University of Connecticut
Mathematics
Fractional Laplacians and Integro-Differential Equations
Mingfeng Zhao
1
Email: mingfeng.zhao@uconn.edu
May 28, 2012
1
Department of Mathematics, University of Connecticut, 196 Auditorium Road, Unit 3009, Storrs, CT 06269-3009
1
Contents
1.
Introduction to the Fractional Laplacians
1
1.1.
Basics on Fourier Analysis
1
1.2.
Definitions of Fractional Laplacians
6
1.3.
Caffarelli-Silvestre’s s-Harmonic Extension
37
1.4.
Poisson Kernels of the Balls for Fractional Laplacians
42
2.
Regularity for Fully Nonlinear Integro-Differential Equations
51
2.1.
Fully Nonlinear Integro-Differential Equations and Viscosity Solutions
51
2.2.
Nonlocal ABP Estimates and Harnack Inequality
64
References
66
1. Introduction to the Fractional Laplacians
1.1. Basics on Fourier Analysis.
Definition 1.1.1. For any f ∈ L1 (Rn ), then
a. The Fourier transform fˆ of f is defined as:
Z
fˆ(x) =
e−2πix·y f (y) dy,
for all x ∈ Rn .
Rn
b. The inverse Fourier transform fˇ of f is defined as:
Z
fˇ(x) =
e2πix·y f (y) dy,
for all x ∈ Rn .
Rn
Proposition 1.1.1. The followings hold:
a. For any f ∈ L1 (Rn ), then fˆ, fˇ ∈ C0 (Rn ).
b. For any f ∈ L1 (Rn ) and any a ∈ Rn , let τa f (x) = f (x − a) for all x ∈ Rn , then τa f ∈ L1 (Rn ), and
−2πix·a ˆ
τd
f (x),
a f (x) = e
for all x ∈ Rn .
c. For any f ∈ L1 (Rn ) and any δ > 0, let Dδ f (x) = f (δx) for all x ∈ Rn , then Dδ f ∈ L1 (Rn ), and
−n ˆ x
d
D
f
,
δ f (x) = δ
δ
for all x ∈ Rn .
d. For any f, g ∈ L1 (Rn ), then f ∗ g ∈ L1 (Rn ), and
f[
∗ g(x) = fˆ(x)ĝ(x),
for all x ∈ Rn .
2
e. For any f, g ∈ L1 (Rn ), then
Z
Z
fˆ(x)g(x) dx =
Rn
Z
f (x)ĝ(x) dx
fˇ(x)g(x) dx =
and
Rn
Z
Rn
f (x)ǧ(x) dx.
Rn
Proof. Exercise.
To study the Fourier transform, it is natural to introduce the Schwartz class S(Rn ) of rapidly decreasing
functions.
Definition 1.1.2. Let f ∈ C ∞ (Rn ), we say that f ∈ S(Rn ) if for all multi-index α and β, we have
sup xα Dβ f (x) < ∞.
x∈Rn
2
Example 1.1.1. Let f (x) = e−π|x| for all x ∈ Rn , then f ∈ S(Rn ) (Exercise) and
Z
2
fˆ(x) =
e−2πix·y e−π|y| dy
Rn
Z
Pn
2
2
j=1 [(yj +ixj ) +xj ]
e−π
=
dy
Rn
2
= e−π|x|
(1.1.1)
n Z
Y
j=1
Z
2
e−π(yj +ixj ) dyj .
R
2
e−π(y+ia) dy = 1.
Claim I: For all a ∈ R, then
R
2
Consider the entire function g(z) = e−πz for all z ∈ C = R2 . For any a ∈ R and fix, by Residue Theorem in
complex analysis, then for any R > 0, we have
Z R
Z a
Z
2
2
e−πy dy +
e−π(R+it) dy +
−R
Z
a
0
−R
2
e−π(y+ia) dy +
Z
R
0
a
2
e−π(R+it) dy, then
For
0
Z
a
0
2
e−π(R+it) dy ≤
=
≤
Z a
−π(R+it)2
|e
|
dy
0
Z a
2
2
e−π(R −t ) dy 0
Z a
−πR2
e
dy
0
=
2
e−πR |a|
→ 0,
Z
For
0
as R → ∞.
2
e−π(−R+it) dy, then
a
Z
a
0
e
−π(−R+it)2
dy ≤
Z
a
0
|e
−π(−R+it)2
2
e−π(−R+it) dy = 0.
dy|
3
a
Z
=
e
−π(R2 −t2 )
0
→ 0,
dy as R → ∞.
Hence, we obtain
Z
2
e−π(y+ia) dy
R
Z
=
R
R→∞
−R
R
Z
=
2
e−πy dy
lim
R→∞
−R
Z
2
=
2
e−π(y+ia) dy
lim
e−πy dy
By Dominated Convergence Theorem
R
=
1 (Exercise).
2
By Claim I and (1.1.1), then fˆ(x) = e−π|x| for all x ∈ Rn .
n
n
n
Definition 1.1.3. Let {fk }∞
k=1 ⊂ S(R ) and f ∈ S(R ), we say that fk → f in S(R ) as k → ∞ if for all
multi-index α and β, we have
sup xα Dβ (fk − f )(x) → 0,
as k → ∞.
x∈Rn
Proposition 1.1.2. The followings hold:
a. Cc∞ (Rn ) ,→ S(Rn ) ,→ C0 (Rn )
T
Lp (Rn ) for all 1 ≤ p < ∞.
b. For any f ∈ S(Rn ), any multi-index α and β, then xα Dβ f (x) ∈ S(Rn ).
c. For any f ∈ S(Rn ) and any s ∈ Rn , then [1 + |x|2 ]s f (x) ∈ S(Rn ).
d. For any f ∈ S(Rn ), then fˆ, fˇ ∈ S(Rn ).
e. For any f ∈ S(Rn ) and any multi-index α, then
α f (x) = (−2πi)|α| xα fˆ(x),
d
D
for all x ∈ Rn .
Proof. Exercise.
Lemma 1.1.1. For any φ ∈ L1 (Rn ) with
Z
Rn
φ(x) dx = 1 and > 0, let φ (x) = −n φ
x
for all x ∈ Rn .
Then
a. If f ∈ Lp (Rn ) for some 1 ≤ p < ∞, then f ∗ φ → f in Lp (Rn ), as & 0.
T
T
b. If f ∈ L∞ (Rn ) C(Rn ), then for any > 0, f ∗ φ ∈ L∞ (Rn ) C(Rn ), and f ∗ φ (x) → f (x) for all
x ∈ Rn , as & 0.
Proof. Exercise.
4
ˇ
Theorem 1.1.1 (Inversion Formula on S(Rn )). For any f ∈ S(Rn ), then fˆ(x) = fˆˇ(x) = f (x) for all x ∈ Rn .
Proof. For any f ∈ S(Rn ), any x ∈ Rn and fix, by Dominated Convergence Theorem (Exercise), we know that
ˇ
fˆ(x)
Z
2
lim e2πix·y e−δπ|y| fˆ(y) dy
=
δ&0
Rn
Z
=
2
e2πix·y e−δπ|y| fˆ(y) dy.
lim
δ&0
Rn
By Fubini-Tonelli’s Theorem (Exercise), then
Z
e
2πix·y −δπ|y|2
e
fˆ(y) dy
Z
=
e
Rn
2πix·y −δπ|y|2
e
e
Rn
−2πiy·z
f (z) dz
dy
Rn
Z
Z
=
2
e−2πi(z−x)·y e−δπ|y| dy
f (z)
dz.
Rn
R
Z
Z
2
e−2πi(z−x)·y e−δπ|y| dy, then
For
Rn
Z
2
e−2πi(z−x)·y e−δπ|y| dy
Z
e−2πi(z−x)·(δ
=
Rn
−1
2
w) −π|w|2 − n
2
e
δ
√
dw
Let w =
δy
Rn
=
n
δ− 2
Z
e−2πiδ
−1
2
(z−x)·w −π|w|2
e
dw
Rn
=
1
n
−π|·|2 (δ − 2 (z − x))
δ − 2 e\
=
δ − 2 e−π|δ
n
=
n
δ− 2 e
−π
−1
2
(z−x)|2
z−x
1
δ2
By Example 1.1.1
2
.
2
Let P (x) = e−π|z| for all z ∈ Rn , then
Z
2
e−2πi(z−x)·y e−δπ|y| dy = P
Rn
1
δ2
(z − x) = P
1
δ2
(x − z),
for all z ∈ Rn .
Hence we have
Z
Z
2
e2πix·y e−δπ|y| fˆ(y) dy
=
Rn
f (z)P
Rn
= f ∗P
Z
P (z) dz = 1, by Lemma 1.1.1, then f ∗ P
Since
Rn
1
δ2
1
δ2
1
δ2
(x − z) dz
(x).
(x) → f (x) for all x ∈ Rn as δ & 0. Therefore, we get
ˇ
fˆ(x) = f (x) for all x ∈ Rn .
By the same argument, we can show that fˆˇ(x) = f (x) for all x ∈ Rn .
ˇ
Corollary 1.1.1 (Inversion Formula on L1 (Rn )). Let f, fˆ ∈ L1 (Rn ), then fˆ(x) = fˆˇ(x) = f (x) for a.e. x ∈ Rn .
5
Proof. For any g ∈ S(Rn ), by Proposition 1.1.2, then g ∈ L1 (Rn ), and ĝ ∈ L1 (Rn ). Since f, fˆ ∈ L1 (Rn ), then
Z
Z
ˇ
fˆ(x)g(x) dx =
fˆ(x)ǧ(x) dx By Proposition 1.1.2
Rn
Rn
Z
=
f (x)ǧˆ(x) dx
By Proposition 1.1.2
f (x)g(x) dx
By Theorem 1.1.1.
Rn
Z
=
Rn
ˇ
Since Cc∞ (Rn ) ⊂ S(Rn ) is dense in L1 (Rn ), then fˆ(x) = f (x) for a.e x ∈ Rn .
Since fˆ ∈ L1 (Rn ), then fˇ(x) = fˆ(−x) ∈ L1 (Rn ). By the same argument as above, we can show that
ˇ(x) = f (x) for a.e. x ∈ Rn .
fˆ
Corollary 1.1.2 (Plancherel’s Identity). There exists a unique bounded linear operator F : L2 (Rn ) → L2 (Rn )
such that
a. For all f ∈ L2 (Rn ), then kf kL2 (Rn ) = kF(f )kL2 (Rn ) .
b. For all f ∈ S(Rn ), then F(f )(x) = fˆ(x) in L2 (Rn ).
Proof. Exercise.
Definition 1.1.4. Let l : S(Rn ) → C be a linear functional, we say that l is a tempered distribution on Rn ,
that is, l ∈ S 0 (Rn ) if l is continuous in the sense that for any fk → f in S(Rn ) as k → ∞, then l(fk ) → l(f ) as
k → ∞.
Remark 1.1.1. Any tempered distribution on Rn is a distribution on Rn , in fact, S 0 (Rn ) ( D0 (Rn ) (Exercise).
p
n
Z
Example 1.1.2. For any 1 ≤ p ≤ ∞ and any f ∈ L (R ), define lf (g) =
then lf ∈ S 0 (Rn ) (Exercise).
f (x)g(x) dx for all g ∈ S(Rn ),
Rn
Example 1.1.3. For any s ∈ R, then |x|s ∈ S 0 (Rn ) if and only if s > − n2 (Exercise).
Lemma 1.1.2. For any l ∈ S 0 (Rn ), define ˆl and ˇl on S(Rn ) as: for any f ∈ S(Rn ), we have ˆl(f ) = l(fˆ) and
ˇl(f ) = l(fˇ). Then
a. ˆl, ˇl ∈ S 0 (Rn ).
b. Let g ∈ L1 (Rn ), then lbg = lĝ and lˇg = lǧ .
Proof. Exercise.
ˇ
Theorem 1.1.2 (Inversion Formula on S 0 (Rn )). For any l ∈ S 0 (Rn ), then ˆl = ˆˇl = l on S(Rn ).
6
Proof. In fact, for all f ∈ S(Rn ), then
ˇ
ˆl(f )
ˆ
ˇl(f )
= ˆl(fˇ)
=
l(fˆˇ)
=
l(f )
By Theorem 1.1.1
= ˇl(fˆ)
=
ˇ
l(fˆ)
=
l(f )
By Theorem 1.1.1.
Remark 1.1.2. References: Duoandikoetxea[11], Folland[14], Folland[15], Grafakos[17], Miao[26], Stein[38],
Stein and Weiss[39], Taylor[40], Torchinsky[41], Vretblad[42], etc.
1.2. Definitions of Fractional Laplacians.
For any f ∈ S(Rn ), by Proposition 1.1.2, then for any k ∈ N, we have
\
k f (x) = (2π|x|)2k fˆ(x),
(−∆)
(1.2.1)
for all x ∈ Rn .
In the view of (1.2.1), we can define the Fractional Laplacian via the Fourier transform.
Definition 1.2.1. For any s > − n2 and any f ∈ S(Rn ), define the Fractional Laplacian (−∆)s f of f as:
Z
s
e2πix·y (2π|y|)2s fˆ(y) dy,
(−∆) f (x) =
for all x ∈ Rn .
Rn
Remark 1.2.1. It is easy to verify that (2π|y|)2s fˆ(y) ∈ L1 (Rn ), (−∆)s f ∈ C0∞ (Rn )
T
L1 (Rn ) and for any
multi-index α, we have Dα (−∆)s f (x) = (−∆)s Dα f (x) for all x ∈ Rn (Exercise). By Corollary 1.1.1, then
\s f (x) = (2π|x|)2s fˆ(x) for all x ∈ Rn .
(−∆)
Lemma 1.2.1. For any 0 < s <
n
2,
then | \
· |2s−n (x) = γn (s)(2π|x|)−2s in S 0 (Rn ), that is, for all f ∈ S(Rn ),
then
Z
|x|
Rn
2s−n
fˆ(x) dx =
Z
γn (s)(2π|x|)−2s f (x) dx.
Rn
n
22s π 2 Γ(s)
.
Where γn (s) =
Γ n−2s
2
Remark 1.2.2. When n ≥ 3 and s = 1, then γn (1) = n(n − 2)αn (Exercise).
7
2
2
2
−π|·|2 (x) = e−π|x| for all x ∈ Rn . Now for any δ > 0, then e−πδ|x| ∈ S(Rn ).
Proof. Recall e−π|x| ∈ S(Rn ) and e\
By Proposition 1.1.1, then
−πδ|·|2 (x)
e\
n
= δ− 2 e
1 2
−π δ − 2 x
n
= δ − 2 e−
π|x|2
δ
for all x ∈ Rn .
,
By the Plancherel’s Theorem, Corollary 1.1.2, we know that the Fourier transform is an isometry of L2 (Rn ),
in particular, the Fourier transform is unitary on L2 (Rn ), that is
Z
Z
ϕ(x)ψ̂(x) dx =
ϕ̂(x)ψ(x) dx, for all ϕ, ψ ∈ L2 (Rn ).
Rn
Rn
Now for all f ∈ S(Rn ), then
Z
2
fˆ(x)e−πδ|x| dx
Z
−πδ|·|2 (x) dx
f (x)e\
=
Rn
Rn
=
δ
−n
2
Z
e−
π|x|2
δ
f (x) dx.
Rn
n−2s
Multiply both sides by δ 2 −1 and integrate over (0, ∞), by Fubini’s Theorem (Exercise), then
Z ∞
Z ∞
Z
Z
π|x|2
n−2s
−1 −πδ|x|2
−s−1 − δ
ˆ
2
f (x)
δ
e
dδ dx =
f (x)
δ
e
dδ dx.
Rn
Rn
0
Notices that for all x ∈ Rn \{0}, we have
Z ∞
n−2s
n−2s
2
n − 2s
−1
−πδ|x|
−
e
|x|2s−n
dδ = π 2 Γ
δ 2
2
0
Z
and
0
∞
e−
π|x|2
δ
δ −s−1 dδ = π −s Γ(s)|x|−2s (Exercise).
0
Hence, we have
Z
n−2s
fˆ(x)π − 2 Γ
Rn
n − 2s
2
|x|2s−n dx =
Z
f (x)π −s Γ(s) · |x|−2s dx.
Rn
That is,
Z
|x|2s−n fˆ(x) dx
=
Rn
=
Z
n−2s
π −s+ 2 Γ(s)
|x|−2s f (x) dx
n
Γ n−2s
R
2
Z
γn (s)(2π|x|)−2s f (x) dx.
Rn
Remark 1.2.3. For Lemma 1.2.1, f needs not be in S(Rn ), for example, Lemma 1.2.1 is still true, if assume
that f ∈ L1 (Rn ) such that there exists some A > 0 such that
|f (x)| + |fˆ(x)| ≤
A
,
1 + |x|n
for all x ∈ Rn .
8
Theorem 1.2.1 (Riesz’s Characterizations of Negative Fractional Laplacians). For any 0 < s <
n
2
and any
f ∈ S(Rn ), then
−s
(−∆)
1
f (x) =
γn (s)
Z
f (y)
dy,
|x − y|n−2s
Rn
for all x ∈ Rn .
1
|x|2s−n is the fundamental solution of (−∆)s in Rn , that is, (−∆)s γn1(s) | · |2s−n = δ.
γn (s)
Z
1
f (y)
dy for all x ∈ Rn , it is easy to verify that Is f ∈ C ∞ (Rn ). For any
Proof. Define Is f (x) =
γn (s) Rn |x − y|n−2s
x ∈ Rn and fix, let hx (y) = f (x − y) for all y ∈ Rn , since f ∈ S(Rn ), then h ∈ S(Rn ). By Lemma 1.2.1, then
Z
Z
|y|2s−n f (x − y) dy =
|y|2s−n hx (y) dy
In particular,
Rn
Rn
Z
=
cx (y) dy.
γn (s)(2π|y|)−2s h
Rn
cx (y), then
For h
Z
b
h(y)
e−2πiy·z f (x − z) dz
=
Rn
Z
e−2πiy·(x−w) f (w) dw
=
Let w = x − z
Rn
= e−2πix·y fˆ(−y).
Then
Z
|y|2s−n f (x − y) dy =
Z
Rn
γn (s)(2π|y|)−2s e−2πix·y fˆ(−y) dy,
for all x ∈ Rn .
Rn
Both sides are multiplied by ĝ and integrating over Rn , by Fubini’s Theorem (Exercise), then
Z
Z
Z
Z
f (x − y)
−2s ˆ
−2πix·y
dx = γn (s)
(2π|y|) f (−y)
e
ĝ(x) dx dy
ĝ(x)
n−2s
Rn
Rn
Rn
Rn |y|
Z
ˇ
= γn (s)
(2π|y|)−2s fˆ(−y)ĝ(−y)
dy
Rn
Z
(2π|y|)−2s fˆ(−y)g(−y) dy
= γn (s)
By Theorem 1.1.1
Rn
Z
(2π|y|)−2s fˆ(x)g(x) dy
= γn (s)
Let x = −y.
Rn
That is, we have
Z
Z
Is f (x)ĝ(x) dx =
Rn
(2π|x|)−2s fˆ(x)g(x) dx.
Rn
−2s ˆ
\s f (x) in S 0 (Rn ). By Theorem 1.1.2, then Is f (x) =
Which implies that Id
f (x) = (−∆)
s f (x) = (2π|x|)
−2s f ∈ L1 (Rn ), and (−∆)−s f ∈ C ∞ (Rn ), by Corollary 1.1.1,
\
(−∆)s f (x) in S 0 (Rn ). Since Is f ∈ C ∞ (Rn ), (−∆)
then
(−∆)
−s
1
f (x) =
γn (s)
Z
Rn
f (y)
dy,
|x − y|n−2s
for all x ∈ Rn (Exercise).
9
n
, then there exists some constant
Lemma 1.2.2 (Hedberg’s inequality). Let 0 < α < n and 1 ≤ p < α
Z
f
(y)
A(n, α, p) > 0 such that for all f ∈ Lp (Rn ), define Iα f (x) =
dy for all x ∈ R, we have
n−α
Rn |x − y|
αp
|Iα f (x)| ≤ Akf kLnp (Rn ) (M f (x))1−
αp
n
for all x ∈ Rn .
,
Where M is the Hardy-Littlewood maximal operator.
