ s,2 (Ω ) ↩→ L2 (Ω ) is compact for any 0 < s < 1. Step 3 The embedding H ˜ s,2 (Ω ). Let ϕ, ϕℓ be as in Step 2 (here ℓ ∈ N). Then for Let us consider a bounded sequence {uk }∞ k=1 in H ∞ s,2 ˜ (Ω ) (the proof is very similar to the estimate of I2 in a fixed ℓ the sequence {ϕℓ uk }k=1 is bounded in H Step 2). ˜ s,2 (Br ) ↩→ L2 (Br ) is compact (see e.g. [5, Theorem 7.1]), there exists a subsequence Since the embedding H 1 ∞ such that ϕ1 uk → u1 in L2 (B2 ). Inductively we will have ϕℓ uℓk → uℓ in L2 (B2ℓ ) where {uℓ+1 k }k=1 ℓ ∞ ℓ+1 ℓ ℓ is a subsequence of {uk }k=1 for ℓ ≥ 1. Moreover, we have u = u on Bℓ . Setting u = limℓ→∞ u it follows k 2 that uk converges to u in L (Ω ), thanks to Step 2. {u1k }∞ k=1
˜ s,2 (Ω ) ↩→ H ˜ r,2 (Ω ) is compact for any 0 ≤ r < s. Step 4 The embedding H Since the composition of two compact operators is compact, we can assume that s − r < 1.
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r
˜ s,2 (Ω ). Setting vk = (−∆) 2 uk we see that {vk }∞ is a bounded Let {uk }∞ k=1 be a bounded sequence in H k=1 s−r,2 ˜ sequence in H (Ω ). Then by Step 3 (up to a subsequence) vk converges to some v in L2 (Ω ) which is ˜ r,2 (Ω ). equivalent to saying that (up to a subsequence) uk converges to some u in H ˜ n2 ,2 (Ω ) ↩→ Lp (Ω ) follows from the compactness of H ˜ n2 ,2 (Ω ) ↩→ Finally, compactness of the embedding H L2 (Ω ), Theorem B and Lemma A.9. Lemma A.8 (Exact Constant). We set f (x) = log
1 , |x|
x ∈ Rn .
Then n
(−∆)σ f (x) = γn 22σ−n π − 2 where Γ is the gamma function and γn =
Γ
Γ (σ) n−2σ 2
1 , |x|2σ
0<σ<
n , 2
(n−1)! |S n |. 2
Proof. Using a rescaling argument one can get (see for e.g. [8, Lemma A.5]) (−∆)σ f (x) = (−∆)σ f (e1 )
1 . |x|2σ n
1 To compute the value of (−∆)σ f (e1 ) we use the fact that γ1n log |x| is a fundamental solution of (−∆) 2 (see for instance [8, Lemma A.2]) i.e., n 1 log (−∆) 2 ϕ(x)dx = γn ϕ(0), ϕ ∈ S(Rn ). |x| n R
Using integration by parts, which can be verified, we obtain n γn ϕ(0) = f (x)(−∆) 2 ϕ(x)dx n R n = (−∆)σ f (x)(−∆) 2 −σ ϕ(x)dx n R (−∆)σ f (e1 ) n−2σ ∨ |ξ| ϕ (x)dx = |x|2σ Rn ∨ 1 σ = (−∆) f (e1 ) (ξ) |ξ|n−2σ ϕ dξ 2σ |x| n R n−2σ Γ n−2σ 1 σ n−2σ− n 2 2 2σ = (−∆) f (e1 )2 |ξ| ϕ dξ n−2σ Γ 2 Rn |ξ| Γ n−2σ n σ n−2σ− n 2 2 = (−∆) f (e1 )2 (2π) 2 ϕ(0), Γ 2σ 2 where in the 4th equality we have used that Γ α2 1 1 α− n 2 α, F =2 |x|n−α |x| Γ n−α 2
0 < α < n,
(22)
in the sense of tempered distribution. Since in our case F is the normalized Fourier transform, the constant in the right hand side of (22) appears slightly different from [14, Section 5.9]. Hence we have the lemma.
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The following lemma is the Vitali’s convergence theorem. Lemma A.9 (Vitali’s Convergence Theorem). Let Ω be a measure space with finite measure µ i.e., µ(Ω ) < ∞. Let fk be a sequence of measurable function on Ω be such that k→∞
(i) fk −−−−→ f almost everywhere in Ω . (ii) For ε > 0 there exists δ > 0 such that ˜ Ω
˜ ⊂ Ω with µ(Ω ˜ ) < δ. |fk |dµ < ε for every Ω
Or, (ii′ ) There exists p > 1 such that sup k∈N
|fk |p dµ < ∞.
Ω
Then fk → f in L1 (Ω ). References [1] D. Adams, A sharp inequality of J. Moser for higher order derivatives, Ann. of Math. 128 (1988) 385–398. [2] Adimurthi, Existence of positive solutions of the semilinear Dirichlet problem with critical growth for the n-Laplacian, Ann. Sc. Norm. Super. Pisa Cl. Sci. (4) 17 (3) (1990) 393–413. [3] M.Struwe Adimurthi, Global compactness properties of semilinear elliptic equations with critical growth, J. Funct. Anal. 175 (2000) 125–167. [4] J. Bao, N. Lam, G. Lu, polyharmonic equations with critical exponential growth in the whole space Rn , Discrete Contin. Dyn. Syst. 36 (2) (2016) 577–600. [5] E. Di Nezza, G. Palatucci, E. Valdinoci, Hitchhiker’s guide to the fractional Sobolev spaces, Bull. Sci. Math. 136 (5) (2012) 521–573. [6] O. Druet, Multibumps analysis in dimension 2: quantification of blow-up levels, Duke Math. J. 132 (2006) 217–269. [7] L. Fontana, C. Morpurgo, Sharp Adams and Moser–Trudinger inequalities on Rn and other spaces of infinite measure, 2015, arXiv:1504.04678. [8] A. Hyder, Structure of conformal metrics on Rn with constant Q-curvature, 2015, arXiv:1504.07095. [9] A. Iannizzotto, M. Squassina, 1/2-Laplacian problems with exponential nonlinearity, J. Math. Anal. Appl. 414 (2014) 372–385. [10] S. Iula, A. Maalaoui, L. Martinazzi, A fractional Moser–Trudinger type inequality in one dimension and its critical points, Differential Integral Equations 29 (5/6) (2016) 455–492. [11] O. Lakkis, Existence of solutions for a class of semilinear polyharmonic equations critical exponential growth, Adv. Difference Equ. 4 (6) (1999) 877–906. [12] N. Lam, G. Lu, Existence of nontrivial solutions to polyharmonic equations with subcritical and critical exponential growth, Discrete Contin. Dyn. Syst. 32 (6) (2012) 2187–2205. [13] N. Lam, G. Lu, A new approach to sharp Moser–Trudinger and Adams type inequalities: a rearrangement-free argument, J. Differential Equations 255 (2013) 298–325. [14] E.H. Lieb, M. Loss, Analysis, second ed., in: Graduate Studies in Mathematics, vol. 14, American Mathematical Society, Providence, RI, ISBN: 0-8218-2783-9, 2001. [15] P.L. Lions, The concentration-compactness principle in the calculus of variations. The limit case. I, Rev. Mat. Iberoam. 1 (1) (1985) 145–201. [16] A. Maalaoui, L. Martinazzi, A. Schikorra, Blow-up behaviour of a fractional Adams–Moser–Trudinger type inequality in odd dimension, Comm. Partial Differential Equations (2016) http://dx.doi.org/10.1080/03605302.2016.1222544. [17] A. Malchiodi, L. Martinazzi, Critical points of the Moser–Trudinger functional on a disk, J. Eur. Math. Soc. (JEMS) 16 (2014) 893–908. [18] L. Martinazzi, A threshold phenomenon for embeddings of H0m into Orlicz spaces, Calc. Var. Partial Differential Equations 36 (2009) 493–506.
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[19] L. Martinazzi, Fractional Adams–Moser–Trudinger type inequalities, Nonlinear Anal. 127 (2015) 263–278. [20] L. Martinazzi, M. Struwe, Quantization for an elliptic equation of order 2m with critical exponential non-linearity, Math Z. 270 (2012) 453–487. [21] N. Masmoudi, F. Sani, Trudinger–Moser inequalities with the exact growth condition in Rn and applications, Comm. Partial Differential Equations 40 (8) (2015) 1408–1440. [22] J. Moser, A sharp form of an inequality by N. Trudinger, Indiana Univ. Math. J. 20 (1970/71) 1077–1092. [23] S.I. Pohozaev, The Sobolev embedding in the case pl = n, in: Proc. Tech. Sci. Conf. on Adv. Sci. Research 1964–1965, Mathematics Section, Moskov. Energet. Inst, Moscow, 1965, pp. 158–170. [24] F. Robert, M. Struwe, Asymptotic profile for a fourth order PDE with critical exponential growth in dimension four, Adv. Nonlinear Stud. 4 (2004) 397–415. [25] B. Ruf, A sharp Trudinger–Moser type inequality for unbounded domains in R2 , J. Funct. Anal. 219 (2) (2005) 340–367. [26] L. Silvestre, Regularity of the obstacle problem for a fractional power of the Laplace operator, Comm. Pure Appl. Math. 60 (1) (2007) 67–112. [27] E.M. Stein, Singular Integrals and Differentiable Properties of Functions, in: Princeton Math. Ser., vol. 30, Princeton University Press, Princeton, NJ, 1970, p. xiv+290. [28] M. Struwe, Quantization for a fourth order equation with critical exponential growth, Math. Z. 256 (2007) 397–424. [29] N.S. Trudinger, On embedding into Orlicz spaces and some applications, J. Math. Mech. 17 (1967) 473–483. [30] V.I. Yudovich, On certain estimates connected with integral operators and solutions of elliptic equations, Dokl. Akad. Nank. SSSR 138 (1961) 805–808.
