A note on truncations in fractional Sobolev spaces

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by  Roberta Musina

A note on truncations in fractional Sobolev spaces
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Roberta Musina∗ and Alexander I. Nazarov†
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1 We study the Nemytskii operators u 7→ |u| and u 7→ u± in fractional Sobolev spaces

Hs(Rn), s>1.
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Keywords: Fractional Laplacian - Sobolev spaces - Truncation operators
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t 2010 Mathematics Subject Classification: 46E35,47H30.
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1 Introduction. Main result
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In this paper we discuss the relation between the map u 7→ |u| and the Dirichlet
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4 s
Laplacian. Recall that the Dirichlet Laplacian (−∆Rn) u of order s > 0 of a function
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0 u ∈ L2(Rn), n ≥ 1, is the distribution
.
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7 h(−∆Rn)su,ϕi ≡ u (−∆Rn)sϕdx := |ξ|2sF[ϕ]F[u]dξ , ϕ ∈ C∞(Rn),
Z Z 0
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Rn Rn
:
v
i where
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r F[u](ξ) = (2π)−n2 e−iξ·xu(x)dx
a Z
Rn
∗Dipartimento di Scienze Matematiche, Informatiche e Fisiche, Universit`a di Udine, via delle
Scienze, 206 – 33100 Udine, Italy. Email: [email protected]. Partially supported by
miur-prin 2015233N54.
†St.Petersburg Department of Steklov Institute, Fontanka, 27, St.Petersburg, 191023, Russia
and St.Petersburg State University, Universitetskii pr. 28, St.Petersburg, 198504, Russia. E-mail:
[email protected].
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is the Fourier transform in Rn. The Sobolev–Slobodetskii space
s
Hs(Rn) = {u ∈ L2(Rn) | (−∆Rn)2u ∈ L2(Rn) }
naturally inherits an Hilbertian structure from the scalar product
s
(u,v) = h(−∆Rn) u,vi+ uvdx.
Z
Rn
The standard reference for the operator (−∆Rn)s and functions in Hs(Rn) is the
monograph [8] by Triebel.
For any positive order s ∈/ N we introduce the constant
22ss Γ n +s
C = 2 . (1)
n,s n (cid:0) (cid:1)
π2 Γ 1−s
(cid:0) (cid:1)
Notice that
C > 0 if ⌊s⌋ is even; C < 0 if ⌊s⌋ is odd, (2)
n,s n,s
where ⌊s⌋ stands for the integer part of s. It is well known that for s ∈ (0,1) and
u,v ∈ Hs(Rn) one has
C (u(x)−u(y))(v(x)−v(y))
s n,s
h(−∆Rn) u,vi = dx. (3)
2 Z |x−y|n+2s
Rn
Let us recall some known facts about the Nemytskii operator |·| : u 7→ |u|.
1. |·| is a Lipschitz transform of H0(Rn) ≡ L2(Rn) into itself.
2. Let 0 < s ≤ 1. Then | · | is a continuous transform of Hs(Rn) into itself, by
general results about Nemytskii operators in Sobolev/Besov spaces, see [7, Theorem
5.5.2/3]. Also it is obvious that for u ∈ H1(Rn)
h−∆|u|,|u|i = h−∆u,ui = |∇u|2dx , h−∆u+,u−i = ∇u+ ·∇u−dx = 0.
Z Z
Rn Rn
Hereandelsewhereu± = max{±u,0} = 1(|u|±u),sothatu = u+−u−,|u| = u++u−.
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On the other hand, for s ∈ (0,1) and u ∈ Hs(Rn) formula (3) gives
u+(x)u−(y)
h(−∆Rn)su+,u−i = −Cn,sZZ |x−y|n+2s dxdy. (4)
Rn×Rn
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From (4) we infer by the polarization identity
4h(−∆Rn)su+,u−i = h(−∆Rn)s|u|,|u|i−h(−∆Rn)su,ui
that if u changes sign then
s s
h(−∆Rn) |u|,|u|i < h(−∆Rn) u,ui, s ∈ (0,1). (5)
We mention also [4, Theorem 6] for a different proof and explanation of (5), that
s
includes thecasewhen (−∆Rn) is replacedby theNavier(orspectral Dirichlet)Lapla-
cian on a bounded Lipschitz domain Ω ⊂ Rn.