Proof. For any r > 0, for any x ∈ Rn , we have
Z
|Iα f (x)|
≤
Rn
|f (y)|
dy
|x − y|n−α
Z
=
|x−y|≤r
=
|f (y)|
dy +
|x − y|n−α
Z
|x−y|>r
|f (y)|
dy
|x − y|n−α
I + II.
For I, we have
I
=
∞ Z
X
k=0
≤
∞
X
2−k−1 r<|x−y|≤2−k r
(2−k−1 r)n−α
∞
X
Z
|f (y)|dy
2−k−1 r<|x−y|≤2−k r
k=0
≤
2(k+1)(n−α) rα−n
Z
=
k=0
≤
∞
X
|f (y)|dy
|x−y|≤2−k r
k=0
∞
X
|f (y)|
dy
|x − y|n−α
αn 2−kn rn
2(k+1)(n−α) rα−n
αn 2−kn rn
Z
2(k+1)(n−α) rα−n αn 2−kn rn M f (x)
k=0
= αn rα
∞
X
2(k+1)(n−α)−kn M f (x)
k=0
= αn rα
∞
X
2−kα+n−α M f (x)
k=0
= αn rα
=
|f (y)|dy
|x−y|≤2−k r
2n−α
M f (x)
1 − 2−α
αn 2n−α α
r M f (x)
(1 − 2−α )
10
For II, when p = 1, we have
Z
II
=
|x−y|>r
rα−n
≤
|f (y)|
dy
|x − y|n−α
Z
|f (y)|dy
|x−y|>r
≤
r
α−n
Z
|f (y)|dy
Rn
rα−n kf kL1 (Rn )
=
n
α,
For II, when 1 < p <
Z
II =
we can get
|x−y|>r
Z
|f (y)|
dy
|x − y|n−α
! p1
Z
|f (y)|p dy
≤
|x−y|>r
|x−y|>r
By Holder’s inequality,
Z
≤
p1
p
Z
|f (y)| dy
Rn
|x−y|>r
Z
|z|q(n−α)
|z|>r
Z
= kf kLp (Rn ) nαn
=
1
q
∞
! q1
1 1
+ =1
p q
1
dy
|x − y|q(n−α)
! q1
! q1
1
= kf kLp (Rn )
1
dy
|x − y|q(n−α)
dz
Let z = x − y
q1
n−1
t
dt
q(n−α)
1
r
t
Z
∞
(nαn ) kf kLp (Rn )
t
n−1−q(n−α)
q1
dt
r
=
q1
1
n−q(n−α)
(nαn ) kf kLp (Rn )
r
q(n − α) − n
n−α
np − pα
pα − n
Since n − q(n − α) = n −
1 =n− p−1 = p−1 <0
1− p
1
q
1
=
n
(nαn ) q
−n+α
q1 kf kLp (Rn ) r q
n−pα
p−1
1
=
(nαn ) q
α− n
q1 kf kLp (Rn ) r p
n−pα
p−1
Hence for all 1 ≤ p <
n
α,
Since
n
1
n
−n+α=n 1−
−n+α=α−
q
p
p
there exists some constant A(n, α, p) > 0 such that
n
|Iα f (x)| ≤ A rα M f (x) + rα− p kf kLp (Rn ) .
11
n
Set rα M f (x) = rα− p kf kLp (Rn ) , then r =
kf kLp (Rn )
M f (x)
α
r M f (x)
=
np
, we can get
kf kLp (Rn )
M f (x)
np α
M f (x)
αp
αp
n
= kf kLnp (Rn ) (M f (x))1−
Therefore, we get
αp
|Iα f (x)| ≤ Akf kLnp (Rn ) (M f (x))1−
αp
n
.
n
2,
Theorem 1.2.2 (Hardy-Littlewood-Sobolev’s Inequality). Let 0 < s <
1<p<
n
2s
and
1
q
=
1
p
−
2s
n,
then
there exists some constant C(n, s, p) > 0 such that for all f ∈ Lp (Rn ), we have
k(−∆)−s f kLq (Rn ) ≤ Ckf kLp (Rn ) .
Proof. Since
1
q
=
1
p
−
2s
n
=
n−2sp
np ,
then
2sp
n − 2sp
q 1−
= p.
=q
n
n
By Lemma 1.2.2, then
k(−∆)−s f kLq (Rn )
p
2sp
≤ Akf kLnp (Rn ) kM f kLq p (Rn )
2sp
p
≤ Ckf kLnp (Rn ) kf kLq p (Rn )
Since M is (p, p)-type
= Ckf kLp (Rn ) .
Theorem 1.2.3. Let 0 < s < 1 and f ∈ S(Rn ), then
Z
f (x) − f (y)
s
(−∆) f (x) = Cn,s P.V.
dy,
n+2s
Rn |x − y|
Z
Remark 1.2.4. For any R > 0, it is easy to see that P.V.
Rn
(Exercise), then
Z
f (x) − f (y)
P.V.
dy
n+2s
n
R |x − y|
Z
=
P.V.
Rn
Z
=
P.V.
Rn
Z
=
P.V.
Rn
for all x ∈ Rn .
1BR (x) (y)∇f (x) · (x − y)
dy = 0 for all x ∈ Rn
|x − y|n+2s
f (x) − f (y)
dy − P.V.
|x − y|n+2s
Z
Rn
1BR (x) (y)∇f (x) · (x − y)
dy
|x − y|n+2s
f (x) − f (y) − 1BR (x) (y)∇f (x) · (x − y)
dy
|x − y|n+2s
f (x) − f (x − z) − 1BR (0) (z)∇f (x) · z
dy
|z|n+2s
Let z = x − y
12
On the other hand, we know that
Z
f (x) − f (y)
P.V.
dy
n+2s
n
R |x − y|
Z
=
P.V.
Rn
Z
=
P.V.
Rn
Hence, we know that
Z
f (x) − f (y)
dy
P.V.
n+2s
n
R |x − y|
Let z = x − y
f (x) − f (x − y)
dy
|y|n+2s
Let y = −z.
Z
Z
1
f (x) − f (x − y)
f (x) − f (x + y)
dy
+
P.V.
dy
P.V.
2
|y|n+2s
|y|n+2s
Rn
Rn
Z
1
f (x + y) + f (x − y) − 2f (x)
−
dy, for all x ∈ Rn (Exercise).
2 Rn
|y|n+2s
=
=
Proof. Approach I: For any max
f (x) − f (x + z)
dz
|z|n+2s
1 − n2 , 0 < s < 1, then − n2 < s − 1 < 0, and
s−1 (−∆)f (x)
(−∆)\
=
\ (x)
(2π|x|)2(s−1) (−∆)f
=
(2π|x|)2(s−1) (2π|x|)2 fˆ(x)
=
(2π|x|)2s fˆ(x)
in S 0 (Rn ).
Which implies that
(−∆)s f (x)
=
=
=
=
(−∆)s−1 (−∆)f (x) (Exercise)
Z
−∆f (y)
1
dy By Theorem 1.2.1
γn (1 − s) Rn |x − y|n−2(1−s)
Z
1
∆[f (x) − f (y)]
dy
γn (1 − s) Rn |x − y|n−2(1−s)
Z
1
∆[f (x) − f (y)]
lim lim
dy for all x ∈ Rn (Exercise).
γn (1 − s) →0 R→∞ <|x−y|<R |x − y|n−2(1−s)
∆[f (x) − f (y)]
dy, by Green’s Theorem, then
n−2(1−s)
<|x−y|<R |x − y|
Z
∆[f (x) − f (y)]
dy
|x
− y|n−2(1−s)
<|x−y|<R
Z
1
=
[f (x) − f (y)]∆
dy
n−2(1−s)
|x − y|
<|x−y|<R
!
Z
1
∂ |x−y|n−2(1−s)
1
∂[f (x) − f (y)]
−
− [f (x) − f (y)]
dS(y)
∂n
∂n
|x − y|n−2(1−s)
|x−y|=
!
Z
1
∂ |x−y|n−2(1−s)
1
∂[f (x) − f (y)]
+
− [f (x) − f (y)]
dS(y)
∂n
∂n
|x − y|n−2(1−s)
|x−y|=R
Z
For
=
I,R + II + IIIR .
13
For ∆
1
, then
|x − y|n−2(1−s)
∆
1
|x −
y|n−2(1−s)
= 2s[n − 2(1 − s)]
1
(Exercise).
|x − y|n+2s
By Dominated Convergence Theorem (Exercise), we have
Z
lim I,R = 2s[n − 2(1 − s)]
R→∞
For
1
∂ |x−y|n−2(1−s)
∂n
<|x−y|
f (x) − f (y)
dy.
|x − y|n+2s
, then
1
∂ |x−y|n−2(1−s)
∂n
=
[−n + 2(1 − s)]
(Exercise).
|x − y|n−1+2s
Since f ∈ S(Rn ), then there exists some constant A > 0 such that
|y||∇f (y)| ≤ A,
for all |y| ≥ 1.
Hence, for all R > |x| + 1, then we know that
Z
|III,R |
≤
|x−y|=R
≤
≤
=
Z
1
A
Rn−2+2s
A
Rn−2+2s
Rn−2(1−s)
Z
|x−y|=R
Z
|x−y|=R
|∇f (y)| dS(y) +
2kf kL∞ (Rn )
|x−y|=R
|n − 2(1 − s)|
dS(y)
Rn−1+2s
1
nαn
dS(y) + 2[n + 2(1 − s)]kf kL∞ (Rn ) 2s
|y|
R
1
nαn
dS(y) + 2[n + 2(1 − s)]kf kL∞ (Rn ) 2s
R − |x|
R
Anαn
nαn
1
+ 2[n + 2(1 − s)]kf kL∞ (Rn ) 2s
R2s−1 R − |x|
R
→ 0,
as R → ∞.
On the other hand, we know that
Z
∂[f (x) − f (y)]
1
dS(y)
|x−y|= |x − y|n−2(1−s)
∂n
Z
∂f (y)
dS(y)
= 2(1−s)−n |x−y|= ∂n
Z
= 2(1−s)−n ∆f (y) dy By Divergence Theorem
|x−y|<
≤
αn k∆f kL∞ (Rn ) 2(1−s)−n+n
=
αn k∆f kL∞ (Rn ) 2(1−s)
→ 0,
as → 0.
14
And also we have
Z
Z
1
∂ |x−y|n−2(1−s)
dS(y) = [−n + 2(1 − s)]1−2s−n
[f (x) − f (y)]
[f (x) − f (y)] dS(y).
∂n
|x−y|=
|x−y|=
Notice that
Z
1−2s−n
∇f (y) · (y − x) dS(y)
|x−y|=
Z
y−x
2−2s−n
∇f (y) ·
dS(y)
|x−y|=
Z
∂f (y)
2−2s−n dS(y)
|x−y|= ∂n
Z
2−2s−n ∆f (y) dy By Divergence Theorem
|x−y|<
=
=
=
≤
αn k∆f kL∞ (Rn ) 2−2s−n+n
=
αn k∆f kL∞ (Rn ) 2(1−s)
→ 0,
as → 0.
Hence, we know that
Z
1−2s−n
[f (x) − f (y)] dS(y)
|x−y|=
Z
Z
1−2s−n
1−2s−n
∇f (y) · (y − x) dS(y) .
[f (x) − f (y) − ∇f (y) · (x − y)] dS(y) + ≤ |x−y|=
|x−y|=
At the mean time, we know that
Z
1−2s−n
[f (x) − f (y) − ∇f (y) · (x − y)] dS(y)
|x−y|=
Z
Z 1Z t
1−2s−n
= (x − y)T D2 f (y + r(x − y))(x − y)drdt dS(y)
|x−y|= 0
0
≤
nαn kD2 f kL∞ (Rn ) 1−2s−n+2+n−1
=
nαn kD2 f kL∞ (Rn ) 2(1−s)
→ 0,
(Exercise)
as → 0.
Therefore, we can conclude that
(−∆)s f (x)
Z
2s[n − 2(1 − s)]
f (x) − f (y)
P.V.
dy
n+2s
γn (1 − s)
Rn |x − y|
Z
f (x) − f (y)
= Cn,s P.V.
dy, for all x ∈ Rn (Exercise).
n+2s
n
R |x − y|
=
15
f (x + y) + f (x − y) − 2f (x)
dy for all x ∈ Rn , then g ∈ L1 (Rn )
|y|n+2s
Z
Approach II: For all 0 < s < 1, let g(x) =
Rn
(Exercise), and
Z
ĝ(x)
=
e
−2πix·z
Rn
Z
=
Rn
1
|y|n+2s
f (z + y) + f (z − y) − 2f (z)
dy dz
|y|n+2s
Z
e−2πix·z [f (z + y) + f (z − y) − 2f (z)] dz dy
By Fubini’s Theorem (Exercise)
Rn
e2πix·y + e−2πix·y − 2]fˆ(x)
dy
|y|n+2s
Rn
Z
|e2πix·y − 1|2
ˆ
−f (x)
dy.
|y|n+2s
Rn
Z
=
=
By Theorem 1.1.1 (Exercise)
It is easy to see that ĝ(0) = 0, and for x 6= 0, we have
ĝ(x)
x
|ei |x| ·z − 1|2
dz
|z|n+2s
Z
= −(2π|x|) fˆ(x)
2s
Rn
Z
It is easy to verify that
Rn
Let z = 2π|x|y (Exercise).
x
|ei |x| ·z − 1|2
dz = An,s is a constant function on Rn \{0} (Exercise), which implies
|z|n+2s
that
g(x) = −An,s (−∆)s f (x),
for all x ∈ Rn .
By Remark 1.2.4 and Approach I, then
−
1
Cn,s
,
=−
An,s
2
for all max
n
n o
1 − , 0 < s < 1.
2
By Complex Unique Analytic Continuation Theorem, then
−
1
Cn,s
,
=−
An,s
2
for all 0 < s < 1.
Which implies that
(−∆)s f (x) = Cn,s P.V.
Z
Rn
f (x) − f (y)
dy,
|x − y|n+2s
for all 0 < s < 1.
Remark 1.2.5. In a summary, we know that for any R > 0 and all f ∈ S(Rn ), then for all x ∈ Rn , we have
(1.2.2)
s
(−∆) f (x)
Z
= Cn,s P.V.
Rn
(1.2.3)
(1.2.4)
= −
Cn,s
2
Z
Z
= Cn,s
Rn
Rn
f (x) − f (y)
dy
|x − y|n+2s
f (x + y) + f (x − y) − 2f (x)
dy
|y|n+2s
f (x) − f (y) − 1BR (x) (y)∇f (y) · (x − y)
dy.
|x − y|n+2s
16
For (1.2.2), the principle value is needed whenever
1
2
≤ s < 1, and the principle value can be ignored whenever
0 < s < 12 . For (1.2.3) and (1.2.4), there is no principle value.
Theorem 1.2.4 (Classical Solution). Let 0 < s < 1, 0 < α ≤ 1, f ∈ L∞ (Rn ) and x ∈ Rn , then
a. If 0 < s < 21 , 2s < α ≤ 1 and f is C 0,α (BR (x)) for some R > 0, then (−∆)s f (x) is well defined, and
there is no principal value.
b. If
1
2
≤ s < 1, α > 2s − 1 and f is C 1,α (BR (x)) for some R > 0, then (−∆)s f (x) is well defined.
Proof. a. By (1.2.2), then
|(−∆)s f (x)|
Z
1
|f (x) − f (y)|
∞
n
dy
+
2C
kf
k
dy
n,s
L (R )
n+2s
n+2s
|y−x|>R |x − y|
|y−x|<R |x − y|
Z
Z ∞
|x − y|α
≤ Cn,s [f ]C α (BR (x))
dy + 2Cn,s nαn kf kL∞ (Rn )
r−n−2s+n−1 dr
n+2s
|y−x|<R |x − y|
R
Z R
Z ∞
α
= Cn,s nαn [f ]C (BR (x))
rα−n−2s+n−1 dr + 2Cn,s nαn kf kL∞ (Rn )
r−2s−1 dr
Z
≤ Cn,s
0
R
R
Z
rα−2s−1 dr +
= Cn,s nαn [f ]C α (BR (x))
0
=
2Cn,s nαn
kf kL∞ (Rn ) R−2s
2s
Cn,s nαn
Cn,s nαn
[f ]C α (BR (x)) Rα−2s +
kf kL∞ (Rn ) R−2s
α − 2s
s
Since α > 2s
≤ Cn,s,R [kf kC 0,α (BR (x)) + kf kL∞ (Rn ) ]
< ∞.
b. By (1.2.4), then
s
|(−∆) f (x)|
Z
≤ Cn,s
|y−x|<R
|f (x) − f (y) − ∇f (y) · (x − y)|
dy + 2Cn,s kf kL∞ (Rn )
|x − y|n+2s
By the proof of the part a, we know that
Z
2Cn,s kf kL∞ (Rn )
|y−x|>R
|f (x) − f (y) − ∇f (y) · (x − y)|
Z 1
= [∇f (y + t(x − y)) − ∇f (y)] · (x − y) dt
0
Z 1
≤ |x − y|
|∇f (y + t(x − y)) − ∇f (y)| dt
0
≤
|x − y|
|y−x|>R
Cn,s nαn
1
dy ≤
kf kL∞ (Rn ) R−2s .
n+2s
|x − y|
s
For |f (x) − f (y) − ∇f (y) · (x − y)|, then
Z
Z
1
[∇f ]C α (BR (x)) (t|x − y|)α dt
0
≤ [∇f ]C α (BR (x)) |x − y|1+α .
1
dy.
|x − y|n+2s
17
Hence, we can obtain
Z
|f (x) − f (y) − ∇f (y) · (x − y)|
dy
|x − y|n+2s
|y−x|<R
Z
≤ [∇f ]C α (BR (x))
|x−y|<R
Z
|x − y|1+α
dy
|x − y|n+2s
R
r1+α−n−2s+n−1 dr
= nαn [∇f ]C α (BR (x))
0
Z
R
= nαn [∇f ]C α (BR (x))
rα+1−2s−1 dr
0
nαn
[∇f ]C α (BR (x)) Rα+1−2s
α + 1 − 2s
=
Since α + 1 > 2s.
Therefore, we know that
|(−∆)s f (x)|
Cn,s nαn
Cn,s nαn
[∇f ]C α (BR (x)) Rα+1−2s +
kf kL∞ (Rn ) R−2s
α + 1 − 2s
s
≤
≤ Cn,s,α,R [kf kC 1,α (BR (x)) + kf kL∞ (Rn ) ]
< ∞.
Remark 1.2.6. The fractional Laplaicans also can be defined for some unbounded functions, for example, let
1
for all x ∈ Rn \{0}, then for any > 0, we know that
f (x) =
n−2s
|x|
Z
f (x) − f (y)
dy
n+2s
|x−y|> |x − y|
Z
Z
Z
f (x) − f (y)
f (x) − f (y)
f (x) − f (y)
=
dy +
dy +
dy
n+2s
n+2s
n+2s
|x|
|x|
|x − y|
|x−y|≥2|x| |x − y|
2|x|>|x−y|≥ 2 |x − y|
2 >|x−y|≥
= I + II + III.
For I , since f is smooth in the region of
n
y:
|x|
2
o
> |x − y| > 0 , by the proof of Theorem 1.2.3, then
Z
lim I = P.V.
&0
|x|
2 >|x−y|
f (x) − f (y)
dy.
|x − y|n+2s
For II , since f is bounded in the region of {y : |x − y| ≥ 2|x|}, then II ∈ C. Now it suffices to show that
III ∈ C. In fact, we know that
Z
III
= f (x)
|x|
2|x|>|x−y|≥ 2
1
dy −
|x − y|n+2s
Z
|x|
2|x|>|x−y|≥ 2
f (y)
dy
|x − y|n+2s
= III1 + III2
Since f is smooth in the region of
n
y : 2|x| > |x − y| ≥
|x|
2
Z
III2
=
2|x|>|x−y|≥
|x|
2
|x
o
, then III1 ∈ C. For III2 , we have
1
|y|n−2s
− y|n+2s
dy
18
≤
=
2
|x|
n+2s Z
2
|x|
n+2s
2|x|>|x−y|≥
1
dy
|y|n−2s
|x|
Z
nαn
|x|
2
|x|
2
r2s−n+n−1 dr
=
"
2s # n+2s
nαn
|x|
2
2s
(2|x|) −
2s
2
|x|
<
∞.
f (x) − f (y)
dy exists, that is, (−∆)s f (x) is well defined for all x ∈
n+2s
|x−y|> |x − y|
Rn \{0}. Furthermore, notice that Theorem 1.2.1 says that (−∆)s γn1(s) | · |2s−n = δ, also by direct computation,
Z
Therefore, we know that P.V.
one can show that (−∆)s f (x) = 0 for all x ∈ Rn \{0} (Exercise).
Definition 1.2.2. Let β > 0 such that β ∈
/ N, and Ω be an open subset of Rn , we say that u ∈ C β (Ω) if
u ∈ C [β],{β} (Ω).
Remark 1.2.7. Using Definition 1.2.2 and Theorem 1.2.3, we know that if u ∈ C 2s+α (BR (x)) for some
0 < α < 1, then (−∆)s f (x) is well defined. Moreover, there exists some constant C(n, s, α, R) > 0 such
that
|(−∆)s f (x)| ≤ C[kf kC 2s+α (BR (x)) + kf kL∞ (Rn ) ].
Remark 1.2.8. Recall Schauder’s Estimate for Laplacian: Let 0 < α < 1, Ω be a bounded smooth domain in
Rn , f ∈ C α (Ω), and u ∈ C 2 (Ω) be the solution of the PDE:

 ∆u(x) = f (x), in Ω
 u(x) = 0, on ∂Ω.
2,α
Then u ∈ Cloc
(Ω) and for any subdomain Ω0 ⊂ Ω, there exists some constant C(n, s, Ω, Ω0 ) > 0 such that
kukC 2,α (Ω0 ) ≤ Ckf kC α (Ω) .
It is natural to expect that if f ∈ C α (Rn ) and u be a classical solution of the PDE:
(−∆)s u(x) = f (x), in Rn .
Then u ∈ C 2s+α (Rn ), and there exists some constant C(n, s, α) > 0 such that
kukC 2s+α (Rn ) ≤ Ckf kC α (Rn ) .
19
Definition 1.2.3. Let Ω be a domain in Rn , f ∈ L∞ (Rn \Ω), we say that u is a classical solution of the follow
Dirichlet Problem:

 (−∆)s u(x) = 0, in Ω
 u(x) = f (x), in Rn \Ω.
(1.2.5)
2s+α
If u ∈ Cloc(Ω)
T
C(Ω)
T
L∞ (Rn ) for some 0 < α < 1.
Theorem 1.2.5 (Strong Maximum Principle). Let Ω be an open subset of Rn (not necessary connected),
T
b : Ω → Rn , c : Ω → R, and u ∈ L∞ (Rn ) C 1 (Ω) such that
(−∆)s u(x) + b(x) · ∇u(x) + c(x)u(x) ≤ 0,
for all x ∈ Ω.
Assume there exists some x0 ∈ Ω such that u(x0 ) = sup u(x), then
x∈Rn
0
a. If c(x) ≥ 0 for all x ∈ Ω and u(x ) ≥ 0, then u(x) ≡ u(x0 ) for all x ∈ Rn .
b. If u(x0 ) = 0, then u(x) ≡ 0 for all x ∈ Rn .
Proof. If assume u(x) 6≡ u(x0 ) in Rn . Since u(x0 ) = sup u(x) and u ∈ C 1 (Ω), then ∇u(x0 ) = 0, and
x∈Rn
s
0
(−∆) u(x )
Z
=
Cn,s P.V.
Rn
>
u(x0 ) − u(y)
dy
|x0 − y|n+2s
0.
Hence, we get
c(x0 )u(x0 ) < 0.
Which contradicts with the assumption c(x0 )u(x0 ) ≥ 0.
Remark 1.2.9. Because the fractional Laplacian is a nonlocal integral operator, then the global maximum
assumption is needed.
Corollary 1.2.1. Let Ω be a proper open subset of Rn , then the solution to the PDE (1.2.5) is unique.
Proof. Exercise.
Proposition 1.2.1. Let 0 < s < 1 and f ∈ C 2 (Rn ), then the followings hold:
a. For any a ∈ Rn , then
(−∆)s τa f (x) = (−∆)s f (x − a),
for all x ∈ Rn .
20
b. For any δ > 0, let g(x) = f (δx) for all x ∈ Rn , then
(−∆)s g(x) = δ 2s (−∆)s f (δx),
for all x ∈ Rn .
c. For any A ∈ O(n), let h(x) = f (Ax) for all x ∈ Rn , then
(−∆)s h(x) = (−∆)s f (Ax),
for all x ∈ Rn .
Proof. Exercise.
Theorem 1.2.6. For any f ∈ S(Rn ), then for any x ∈ Rn , we have
lim (−∆)s f (x) = f (x),
s&0
lim (−∆)s f (x) = −∆f (x).
and
s%1
Proof. Approach I: a. For any x ∈ Rn and fix, then for any R > 1, we know that
Z
f (x + y) + f (x − y) − 2f (x)
Cn,s
s
dy
(−∆) f (x) = −
2
|y|n+2s
n
R
Z
Z
Cn,s
f (x + y) + f (x − y) − 2f (x)
Cn,s
f (x + y) + f (x − y) − 2f (x)
= −
dy −
dy
n+2s
2
|y|
2
|y|n+2s
|y|≥R
|y|<R
= I + II.
For II, since
Z
|f (x + y) + f (x − y) − 2f (x)| = ≤
1
1
ty T · D2 f (x + rty) · y drdt (Exercise)
0
−1
Z 1Z 1
kD2 f kL∞ (Rn )
t|y|2 drdt
Z
−1
0
= |y|2 kD2 f kL∞ (Rn ) .
Hence, we get
Z
f (x + y) + f (x − y) − 2f (x) dy |y|<R
|y|n+2s
≤
kD2 f kL2 (Rn )
Z
|y|<R
2
Z
= nαn kD f kL2 (Rn )
|y|2
dy
|y|n+2s
R
r−n−2s+2+n−1 dr
0
=
nαn
kD2 f kL2 (Rn ) R2−2s .
2 − 2s
s22s−1 Γ n+2s
C(n, s)
2
Since −
=−
, then
n
2
π 2 Γ(1 − s)
s22s−1 Γ n+2s Z
f (x + y) + f (x − y) − 2f (x) 2
dy −
n
|y|n+2s
π 2 Γ(1 − s)
|y|<R
s22s−1 Γ n+2s
nαn
2
≤
kD2 f kL2 (Rn ) R2−2s
n
π 2 Γ(1 − s) 2 − 2s
21
→ 0,
as s & 0.
For I, since f ∈ S(Rn ), then lim
|y|→∞
f (y) = 0, and for any > 0, there exists some R0 > 0 such that
|f (x + y)| ≤ ,
for all R ≥ R0 .
Now we find a large R ≥ R0 ≥ 1 and fix such that
22s Γ n+2s
nαn
2
≤ 1,
n
2
π Γ(1 − s) 2
for all 0 ≤ s ≤
1
.
2
Hence, we know
C Z
f (x + y) + f (x − y) n,s
dy 2
|y|n+2s
|y|≥R
Z
|y|−n−2s dy
≤ Cn,s |y|≥R
Z
∞
= Cn,s nαn r−n−2s+n−1 dr
R
nαn −2s
R 2s
s22s Γ n+2s
nαn −2s
2
R n
π 2 Γ(1 − s) 2s
22s Γ n+2s
nαn −2s
2
R n
π 2 Γ(1 − s) 2
= Cn,s
=
=
≤ .
Also we know that
−
Cn,s
2
Z
|y|≥R
−2f (x)
dy
|y|n+2s
Z
=
|y|−n−2s dy
Cn,s f (x)
|y|≥R
Z
=
∞
r−n−2s+n−1 dr
Cn,s nαn f (x)
R
=
=
22s Γ n+2s
nαn −2s
2
R f (x)
n
π 2 Γ(1 − s) 2
n
22s Γ n+2s
π2
2
R−2s f (x)
n
π 2 Γ(1 − s) Γ n2
→ f (x),
as s & 0.
Hence, we get
lim (−∆)s f (x)
s&0
=
lim −
s&0
Cn,s
2
Z
|y|≥R
= f (x).
Therefore, we obtain (−∆)s f (x) → f (x) for all x ∈ Rn , as s & 0.
−2f (x)
dy
|y|n+2s
22
b. For any x ∈ Rn and fix, then
Z
Cn,s
f (x + y) + f (x − y) − 2f (x)
s
(−∆) f (x) = −
dy
2
|y|n+2s
n
R
Z
Z
Cn,s
Cn,s
f (x + y) + f (x − y) − 2f (x)
f (x + y) + f (x − y) − 2f (x)
dy
−
dy.
= −
n+2s
2
|y|
2
|y|n+2s
|y|≥1
|y|<1
Z
f (x + y) + f (x − y) − 2f (x)
For
dy, then
|y|n+2s
|y|≥1
Z
Z
f (x + y) + f (x − y) − 2f (x) |f (x + y) + f (x − y) − 2f (x)|
dy ≤
dy
n+2s
|y|≥1
|y|
|y|n+2s
|y|≥1
Z
≤ 4kf kL∞ (Rn )
|y|−n−2s dy
|y|≥1
Z
4kf kL∞ (Rn ) nαn
=
∞
r−n−2s+n−1 dr
1
2nαn
kf kL∞ (Rn ) .
s
=
s22s−1 Γ n+2s
C(n, s)
2
Since −
=−
, then
n
2
π 2 Γ(1 − s)
s22s−1 Γ n+2s Z
f
(x
+
y)
+
f
(x
−
y)
−
2f
(x)
2
dy
−
n
n+2s
2
|y|
π Γ(1 − s)
|y|≥1
s22s−1 Γ n+2s
2nαn
2
≤
kf kL∞ (Rn )
n
s
π 2 Γ(1 − s)
22s Γ n+2s
2
=
nαn kf kL∞ (Rn )
n
π 2 Γ(1 − s)
22s Γ n+2s
nαn
1−s
2
=
kf kL∞ (Rn )
n
(1 − s)Γ(1 − s)
π2
1 − s 22s Γ n+2s
2
=
kf kL∞ (Rn )
n
Γ(2 − s)
π2
→ 0,
as s % 1.
Which implies that
s22s−1 Γ n+2s
2
lim (−∆) f (x) = lim −
n
s%1
s%1
π 2 Γ(1 − s)
s
Z
|y|<1
f (x + y) + f (x − y) − 2f (x)
dy.
|y|n+2s
For |f (x + y) + f (x − y) − 2f (x)|, we know that
Z 1Z 1
f (x + y) + f (x − y) − 2f (x)
ty T · D2 f (x + rty) · y drdt (Exercise).
0
−1
Then we get
|f (x + y) + f (x − y) − 2f (x) − y T · D2 f (x) · y|
23
Z 1 Z 1
ty T
= −1
0
Z 1 Z 1
ty T
= −1
0
Z 1 Z 1
ty T
= · D f (x + rty) · y drdt − y · D f (x) · y Z
2
2
· [D f (x + rty) − D f (x)] · y drdt Since
2
T
2
1
1
Z
0
t drdt = 1
−1
1
dD2 f (x + wrty)
·
dw · y drdt
dw
0
−1
0
Z 1 Z 1
Z 1
T
2
ty ·
∇D f (x + wrty) · rty dw · y drdt
= −1
0
Z
1
Z
1
0
Z
≤
0
−1
Z
1
t|y||∇D2 f (x + wrty)||r|t|y||y| dwdrdt
0
Z
≤
kD3 f kL∞ (Rn ) |y|3
≤
kD3 f kL∞ (Rn ) |y|3 .
0
1
Z
1
−1
Z
1
t2 |r| dwdrdt
0
Hence, we obtain
Z
f (x + y) + f (x − y) − 2f (x) − y T · D2 f (x) · y
dy
|y|<1
|y|n+2s
≤
kD3 f kL∞ (Rn )
Z
|y|<1
3
Z
= kD f kL∞ (Rn ) nαn
|y|3
dy
|y|n+2s
1
r3−n−2s+n−1 dr
0
=
nαn
kD3 f kL∞ (Rn )
3 − 2s
≤ nαn kD3 f kL∞ (Rn ) .
Which implies that
s22s−1 Γ n+2s Z
f (x + y) + f (x − y) − 2f (x) − y T · D2 f (x) · y
2
dy
−
n
n+2s
|y|
π 2 Γ(1 − s)
|y|<1
s22s−1 Γ n+2s
2
≤
nαn kD3 f kL∞ (Rn )
n
π 2 Γ(1 − s)
nαn s22s−1 Γ n+2s
1
2
=
kD3 f kL∞ (Rn )
n
Γ(1 − s)
π2
nαn s22s−1 Γ n+2s
1−s
2
=
kD3 f kL∞ (Rn )
n
(1 − s)Γ(1 − s)
π2
1 − s nαn s22s−1 Γ n+2s
2
=
kD3 f kL∞ (Rn )
n
Γ(2 − s)
π2
→ 0,
as s % 1.
Hence, we know that
s
lim (−∆) f (x)
s%1
s22s−1 Γ n+2s
2
= lim −
n
s%1
π 2 Γ(1 − s)
Z
|y|<1
y T · D2 f (x) · y
dy
|y|n+2s
24
n
X
s22s−1 Γ n+2s
2
lim
−
n
2 Γ(1 − s)
s%1
π
i,j=1
Z
Dij f (x) · yi yj
dy
|y|n+2s
Z
n
X
s22s−1 Γ n+2s
yi yj
2
Dij f (x) lim −
dy.
n
n+2s
2 Γ(1 − s)
s%1
|y|
π
|y|<1
i,j=1
=
=
|y|<1
Z
For
yi yj
dy, if i 6= j, then
n+2s
|y|
|y|<1
Z
Z
yi yj
zi zj
dy = −
dz
n+2s
n+2s
|y|
|z|
|y|<1
|z|<1
Let zk = yk for all 1 ≤ k ≤ n with k 6= i, and zi = −yi
Z
−
=
|y|<1
Z
Which implies that
|y|<1
yi yj
dy
|y|n+2s
Let y = z.
yi yj
dy = 0. Hence, we obtain
|y|n+2s
n
X
s22s−1 Γ n+2s
2
Dii f (x) lim −
lim (−∆) f (x) =
n
2 Γ(1 − s)
s%1
s%1
π
i=1
s
Z
For
|y|<1
Z
|y|<1
yi yi
dy.
|y|n+2s
yi yi
dy, then
|y|n+2s
Z
|y|<1
yi yi
dy
|y|n+2s
Z
=
|z|<1
Z
=
|y|<1
z1 z1
dz
|z|n+2s
Let z1 = yi , zi = y1 , and zk = yk for all k 6= 1, i
y1 y1
dy,
|y|n+2s
Let y = z.
So we get
Z
|y|<1
yi yi
dy
|y|n+2s
=
=
=
n Z
yj yj
1X
dy
n j=1 |y|<1 |y|n+2s
Z
|y|2
1
dy
n |y|<1 |y|n+2s
αn
.
2 − 2s
Therefore, we get
n
X
s22s−1 Γ n+2s
αn
2
lim (−∆) f (x) =
Dii f (x) lim −
n
2
s%1
s%1
2
−
2s
π Γ(1 − s)
i=1
s22s−1 Γ n+2s
αn
2
= ∆f (x) lim −
.
n
s%1
π 2 Γ(1 − s) 2 − 2s
s
s22s−1 Γ n+2s
αn
2
For −
, then
n
2
π Γ(1 − s) 2 − 2s
−
s22s−1 Γ n+2s
αn
2
n
π 2 Γ(1 − s) 2 − 2s
=
−
s22s−1 Γ n+2s
αn
2
n
π 2 Γ(1 − s) 2 − 2s
25
=
=
→
=
=
=
αn s22s−1 Γ
1
−
n
(1 − s)Γ(1 − s)
π2
αn s22s−2 Γ n+2s
1
2
−
n
Γ(2 − s)
π2
αn 22−2 Γ n+2
1
2
−
n
Γ(2 − 1)
π2
αn Γ n2 + 1
−
n
π2
n
2π 2 n
n
Γ
n
2
2
nΓ( 2 )
−
n
π2
n+2s
2
−1.
Therefore, we can conclude that
lim (−∆)s f (x) = −∆f (x),
s%1
for all x ∈ Rn .
Approach II: Since f ∈ S(Rn ), then for all multi-index α, we know that
s D α f (x) = (2π|x|)2s D
α f (x),
\
[
(−∆)
for all x ∈ Rn .
By Plancherel’s Theorem, Corollary 1.1.2, then
s (D α f ) − D
α f k 2 n = k(2π|x|)2s D
α f (x) − D
α f (x)k 2 n .
\
[
[
[
k(−∆)s (Dα f ) − Dα f kL2 (Rn ) = k(−∆)
L (R )
L (R )
By Dominated Convergence Theorem (Exercise), then
lim k(−∆)s (Dα f ) − Dα f kL2 (Rn ) = 0.
s&0
That is, for all k ∈ N, we have (−∆)s f → f in H k (Rn ) as s & 0. By Sobolev Embedding Theorem, then