Contents lists available at ScienceDirect
Nonlinear Analysis www.elsevier.com/locate/na
Moser functions and fractional Moser–Trudinger type inequalities Ali Hyder Universit¨ at Basel, Switzerland
article
info
abstract
Article history: Received 22 October 2015 Accepted 27 August 2016 Communicated by Enzo Mitidieri
We improve the sharpness of some fractional Moser–Trudinger type inequalities, particularly those studied by Lam–Lu and Martinazzi. As an application, improving upon works of Adimurthi and Lakkis, we prove the existence of weak solutions to
Keywords: Fractional Moser–Trudinger inequalities Nonlocal equations
the problem n
(−∆) 2 u = λuebu
2
in Ω , 0 < λ < λ1 , b > 0,
with Dirichlet boundary condition, forn any domain Ω in Rn with finite measure. Here λ1 is the first eigenvalue of (−∆) 2 on Ω . © 2016 Elsevier Ltd. All rights reserved.
1. Introduction to the problem Let n ≥ 2 and let Ω be a bounded domain in Rn . The Sobolev embedding theorem states that W0k,p (Ω ) np and kp < n. However, it is not true that W0k,p (Ω ) ⊂ L∞ (Ω ) for kp = n. In ⊂ Lq (Ω ) for 1 ≤ q ≤ n−kp the borderline case, as shown by Yudovich [30], Pohozaev [23] and Trudinger [29], W01,n (Ω ) embeds into an Orlicz space and in fact n n−1 sup eα|u| dx < ∞, (1) u∈W01,n (Ω), ∥∇u∥Ln (Ω) ≤1
Ω
for some α > 0. Moser [22] found the best constant α in the inequality (1), obtaining the so called Moser–Trudinger inequality: n 1 1 n−1 sup eαn |u| dx < ∞, αn = n|S n−1 | n−1 . (2) 1,n |Ω | Ω u∈W (Ω), ∥∇u∥Ln (Ω) ≤1 0
The constant αn in (2) is the best constant in the sense that for any α > αn , the supremum in (1) is infinite. A generalized version of Moser–Trudinger inequality is the following theorem of Adams [1]: E-mail address: [email protected]. http://dx.doi.org/10.1016/j.na.2016.08.024 0362-546X/© 2016 Elsevier Ltd. All rights reserved.
A. Hyder / Nonlinear Analysis 146 (2016) 185–210
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Theorem A ([1]). If k is a positive integer less than n, then there is a constant C = C(k, n) such that n n−k sup eα|u| dx ≤ C|Ω |, u∈Cck (Ω), ∥∇k u∥
n ≤1 L k (Ω)
Ω
where n k k+1 n n−k π22 Γ 2 n−k+1 , k = odd, n Γ 2 α = α(k, n) = n−1 n n k |S | π 2 2 Γ k2 n−k , k = even, Γ n−k 2 and ∇k := ∇∆
k−1 2
k
for k odd and ∇k = ∆ 2 for k even. Moreover the constant α is sharp in the sense that n n−k dx = ∞, (3) sup f (|u|)eα|u| u∈Cck (Ω), ∥∇k u∥
n ≤1 L k (Ω)
Ω
for any f : [0, ∞) → [0, ∞) with limt→∞ f (t) = ∞.1 In a recent work Martinazzi [19] has studied the Adams inequality in a fractional setting. In order to state its result first we recall the space |u(x)| n 1 n Ls (R ) := u ∈ Lloc (R ) : dx < ∞ . n+2s Rn 1 + |x| The operator (−∆)s can be defined on the space Ls (Rn ) via the duality s ⟨(−∆) u, ϕ⟩ := u(−∆)s ϕdx, ϕ ∈ S(Rn ),
(4)
Rn
where (−∆)s ϕ = F −1 |ξ|2s ϕˆ ,
ϕ ∈ S(Rn ),
F is the normalized Fourier transform and S(Rn ) is the Schwartz space. Notice that the integral in (4) is well-defined thanks to [8, Proposition 2.1]. Now for an open set Ω ⊆ Rn (possibly Ω = Rn ), s > 0 and 1 ≤ p ≤ ∞ we define the fractional Sobolev ˜ s,p (Ω ) by space H ˜ s,p (Ω ) := u ∈ Lp (Ω ) : u = 0 on Rn \ Ω , (−∆) 2s u ∈ Lp (Rn ) . H Theorem B ([19]). For any open set Ω ⊂ Rn with finite measure and for any p ∈ (1, ∞) we have p′ sup eαn,p |u| dx ≤ Cn,p |Ω |, n ,p
˜p u∈H
n
(Ω), ∥(−∆) 2p u∥Lp (Ω) ≤1
Ω
where the constant αn,p is given by αn,p =
Γ
n 2p
p′ n n 2p π 2 .
n |S n−1 | Γ np−n 2p
(5)
Moreover, the constant αn,p is sharp in the sense that we cannot replace it with any larger one without making the above supremum infinite. 1 Identity (3) is proven in [1], although not explicitly stated.
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Notice that condition (3) in Theorem A is sharper than only requiring that the constant α in the exponential is sharp, as done in Theorem B. In fact Martinazzi asked whether it is true that p′ sup f (|u|)eαn,p |u| dx = ∞, (6) n ,p
˜p u∈H
n
Ω
(Ω), ∥(−∆) 2p u∥Lp (Ω) ≤1
for any f : [0, ∞) → [0, ∞) with lim f (t) = ∞,
t→∞
f is Borel measurable,
(7)
and αn,p is given by (5). The point here is that Adams constructs smooth and compactly supported test functions similar to the standard Moser functions (constant in a small ball, and decaying logarithmically on an annulus), and then k, n k he estimates their H0 -norms in a very precise way. This becomes much more delicate when k is not integer because instead of computing partial derivatives, one has to estimate the norms of fractional Laplacians (the n term ∥(−∆) 2p u∥Lp (Ω) in (6)). This is indeed done in [19], but the test functions introduced by Martinazzi are not efficient enough to prove (6). As we shall see this has consequences for applications to PDEs. We shall prove that the answer to Martinazzi’s question is positive, indeed in a slightly stronger form, n namely the supremum in (6) is infinite even if we consider the full H p ,p -norm on the whole space. More precisely we have: Theorem 1.1. Let Ω be an open set in Rn with finite measure and let f : [0, ∞) → [0, ∞) satisfy (7). Then p′ sup f (|u|)eαn,p |u| dx = ∞, 1 < p < ∞, n ,p
˜p u∈H
n
≤1 (Ω), ∥u∥p +∥(−∆) 2p u∥p Lp (Rn ) Lp (Ω)
Ω
where the constant αn,p is given by (5). The main difficulty in the proof of Theorem 1.1 is to construct test and cut-off functions in a way that their fractional Laplacians of suitable orders can be estimated precisely. This will be done in Section 2. Here we mention that using a Green’s representation formula, Iula–Maalaoui–Martinazzi [10] proved a ˜ np ,p (Ω ) particular case of Theorem 1.1 in one dimension. Their proof, though, does not extend to spaces H when np > 1 because the function constructed using the Green representation formula do not enjoy enough smoothness at the boundary. Trying to solve this with a smooth cut-off function at the boundary allows to prove (6) only when f grows fast enough at infinity (for instance f (t) ≥ ta for some a > p′ ). Now we move to Moser–Trudinger type inequalities on domains with infinite measure. In this direction we refer to [25,13,21] and the references there in. For our purpose, here we only state the work of Lam–Lu [13]. Theorem C ([13]). Let p ∈ (1, ∞) and τ > 0. Then for every domain Ω ⊂ Rn with finite measure, there exists C = C(n, p, τ ) > 0 such that p′ eαn,p |u| dx ≤ C(|Ω | + 1), sup n ,p
u∈H p
n
(Rn ), ∥(τ I−∆) 2p u∥Lp (Rn ) ≤1
Ω
and sup u∈H
n ,p n p (Rn ), ∥(τ I−∆) 2p
u∥Lp (Rn ) ≤1
′
Φ(αn,p |u|p )dx < ∞,
Rn
where αn,p is given by (5) and Φ(t) := et −
jp −2 j j=0
t , j!
jp := min{j ∈ N : j ≥ p}.
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Furthermore, the constant αn,p is sharp in the above inequalities, i.e., if αn,p is replaced by any α > αn,p , then the supremums are infinite. In the spirit of Theorem 1.1 we prove a stronger version of the sharpness of the constant in Theorem C, in the sense that, even without increasing the constant αn,p we can make the two supremums in Theorem C infinite by multiplying the exponential by a function f going to infinity arbitrarily slow. Moreover it is sufficient to consider functions with compact support. Theorem 1.2. Let Ω ⊂ Rn be a domain with finite measure and let f : [0, ∞) → [0, ∞) satisfy (7). Then for any τ > 0 and for any p ∈ (1, ∞) we have (with the notations as in Theorem C) p′ sup f (|u|)eαn,p |u| dx = ∞, n ,p
˜p u∈H
n
(Ω), ∥(τ I−∆) 2p u∥Lp (Rn ) ≤1
Ω
and sup n ,p
˜p u∈H
n
(Ω), ∥(τ I−∆) 2p u∥Lp (Rn ) ≤1
′
f (|u|)Φ(αn,p |u|p )dx = ∞.