3. Let 1 < s < 3. The results in [2] and [6] (see also Section 4 of the exhaustive
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survey [3]) imply that |·| is a bounded transform of Hs(Rn) into itself. That is, there
exists a constant c(n,s) such that
h(−∆Rn)s|u|,|u|i ≤ c(n,s)h(−∆Rn)su,ui, u ∈ Hs(Rn).
In particular, |·| is continuous at 0 ∈ Hs(Rn).
It is easy to show that the assumption s < 3 can not be improved, see Example
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1 below and [2, Proposition p. 357], where a more general setting involving Besov
spaces Bs,q(Rn), s ≥ 1+ 1, is considered.
p p
At our knowledge, the continuity of | · | : Hs(Rn) → Hs(Rn), s ∈ (1, 3), is an
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open problem. We can only point out the next simple result.
Proposition 1 Let 0 < τ < s < 3. Then |·| : Hs(Rn) → Hτ(Rn) is continuous.
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Proof. Recall that Hs(Rn) ֒→ Hτ(Rn) for 0 < τ < s. Actually, the Cauchy-
Bunyakovsky-Schwarz inequality readily gives the well known interpolation inequal-
ity
τ s−τ
h(−∆Rn)τv,vi = |ξ|2τ|F[v]|2dξ ≤ h(−∆Rn)sv,vi s |v|2dx s, v ∈ Hs(Rn).
Z (cid:16) (cid:17) (cid:16)Z (cid:17)
Rn Rn
Since | · | is continuous L2(Rn) → L2(Rn) and bounded Hs(Rn) → Hs(Rn), the
(cid:3)
statement follows immediately.
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Now we formulate our main result. It provides the complete proof of [5, Theorem
1] for s below the threshold 3 and gives a positive answer to a question raised in [1,
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Remark 4.2] by Nicola Abatangelo, Sven Jahros and Albero Saldan˜a.
Theorem 1 Let s ∈ (1, 3) and u ∈ Hs(Rn). Then formula (4) holds. In particular,
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if u changes sign then
s s
h(−∆Rn) |u|,|u|i > h(−∆Rn) u,ui.
Our proof is deeply based on the continuity result in Proposition 1. The knowl-
edge of continuity of |·| : Hs(Rn) → Hs(Rn) could considerably simplify it.
We denote by c any positive constant whose value is not important for our pur-
poseses. Itsvaluemaychangelinetoline. Thedependance ofconcertainparameters
is shown in parentheses.
2 Preliminary results and proof of Theorem 1
We begin with a simple but crucial identity that has been independently pointed out
in [5, Lemma 1] and [1, Lemma 3.11] (without exact value of the constant). Notice
that it holds for general fractional orders s > 0.
Theorem 2 Let s > 0, s ∈/ N. Assume that v,w ∈ Hs(Rn) have compact and
disjoint supports. Then
v(x)w(y)
s
h(−∆Rn) v,wi = −Cn,s ZZ |x−y|n+2s dxdy. (6)
Rn×Rn
Proof. Let ρ be a sequence of mollifiers, and put w := w∗ρ . Formula (3) gives
h h h
h(−∆Rn)sv,whi = h(−∆Rn)s−⌊s⌋v,(−∆)⌊s⌋whi
C v(x)−v(y) (−∆)⌊s⌋w (x)−(−∆)⌊s⌋w (y)
n,s−⌊s⌋ h h
= dxdy.
2 ZZ (cid:0) (cid:1)(cid:0)|x−y|n+2(s−⌊s⌋) (cid:1)
Rn×Rn
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Since for large h the supports of v and w are separated, we have
h
v(x) (−∆)⌊s⌋w (y)
s h
h(−∆Rn) v,whi = −Cn,s−⌊s⌋ZZ |x−y|n+2(s−⌊s⌋) dydx.