(−∆)s f (x) → f (x) uniformly in Rn as s & 0. By the same argument, we can show that (−∆)s f (x) → −∆f (x)
uniformly in Rn as s % 1.
Remark 1.2.10. By proof of Approach I in Theorem 1.2.5, the conditions on f can be general, such as, if
T
f ∈ C 2 (Rn ) C0 (Rn ), then lim (−∆)s f (x) = f (x) for all x ∈ Rn , in this case, the condition lim f (x) = 0
|x|→∞
s&0
n
must be needed. If f ∈ C (R ), then lim (−∆)s f (x) = −∆f (x) for all x ∈ Rn ,
3
s%1
Definition 1.2.4. For any s ∈ R, we define
n
o
s
H s (Rn ) = f ∈ S 0 (Rn ) : [1 + |x|2 ] 2 fˆ(x) ∈ L2 (Rn ) .
26
Define the norm of f ∈ H s (Rn ) as:
s
kf kH s (Rn ) = k[1 + |x|2 ] 2 fˆ(x)kL2 (Rn ) .
Lemma 1.2.3. For any s ∈ R, then H s (Rn ) is a Hilbert space.
Proof. It is clear that H s (Rn ) is a pre-Hilbert space under the inner product defined by: for any φ, ψ ∈ H s (Rn ),
we have
Z
fˆ(x)ĝ(x)[1 + |x|2 ]s dx.
< f, g >=
Rn
Let {fk }k≥1 be a Cauchy Sequence in H s (Rn ), that is, we have
lim kfk − fl kH s (Rn )
k,l→∞
=
=
=
s
lim k[1 + |x|2 ] 2 [fˆk (x) − fˆl (x)]kL2 (Rn )
k,l→∞
lim kfˆk − fˆl kL2 Rn ,[1+|x|2 ] 2s dx
k,l→∞
0.
s
s
That is, the sequence {fˆk }k≥1 is Cauchy in L2 Rn , [1 + |x|2 ] 2 dx . Since L2 Rn , [1 + |x|2 ] 2 dx is a Banach
s
space, then there exists some g ∈ L2 Rn , [1 + |x|2 ] 2 dx such that
lim kfˆk − gkL2 Rn ,[1+|x|2 ] 2s dx = 0.
k→∞
s
Since g ∈ L2 Rn , [1 + |x|2 ] 2 dx , then g ∈ S 0 (Rn ) (Exercise). Let f = ǧ ∈ S 0 (Rn ), then
fˆ = g.
Hence, we get
lim kfˆk − fˆkL2 Rn ,[1+|x|2 ] 2s dx
k→∞
=
=
=
s
lim k[1 + |x|2 ] 2 [fˆk (x) − fˆ(x)]kL2 (Rn )
k→∞
lim kfk − f kH s (Rn )
k→∞
0.
That is, fk → f in H s (Rn ), as k → ∞. Therefore, H s (Rn ) is a Hilbert space.
Lemma 1.2.4. For any s ∈ Rn , then S(Rn ) is dense in H s (Rn ).
Proof. By Lemma 1.2.2, we know that H s (Rn ) is a Hilbert space, then it suffices to show that for any g ∈ H s (Rn )
such that (f, g) = 0 for all f ∈ S(Rn ), then g ≡ 0. In fact, let g ∈ H s (Rn ) such that (f, g) = 0 for all f ∈ S(Rn ),
that is,
Z
Rn
fˆ(x)ĝ(x)[1 + |x|2 ]s dx = 0.
27
By Corollary 1.1.1, F : S(Rn ) → S(Rn ) is unitary, and S(Rn ) is dense in L2 (Rn ), then ĝ(x)[1 + |x|2 ]s ≡ 0 in
Rn , that is, ĝ(x) ≡ 0 in Rn . By Theorem 1.1.2, F : S 0 (Rn ) → S 0 (Rn ) is unitary, then g ≡ 0. Therefore, we can
conclude that S(Rn ) is dense in H s (Rn ).
Lemma 1.2.5. For any s ∈ Rn , then (H s (Rn ))∗ = H −s (Rn ).
s
s
Proof. a. For any f ∈ H −s (Rn ) and g ∈ H s (Rn ), then [1 + |x|2 ]− 2 fˆ(x), [1 + |x|2 ] 2 ĝ(x) ∈ L2 (Rn ). By Holder’s
inequality, then
Z
f (g)
ĝ(x)fˆ(x) dx
=
Rn
≤
s
s
k[1 + |x|2 ]− 2 fˆ(x)kL2x (Rn ) k[1 + |x|2 ] 2 · ĝ(x)kL2x (Rn )
By Cauchy-Schwarz’s inequality
=
kf kH −s (Rn ) kgkH s (Rn ) .
Which implies that f ∈ (H s (Rn ))∗ .
b. For any l ∈ (H s (Rn ))∗ and fix. Now for any g ∈ L2 (Rn ), by Plancherel’s Theorem, Corollary 1.1.2,
s
then ĝ ∈ L2 (Rn ) ⊂ S 0 (Rn ), which implies that [1 + |x|2 ]− 2 ĝ(x) ∈ L2x (Rn , [1 + |x|2 ]s dx) ⊂ S 0 (Rn ). Let gs =
s
F −1 [1 + |x|2 ]− 2 ĝ(x) ∈ S 0 (Rn ), then we know that gs ∈ H s (Rn ). Now we define
ls (g)
= l (gs )
≤
klkkgs kH s (Rn )
s
= klkk[1 + |x|2 ]− 2 ĝ(x)kL2x (Rn ,[1+|x|2 ]s dx)
=
klkkĝkL2 (Rn )
=
klkkgkL2 (Rn ) ,
By Plancherel’s Theorem, Corollary 1.1.2.
That is, ls ∈ (L2 (Rn ))∗ . By Riesz Representation Theorem, then there exists some f ∈ L2 (Rn ) ⊂ S 0 (Rn )
such that for all g ∈ L2 (Rn ), we have
ls (g)
=
=
(g, f )
Z
g(x)f (x) dx
Rn
Z
=
ĝ(x)fˆ(x) dx,
By Plancherel’s Theorem, Corollary 1.1.2.
Rn
s
s
Let fs = F −1 [1 + |x|2 ] 2 fˆ(x) ∈ S 0 (Rn ), then fbs (x) = [1 + |x|2 ] 2 fˆ(x), and
s
[1 + |x|2 ]− 2 fbs (x) = fˆ(x) ∈ L2x (Rn ).
28
Which implies that fs ∈ H −s (Rn ). Now it suffices to show that l(φ) = (φ, fs ) for all φ ∈ H s (Rn ). Since
s
s
φ ∈ H s (Rn ), then [1 + |x|2 ] 2 φ̂(x) ∈ L2x (Rn ). Hence, we obtain g = F −1 [1 + | · |2 ] 2 · φ̂ ∈∈ L2 (Rn ). Hence,
s
s
gs = F −1 [1 + | · |2 ]− 2 [1 + | · |2 ] 2 φ̂ = F −1 (φ̂) = φ, which implies that
s
ls F −1 [1 + | · |2 ] 2 φ̂
=
=
=
=
l(φ)
s
F −1 [1 + | · |2 ] 2 φ̂ , f
s
[1 + | · |2 ] 2 φ̂, fˆ
By Plancherel’s Theorem, Corollary 1.1.2
Z
s
[1 + |x|2 ] 2 φ̂(x)fˆ(x) dx
Rn
Z
=
s
[1 + |x|2 ] 2 φ̂(x)fˆ(x) dx
Rn
Z
=
s
s
[1 + |x|2 ] 2 φ̂(x)[1 + |x|2 ]− 2 fbs (x) dx
Rn
Z
=
φ̂(x)fbs (x) dx
Rn
=
fs (φ).
Lemma 1.2.6. For any s1 < s2 , then H s2 (Rn ) ,→ H s1 (Rn ).
Proof. Exercise.
Lemma 1.2.7. Let s >
n
, then H s (Rn ) ,→ C0 (Rn ).
2
n
Proof. Let f ∈ H s (Rn ), since s >
> 0, then f ∈ L2 (Rn ). By Plancherel’s Identity, Corollary 1.1.1, then
2
ˇ
fˆ ∈ L2 (Rn ) and fˆ = f in L2 (Rn ). Notice that F −1 : L1 (Rn ) → C0 (Rn ). So it suffices to show that fˆ ∈ L1 (Rn ).
s
s
Since f ∈ H s (Rn ), then |fˆ(x)|[1 + |x|2 ] 2 ∈ L2x (Rn ). Since [1 + |x|2 ]− 2 ∈ L2 (Rn ) (Exercise), then
Z
|fˆ(x)| dx
Z
=
Rn
s
s
|fˆ(x)|[1 + |x|2 ] 2 · [1 + |x|2 ]− 2 dx
Rn
≤
s
s
k|fˆ(x)|[1 + |x|2 ] 2 kL2x (Rn ) k[1 + |x|2 ]− 2 kL2 (Rn )
By the Holder’s Inequality
= C(n, s)kf kH s (Rn ) .
Therefore, fˆ ∈ L1 (Rn ), which implies that f ∈ C0 (Rn ), and
kf kC(Rn ) ≤ kfˆkL1 (Rn ) ≤ C(n, s)kf kH s (Rn ) .
29
Lemma 1.2.8. For any multi-index α and s ∈ R, then Dα : H s (Rn ) → H s−|α| (Rn ).
Proof. Exercise.
Theorem 1.2.7. Let s =
n
+ α for some 0 < α < 1, then H s (Rn ) ,→ C α (Rn ).
2
Proof. For any f ∈ H s (Rn ), by the proof of Lemma 1.2.6, then
ˇ
f (x) = fˆ(x),
for all x ∈ Rn .
So for all x, y ∈ Rn , then
|f (x + y) − f (x)|
=
=
=
ˇ
ˇ
|fˆ(x + y) − fˆ(x)|
Z
Z
2πi(x+y)·z ˆ
2πix·z ˆ
f (z) dz −
e
f (z) dz ne
R
Rn
Z
2πi(x+y)·z
2πx·z ˆ
[e
−e
]f (z) dz Rn
Z
≤
|e2πi(x+y)·z − e2πx·z ||fˆ(z)| dz
Rn
Z
=
|e2πiy·z − 1||fˆ(z)| dz
Rn
Z
=
s
s
|e2πiy·z − 1|[1 + |z|2 ]− 2 [1 + |z|2 ] 2 |fˆ(z)| dz
Rn
≤
=
s
s
k[1 + |z|2 ] 2 |fˆ(z)|kL2z (Rn ) k|e2πiy·z − 1|[1 + |z|2 ]− 2 kL2 (Rn )
Z
12
2πiy·z
2
2 −s
kf kH s (Rn )
|e
− 1| [1 + |z| ] dz .
By Holder’s Inequality
Rn
Notice that |e2πy·z − 1| = | sin(πy · z)| ≤ π|y||z| for all y, z ∈ Rn . If |y| > 1, by the proof of Lemma 1.2.6, then
|f (x + y) − f (x)|
≤ 2kf kL∞ (Rn ) ≤ 2kf kH s (Rn ) .
|y|α
sup
x∈Rn
Now assume 0 < |y| ≤ 1, then
Z
|e2πiy·z − 1|2 [1 + |z|2 ]−s dz
Rn
Z
|e
=
2πiy·z
2 −s
2
− 1| [1 + |z| ]
Z
1
|z|> |y|
Z
2
≤
2
2
2 −s
π |y| |z| [1 + |z| ]
1
|z|≤ |y|
2
= π |y|
2
|e2πiy·z − 1|2 [1 + |z|2 ]−s dz
dz +
1
|z|≤ |y|
Z
4[1 + |z|2 ]−s dz
dz +
1
|z|> |y|
Z
2
2 −s
|z| [1 + |z| ]
1
|z|≤ |y|
Z
dz + 4
1
|z|> |y|
[1 + |z|2 ]−s dz.
30
Z
For
|z|2 [1 + |z|2 ]−s dz, then
1
|z|≤ |y|
Z
Z
|z|2 [1 + |z|2 ]−s dz
|z|2 [|z|2 ]−s dz
≤
1
|z|≤ |y|
1
|z|≤ |y|
Z
|z|2−2s dz
=
1
|z|≤ |y|
Z
=
1
|y|
nαn
rn−2s+1 dr
0
Z
=
1
|y|
nαn
rn−n−2α+1 dr
0
Z
1
|y|
r−2α+1 dr
=
nαn
=
nαn
|y|2α−2 .
2 − 2α
0
Z
For
[1 + |z|2 ]−s dz, then
1
|z|> |y|
Z
2 −s
[1 + |z| ]
Z
dz
=
∞
[1 + r2 ]−s rn−1 dr
nαn
1
|z|> |y|
1
|y|
Z
≤
∞
r−2s+n−1 dr
nαn
Since
1
|y|
nαn
|y|2s−n
2s − n
nαn
|y|2α .
2s − n
=
=
1
>1
|y|
Since n − 2s < 0
Hence, we get
Z
|e
2πiy·z
2
2 −s
− 1| [1 + |z| ]
Rn
π 2 nαn
4nαn
+
dz ≤
|y|2α .
2 − 2α 2s − n
That is, for all x ∈ Rn and 0 < |y| ≤ 1, we have
|f (x + y) − f (x)| ≤ C(n, s)|y|α kf kH s (Rn ) .
That is, we have
sup
x,y∈Rn
α
|f (x + y) − f (x)|
≤ C(n, s)kf kH s (Rn ) .
|y|2α
n
Therefore, f ∈ C (R ), and
kf kC α (Rn ) ≤ C(n, s)kf kH s (Rn ) .
Theorem 1.2.8. Let 0 < s <
2n
n
, then H s (Rn ) ,→ L n−2s (Rn ).
2
31
Proof. Let p =
2n
> 2, for any f ∈ S(Rn ), then
n − 2s
Z ∞
p
tp−1 |{x : |f (x)| > t}| dt (Exercise).
kf kLp (Rn ) = p
0
T
Since f ∈ S(Rn ), then fˆ ∈ S(Rn ), in particular, f, fˆ ∈ L1 (Rn ) L2 (Rn ). For any R > 0, let
Z
f1,R (x) =
e2πix·y fˆ(y)1BR (0) (y) dy, for all x ∈ Rn .
Rn
ˆ
Then f1,R ∈ C0 (Rn ), fd
1,R = f 1BR (0) , and
kf1,A kL∞ (Rn )
kfˆ1BR (0) kL1 (Rn )
Z
=
|fˆ(x)|1BR (0) (x) dx
≤
Rn
Z
=
Rn
s
s
[1 + |x|2 ] 2 |fˆ(x)|[1 + |x|2 ]− 2 1BR (0) (x) dx
# 21
"Z
≤
2 −s
[1 + |x| ]
kf kH s (Rn )
dx
By Cauchy-Schwarz’s inequality.
BR (0)
Z
For
[1 + |x|2 ]−s dx, then
BR (0)
Z
[1 + |x|2 ]−s dx
Z
R
= nαn
BR (0)
0
Z
rn−1
dr
[1 + r2 ]s
R
≤ nαn
rn−1−2s dr
0
nαn n−2s
R
n − 2s
=
Since n − 2s > 0.
That is, we get
r
n−2s
nαn
R 2 kf kH s (Rn ) .
n − 2s
T
d
Let f2,R (x) = f (x) − f1,R (x) for all x ∈ Rn , then f2,R ∈ L1 (Rn ) L2 (Rn ), and fˆ = fd
1,R + f2,R , that is,
kf1,R kL∞ (Rn ) ≤
n
ˆ
fd
2,R (x) = f (x)1{x:|x|>R} (x) for all x ∈ R . Then for any t > 0, we have
t [
t
x : |f2,R (x)| >
.
{x : |f (x)| > t} ⊂ x : |f1,R (x)| >
2
2