Ω
As an application of Theorem 1.1 (in the case p = 2 and f (t) = t2 , compare to (19)) we prove the existence of (weak) solution to a semilinear elliptic equation with exponential nonlinearity. In order to state the theorem first we need the following definition. Definition 1.1. Let Ω be an open set in Rn with finite measure. Let f ∈ Lp (Ω ) for some p ∈ (1, ∞). We say that u is a weak solution of n
(−∆) 2 u = f ˜ if u ∈ H
n 2 ,2
(Ω ) satisfies
n 4
n 4
(−∆) u(−∆) vdx = Rn
in Ω ,
˜ n2 ,2 (Ω ). f vdx for every v ∈ H
Ω
Theorem 1.3. Let Ω be an open set in Rn with finite measure. Let 0 < λ < λ1 and b > 0. Then there exists a nontrivial weak solution to the problem n
(−∆) 2 u = λuebu
2
in Ω .
n
(8)
˜ 2 ,2 (Ω ) ↩→ L2 (Ω ) is compact for any open set Ω with finite measure Due to the fact that the embedding H (see Lemma A.7 in the Appendix), we do not need any regularity assumption or boundedness assumption on the domain Ω . Eq. (8) has been well studied by several authors in even and odd dimensions, with emphasis both on existence and compactness properties see e.g. [3,6,9,11,16–18,20,24,28]. For instance, Lakkis [11], extending a work of Adimurthi [2], proved the existence of solution to (8) in any even dimension. In [12], Lam–Lu have studied Eq. (8) in even dimension with more general right hand side, namely, equation of the form n (−∆) 2 u = f (x, u), where the function f may not satisfy the Ambrosetti–Rabinowitz condition. In a recent work Iannizzotto–Squassina [9] have proven existence of nontrivial weak solution of (8) with Ω = (0, 1) under an assumption, which turns out to be satisfied thanks to our Theorem 1.1, applied with p = 2 (see Lemma 3.5). Finally, we mention that in a recent work Bao–Lam–Lu [4] have studied the existence of positive solutions to a polyharmonic equation on the whole space R2m , more precisely (I − ∆)m u = f (x, u),
in R2m , m ≥ 1,
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where the function f has critical growth at infinity. Moreover, under certain assumptions on f , they also discussed radial symmetry and regularity of solutions. 2. Moser type functions and proof of Theorems 1.1 and 1.2 We construct Moser type functions as follows: First we fix two smooth functions η and ϕ such that 0 ≤ η, ϕ ≤ 1, 3 3 ∞ η ∈ Cc ((−1, 1)), η = 1 on − , , 4 4 and ϕ ∈ Cc∞ ((−2, 2)),
ϕ = 1, on (−1, 1).
For ε > 0, we set ψε (t) =
1 − ϕε (t) η(t)
1 if 0 ≤ t ≤ 2 1 if t ≥ , 2
and − p1 1 1 1 vε (x) = log ϕε (|x|) + log ψε (|x|) log ε ε |x|
x ∈ Rn ,
where ϕε (t) = ϕ
t . ε
Our aim is to show that the supremums (in Theorems 1.1 and 1.2) taken over the functions {vε }ε>0 (up to a proper normalization) are infinite. The following proposition is crucial in the proof of Theorem 1.1. Proposition 2.1. Let uε (x) := |S
1 n−1 − p
|
n p′
n 2
2 π Γ
n 2p′
1 Γ
n 2p
vε (x). γn
Then for 1 < p < ∞ there exists a constant C > 0 such that −1 p1 n 1 ∥(−∆) 2p uε ∥Lp (Rn ) ≤ 1 + C log . ε Proof. Since the proof of above proposition is quite trivial if n is not an integer. the case when 2p
n 2p
is an integer, from now on we only consider
From Lemmas 2.2 and 2.4 we have −1 n 1 ∥(−∆) 2p uε ∥pLp (B3ε ∪B c ) ≤ C log . 2 ε In order to estimate (−∆)σ vε on the domain {x : 3ε < |x| < 2} we consider the function − p1 1 1 Rε (x) = vε (x) − log log =: fε (x) + gε (x) x ∈ Rn , ε |x|
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where
− p1 1 1 vε (x) − log log if |x| < 2ε fε (x) := ε |x| 0 if |x| ≥ 2ε, − p1 1 1 1 log − log ϕε (|x|) = log ε ε |x| and − p1 1 1 vε (x) − log 1 log if |x| > ε |x| 2 gε (x) := 1 0 if |x| ≤ 2 − p1 1 1 = log (η(|x|) − 1) log . ε |x| It is easy to see that for any σ > 0
1 sup |(−∆) gε (x)| ≤ C log ε x∈Rn σ
− p1 .
(9)
With the help of Lemma A.8 and the triangle inequality we bound − p1 n n n 1 1 1 2p 2p 2p (−∆) log |(−∆) uε (x)| = (−∆) Rε (x) + log 1 ε |x| n |S n−1 | p βn, 2p − p1 n 1 1 1 2p ≤ C|(−∆) Rε (x)| + log n . 1 n−1 ε |S | p |x| p Using the elementary inequality (a + b)q ≤ aq + Cq (bq + aq−1 b), we get |(−∆) 3ε<|x|<2
n 2p
1 ≤ q < ∞, a ≥ 0, b ≥ 0,
−1 n 1 1 1 dx + C |(−∆) 2p Rε (x)|p dx uε (x)| dx ≤ log n−1 n ε |S | |x| 3ε<|x|<2 3ε<|x|<2 − 1′ p n 1 1 2p R (x)|dx + C log n |(−∆) ε ′ ε 3ε<|x|<2 |x| p −1 − 1′ p n 1 1 1 2p R (x)|dx, ≤ 1 + C log + C log n |(−∆) ε ′ p ε ε 3ε<|x|<2 |x|
p
where the last inequality follows from Lemma 2.3. Using the pointwise estimate in Lemma 2.3 and (9) one can show that − p1 n 1 1 2p Rε (x)|dx ≤ C log , n |(−∆) ′ ε 3ε<|x|<2 |x| p which completes the proof.
Lemma 2.2. Let p ∈ (1, ∞). Then there exists a constant C = C(n, p, σ) > 0 such that − p1 1 n |(−∆)σ vε (x)| ≤ C log ε−2σ for |x| ≤ 3ε, 0 < σ < . ε 2
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Moreover, ∥(−∆)
n 2p
vε ∥pLp (B3ε )
1 ≤ C log ε
−1 .
Proof. We claim that for every nonzero multiindex α ∈ Nn there exists C = C(n, α) > 0 such that − p1 1 α |D vε (x)| ≤ C log ε−|α| , x ∈ Rn . ε
(10)
In order to prove (10), first notice that 1 1 p′ if |x| ≤ ε log ε vε (x) = − p1 1 1 1 log η(|x|) if |x| ≥ , log ε |x| 2 and that η ∈ Cc∞ ((−1, 1)). Therefore, the estimate in (10) holds for x ∈ Bε ∪ B c1 . For x ∈ B 21 \ Bε and for 2 a nonzero multiindex α ∈ Nn we compute 1 1 p α 1 1 1 log D vε (x) = Dα ϕε (|x|) log + Dα ψε (|x|) log + Cα,β Dβ log Dα−β ψε (|x|) ε ε |x| |x| 0<β≤α 1 1 1 + Cα,β Dβ log Dα−β ψε (|x|), = Dα ϕε (|x|) log − log ε |x| |x| 0<β≤α
where Cα,β ’s are some constants, and in the second equality we have used that Dα ψε (|x|) = −Dα ϕε (|x|), which follows from the fact that ϕε + ψε = 1 on B 12 . Moreover, from the definition of ϕ and ϕε we have that Cε−|α| if x ∈ B2ε \ Bε α |D ϕε (|x|)| ≤ c 0 if x ∈ B2ε ∪ Bε . Therefore, 1 1 |D ϕε (|x|)| log − log ≤ Cε−|α| . ε |x| α
It is easy to see that for 0 < β ≤ α β D log(|x|) Dα−β ψε (|x|) ≤ C(α, β)|x|−|β| ε−|α−β| ≤ C(α, β)ε−|α| ,
for x ∈ B 12 \ Bε .