Rn×Rn
Here we can integrate by parts. Using (1) one computes for a > 0
C C (n+2a)(2a+2) C
n,a n,a n,a+1
∆ = = −
|x−y|n+2a |x−y|n+2a+2 |x−y|n+2(a+1)
and obtains (6) with w instead of w.
h
Since the supports of v and w are separated, it is easy to pass to the limit as
h → ∞ and to conclude the proof. (cid:3)
Remark 1 Motivated by (6) and (2), A.I. Nazarov conjectured in [5] that
s s
h(−∆Rn) |u|,|u|i−h(−∆Rn) u,ui < 0 if ⌊s⌋ is even;
s s
h(−∆Rn) |u|,|u|i−h(−∆Rn) u,ui > 0 if ⌊s⌋ is odd
for any not integer exponent s > 0 and for any changing sign function u ∈ Hs(Rn)
such that u± ∈ Hs(Rn).
Lemma 1 Let s ∈ (1, 3) and ε > 0. If a function u ∈ Hs(Rn) has compact support
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then (u−ε)+ ∈ Hs(Rn), and
h(−∆Rn)s(u−ε)+,(u−ε)+i ≤ c(n,s)h(−∆Rn)su,ui+c(n,s,supp(u))ε2.
Proof. Take a nonnegative function η ∈ C∞(Rn) such that η ≡ 1 on supp(u).
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Clearly u − εη ∈ Hs(Rn). Hence, by Item 3 in the Introduction we have that
(u−εη)+ = (u−ε)+ ∈ Hs(Rn) and
h(−∆Rn)s(u−ε)+,(u−ε)+i ≤ c(n,s)h(−∆Rn)s(u−εη),u−εηi
≤ c(n,s) h(−∆Rn)su,ui+ε2h(−∆Rn)sη,ηi .
(cid:0) (cid:1)
(cid:3)
The proof is complete.
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In order to simplify notation, for u : Rn → R and s > 0 we put
u+(x)u−(y)
Φs(x,y) = .
u |x−y|n+2s
Lemma 2 Let s ∈ (1, 3) and u ∈ Hs(Rn)∩C0(Rn). Then (4) holds, and in particular
2 0
Φs ∈ L1(Rn ×Rn).
u
Proof. Thanks to Lemma 1 we have that (u−−ε)+ ∈ Hs(Rn)∩C0(Rn) for anyε > 0.
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Next, the supports of the functions u+ and (u−−ε)+ are compact and disjoint. Thus
we can apply Theorem 2 to get
u+(x)(u(y)− −ε)+
h(−∆Rn)su+,(u− −ε)+i = −Cn,sZZ |x−y|n+2s dxdy. (7)
Rn×Rn
Take a decreasing sequence ε ց 0. From Lemma 1 we infer that (u− − ε)+ → u−
weakly in Hs(Rn), as (u− −ε)+ → u− in L2(Rn). Hence the duality product in (7)
converges to the the duality product in (4). Next, the integrand in the right-hand
side of (7) increases to Φs a.e. on Rn ×Rn. By the monotone convergence theorem
u
(cid:3)
we get the convergence of the integrals, and the conclusion follows immediately.
Lemma 3 Let s ∈ (1, 3) and u ∈ Hs(Rn). Then Φs ∈ L1(Rn ×Rn).
2 u
Proof. Take a sequence of functions u ∈ C∞(Rn) such that u → u in Hs(Rn) and
h 0 h
almost everywhere. Since Φs → Φs a.e. on Rn×Rn, Fatou’s Lemma, Lemma 2 for
uh u
u and the boundeness of v 7→ v± in Hs(Rn) give
h
ZZ Φsu(x,y)dxdy ≤ lihm→i∞nf ZZ Φsuh(x,y)dxdy = c(n,s)lihm→i∞nfh(−∆Rn)su+h,u−hi
Rn×Rn Rn×Rn
s s
≤ c(n,s) limh(−∆Rn) uh,uhi = c(n,s)h(−∆Rn) u,ui,
h→∞
(cid:3)
that concludes the proof.