 np
t
 , which implies that
Now for any t > 0, let R(t) =  q
nαn
4 n−2s kf kH s (Rn )
r
kf1,R(t) kL∞ (Rn )
n
nαn
[R(t)] 2 −s kf kH s (Rn )
n − 2s

 np n−2s
2
r
t
nαn 

q
≤
kf kH s (Rn )
n − 2s 4 nαn kf k s n
≤
n−2s
H (R )
32
2n
nαn 
q
n − 2s 4
=
r
nαn
q
n − 2s 4
=
<
x : |f1,R(t) (x)| >
t
2
t
nαn
s
n
n−2s kf kH (R )
t
kf kH s (Rn )

kf kH s (Rn )
nαn
s
n
n−2s kf kH (R )
t
4
t
.
2
=
Hence,
 n−2s
· n−2s
n
2

r
= ∅, which gives us
∞
Z
kf kpLp (Rn )
x : |f2,R(t) (x)| > t dt.
2 p−1
≤ p
t
0
On the other hand, then
x : |f2,R(t) (x)| > t 2 Z
dx
=
|f2,R(t) |> 2t
Z
≤
|f2,R(t) |> 2t
≤
2|f2,R(t) (x)|
t
2
dx
4
kf2,R(t) k2L2 (Rn ) .
t2
Hence,
kf kpLp (Rn )
∞
Z
tp−1
≤ p
0
Z
=
4
kf2,R(t) k2L2 (Rn ) dt
t2
∞
tp−3 kf2,R(t) k2L2 (Rn ) dt
4p
0
Z
=
∞
2
tp−3 kf\
2,R(t) kL2 (Rn ) dt
4p
By Plancherel’s Identity, Corollary1.1.1
0
Z
=
∞
4p
t
p−3
Z
Rn
0
Z
=
∞
4p
tp−3
Z
Z
4p
|fˆ(x)|2 1{x:|x|>R(t)} (x) dx
dt
Rn
0
=
2
\
|f2,R(t) (x)| dx dt
|fˆ(x)|2
Z
Rn
∞
t
p−3
1{x:|x|>R(t)} (x) dt dx
By Tonelli’s Theorem.
0
 np