This completes the proof of (10). In the case when σ is not an integer then we write σ = m + s where 0 < s < 1 and m is a nonnegative integer. Then for |x| ≤ 3ε we have (the following equivalent definition of fractional Laplacian can be found in [26,5]) (−∆)m vε (x + y) + (−∆)m vε (x − y) − 2(−∆)m vε (x) σ (−∆) vε (x) = Cn,s dy. |y|n+2s Rn Splitting Rn into A1 = {x : |x| ≤ 2ε} ,
A2 =
x : 2ε < |x| ≤
1 4
and A3 =
we have σ
(−∆) vε (x) = Cn,s
3 i=1
Ii ,
x : |x| >
1 4
,
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where Ii := Ai
(−∆)m vε (x + y) + (−∆)m vε (x − y) − 2(−∆)m vε (x) dy. |y|n+2s
For y ∈ A1 , using (10) we have |(−∆)m vε (x + y) + (−∆)m vε (x − y) − 2(−∆)m vε (x)| ≤ |y|2 ∥D2 (−∆)m vε ∥L∞ − p1 1 ≤ C|y|2 ε−2m−2 log , ε and hence − p1 − p1 1 1 dy |I1 | ≤ Cε−2m−2 log ≤ C log ε−2σ . n+2s−2 ε |y| ε A1 For m ≥ 1, again by (10) − p1 1 |(−∆)m vε (x + y) − (−∆)m vε (x)| ≤ C log ε−2m . ε Therefore, − p1 − p1 1 1 dy −2m |I2 + I3 | ≤ C log ≤ C log ε ε−2σ . n+2s ε ε |y|>ε |y| Since on A2 |x + y| ≤ 3ε + 41 < 12 , one has 1 1 1 p |vε (x + y) − vε (x)| = log log (ϕε (|x + y|) + ψε (|x + y|) − ϕε (|x|) − ψε (|x|)) ε ε ε ε + log ψε (|x + y|) − log ψε (|x|) |x + y| |x| ε ε = log ψε (|x + y|) − log ψε (|x|) |x + y| |x| ε ≤ C + log ψε (|x + y|) . |x + y| Hence, for m = 0, changing the variable y → εz − p1 1 |I2 | ≤ C log ε − p1 1 ≤ C log ε − p1 1 ≤ C log ε − p1 1 ≤ C log ε
ε − p1 ψε (|x + y|) log |x+y| 1 −2s ε + C log dy ε |y|n+2s ε<|y|< 41 − p1 log | x + z| ψε (ε| x + z|) 1 −2s −2s ε ε ε + C log ε dz n+2s ε |z| |z|>1 − p1 1 log (3 + |z|) −2s −2s ε + C log ε dz ε |z|n+2s |z|>1 ε−2s .
− 1 Finally, for m = 0, using that |vε | ≤ C log 1ε p on B c1 , we bound 8
− p1 − p1 1 dy 1 |I3 | ≤ C log ≤ C log . n+2s ε ε |y|≥ 14 |y| The lemma follows immediately.
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Lemma 2.3. For |x| ≥ 3ε we have
|(−∆)σ fε (x)| ≤ C
1 |x|2σ
log
1 ε
1 −p
n ε |x| n−2m ε |x|
if 0 < σ < 1 if 1 < σ = m + s <
n , 2
where m is a positive integer and 0 < s < 1. In particular − p1 n 1 ∥(−∆) 2p Rε ∥Lp (B2 \B3ε ) ≤ C log . ε Proof. Notice that for every nonzero multiindex α ∈ Nn we have 1 if |x| < ε − p1 |x||α| 1 α 1 |D fε (x)| ≤ C log if ε < |x| ≤ 2ε ε ε|α| 0 if |x| ≥ 2ε. First we consider 0 < σ < 1. Using that |ϕε | ≤ 1, changing the variable y → εy and by H¨older inequality we obtain fε (x) − fε (y) σ dy |(−∆) fε (x)| = C n+2σ Rn |x − y| 1 1 − p1 log − log ϕ (|y|) ε ε |y| 1 = C log dy n+2σ ε |x − y| |y|<2ε p1 1′ − p1 p dy 1 p′ |log |y|| dy ≤ Cε log np+2pσ ε |y|<2 |x − εy| |y|<2 p1 − p1 n |x| 1 dy 1 ≤ Cεn log , x np+2pσ 2ε | ε εn |x|np+2pσ |y|< |x| |x| − y| n − p1 1 ε 1 , ≤ C 2σ log |x| |x| ε n
where in the second last inequality we have used a change of variable y → |x| ε y and the last inequality follows from the uniform bound 1 2ε ≤ C for every |x| ≥ 3ε, |y| ≤ . (11) x np+2pσ | |x| − y| |x| For σ > 1, changing the variable y → |x|y and by (11) we have (−∆)m fε (x) − (−∆)m fε (y) |(−∆)σ fε (x)| = C dy |x − y|n+2s n R (−∆)m fε (y) =C dy |y|<2ε |x − y|n+2s − p1 1 1 1 ≤ C log dy 2m |x − y|n+2s ε |y| |y|<2ε n−2m − p1 1 ε 1 ≤ C 2σ log . |x| |x| ε We conclude the lemma by (9).
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Lemma 2.4. For 0 < σ <
n 2
there exists a constant C = C(n, σ) such that
− p1 1 1 |(−∆)σ vε (x)| ≤ C log ε |x|n+2σ
for every x ∈ B2c .
Moreover, −1 n 1 ∥(−∆) 2p vε ∥pLp (B c ) ≤ C log . 2 ε Proof. If 0 < σ < 1 then
vε (y) dy, |x| > 2 |x − y|n+2σ |y|<1 1 ≤ C n+2σ vε (y)dy |x| |y|<1 − p1 1 1 (log |y| + log 2) dy ≤ C log ε |x|n+2σ |y|<1 − p1 1 1 ≤ C log . n+2σ ε |x|
σ
|(−∆) vε (x)| = C
(12)
Since the integral in the right hand side of (12) is a proper integral, differentiating under the integral sign one can prove the lemma in a similar way. Proof of Theorem 1.1. Without loss of generality we can assume that B1 ⊆ Ω . Let uε be defined as in Proposition 2.1. We set u ¯ε (x) =
uε (x) ∥uε ∥pLp (Ω)
+ ∥(−∆)
n 2p
uε ∥pLp (Rn )
x ∈ Rn .
p1 ,
n ˜ np ,p (Ω ) and ∥¯ Then u ¯ε ∈ H u∥pLp (Ω) + ∥(−∆) 2p u ¯∥pLp (Rn ) = 1. We claim that there exists a constant δ > 0 such that ′ lim sup exp αn,p |¯ uε |p dx =: lim sup Iε ≥ δ. (13)
ε→0
ε→0
Bε
Noticing that (restriction of u ¯ε on Bε is a constant function) lim inf u ¯ε (x) = ∞,
ε→0 x∈Bε
and by (7), we obtain f (|u|)e
sup ˜ u∈H
n n ,p p (Ω), ∥u∥p +∥(−∆) 2p Lp (Ω)
u∥p ≤1 Lp (Rn )
Ω
′
αn,p |u|p
p′
f (|¯ uε |)eαn,p |¯uε | dx
dx ≥ lim sup ε→0
Ω
p′
f (|¯ uε |)eαn,p |¯uε | dx ε→0 Bε ≥ lim sup Iε inf f (|¯ uε (x)|) ≥ lim sup
ε→0
x∈Bε
≥ δ lim sup inf f (|¯ uε (x)|) ε→0
= ∞.
x∈Bε
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To prove (13) we choose ε = e−k . Noticing that − pp′ p′ C = −C , lim −k + k 1 + k→∞ k p −1 1 , ∥uε ∥pLp (Rn ) ≤ C log ε and using Proposition 2.1 we have n n log
Iε ≥ |B1 |ε e for some δ > 0.
1 ε
1+C (log
1 ε
)
−1
− pp′
C
= |B1 |e−kn+kn(1+ k )
−
p′ p
≥ δ,
In order to prove Theorem 1.2, first we prove the following proposition which gives a similar type of estimate as in Proposition 2.1. Proposition 2.5. Let τ > 0 and 1 < p < ∞. Then there exists a constant C > 0 such that −1 p1 n 1 ∥(τ I − ∆) 2p uε ∥Lp (Rn ) ≤ 1 + C log . ε Proof. We set n
n
wε (x) = (τ I − ∆) 2p uε (x) − (−∆) 2p uε (x). We observe that there exists C = C(p) > 0 such that 1
h(t) = (1 + t)p − 1 − C(tp + tp−1 + t 2 ) < 0,
for every t > 0, 1 ≤ p < ∞,
which follows from the fact that h(0) = 0 and h′ (t) < 0 for every t > 0. Therefore, there holds 1
1
(a + b)p ≤ ap + Cp (bp + abp−1 + b 2 ap− 2 ),
a ≥ 0, b ≥ 0, 1 ≤ p < ∞,
for some constant Cp > 0 and using this inequality we bound n n |(τ I − ∆) 2p uε (x)|p dx = |wε (x) + (−∆) 2p uε (x)|p dx n n R R n n ≤ |(−∆) 2p uε (x)|p + C |wε (x)|p dx + C |(−∆) 2p uε (x)||wε (x)|p−1 dx Rn Rn Rn n 1 1 +C |(−∆) 2p uε (x)|p− 2 |wε (x)| 2 dx Rn
=: I1 + I2 + I3 + I4 . From Proposition 2.1 we have
1 I1 ≤ 1 + C log ε
−1 .
To estimate I2 , I3 and I4 we will use pointwise estimates on (−∆)σ uε , (−∆)σ wε and Lp estimates on (−∆)σ wε . For 0 < σ < n2 combining Lemmas 2.2–2.4, A.8, and (9) we get −2σ − p1 if |x| < 3ε ε 1 σ |(−∆) uε (x)| ≤ C log (14) |x|−2σ if 3ε < |x| < 2 ε |x|−n−2σ if |x| > 2.