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Proof of Theorem 1. Take a sequence u ∈ C∞(Rn) such that u → u in Hs(Rn)
h 0 h
and almost everywhere. Consider the nonnegative functions
v := u+ ∧u+ = u+ −(u+ −u+)+ , w := u− ∧u− = u− −(u− −u−)+.
h h h h h h
Then v ,w ∈ Hs(Rn). Next, take any exponent τ ∈ (1,s). By Proposition 1 we
h h
have that u± −u± → 0 in Hτ(Rn); hence (u± −u±)+ → 0 in Hτ(Rn) by Item 3 in
h h
the Introduction. Thus,
v → u+ , w → u− in Hτ(Rn) and almost everywhere, as h → ∞. (8)
h h
Now we take a small ε > 0. Recall that (v − ε)+ ∈ Hτ(Rn) by Lemma 1.
h
Moreover, from 0 ≤ v ≤ u+, 0 ≤ w ≤ u− it follows that
h h h h
supp((v −ε)+) ⊆ {u ≥ ε}; supp(w ) ⊆ supp(u−).
h h h h
In particular, the functions (v −ε)+,w have compact and disjoint supports. Thus
h h
we can apply Theorem 2 to infer
(v (x)−ε)+w (y)
h(−∆Rn)τ(vh −ε)+,whi = −Cn,τ ZZ h |x−y|n+2τh dxdy.
Rn×Rn
We first take the limit as ε ց 0. The argument in the proof of Lemma 2 gives
v (x)w (y)
τ h h
h(−∆Rn) vh,whi = −Cn,τ ZZ |x−y|n+2τ dxdy. (9)
Rn×Rn
Next we push h → ∞. By (8) we get
limh(−∆Rn)τvh,whi = h(−∆Rn)τu+,u−i.
h→∞
Further, since the integrand in the right-hand side of (9) does not exceed Φτ(x,y),
u
Lemma 3, (8) and Lebesgue’s theorem give
v (x)w (y)
lim h h dxdy = Φτ(x,y)dxdy.
h→∞ ZZ |x−y|n+2τ ZZ u
Rn×Rn Rn×Rn
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Thus, we proved (4) with s replaced by τ. It remains to pass to the limit as τ ր s.
By Lebesgue’s theorem, we have
limh(−∆Rn)τu+,u−i = lim |ξ|2τF[u+]F[u−]dξ
τրs τրs Z
Rn
= |ξ|2sF[u+]F[u−]dξ = h(−∆Rn)su+,u−i.
Z
Rn
Now we fix τ ∈ (1,s) and notice that 0 ≤ Φτ ≤ max{Φτ0,Φs} for any τ ∈ (τ ,s).
0 u u u 0
Therefore, Lemma 3 and Lebesgue’s theorem give
lim Φτ(x,y)dxdy = Φs(x,y)dxdy.
τրs ZZ u ZZ u
Rn×Rn Rn×Rn
The proof of (4) is complete. The last statement follows immediately from (4),
(cid:3)
polarization identity and (2).
Example 1 It is easy to construct a function u ∈ C∞(Rn) such that u+ ∈ Hs(Rn)
0
if and only if s < 3.
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Take ϕ ∈ C∞(R) satisfying ϕ(0) = 0,ϕ′(0) > 0 and xϕ(x) ≥ 0 on R. By direct
0
computation one checks that ϕ+ = χ ϕ ∈ Hs(R) if and only if s < 3. If n = 1
(0,∞) 2
we are done. If n ≥ 2 we take u(x ,x ,...,x ) = ϕ(x )ϕ(x )...ϕ(x ).
1 2 n 1 2 n
Acknowledgements. ThefirstauthorwishestothankUniversit´eLibredeBruxelles
for the hospitality in February 2016. She is grateful to Denis Bonheure, Nicola
Abatangelo, Sven Jahros and Albero Saldan˜a for valuable discussion on this subject.
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