t
 , then whenever |x| > R(t), we know that
By the Definition of R(t) =  q
nαn
4 n−2s kf kH s (Rn )
r
t<4
n
nαn
kf kH s (Rn ) |x| p .
n − 2s
33
Hence, we have
kf kpLp (Rn )
Z
≤
4p
Rn
=
Since p =
0
r
p−2
n
nαn
1
4
kf kH s (Rn ) |x| p
dx Since p > 2
p−2
n − 2s
Rn
r
p−2
Z
n(p−2)
4p
nαn
|fˆ(x)|2 |x| p dx.
4
kf kp−2
s (Rn )
H
p−2
n − 2s
Rn
Z
=
 √

n
nαn
Z 4 n−2s
kf kH s (Rn ) |x| p
|fˆ(x)|2 
tp−3 dt dx
|fˆ(x)|2
4p
2n
2n
2n − 2n + 4s
4s
, then p − 2 =
−2=
=
, and
n − 2s
n − 2s
n − 2s
n − 2s
n(p − 2)
p
=
=
4s
n n−2s
2n
n−2s
2s.
Hence, we get
kf kpLp (Rn )
≤
≤
=
=
r
p−2
Z
nαn
kf kp−2
4
|fˆ(x)|2 |x|2s dx
H s (Rn )
n − 2s
n
R
r
p−2
Z
4p
nαn
4
kf kp−2
|fˆ(x)|2 [1 + |x|2 ]s dx
H s (Rn )
p−2
n − 2s
n
R
r
p−2
4p
nαn
2
4
kf kp−2
H s (Rn ) kf kH s (Rn )
p−2
n − 2s
r
p−2
nαn
4p
kf kpH s (Rn ) .
4
p−2
n − 2s
4p
p−2
Therefore, there exists some constant C > 0 which just depends on n and s such that
kf k
2n
L n−2s
≤ Ckf kH s (Rn ) .
Theorem 1.2.9. For any 0 < s <
n
, there exists some constant C(n, s) > 0 such that for all f ∈ S(Rn ), we
2
have
kf k
2n
L n−2s (Rn )
≤ Ck| · |s fˆkḢ s (Rn ) .
T
Proof. For any f ∈ S(Rn ), then fˆ ∈ S(Rn ). Let g(x) = |x|s fˆ(x) ∈ L1 (Rn ) L∞ (Rn ) for all x ∈ Rn , then
g ∈ L2 (Rn ), and fˆ(x) = |x|−s g(x) for all x ∈ Rn . Since g ∈ L2 (Rn ), then there exists some h ∈ L2 (Rn ) such
that ĥ = g. Recall F −1 (| · |−s ) = γn (s)| · |s−n . Consider
Is (h)(x) = | · |
s−n
Z
∗ h(x) =
Rn
h(y)
dy.
|x − y|n−s
34
−s
Then I[
· ĥ(x) = γn (s)|x|−s · g(x) = fˆ(x) for all x ∈ Rn , which implies that In−s (h) = f .
s (h)(x) = γn (s)|x|
On the other hand, by Hardy-Littlewood-Sobolev’s inequality, Theorem 1.2.2 and h ∈ L2 (Rn ), then f =
2n
In−s (h) ∈ L n−2s (Rn ), and there exists some constant C > 0 which just depends on n and s such that
kf k
2n
L n−2s (Rn )
= kIs (h)k
2n
L n−2s (Rn )
≤ CkhkL2 (Rn )
= CkĝkL2 (Rn )
= CkgkL2 (Rn )
By Plancherel’s Theorem, Corollary 1.1.1
= Ck| · |s fˆkL2x (Rn ) .
Theorem 1.2.10. Let 0 < s < 1. Then there exists some constant An,s which just depends on n and s, such
that for all f ∈ H s (Rn ), we have
Z
(2π|x|)2s |fˆ(x)|2 dx = Bn,s
Z
Rn
Where Bn,s
Z
Rn
Rn
|f (x) − f (y)|2
dxdy.
|x − y|n+2s
s22s−1 Γ n+2s
2
=
.
n
π 2 Γ(1 − s)
Proof. Since f ∈ H s (Rn ), then
Z
Rn
Z
Z
Rn
Z
=
Rn
Rn
|f (x) − f (y)|2
dxdy
|x − y|n+2s
|f (z + y) − f (y)|2
dzdy
|z|n+2s
Let z = x − y
|f (z + y) − f (y)|2
dydz By Fubini’s Theorem
|z|n+2s
Rn Rn
Z
Z
=
|z|−n−2s
|f (z + y) − f (y)|2 dy dz
Z
Z
=
Rn
Z
=
Rn
Z
=
Rn
Rn
|z|−n−2s kf (· + z) − f k2L2 (Rn ) dz
|z|−n−2s kf \
(· + z) − fˆk2L2 (Rn ) dz
By Plancherel’s Theorem, Corollary 1.1.2.
By Proposition 1.1.1, then
f\
(· + z)(ξ) = e2πiξ·z fˆ(ξ),
for all z, ξ ∈ Rn .
35
Then we have
|f (x) − f (y)|2
dxdy
n+2s
Rn Rn |x − y|
Z
2
dz
|z|−n−2s e2πiξ·z fˆ(ξ) − fˆ(ξ)
L2 (Rn )
n
R
Z
Z
2πiξ·z
2
−n−2s
e
|z|
− 1 |fˆ(ξ)|2 dξdz
Z
=
=
Z
Rn
Rn
Z
|fˆ(ξ)|2
=
Rn
Z
2πiξ·z
2
e
− 1 |z|−n−2s dz
dξ
By Fubini’s Theorem
Rn
"Z
#
x
2 |x| −n−2s
ξ· |ξ|
−n
=
(2π|ξ|)
dx dξ
− 1
e
2π|ξ|
Rn
Rn
Z Z
2
ξ·x
(2π|ξ|)2s |fˆ(ξ)|2
=
e |ξ| − 1 |x|−n−2s dx dξ.
Z
|fˆ(ξ)|2
Rn
Let z =
x
2π|ξ|
Rn
Hence, we can obtain
Bn,s
=
1
Cn,s
=
An,s
2
=
s22s−1 Γ n+2s
2
.
n
π 2 Γ(1 − s)
Definition 1.2.5. Let s > 0, 1 ≤ p < ∞ and Ω be a domain in Rn , we say that f ∈ W s,p (Ω) if
Z Z
1
|f (x) − f (y)|p p
< ∞.
kf kW s,p (Ω) = kf kW [s],p (Ω) +
n+sp
Ω Ω |x − y|
Remark 1.2.11. The general Sobolev embedding theorems hold for W s,p (Ω), that is,
np
a. If sp < n, then W s,p (Ω) ,→ L n−sp (Ω).
b. If sp > n, then W s,p (Ω) ,→ C 0,α (Ω) for some 0 < α < 1.
Theorem 1.2.11. Let
1
2
< s < 1 and u ∈ C 2 (R) such that lim
x→±∞
Z
Proof. For any R > 0,
Z R
Z
s
(−∆) u(y) dy =
−R
R
|x|→∞
R
lim
R→∞
u(x) = L± and lim u0 (x) = 0, then
(−∆)s u(y) dy = 0.
−R
Z
u(y) − u(z)
Cn,s P.V.
dz dy
1+2s
−R
R |y − z|
Z R Z
u(y) − u(y + w)
= Cn,s
P.V.
dw
dy Let w = z − y
|w|1+2s
−R
R
#
Z R "Z
Z
u(y) − u(y + w)
u(y) − u(y + w)
= Cn,s
dw + P.V.
dw dy
|w|1+2s
|w|1+2s
−R
|w|≥1
|w|<1
36
"Z
R
Z
u(y) − u(y + w)
dw −
|w|1+2s
= Cn,s
−R
R
Z
|w|≥1
Z
= Cn,s
−R
Z
|w|≥1
R
Z
R
Z
For
−R
R
Z
Z
−R
|w|≥1
|w|≥1
|w|<1
u(y + w) − u(y) − u0 (y)w
dw
|w|1+2s
#
dy
u(y) − u(y + w)
dwdy
|w|1+2s
Z
−Cn,s
−R
Z
|w|<1
u(y + w) − u(y) − u0 (y)w
dwdy.
|w|1+2s
u(y) − u(y + w)
dwdy, then
|w|1+2s
u(y) − u(y + w)
dwdy
|w|1+2s
Z
=
R
Z
−
−R
Z
R
|w|≥1
R1
Z
0
=
−R
Z
R
u(y + w) − u(y)
dwdy
|w|1+2s
|w|≥1
Z
Z
1
=
−R
|w|≥1
Z
=
|w|≥1
Z
=
|w|≥1
Z
=
|w|≥1
0
w
|w|1+2s
u0 (y + tw) · w dt
dwdy
|w|1+2s
u0 (y + tw) · w
dtdwdy
|w|1+2s
Z 1Z R
u0 (y + tw) dydtdw
w
|w|1+2s
Z
w
|w|1+2s
Z
By the Fubini’s Theorem
−R
0
1
R
u(y + tw)|−R dtdw
0
1
[u(R + tw) − u(−R + tw)] dtdw
0
Z
→
=
=
Z
R
Z
For
−R
Z
R
−R
Z
|w|<1
|w|<1
w
[L+ − L− ] dtdw
1+2s
|w|
|w|≥1
Z
w
dw
[L+ − L− ]
1+2s
|w|
|w|≥1
By Dominated Convergence Theorem
0.
u(y + w) − u(y) − u0 (y)w
dwdy, then
|w|1+2s
u(y + w) − u(y) − u0 (y)w
dwdy
|w|1+2s
Z
R
Z
=
−R
=
=
=
0
|w|<1
R1
u0 (y + tw)w dt − u0 (y)w
dwdy
|w|1+2s
[u0 (y + tw) − u0 (y)] dt · w
dwdy
|w|1+2s
−R |w|<1
R 1 R 1 00
Z R Z
u (y + rtw) · w2 drdt
0 0
dwdy
|w|1+2s
−R |w|<1
Z R Z
Z 1Z 1
1
· u00 (y + rtw) drdtdwdy
2s−1
−R |w|<1 0
0 |w|
Z
Z 1Z 1Z R
1
u00 (y + rtw) dydrdtdw
2s−1
0
−R
|w|≤1 |w|
0
Z
=
R1
R
Z
0
37
Z
=
|w|≤1
Z
=
|w|≤1
→
Z
0,
By Fubini’s Theorem
Z 1Z 1
1
R
u0 (y + rtw)|−R drdtdw
|w|2s−1 0 0
Z 1Z 1
1
[u0 (R + rtw) − u0 (−R + rtw)] drdtdw
|w|2s−1 0 0
By Dominated Convergence Theorem.
R
Therefore, we can conclude that
(−∆)s u(y) dy → 0, as R → ∞.
−R
Remark 1.2.12. References: Adams and Fournier[1], Applebaum[2], Bahouri, Chemin and Danchin[3], Blumenthal and Getoor[4], Grafakos[17], Haroske and Triebel[18], Maz’ya[24], Maz’ya and Shaposhnikova[23], Mellet, Roquejoffre and Sire[25], Miao[26], Nezza, Palatucci and Valdinoci[30], Riesz[31], Riesz[32], Shakarchi and
Stein[33], Silvestre[36], Stein[38], etc.
1.3. Caffarelli-Silvestre’s s-Harmonic Extension.
as:
Definition 1.3.1. For any 0 < s < 1, we define the s-Poisson kernel in Rn+1
+
Pys (x) = B(n, s)
y 2s
[|x|2 + y 2 ]
n+2s
2
,
for all (x, y) ∈ Rn+1
+ .
Γ n+2s
2
Where B(n, s) = n
.
π 2 Γ(s)
Lemma 1.3.1. For any 0 < s < 1, then
Z
x
a. Pys (x) = y −n P1s
for all (x, y) ∈ Rn+1
and
Pys (x) dx = 1 for all y > 0, that is, Pys is an
+
y
Rn
approximate identity.
b. div[y 1−2s ∇Pys (x, y)] = 0 for all (x, y) ∈ Rn+1
+ .
Proof. Exercise.
Theorem 1.3.1. Let 0 < s < 1, for any f ∈ C 2 (Rn ), let f¯(x, y) = Pys ∗ f (x), then f¯ ∈ C 2 (Rn+1
+ ), and
(1.3.1)


div[y 1−2s ∇f¯(x, y)] = 0,
in Rn+1

+




lim f¯(x, y) = f (x),
in Rn
y&0





 − lim y 1−2s Dy f¯(x, y) = D(n, s)(−∆)s f (x), in Rn
y&0
Where D(n, s) =
21−2s Γ(1 − s)
.
Γ(s)
38
1−2s
¯
Proof. By Lemma 1.3.1, then f¯ ∈ C 2 (Rn+1
∇f¯(x, y)] = 0 for all (x, y) ∈ Rn+1
+ ), div [y
+ , and lim f (x, y) =
y&0
n
f (x) for all x ∈ R . On the other hand, we know that
f¯(x, y) − f (x)
y&0
y 2s
Z
= −2sB(n, s) lim y −2s
− lim y 1−2s Dy f¯(x, y)
= −2s lim
y&0
y&0
y&0
Rn
f (z) − f (x)
[|z − x|2 + y 2 ]
Z
Rn
n+2s
2
[|z − x|2 + y 2 ]
Rn
n+2s
2
dz
n+2s
2
By Lemma 1.3.1
dz.
dz, then
f (z) − f (x)
[|z −
[|z − x|2 + y 2 ]
f (z) − f (x)
Z
Z
y 2s [f (z) − f (x)]
Rn
= −2B(n, s) lim
For
By L’Hospital’s Rue
x|2
+
y2 ]
f (w + x) − f (x)
Z
n+2s
2
dz
=
[|w|2 + y 2 ]
Rn
n+2s
2
f (w − x) − f (x)
Z
=
[|w|2 + y 2 ]
Rn
n+2s
2
dw
Let w = z − x
dw
Let w = x − z.
Hence, we can obtain
− lim y 1−2s Dy f¯(x, y)
y&0
f (w + x) + f (w − x) − 2f (x)
Z
=
−sB(n, s) lim
y&0
Z
= −sB(n, s)
Rn
Rn
[|w|2 + y 2 ]
n+2s
2
dw
f (w + x) + f (w − x) − 2f (x)
dw
|w|n+2s
By Dominated Convergence Theorem
=
2sB(n, s)
(−∆)s f (x)
C(n, s)
=
(−∆)s f (x) (Exercise).
Theorem 1.3.2. Let −1 < a < 1, and u ∈ C 2 (Rn+1
+ ) be a solution of the PDE:
div [y a ∇u(x, y)] = 0,
uniformly for all (x, y) ∈ Rn+1
+ .
And such that for some R > 0, we have
lim y a Dy u(x, y) = 0,
y&0
for all |x| < R.
Consider the even extension of ũ on Rn+1 given by:

 u(x, y),
if y > 0
ũ(x, y) =
 u(x, −y), if y < 0.
39
Then ũ is a weak solution of the PDE:
div [|y|a ∇ũ(x, y)] = 0,
for all (x, y) ∈ BR (0).
That is, for any φ ∈ C0∞ (BR (0)), we have
Z
∇ũ(x, y) · ∇φ(x, y)|y|a dxdy = 0.
BR (0)
Proof. For any φ ∈ C0∞ (BR (0)) and fix, for any small > 0, we have
Z
∇ũ(x, y) · ∇φ(x, y)|y|a dxdy
BR (0)
Z
∇ũ(x, y) · ∇φ(x, y)|y|a dxdy
=
{(x,y): |(x,y)|<R, |y|<}
Z
∇ũ(x, y) · ∇φ(x, y)|y|a dxdy
+
{(x,y): |(x,y)|<R, |y|>}
= I + II .
For I , we have
|I |
=
Z
{(x,y):
Z
∇ũ(x, y) · ∇φ(x, y)|y| dxdy |y|<}
a
|(x,y)|<R,
|∇ũ(x, y) · ∇φ(x, y)||y|a dxdy
≤
{(x,y): |(x,y)|<R, |y|<}
Z
|∇ũ(x, y)||∇φ(x, y)||y|a dxdy
≤
{(x,y): |(x,y)|<R, |y|<}
Z
≤
k∇ukL∞ (BR (0)) k∇φkL∞ (BR (0)) 2αn Rn
=
2αn Rn
k∇ukL∞ (BR (0)) k∇φkL∞ (BR (0)) 1+a ,
1+a
→
0,
|y|a dy
0
Since a + 1 > 0
as & 0.
On the other hand, for II , since (x, y) ∈ BR (0), and |y| > , then ũ is C 2 in {(x, y) : |(x, y)| < R, |y| > },
and ũ satisfies the PDE:
div [|y|a ∇ũ(x, y)] = 0,
for all (x, y) ∈ {(x, y) : |(x, y)| < R, |y| > }.
So we have
div [|y|a φ(x, y) · ∇ũ(x, y)]
= φ(x, y)div [|y|a ∇ũ(x, y)] + |y|a ∇ũ(x, y) · ∇φ(x, y)
= |y|a ∇ũ(x, y) · ∇φ(x, y).
40
Then we get
Z
II
div [|y|a φ(x, y) · ∇ũ(x, y)] dxdy
=
{(x,y): |(x,y)|<R, |y|>}
Z
[|y|a φ(x, y)∇ũ(x, y)] · n(x, y) dy
=
By Divergence Theorem
{(x,y): |(x,y)|<R, |y|=}
Z
[a φ(x, y)∇ũ(x, y)] · n(x, y) dy
=
{(x,y): |(x,y)|<R, |y|=}
Z
Z
[a φ(x, )Dy ũ(x, )] dy +
= −
{(x,y): |(x,y)|<R, y=}
[a φ(x, −)Dy ũ(x, −)] dy
{(x,y): |(x,y)|<R, y=−}
Z
[a φ(x, )Dy u(x, )] dy
= −2
Since ũ(x, y) = ũ(−x, y) = u(x, y) for all x > 0
{(x,y): |(x,y)|<R, x=}
Z
[a φ(x, )Dy u(x, )] dy.
= −2
{(x,y): |(x,y)|<R, y=}
Hence, we have
|II |
Z
= −2
{(x,y):
[ φ(x, )Dy u(x, )] dy y=}
a
|(x,y)|<R,
Z
≤ 2kφkL∞ (BR (0))
a |Dy u(x, )| dx
|y|<R
But since
lim y a Dy u(x, y) = 0,
y&0
for all |x| < R.
So we get
lim II = 0.
&0
Therefore, we get
Z
∇ũ(x, y) · ∇φ(x, y)|y|a dxdy = 0.
BR (0)
Remark 1.3.1. Theorem 1.3.2 still holds if we just assume
Z
lim
y&0
y a φ(x, y)Dy u(x, y) dx = 0,
for all φ ∈ C0∞ (BR (0)).
|x|<R
Remark 1.3.2. Let 0 < s < 1, w(x, y) = |y|1−2s for all x ∈ Rn and all y ∈ R\{0}, then div [w(x, y)∇u(x, y)] = 0
is a degenerate but not uniformly elliptic PDE, however, the good thing is that it has divergent form and the
weight w is a special kind of functions.
41
Definition 1.3.2. Let 1 < p < ∞ and w be a nonnegative locally integrable function on Rn , we say that
w ∈ Ap , if there exists some constant C > 0 such that for all cube Q in Rn , we have
p−1
Z
Z
1
1
1
w(x)dx
[w(x)]− p−1 dx
≤ C < ∞.
|Q| Q
|Q| Q
Example 1.3.1. Let a ∈ R and w(y) = |y1 |a for all y ∈ Rn \{0}, then w ∈ A2 if and only if −1 < a < 1
(Exercise).
Definition 1.3.3. Let Ω be a Liplichitz domain in Rn and w be a nonnegative locally integrable function on
Ω, define the weighted Sobolev space H 1 (Ω, w) as the completion of C ∞ (Ω) under the norm:
Z
21
[[ϕ(x)]2 + |∇ϕ(x)|2 ] w(x) dx .
kϕkH 1 (Ω,w) =
Ω
Define the space
H01 (Ω, w)
as the completion of Cc∞ (Ω) under the norm:
Z
21
kϕkH01 (Ω,w) =
|∇ϕ(x)|2 w(x) dx .
Ω
Theorem 1.3.3 (Weighted Sobolev Imbedding Theorem). Let w ∈ A2 and Ω be a bounded smooth domain in
Rn , then there exists some constant C(n, Ω) > 0 and δ(n) > 0 such that u ∈ Cc∞ (Ω), and 1 ≤ k ≤
n
n−1
+ δ, we
have
kukL2k (Ω,w) ≤ Ck∇ukL2 (Ω,w) .
Definition 1.3.4. Let Ω be a bounded smooth domain in Rn , w ∈ A2 , and
|f |
w
∈ L2 (Ω, w). We say that
u ∈ H 1 (Ω, w) is a weak solution of div [w(x)∇u(x)] = f (x) in Ω if for any ϕ ∈ Cc∞ (Ω), we have
Z
Z
∇u(x) · ∇ϕ(x) w(x)dx =
f (x)ϕ(x) dx.
Ω
Ω
Theorem 1.3.4. Let Ω be a domain in Rn and u be a local subsolution of div [w(x)∇u(x)] ≥ 0 in Ω. Then
there exists some constant C(w, Ω) such that for all bass BR ⊂ Ω, we have
21
Z
1
|u(x)|2 w(x) dx
,
sup u ≤ C
w(BR ) BR
BR
2
Z
where w(BR ) =
w(x) dx.
BR
Theorem 1.3.5 (Harnack Inequality). Let Ω be a domain in Rn and u be a positive local solution of
div [w(x)∇u(x)] = 0 in Ω. Then for any compact subdomain K of Ω, there exists some constant C(K, Ω, w)
such that
max u ≤ C inf u.
K
K
42
Remark 1.3.3. References: Caffarelli and Silvestre[6], Coifman and Fefferman[8], Ding, Lu and Yan[9], Fabes,
Jerison and Kenig[12], Fabes, Kenig and Serapioni[13], Muckenhoupt and Wheeden[28], Muckenhoupt[27], Muckenhoupt and Wheeden[29], Stein[37], Torchinsky[41], etc..
1.4. Poisson Kernels of the Balls for Fractional Laplacians.
Theorem 1.4.1. For any x ∈ B1 (0), consider the function
Z
1 − |x|2
1
dS(y).
u(x) =
nαn
|x
−
y|n
|y|=1
Then u(x) ≡ 1 for all x ∈ B1 (0).
Proof. See Theorem 2.6, in Page 20, in Gilbarg and Trudinger[16].
Definition 1.4.1. Let 0 < s < 1, Ω be an open subset of R, we say that the function P : Rn × Rn → R
is a Poisson’s kernel of Ω with respect to (−∆)s if P has the property: for any f ∈ C(Rn \Ω), let u(x) =
Z
P (x, y)f (y) dy for all x ∈ Ω, then (−∆)s u(x) = 0 for all x ∈ Ω.
Rn \Ω
Definition 1.4.2. For any a ∈ Rn and R > 0, define Ia,R : Rn \{a} → Rn \{a} as:
Ia,R (x) = a +
R2
(x − a),
|x − a|2
for all x ∈ Rn \{a}.
We say that Ia,R is an sphere inversion transform with respect to the sphere ∂BR (a).
Lemma 1.4.1. Let z ∈ S n−1 , then det (δij − 2zi zj ) = −1.
Proof. Let f (z) = det (δij − 2zi zj ) for all z ∈ S n−1 . Obviously, f is a continuous function on S n−1 . Look at
f (e1 ), we know that f (e1 ) = (aij ), where a11 = 1 − 2 · 12 = −1, if i 6= j, then aij = 0 − 2 · 0 = 0, and if i = j 6= 1,
then aii = 1 − 2 · 02 = 1, that is,

f (e1 )
=
 −1

 0

det 
 ···


0
0
···
1
···
···
···
···0
···0

0 

0 


··· 


1
= −1.
Claim I: Let A = (δij − 2zi zj ), then A2 = I.
In fact, for all 1 ≤ i, ≤ j ≤ n, we have
(A2 )ij
=
n
X
k=1
[(δik − 2zi zk ) · (δkj − 2zk zj )]
43
=
n
X
[δik δkj − 2δkj zi zk − 2δik zk zj + 4zi zk zk zj ]
k=1
=
δij − 2zi zj − 2zi zj + 4zi zj |z|2
=
δij − 2zi zj − 2zi zj + 4zi zj
=
δij .
Since |z| = 1
That is, A2 = I. In particular, we get det A = ±1, so f (z) = ±1 for all z ∈ S n−1 . Since f is continuous on
S n−1 and f (e1 ) = −1, hence we get f (z) = −1 for all z ∈ S n−1 .
Lemma 1.4.2. For any x, y ∈ S n−1 and any k, l ∈ R, then |kx + ly| = |lx + ky|.
Proof. Exercise.
Lemma 1.4.3. For any a, b ∈ Rn \{0}, then
a
b |b − a|
|a|2 − |b|2 = |a||b| .
Proof. In fact, we know that
a
b |a|2 − |b|2 =
=
=
=
1 a
|a| b −
|a| |a|
|b| |b| b 1 |a| a
−
+
|a| |b| |a| |b| 1 a
b − +
|a| |b| |b| By Lemma 1.4.2
|b − a|
.
|a||b|
Proposition 1.4.1. For any a ∈ Rn and R > 0, then the followings hold:
a. For any x ∈ Rn \{a}, then |Ia,R (x) − a| =
R2
.
|x − a|
2
b. Ia,R
= Id on Rn \{a}.
c. For any x ∈ ∂BR (a), then Ia,R (x) = x.
d. For all x ∈ Rn \{a}, then det DIa,R (x) = −
R
|x − a|
2n
.
e. For any x, y ∈ Rn \{a}, then
|Ia,R (x) − Ia,R (y)| =
R2 |x − y|
.
|x − a||y − a|
44
f. For any x, y ∈ Rn \{a}, then
|y − a|
|Ia,R (x) − y|
=
.
|x − Ia,R (y)|
|x − a|
Proof. a. By Definition 1.4.2, then
|Ia,R (x) − a| =
=
R2
|x − a|
|x − a|2
R2
.
|x − a|
b. For any x ∈ Rn \{0}, by Definition 1.4.2, then
2
Ia,R
(x)
a+
R2
[Ia,R (x) − a]
|Ia,R (x) − a|2
= a+
R2 |x − a|2 R2
(x − a)
R4
|x − a|2
=
=
a+x−a
=
x.
By the result of part a
c. Since x ∈ ∂BR (a), then |x − a| = R. By Definition 1.4.2, then
Ia,R (x) = a +
R2
(x − a) = a + x − a = x.
R2
d. Let J(x) = Ia,R (x) for all x ∈ Rn \{a}, that is, Ji (x) = ai +
R2
|x−a|2 (xi
− ai ) for all 1 ≤ i ≤ n, and for all
1 ≤ j ≤ n, we have
Dj Ji (x)
=
−
=
−
=
2R2 xj − aj
R2
(x
−
a
)
+
δij
i
i
|x − a|3 |x − a|
|x − a|2
2R2 xi − ai xj − aj
R2
+
δij
2
|x − a| |x − a| |x − a|
|x − a|2
xi − ai xj − aj
R2
δij − 2
|x − a|2
|x − a| |x − a|
By Lemma 1.4.1, then we know that
det DIa,R (x) = −
R
|x − a|
2n
,
for all x ∈ Rn \{a}.
e. In fact, we have
R2
R2
|Ia,R (x) − Ia,R (y)| = a +
(x − a) − a −
(y − a)
2
2
|x − a|
|y − a|
2
2
R
R
(x − a) −
(y − a)
= 2
2
|x − a|
|y − a|
y − a 2 x−a
= R −
|x − a|2
|y − a|2 45
= R2
|(x − a) − (y − a)|
|x − a||y − a|
By the Lemma 1.4.3
R2 |x − y|
.
|x − a||y − a|
=
f. By the results of part e and part b, then
|Ia,R (x) − y| =
R2 |x − Ia,R (y)|
.
|x − a||Ia,R (y) − a|
By the result of part a, then
|Ia,R (x) − y|
|x − Ia,R (y)|
=
R2 |y − a|
|x − a| R2
=
|y − a|
.
|x − a|
Lemma 1.4.4. For all x, y ∈ Rn , define
P (x, y) =
1 − |x|2 s
Γ n2
1
.
n
π 2 B(s, 1 − s) 1 − |y|2 |x − y|n
Z
P (x, y) dy = 1 for all |x| < 1.
Then
|y|>1
Proof. In fact, we know that
Z
Z
P (x, y) dy
=
|y|>1
|y|≥1
=
=
=
=
1 − |x|2 s
Γ n2
1
dy
n
2
|x − y|n
π 2 B(s, 1 − s) 1 − |y|
Z
n
Γ 2
[1 − |x|2 ]s
[|y|2 − 1]−s |x − y|−n dy
π B(s, 1 − s)
|y|≥1
Z
Z ∞
Γ n2
2 s
2
−s n−1
−n
[1
−
|x|
]
[r
−
1]
r
|x
−
rz|
dS(z)
dr
n
π 2 B(s, 1 − s)
1
S n−1
n
2
Let y = rz
Z
Z ∞
−n
Γ n2
2 s
2
−s n−1
−n x
[1 − |x| ]
[r − 1] r
r − z
dS(z) dr
n
r
π 2 B(s, 1 − s)
1
S n−1
Z
Z ∞
x
−n
Γ n2
2 s
2
−s −1
[1
−
|x|
]
[r
−
1]
r
−
z
dS(z)
dr.
n
π 2 B(s, 1 − s)
1
S n−1 r
Since |x| < 1, then xr < 1. By Theorem 1.4.1, then
x
−n
dS(z)
− z
S n−1 r
Z
=
nαn
2
1 − x
r
=
nαn r2
.
r2 − |x|2
46
Hence, we have
Z
P (x, y) dy
Z ∞
Γ n2
nαn r2
2 s
dr
[1
−
|x|
[r2 − 1]−s r−1 2
]
n
r − |x|2
π 2 B(s, 1 − s)
1
Z ∞
Γ n2
2 s
[r2 − 1]−s r[r2 − |x|2 ]−1 dr
nα
[1
−
|x|
]
n
n
π 2 B(s, 1 − s)
1
Z ∞
n
2
Γ n2
α
1
2π
2 s
[t − 1]− 2 [t − |x|2 ]−1 dt
[1
−
|x|
]
n
2
π 2 B(s, 1 − s) Γ n2
1
=
|y|>1
=
=
Let t = r2
=
=
=
1
[1 − |x|2 ]s
B(s, 1 − s)
Z
1
[1 − |x|2 ]s
B(s, 1 − s)
Z
1
[1 − |x|2 ]s
B(s, 1 − s)
Z
∞
r−s [r + 1 − |x|2 ]−1 dr
Let r = t − 1
0
∞
r
−s
2 −1
[1 − |x| ]
0
r
+1
1 − |x|2
−1
dr
∞
[(1 − |x|2 )t]−s [1 − |x|2 ]−1 [t + 1]−1 [1 − |x|2 ] dt
0
Let r = (1 − |x|2 )t
Z ∞
1
t−s [1 + t]−1 dt
B(s, 1 − s) 0
1
B(s, 1 − s) (Exercise)
B(s, 1 − s)
=
=
=
1.
Lemma 1.4.5. For any 0 < s < 1 and |y| > 1 and fix, let
f (x, y) =
|1 − |x|2 |s
,
|x − y|n
for all x ∈ Rn .
Then (−∆)s f (x, y) = 0 for all x ∈ B1 (0).
Proof. Exercise.
Theorem 1.4.2 (The Poisson’s Kernel). Let f ∈ C(Rn \B1 (0)) and u(x) =
Z
P (x, y) dy in B1 (0) and
|y|>1
u(x) ≡ f (x) in Rn \B1 (0), then
(−∆)s u(x) = 0,
∀x ∈ B1 (0).
Proof. Exercise.
Theorem 1.4.3 (The Generalized Harnack Inequality). Let u be a nonnegative bounded function on Rn and
be s-harmonic in B1 (0), then
1 − |y|2 s 1 − |y −n
u(x),
u(y) ≤ 1 − |x|2 1 + |x| ∀x, y ∈ B1 (0).
47
Proof. Exercise.
Theorem 1.4.4. For any 0 < s < 1, then there exists a function u ∈ C 2 (B1 (0))
T
L∞ (Rn ) such that