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With the help of (14) one can verify that − p1
σ
∥(−∆) uε ∥Lp (Rn )
1 ≤ C(n, p, σ) log ε
n , 2p
(15)
p2 . p+1
(16)
1 ≤ p < ∞, 0 ≤ σ <
,
and together with Lemma A.2 −1
1 I2 ≤ C log ε
.
We conclude the proposition by showing that −1 n 1 p−q q 2p |wε | |(−∆) vε | dx ≤ C(n, p, q) log , ε Rn
0
It follows from Lemma A.1 that
1 |wε (x)| ≤ C log ε and for
n 2p
− p1
x ∈ Rn ,
,
n < 1, 2p
>1 n − +2 − p1 ε p n 1 |wε (x)| ≤ C log |x|− p +2 ε 1
if |x| < 3ε if 3ε < |x| < 2 if |x| > 2,
thanks to (14) and (15). Splitting Rn into A1 = {x : |x| ≤ 2}
and A2 = {x : |x| > 2} ,
we have
n
|wε |q |(−∆) 2p vε |p−q dx =
Rn
2
Ji ,
n
|wε |q |(−∆) 2p vε |p−q dx,
Ji :=
i = 1, 2.
Ai
i=1
−1 Using (14) one can show that J1 ≤ C log 1ε and together with q < gives (16).
p2 p+1
−1 one has J2 ≤ C log 1ε , which
Proof of Theorem 1.2. Here also we can assume that B1 ⊆ Ω . We choose M > 0 large enough such that p′ ′ 1 Φ(αn,p tp ) ≥ eαn,p t , t ≥ M. 2 We set uε u ¯ε = . n ∥(τ I − ∆) 2p uε ∥Lp (Rn ) Then we have
′ f (|¯ uε |)Φ αn,p |¯ uε |p dx ≥
Rn
′ f (|¯ uε |)Φ αn,p |¯ uε |p dx
uε ≥M
1 ≥ 2
p′
f (|¯ uε |)eαn,p |¯uε | dx,
Bε
for ε > 0 small enough. Now the proof follows as in Theorem 1.1, thanks to Proposition 2.5.
Remark. From the point of view of conductor capacity estimate (see e.g. [1, p. 393], [12, p. 2193]), it would be interesting to know whether n
∥(−∆) 2p uε ∥Lp (Rn ) ≥ 1,
n
and ∥(τ I − ∆) 2p uε ∥Lp (Rn ) ≥ 1,
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or not. Here we recall that the “original” Moser functions (see [22]), that is 0 for x ∈ R2 \ B1 1 u(ε, x) := √ 1/ log(1/ε) log(1/|x|) for x ∈ B1 \ Bε 2π log(1/ε) for x ∈ Bε , satisfy ∥∇u(ε, ·)∥L2 (B1 ) = 1. Lemma 2.6. Let uε be as in Proposition 2.1. Then there exists C > 0 (depends on n, p, τ, ϕ, η) such that −1 −1 n n 1 1 p p 2p 2p ∥(−∆) uε ∥Lp (B1 ) ≥ 1 − C log , ∥(τ I − ∆) uε ∥Lp (B1 ) ≥ 1 − C log . ε ε Moreover, if
n 2p
n
is an integer, then ∥(−∆) 2p uε ∥Lp (B1 ) > 1.
Proof. Using the inequality ||a| − |b||p ≥ |a|p − p|a|p−1 |b|,
p ≥ 1, a, b ∈ R,
we obtain
n
|(−∆) 2p uε |p dx
3ε<|x|<1
−1 − 1′ p n 1 1 1 1 1 2p R |dx, ≥ log dx − C log n |(−∆) ε n−1 n ′ p ε |S | |x| ε 3ε<|x|<1 3ε<|x|<1 |x|
and
n
|(τ I − ∆) 2p uε |p dx ≥
3ε<|x|<1
n
|(−∆) 2p uε |p dx − C
3ε<|x|<1
n
|(−∆) 2p uε |p−1 |wε |dx,
3ε<|x|<1
where Rε and wε are defined in the proof of Propositions 2.1 and 2.5 respectively. First part of the lemma follows as in Propositions 2.1 and 2.5. n
n We choose r < 1 so that uε ∈ Cc∞ (Br ). If 2p is an integer, then the support of (−∆) 2p uε is a subset of Br \ Bε . Therefore, by H¨ older inequality n 1 uε (0) = Kn,p (−∆) 2p uε (x) n dx |x| p′ ε<|x|
where the first identity follows from the fact that the function n Γ 2p′ 1 1 −n −n p 2 Kn,p π n := 2 n , ′ ′ n p |x| Γ 2p |x| p n
is a fundamental solution of the operator (−∆) 2p . Since 1′ p n 1 n n 1 1 n−1 − p ′ uε (0) = |S | 2p π 2 Γ log ′ 2p Γ n γ ε 2p
1
= Kn,p |S n−1 | p′ n
we have ∥(−∆) 2p uε ∥Lp (B1 ) > 1.
log
1 ε
1′ p
,
n
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3. Proof of Theorem 1.3 n ˜ n2 ,2 (Ω ) and α0 = αn,2 . Throughout this section we use the notation ∥u∥ = ∥(−∆) 4 u∥L2 (Rn ) , H = H To prove Theorem 1.3 we follow the approach in [2,11]. First we prove that λ1 > 0, which makes the statement of Theorem 1.3 meaningful.
Lemma 3.1. Let Ω be an open set in Rn with finite measure. Then λ1 > 0 and there exists a function u ∈ H such that ∥u∥L2 (Ω) = 1,
and
∥u∥2 = λ1 .
Proof. We recall that λ1 = inf ∥u∥2 : u ∈ H, ∥u∥L2 (Ω) = 1 . Let {uk }∞ k=1 ⊂ H be a sequence such that lim ∥uk ∥2 = λ1 ,
∥uk ∥L2 (Ω) = 1
uk ⇀ u0
u k → u0
k→∞
for every k.
Then up to a subsequence in H,
in L2 (Ω ),
where the latter one follows from the compact embedding H ↩→ L2 (Ω ) (see Lemma A.7). Therefore, λ1 ≤ ∥u0 ∥2 ≤ lim inf ∥uk ∥2 = λ1 ,
∥u0 ∥L2 (Ω) = 1.
k→∞
Let us now define the functional 1 J(u) = ∥u∥2 − 2
u ∈ H,
G(u)dx, Ω
where G(t) =
t
2
g(r) := λrebr ,
g(r)dr,
0 < λ < λ1 , b > 0.
0
Then J is C 2 and the Fr´echet derivative of J is given by n n 4 4 g(u)vdx, DJ(u)(v) = (−∆) u(−∆) vdx − Rn
v ∈ H.
Ω
We also define 2
F (u) = DJ(u)(u) = ∥u∥ −
g(u)udx, Ω
1 I(u) = J(u) − F (u), 2
S = {u ∈ H : u ̸= 0, F (u) = 0} . Observe that if u ∈ H is a nontrivial weak solution of (8) then u ∈ S. With the above notations we have: Lemma 3.2. The set S is closed in the norm topology and α0 0 < s2 < , s := 2 inf J(u). u∈S b Proof. Since F is continuous (actually F is C 1 as J is C 2 ), it is enough to show that 0 is an isolated point of S. If not, then there exists a sequence {uk } ⊂ S such that ∥uk ∥ → 0 as k → ∞. We set vk = ∥uukk ∥ . From the compactness of the embedding H ↩→ Lq (Ω ) for any 1 ≤ q < ∞, we can assume that (up to a subsequence)
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vk ⇀ v in H and vk → v almost everywhere in Ω . By Lemma 3.4 we get 1 k→∞ bu2k 2 1=λ e vk dx −−−−→ λ v 2 dx ≤ λ ∥v∥2 < 1, λ1 Ω Ω which is a contradiction. Hence S is closed. Since, 1 bt2 1 2 e + > 0, f (t) := t − b b
for t > 0, b > 0,
which follows from f (0) = 0 and f ′ (t) > 0 for t > 0, we have λ 1 bu2 1 I(u) = u2 − e + dx > 0, 2 Ω b b
if u ∈ H \ {0},
(17)
and in particular J(u) = I(u) > 0 for u ∈ S. If possible, we assume that s = 0. Then there exists a sequence {uk } ⊂ S such that J(uk ) → 0 as k → ∞. Moreover, 2 2 2 u2k ebuk dx + λ u2k ebuk dx ∥uk ∥2 = λ u2k ebuk dx = λ u2k ≤ 2b
u2k > 2b
Ω
λ ≤4 2
u2k > 2b
u2k
1 − b
e
bu2k
1 + b
dx + λ u2k ≤ 2b
2
u2k ebuk dx,
≤ 4J(uk ) + λ u2k ≤ 2b
2
u2k ebuk dx (18)
and hence uk is bounded in H. Then up to a subsequence uk → u, a.e. in Ω and uk ⇀ u. Using Fatou lemma and (ii) in Lemma 3.4 we obtain λ 1 bu2 1 I(u) = u2 − e + dx ≤ lim inf I(uk ) = lim inf J(uk ) = 0, k→∞ k→∞ 2 Ω b b and hence u = 0, thanks to (17). It follows from (18) that uk → 0 in H which is a contradiction as S is closed. We prove now s2 < α0 b−1 . First we fix u ∈ H with ∥u∥ = 1. We consider the function 2 2 2 Fu (t) := F (tu) = ∥tu∥ − λ t2 u2 ebt u dx, t ≥ 0. Ω
Then 2 2 Fu (t) ≥ t2 λ1 u2 dx − λ u2 ebt u dx > 0, Ω
Ω
for t > 0 sufficiently small and limt→∞ Fu (t) = −∞. Hence, the continuity of Fu implies that there exists tu > 0 such that Fu (tu ) = 0, i.e., tu u ∈ S. Thus 1 1 s2 ≤ J(tu u) ≤ ∥tu u∥2 = t2u . 2 2 2 Again using that tu u ∈ S we have 2 1 1 1 2 bs2 u2 u e dx ≤ 2 λ (tu u)2 eb(tu u) dx = 2 ∥tu u∥2 = , λt λt λ Ω Ω u u which implies that sup ∥u∥≤1, u∈H
and by Theorem 1.1 we deduce that s2 < α0 b−1 .
2
u2 ebs
Ω
u2
dx < ∞,
(19)
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Lemma 3.3. Let u ∈ S be a minimizer of J on S. Then DJ(u) = 0. Proof. We fix a function v ∈ H \ {0} and consider the function γ > 0, t ∈ R.
Fu,v (γ, t) := F (γu + tv),
Differentiating Fu,v with respect to γ and using that F (u) = 0, we get 2 ∂Fu,v (1, 0) = −2bλ u4 ebu dx < 0. ∂γ n R Hence, by implicit function theorem, there exists δ > 0 such that we can write γ = γ(t) as a C 1 function of t on the interval (−δ, δ) which satisfies γ(0) = 1,
Fu,v (γ(t), t) = 0,
for every t ∈ (−δ, δ).
Moreover, choosing δ > 0 smaller if necessary, we have γ(t)u + tv ∈ S for every t ∈ (−δ, δ). We write J(u + tv) − J(u) t J(γ(t)u + tv) − J(u) J(γ(t)u + tv) − J(u + tv) − = lim . t→0 t t
DJ(u)(v) = lim
t→0
Since J is C 1 , a first order expansion of J yields J(γ(t)u + tv) − J(u + tv) = J((u + tv) + (γ(t) − 1)u) − J(u + tv) = DJ(u + tv)((γ(t) − 1)u) + o ((γ(t) − 1)∥u∥) = (γ(t) − 1)DJ(u + tv)(u) + (γ(t) − 1)∥u∥o(1). Therefore, using that F (u) = 0, J(γ(t)u + tv) − J(u + tv) = γ ′ (0)DJ(u)(u) = 0. t On the other hand, since u is a minimizer of J on S and γ(t)u + tv ∈ S, J(γ(t)u + tv) − J(u) ≥ 0 if t ≥ 0 = t ≤ 0 if t ≤ 0, lim
t→0
implies that (since it exists) J(γ(t)u + tv) − J(u) = 0. t This shows that DJ(u)(v) = 0 for every v ∈ H, i.e., DJ(u) = 0. lim
t→0
Proof of Theorem 1.3. Let {uk } be a sequence in S such that limk→∞ J(uk ) → bounded sequence in H and consequently, up to a subsequence uk → u, a.e.
uk ⇀ u,
s2 2 .
ℓ := lim ∥uk ∥,
in Ω ,
k→∞
for some u ∈ H. First we claim that u ̸= 0. Assuming u = 0, by (ii) in Lemma 3.4 we get λ α0 2 bu2k (e − 1)dx = s2 < , lim ∥uk ∥ = lim 2 J(uk ) + k→∞ k→∞ 2b Ω b and hence by (i) in Lemma 3.4 lim ∥uk ∥2 = lim λ
k→∞
a contradiction as S is closed.
k→∞
Ω
2
u2k ebuk dx = 0,
Then by (18) uk is a
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We claim that ℓ = ∥u∥. Then uk → u in H and applying Lemmas 3.2 and 3.3 we have Theorem 1.3. If the claim is false then necessarily we shall have ℓ > ∥u∥. One has
λ bu2k lim ∥uk ∥ = lim 2 J(uk ) + (e − 1)dx k→∞ k→∞ 2b Ω 2 2 s λ =2 + (ebu − 1)dx, 2 2b Ω 2
= s2 − 2J(u) + ∥u∥2 . We divide the proof in two cases, namely J(u) ≤ 0 and J(u) > 0. Case 1. We consider that J(u) ≤ 0. Since u ̸= 0, 2 2 λ (ebu − 1)dx < λ u2 ebu dx, ∥u∥2 ≤ b Ω Ω where the second inequality follows from (17). It is easy to see that we can choose 0 < t0 < 1 such that 2 ∥t0 u∥2 = λ (t0 u)2 eb(t0 u) dx, Ω
that means t0 u ∈ S. Using that I(tu) is strictly monotone increasing in t, which follows from the expression in (17), we obtain s2 s2 ≤ J(t0 u) = I(t0 u) < I(u) ≤ lim inf J(uk ) = , k→∞ 2 2 a contradiction. Case 2. Here we assume that J(u) > 0. Then ℓ2 = lim ∥uk ∥2 = s2 − 2J(u) + ∥u∥2 < s2 + ∥u∥2 < k→∞
Taking vk =
uk ∥uk ∥
α0 + ∥u∥2 . b
(20)
we see that (up to a subsequence) vk ⇀ v :=
u , ℓ
vk → v, a.e.
in Ω ,
and by Lemma A.5, for every p < (1 − ∥v∥2 )−1 2 sup epα0 vk dx < ∞. k∈N
Ω
Taking (20) into account we have 0 < ℓ2 − ∥u∥2 = s2 − 2J(u) <
α0 , b
and therefore, we can choose ε0 > 0 such that 1 + ε0 =
α0 1 , 2 b ℓ − ∥u∥2
i.e., ℓ2 (1 + ε0 ) =
α0 b
1−
∥u∥2 ℓ2
−1 .
For k large enough such that ∥uk ∥2 ≤ ℓ2 (1 + ε20 ) holds, we observe that b∥uk ∥2 ≤ p0 α0 for some 1 < p0 < (1 − ∥v∥2 )−1 . Thus, for some p1 > 1, p2 > 1 with p1 p2 p0 < (1 − ∥v∥2 )−1 we obtain 2 2 p1 p1 p0 α0 vk 1 ′ u2k ebuk dx ≤ sup ∥u2p ∥Lp2 (Ω) < ∞, sup k ∥Lp2 (Ω) ∥e k∈N
k∈N
Ω
and together with Lemma A.9 lim
k→∞
Ω
2
u2k ebuk dx =
Ω
2
u2 ebu dx.
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Indeed, 2
2
2
∥u∥ < ℓ = lim ∥uk ∥ = λ lim k→∞
k→∞
2 u2k ebuk dx
=λ
Ω
2
u2 ebu dx,
Ω
and we can now proceed as in Case 1.
Lemma 3.4. Let uk , vk ∈ H such that uk ⇀ u in H, uk → u, a.e. in Ω , vk ⇀ v in H and vk → v, a.e. in Ω . Then (i) If lim sup ∥uk ∥2 < k→∞
α0 , b
then for every integer ℓ ≥ 1
2
ebuk vkℓ dx =
lim
k→∞
Ω
2
ebu v ℓ dx.
Ω
(ii) If
2
u2k ebuk dx < ∞,
lim sup k→∞
Ω
then
2
ebuk dx =
lim
k→∞
Ω
2
ebu dx.
Ω
Proof. We prove the lemma with the help of Lemma A.9 (in the Appendix). We choose p > 1 such that for k large enough p∥uk ∥2 < αb0 holds and together with Theorem C we have 2 sup epbuk dx < ∞. k∈N
˜ Since the embedding H
n 2 ,2
Ω
(Ω ) ↩→ Lq (Ω ) is compact (see Lemma A.7) for every 1 ≤ q < ∞, we have vkq → v q
in L1 (Ω ).
Indeed, 2
2
sup ∥ebuk vkℓ ∥Lp (Ω) ≤ ∥vkℓ ∥Lp′ (Ω) ∥ebuk ∥Lp (Ω) < ∞, k∈N
and we conclude (i). Now (ii) follows from e
bu2k
u2k >M
1 dx ≤ M
2
u2k >M
2
u2k ebuk dx ≤
C , M
which implies that the function fk := ebuk satisfies the condition in Lemma A.9.
In the following lemma we prove that the assumption H ′ (v) in [9] is true under certain conditions. 2
Lemma 3.5. Let α0 > 0. Let f (t) = eα0 t h(t) satisfies H(i)–(iii) in [9]. Let h ≥ 0 on [0, ∞) and h(−t) = −h(t). Let s f (st) be a monotone increasing function with respect to t on (0, ∞), s ̸= 0. If t
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√ ˜ 21 ,2 ((0, 1)) such that 2π∥(−∆) 14 u∥L2 (R) = 1 and limt→∞ h(t)t = ∞ then there exists u ∈ H 2 1 ω t F (tu)dx < sup Φ(tu) := sup − , 4π 2α0 t>0 t>0 0 where
t
f (s)ds,
F (t) = 0
and ω is as in [9]. ˜ 21 ,2 ((0, 1)) such that Proof. For a given M > 0 we can choose u ∈ H 1 2 √ 1 2π f u udx > M, 2π∥(−∆) 4 u∥L2 (R) = 1, α0 0 thanks to Theorem 1.1. Differentiating with respect to t one has 1 f (tu) 1 − udx . Φ ′ (tu) = t 2π t 0 2 Hence, for t ≥ 2π α0 =: t0 and 2πM > t0 1 f (t0 u) 1 ′ − udx < 0. Φ (tu) ≤ t 2π t0 0 Thus Φ ′ (tu) ≤ 0 on (t0 − ε, ∞) for some ε > 0 and therefore, sup Φ(tu) = t>0
Φ(tu) ≤
sup t∈(0, t0 −ε)
t2 π < . 4π 2α 0 t∈(0, t0 −ε) sup
Since ω = π, thanks to Theorem B, we conclude the lemma.
Acknowledgments I would like to thank my advisor Prof. Luca Martinazzi for all the useful discussion and encouragement. I would also like to thank the reviewers for carefully reading the manuscript and many useful comments and suggestions. The author is supported by the Swiss National Science Foundation, project no. PP00P2-144669. Appendix Lemma A.1 (Pointwise Estimate). Let s > 0 and not an integer. Let m be the smallest integer greater than s. Then for any τ > 0 |(τ I − ∆)s u(x) − (−∆)s u(x)| ≤ C
m−1
|(−∆)s−j u(x)| + C∥(−∆)σ u∥L1 (Rn ) ,
u ∈ S(Rn ),
j=1
where σ ∈ max{ n2 − m + s, 0}, can be interpreted as zero.
n 2
, the constant C depends only on n, s, σ, τ and for m = 1 the above sum
Proof. We set f (t) = ts on R+ . By Taylor’s expansion we have f (t + τ ) = f (t) + τ f ′ (t) + · · · +
τ m−1 m−1 τm m f (t) + f (ξt ), (m − 1)! m!
for some t < ξt < t + τ.
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In particular (τ + t2 )s = t2s + c1 t2s−2 + c2 t2s−4 + · · · + cm−1 t2s−2m+2 + E(t), where the function E satisfies the estimate |E(t)| ≤ C(1 + t)2s−2m ,
t > 0.
Therefore, for u ∈ S(Rn ) F((τ I − ∆)s u)(ξ) = (τ + |ξ|2 )s u ˆ 2s = |ξ| + c1 |ξ|2s−2 + · · · + cm−1 |ξ|2s−2m+2 + E(|ξ|) u ˆ =
m−1
cj |ξ|2s−2j u ˆ + E(|ξ|)ˆ u(ξ)
j=0
=
m−1
cj F((−∆)s−j u) + E(|ξ|)ˆ u(ξ),
j=0
and hence (τ I − ∆)s u(x) =
m−1
cj (−∆)s−j u(x) + F −1 (E u ˆ)(x).
j=0
To estimate the term F observe that
−1
(E u ˆ) (uniformly in x) in terms of L1 (Rn ) norm of (fractional) derivative of u, we 1 σ |E(|ξ|)ˆ u(ξ)| = E(|ξ|) 2σ (−∆) u(ξ) |ξ| C σ u(ξ) ≤ 2σ (−∆) |ξ| (1 + |ξ|2 )m−s C ≤ 2σ ∥(−∆)σ u∥L1 (Rn ) . |ξ| (1 + |ξ|2 )m−s
Thus −1 F (E u ˆ)(x) ≤ C∥E u ˆ∥L1 (Rn ) ≤ C∥(−∆)σ u∥L1 (Rn ) , and we complete the proof.
Lemma A.2 (Lp Estimate). Let s > 0 be a noninteger. Let τ > 0 be any fixed number. Then for p ∈ (1, ∞) there exists C = C(n, s, p, τ ) > 0 such that ∥u∥Lp (Rn ) if s < 1 ∥(τ I − ∆)s u − (−∆)s u∥Lp (Rn ) ≤ C ∥u + (−∆)s−1 u∥Lp (Rn ) if s > 1. Proof. We have F((τ I − ∆)s u)(ξ) − F((−∆)s u)(ξ) = (τ + |ξ|2 )s − |ξ|2s u ˆ(ξ) ˆ(ξ) if s < 1 (τ + |ξ|2 )s − |ξ|2s u = (τ + |ξ|2 )s − |ξ|2s (1 + |ξ|2s−2 )ˆ u(ξ) if s > 1 1 + |ξ|2s−2 m(ξ)ˆ u(ξ) if s < 1 =: m(ξ)F u + (−∆)s−1 u (ξ) if s > 1. Now the proof follows from the Hormander multiplier theorem (see [27, p. 96]).
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The following lemma appears already in [7, p. 46], but for the reader’s convenience we give a more detailed proof. Lemma A.3 (Equivalence of Norms). Let σ > 0. Then for p ∈ (1, ∞) there exists a constant C > 0 such that for every u ∈ S(Rn ) 1 ∥u∥Lp (Rn ) + ∥(−∆)σ u∥Lp (Rn ) ≤ ∥(I − ∆)σ u∥Lp (Rn ) C ≤ C ∥u∥Lp (Rn ) + ∥(−∆)σ u∥Lp (Rn ) . Proof. We set Gσ (x) =
1 1 σ σ 2 (4π) Γ 2
∞
e−π
|x|2 t
t
e− 4π t
0
−n+σ 2
dt , t
which is the Bessel potential of order σ (see [27, p. 130]). Then 1 ˆ σ (x) = 1 n Gσ (x)dx = 1, G σ . 2 (1 + |x|2 ) 2 n (2π) R Setting f = (I − ∆)σ u we can write u = G2σ ∗ f and by Young’s inequality one has ∥u∥Lp (Rn ) ≤ ∥f ∥Lp (Rn ) . Again writing u = G2σ ∗ f and taking Fourier transform we obtain 1 F((−∆)σ u) = |ξ|2σ u ˆ = |ξ|2σ fˆ =: m(ξ)fˆ, (1 + |ξ|2 )σ and by Hormander multiplier theorem we get ∥(−∆)σ u∥Lp (Rn ) ≤ C∥f ∥Lp (Rn ) . Thus, ∥u∥Lp (Rn ) + ∥(−∆)σ u∥Lp (Rn ) ≤ C∥(I − ∆)σ u∥Lp (Rn ) . To conclude the lemma, it is sufficient to show that ∥(−∆)s u∥Lp (Rn ) ≤ C(n, s, σ, p)(∥u∥Lp (Rn ) + ∥(−∆)σ u∥Lp (Rn ) ),
0 < s < σ,
(21)
thanks to Lemma A.2. In order to prove (21) we fix a function ϕ ∈ Cc∞ (B2 ) such that ϕ = 1 on B1 . Then F((−∆)s u) = |ξ|2s u ˆ = |ξ|2s ϕˆ u + |ξ|2s (1 − ϕ)ˆ u = m1 (ξ)ˆ u + m2 (ξ)F((−∆)σ u), where m1 (ξ) = |ξ|2s ϕ(ξ), m2 (ξ) = |ξ|2s−2σ (1 − ϕ(ξ)) are multipliers and we conclude (21) by Hormander multiplier theorem. Lemma A.4 (Embedding to An Orlicz Space). Let Ω be an open set with finite measure. Then for every ˜ n2 ,2 (Ω ) u∈H 2 eu dx < ∞. Ω n
Proof. We set f = (−∆) 4 u. By [19, Proposition 8] we have u(x) = G(x, y)f (y)dy, 0 ≤ G(x, y) ≤ Ω
Cn n , |x − y| 2
where G is a Greens function. We choose M > 0 large enough such that ∥f˜∥L2 Cn < α0 , where f˜ = f − f χ{|f |≤M } . Then |f˜(y)| ˜ ˜ n n |u(x)| ≤ C(M ) + Cn I 2 f (x), I 2 f (x) := n dy, Ω |x − y| 2 and by [1, Theorem 2] we conclude the proof.
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As a consequence of the above lemma one can prove a higher dimensional generalization of Lions lemma [15] (for a simple proof see e.g. [9, Lemma 2.6]), namely ˜ n2 ,2 (Ω ) such that Lemma A.5 (Lions). Let uk be a sequence in H n
n
n
˜ 2 ,2 (Ω ), uk ⇀ u in H ∥(−∆) 4 uk ∥L2 (Rn ) = 1. 0 < ∥(−∆) 4 u∥L2 (Rn ) < 1, −1 n ∞ Then for every 0 < p < 1 − ∥(−∆) 4 u∥2L2 (Rn ) , the sequence {eα0 puk }1 is bounded in L1 (Ω ). Lemma A.6 (Poincar´e Inequality). Let Ω be an open set with finite measure. Then there exists a constant C > 0 such that s
∥u∥L2 (Ω) ≤ C∥(−∆) 2 u∥L2 (Rn ) ,
˜ s,2 (Ω ). for every u ∈ H
Proof. We have |ˆ u(ξ)| ≤
1 1 1 2 n ∥u∥L1 (Ω) ≤ n |Ω | ∥u∥L2 (Ω) , (2π) 2 (2π) 2
and hence ∥u∥2L2 (Ω) =
|ˆ u|2 dξ =
Rn
|ˆ u|2 dξ +
|ξ|<δ
|ˆ u|2 dξ
|ξ|≥δ
1 2 n −2s |Ω |∥u∥ |B |δ + δ |ξ|2s |ˆ u|2 dξ 2 1 L (Ω) (2π)n |ξ|≥δ s 1 n 2 −2s ≤ |Ω ||B |δ ∥u∥ + δ |F((−∆) 2 u)(ξ)|2 dξ. 2 1 L (Ω) n (2π) Rn
≤
Choosing δ > 0 so that
1 n (2π)n |Ω ||B1 |δ
=
1 2
we complete the proof.
Lemma A.7 (Compact Embedding). Let Ω be an open set in Rn with finite measure. Then the embedding ˜ s,2 (Ω ) ↩→ H ˜ r,2 (Ω ) is compact for any 0 ≤ r < s (with the notation H ˜ 0,2 (Ω ) = L2 (Ω )). Moreover, H n ˜ 2 ,2 (Ω ) ↩→ Lp (Ω ) is compact for any p ∈ [1, ∞). H Proof. We prove the lemma in few steps. ˜ s,2 (Ω ) ↩→ H ˜ r,2 (Ω ) is continuous for any 0 ≤ r < s. Step 1 The embedding H With the notation (−∆)0 u = u we see that r 2 2r 2 2r 2 2 ∥(−∆) u∥L2 (Rn ) = |ξ| |ˆ u| dξ = |ξ| |ˆ u| dξ + |ξ|2r |ˆ u|2 dξ Rn |ξ|≤1 |ξ|>1 s 2 ≤ |ˆ u| dξ + |ξ|2s |ˆ u|2 dξ ≤ ∥u∥2L2 (Ω) + ∥(−∆) 2 u∥2L2 (Rn ) , |ξ|≤1
|ξ|>1
which is Step 1, thanks to Lemma A.6 Step 2 For a given s > 0 and a given ε > 0 there exists R > 0 such that ∥u∥L2 (Ω∩BRc ) ≤ ε∥u∥H˜ s,2 (Ω) ,
˜ s,2 (Ω ). for every u ∈ H
To prove Step 2 it is sufficient to consider 0 < s < 1, thanks to Step 1.
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We fix ϕ ∈ Cc∞ (B2 ) such that ϕ = 1 on B1 and 0 ≤ ϕ ≤ 1. Setting ϕr (x) = ϕ( xr ) we get ∥(1 − ϕr )u∥2L2 (Rn ) = ∥F((1 − ϕr )u)∥2L2 (Rn ) 2 = |F((1 − ϕr )u)| dξ + |F((1 − ϕr )u)|2 dξ |ξ|
1 |BR1 | ≤ (2π)n
|ξ|≥R1 2
|(1 − ϕr )u|dx
+
Rn
R1−2s
|ξ|2s |F((1 − ϕr )u)|2 dξ
|ξ|≥R1
=: I1 + I2 . Using that supp (1 − ϕr )u ⊂ Ω ∩ Brc and by H¨ older inequality we bound 1 1 I1 ≤ |BR1 ||Ω ∩ Brc | |BR1 ||Ω ∩ Brc |∥u∥2L2 (Ω) . |(1 − ϕr )u|2 dx ≤ n (2π)n (2π) c Ω∩Br From [5, Proposition 3.4] we have |ξ|2s |ˆ u|2 dξ = C(n, s) Rn
Rn
Rn
|u(x) − u(y)|2 dxdy, |x − y|n+2s
and hence I2 ≤ =
R1−2s
|ξ|2s |F((1 − ϕr )u)|2 dξ
Rn
C0 R1−2s
Rn ×Rn
((1 − ϕr (x))u(x) − (1 − ϕr (y))u(y))2 dxdy |x − y|n+2s 2
((1 − ϕr (x))(u(x) − u(y)) − u(y)(ϕr (x) − ϕr (y))) dxdy |x − y|n+2s Rn ×Rn u2 (y)(ϕr (x) − ϕr (y))2 (1 − ϕr (x))2 (u(x) − u(y))2 + dxdy ≤ 2C0 R1−2s |x − y|n+2s |x − y|n+2s Rn ×Rn (ϕr (x) − ϕr (y))2 (u(x) − u(y))2 −2s −2s 2 dxdy + 2C R u (y) dxdy ≤ 2C0 R1 0 1 |x − y|n+2s |x − y|n+2s Rn Rn Rn ×Rn = C0 R1−2s
≤ C1 R1−2s (∥(−∆)s u|∥2L2 (Rn ) + ∥u∥2L2 (Ω) ), where in the last inequality we have used that (ϕr (x) − ϕr (y))2 dx ≤ C, |x − y|n+2s Rn c Thus we have Step 2 by choosing R so that |BR1 ||Ω ∩ BR |<
y ∈ Rn , r ≥ 1. ε 2
where C1 R1−2s = 2ε .
˜ s,2 (Ω ) ↩→ L2 (Ω ) is compact for any 0 < s < 1. Step 3 The embedding H ˜ s,2 (Ω ). Let ϕ, ϕℓ be as in Step 2 (here ℓ ∈ N). Then for Let us consider a bounded sequence {uk }∞ k=1 in H ∞ s,2 ˜ (Ω ) (the proof is very similar to the estimate of I2 in a fixed ℓ the sequence {ϕℓ uk }k=1 is bounded in H Step 2). ˜ s,2 (Br ) ↩→ L2 (Br ) is compact (see e.g. [5, Theorem 7.1]), there exists a subsequence Since the embedding H 1 ∞ such that ϕ1 uk → u1 in L2 (B2 ). Inductively we will have ϕℓ uℓk → uℓ in L2 (B2ℓ ) where {uℓ+1 k }k=1 ℓ ∞ ℓ+1 ℓ ℓ is a subsequence of {uk }k=1 for ℓ ≥ 1. Moreover, we have u = u on Bℓ . Setting u = limℓ→∞ u it follows k 2 that uk converges to u in L (Ω ), thanks to Step 2. {u1k }∞ k=1
˜ s,2 (Ω ) ↩→ H ˜ r,2 (Ω ) is compact for any 0 ≤ r < s. Step 4 The embedding H Since the composition of two compact operators is compact, we can assume that s − r < 1.
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208
r
˜ s,2 (Ω ). Setting vk = (−∆) 2 uk we see that {vk }∞ is a bounded Let {uk }∞ k=1 be a bounded sequence in H k=1 s−r,2 ˜ sequence in H (Ω ). Then by Step 3 (up to a subsequence) vk converges to some v in L2 (Ω ) which is ˜ r,2 (Ω ). equivalent to saying that (up to a subsequence) uk converges to some u in H ˜ n2 ,2 (Ω ) ↩→ Lp (Ω ) follows from the compactness of H ˜ n2 ,2 (Ω ) ↩→ Finally, compactness of the embedding H L2 (Ω ), Theorem B and Lemma A.9. Lemma A.8 (Exact Constant). We set f (x) = log
1 , |x|
x ∈ Rn .
Then n
(−∆)σ f (x) = γn 22σ−n π − 2 where Γ is the gamma function and γn =
Γ
Γ (σ) n−2σ 2
1 , |x|2σ
0<σ<
n , 2
(n−1)! |S n |. 2
Proof. Using a rescaling argument one can get (see for e.g. [8, Lemma A.5]) (−∆)σ f (x) = (−∆)σ f (e1 )
1 . |x|2σ n
1 To compute the value of (−∆)σ f (e1 ) we use the fact that γ1n log |x| is a fundamental solution of (−∆) 2 (see for instance [8, Lemma A.2]) i.e., n 1 log (−∆) 2 ϕ(x)dx = γn ϕ(0), ϕ ∈ S(Rn ). |x| n R
Using integration by parts, which can be verified, we obtain n γn ϕ(0) = f (x)(−∆) 2 ϕ(x)dx n R n = (−∆)σ f (x)(−∆) 2 −σ ϕ(x)dx n R (−∆)σ f (e1 ) n−2σ ∨ |ξ| ϕ (x)dx = |x|2σ Rn ∨ 1 σ = (−∆) f (e1 ) (ξ) |ξ|n−2σ ϕ dξ 2σ |x| n R n−2σ Γ n−2σ 1 σ n−2σ− n 2 2 2σ = (−∆) f (e1 )2 |ξ| ϕ dξ n−2σ Γ 2 Rn |ξ| Γ n−2σ n σ n−2σ− n 2 2 = (−∆) f (e1 )2 (2π) 2 ϕ(0), Γ 2σ 2 where in the 4th equality we have used that Γ α2 1 1 α− n 2 α, F =2 |x|n−α |x| Γ n−α 2
0 < α < n,
(22)
in the sense of tempered distribution. Since in our case F is the normalized Fourier transform, the constant in the right hand side of (22) appears slightly different from [14, Section 5.9]. Hence we have the lemma.
A. Hyder / Nonlinear Analysis 146 (2016) 185–210
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The following lemma is the Vitali’s convergence theorem. Lemma A.9 (Vitali’s Convergence Theorem). Let Ω be a measure space with finite measure µ i.e., µ(Ω ) < ∞. Let fk be a sequence of measurable function on Ω be such that k→∞
(i) fk −−−−→ f almost everywhere in Ω . (ii) For ε > 0 there exists δ > 0 such that ˜ Ω
˜ ⊂ Ω with µ(Ω ˜ ) < δ. |fk |dµ < ε for every Ω
Or, (ii′ ) There exists p > 1 such that sup k∈N
|fk |p dµ < ∞.
Ω
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