(−∆)s u(x) = 0, for all x ∈ B1 (0)





 u(x) > 0,
for all x ∈ B1 (0)\{0}
 u(0) = 0.






 |u(x)| ≤ 1,
for all x ∈ Rn .
Proof. For any 1 < R1 < R2 < ∞, we consider the function gR1 ,R2 defined on Rn \B1 (0): for all x ∈ Rn \B1 (0),
we have
gR1 ,R2 (x)
=



1,
if 1 ≤ |x| < R1



−1, if R1 ≤ |x| < R2




 0,
if |x| ≥ R2 .
Consider the s−Poisson integral of gR1 ,R2 :
uR1 ,R2 (x) =



Z
Γ n2
gR1 ,R2 (y)
2 s
[1 − |x| ]
dy, if |x| < R
n
2
s
n
2
π B (s, 1 − s)
Rn \B1 (0) [|y| − 1] |x − y|


gR1 ,R2 (x),
if |x| ≥ 1.
Claim I: |uR1 ,R2 (x)| ≤ 1 for all x ∈ Rn .
For any x ∈ Rn and fix. If |x| ≥ R, then uR1 ,R2 (x) = gR1 ,R2 (x). Since |gR1 ,R2 (x)| ≤ 1, then |uR1 ,R2 (x)| ≤ 1.
If |x| < 1, then
|uR1 ,R2 (x)| =
≤
≤
=
=
Z
Γ n2
g
(y)
R
,R
1
2
[1 − |x|2 ]s
dy
n
2
s
n
π 2 B (s, 1 − s)
Rn \B1 (0) [|y| − 1] |x − y|
Z
Γ n2
|gR1 ,R2 (y)|
2 s
[1
−
|x|
]
dy
n
2 − 1]s |x − y|n
[|y|
n
π 2 B (s, 1 − s)
R \B1 (0)
Z
Γ n2
1
2 s
[1
−
|x|
]
dy
n
2 − 1]s |x − y|n
[|y|
n
π 2 B (s, 1 − s)
R \B1 (0)
Since |gR1 ,R2 (y)| ≤ 1 for all y ∈ Rn
Z
1 − |x|2 s
Γ n2
1
dy
n
2
2
|x − y|n
π B (s, 1 − s) |y|>1 1 − |y|
1.
Therefore, we can conclude that |uR1 ,R2 (x)| ≤ 1 for all x ∈ Rn . In a summary, we have known that uR1 ,R2
is a bounded function in Rn and satisfies (−∆)s uR1 ,R2 (x) = 0 for all x ∈ B1 (0).
48
Claim II: For any T > S ≥ 1, then
2 s
Z
2
[1 − |x| ]
[|y| − 1]
−s
−n
|x − y|
S<|y|<T
In fact, for all |x| < 1 and fix, we have
Z
[1 − |x|2 ]s
nαn
dy =
2
Z
T 2 −1
1−|x|2
S 2 −1
1−|x|2
r−s [1 + r]−1 dr,
for all |x| < 1.
[|y|2 − 1]−s |x − y|−n dy
S<|y|<T
[1 − |x|2 ]s
=
Z
T
[r2 − 1]−s rn−1
Z
|x − rz|−n dS(z) dr
S n−1
S
Let y = rz
Z
Z T
2 s
2
−s n−1
[r − 1] r
[1 − |x| ]
x
−n
r − z
dS(z) dr
r
S
S n−1
Z
Z T
−n
x
2 s
2
−s −1
dS(z) dr.
[1 − |x| ]
[r − 1] r
− z
S
S n−1 r
=
=
−n
x
, then |w| < 1. By Theorem 1.4.1, then
r
Z
x −n
nαn
dS(z) =
z − 2
r
∂B1 (0)
1 − x
Since |x| < R ≤ S ≤ r ≤ T , let w =
r
= nαn r [r2 − |x|2 ]−1 .
2
Hence, we get
[1 − |x|2 ]s
Z
[|y|2 − 1]−s |x − y|−n dy
S<|y|<T
=
[1 − |x|2 ]s
Z
T
[r2 − 1]−s r−1 nαn r2 [r2 − |x|2 ]−1 dr
S
=
nαn [1 − |x|2 ]s
Z
T
[r2 − 1]−s [r2 − |x|2 ]−1 r dr
S
=
Z
T 2 −1
1
dt Let t = r2 − R2
2
−1
Z T 2 −1
nαn
t
2 s
−s
2
2 −1
[1 − |x| ]
t [R − |x| ]
+1
dt
2
1 − |x|2
S 2 −1
Z T 2 −12
1−|x|
nαn
−1
2 s
[1 − |x| ]
[(1 − |x|2 )r]−s [1 − |x|2 ]−1 [r + 1] (1 − |x|2 ) dr
S 2 −1
2
2
nαn [1 − |x|2 ]s
ts [t + 1 − |x|2 ]−1
S 2 −1
=
=
1−|x|
=
nαn
2
Z
T 2 −1
1−|x|2
S 2 −1
1−|x|2
r−s [1 + r]−1 dr.
By the definition of uR1 ,R2 (x), then
"Z
Γ n2
2 s
[1 − |x| ]
[|y|2 − 1]−s |x − y|−n dy
uR1 ,R2 (x) =
n
π 2 B(s, 1 − s)
1≤|y|≤S
Let r =
t
1 − |x|2
49
#
Z
2
−
[|y| − 1]
−s
|x − y|
−n
dy
S≤|y|≤T
=
 R2 −1

Z 1 2
Z R22 −12
1−|x|
1−|x|
Γ n2
nαn 
r−s [1 + r]−1 dr − R2 −1 r−s [1 + r]−1 dr
n
1
π 2 B(s, 1 − s) 2
0
2
1−|x|
=
=
By the Claim II

Z R12 −1
Z R22 −12
n
n
2
1−|x|
1−|x|
Γ 2
Γ 2

r−s [1 + r]−1 dr − R2 −1 r−s [1 + r]−1 dr
n
n
1
π 2 B(s, 1 − s) π 2
0
1−|x|2
 R2 −1

Z 1 2
Z R22 −12
1−|x|
1−|x|
1

r−s [1 + r]−1 dr − R2 −1 r−s [1 + r]−1 dr
1
B(s, 1 − s)
0
2
1−|x|
R12 − 1
R22 − 1
1
2
2
≥
1,
E
=
R
−
1
and
F
=
R
−
1,
then
=
qE,
and
= qF . Consider the
1
2
1 − |x|2
1 − |x|2
1 − |x|2
Z b
function J(a, b) =
r−s [1 + r]−1 dr for all b ≥ a ≥ 0. Then we have
Let q =
a
B(s, 1 − s)uR1 ,R2 (x)
= J(0, qE) − J(qE, qF )
Z qE
Z
=
r−s [1 + r]−1 dr −
0
qF
r−s [1 + r]−1 dr
qE
= K(q, E, F ).
Then we know that
∂K(q, E, F )
∂q
=
[qE]−s [1 + qE]−1 E − [qF ]−s [1 + qF ]−1 F + [qE]−s [1 + qE]−1 E
=
2[qE]−s [1 + qE]−s E − [qF ]−s [1 + qF ]−1 F
1−s
F 1−s
−s 2E
q
−
.
1 + qE
1 + qF
=
We want to find R1 and R2 such that uR1 ,R2 (0) = 0, and uR1 ,R2 (x) > 0 for all 0 < |x| < 1, since q is a
∂K(q, E, F )
>0
strictly increasing function of |x|, then it suffices to find R1 , R2 such that K(1, E, F ) = 0 and
∂q
for all q ≥ 1, that is,
J(0, E) = J(E, F ),
For the condition
and
F 1−s
2E 1−s
>
,
1 + qE
1 + qF
for all q ≥ 1.
2E 1−s
F 1−s
>
for all q ≥ 1, we will find R1 , R2 such that 0 < E ≤ 1 < F . Then we
1 + qE
1 + qF
know that
2E 1−s
F 1−s
>
1 + qE
1 + qF
for all q ≥ 1
⇐⇒
2F s−1 [1 + qF ] > E s−1 [1 + qE]
⇐⇒
2F s−1 − E s−1 > q[E s − 2F s ]
for all q ≥ 1
for all q ≥ 1
50
⇐⇒
E s−1 − 2F s−1 < q[2F s − E s ]
⇐⇒
E s−1 − 2F s−1 < 2F s − E s
⇐⇒
E s−1 + E s < 2[F s−1 + F s ].
for all q ≥ 1
Since 0 < E ≤ 1 < F
Now our goal is to find some R1 , R2 such that 0 < E ≤ 1 < F , J(0, E) = J(E, F ) and E s−1 + E s <
2[F s−1 + F s ]. First, we compare J(0, 1) and J(1, ∞). Notice that
Z 1
J(0, 1) =
r−s [1 + r]−1 dr
0
Z
∞
=
1
Z
−1
1
1
1
dt Let r =
ts 1 +
2
t
t
t
∞
ts+1−2 [1 + t]−1 dt
=
1
Z
=
∞
rs−1 [1 + r]−1 dr.
1
1
, then s − 1 < −s, which implies that J(0, 1) < J(1, ∞). In this case, we let E ∗ = 1, since
2
J(0, 1) < J(1, ∞), then there exists q unique 1 < F ∗ < ∞ such that J(0, 1) = J(0, E ∗ ) = J(1, F ∗ ). On the
If 0 < s <
other hand, we know that (E ∗ )1−s + (E ∗ )s = 2, and 2[(F ∗ )s−1 + (F ∗ )s ] > 2(F ∗ )s > 2 = (E ∗ )1−s + (E ∗ )s .
1
If ≤ s < 1, then s − 1 ≥ −s, which implies that J(0, 1) ≥ J(1, ∞). In this case, for J(0, 1), then there
2
J(0, 1)
exists some 0 < E0 < 1 such that J(0, E0 ) =
. Define M ∗ = sup [E s−1 + E s ], then there exists some
2
E0 ≤E≤1
J(0, 1)
∗
∗
∗ s−1
∗ s
∗
large F > 1 such that M < 2[(F )
+ (F ) ]. For this F , since J(0, E0 ) =
, then there exists some
2
∗
J(0, F )
E0 < E ∗ such that J(0, E ∗ ) =
, which implies that J(0, E ∗ ) = J(E ∗ , F ∗ ). Since J(0, 1) ≥ J(1, ∞),
2
then E∗ < 1. On the other hand, we have (E ∗ )1−s + (E ∗ )s < M ∗ < 2[(F ∗ )s−1 + (F ∗ )s ].
In a summary, we can conclude that we can find some R1 , R2 such that 0 < E ∗ ≤ 1 < F ∗ < ∞, J(0, E ∗ ) =
J(1, F ∗ ) and (E ∗ )1−s + (E ∗ )s < 2[(F ∗ )s−1 + (F ∗ )s ]. Therefore, our statement is true.
Remark 1.4.1. References: Blumenthal, Getoor and Ray[5], Blumenthal and Getoor[4], Kassmann[19],
Landkof[22], Riesz[31], Riesz[32], etc..
51
2. Regularity for Fully Nonlinear Integro-Differential Equations
2.1. Fully Nonlinear Integro-Differential Equations and Viscosity Solutions.
For any r > 0, we define the parabolic cylinders as: Qr = Br (0) × [−r2s , 0].
Lemma 2.1.1. If u satisfies the equation: ut + (−∆)s u = f , then for any λ > 0 and any α ∈ R, we have
v(x, t) = λ−α u(λx, λ2s t) satisfies: vt (x, t) + (−∆)s v(x, t) = λ2s−α f λ (x, t) with f λ (x, t) = f (λx, λ2s t).
Proof. Exercise.
Lemma 2.1.2 (The Point Estimate). For 0 < µ ≤ αn , then there exists some 0 > 0 and θ > 0 which just
depend on s, µ and n such that for function u on Rn × [−2, 0] satisfies:



ut + (−∆)s u ≤ 0 , in B2 (0) × [−2, 0]


 \
≤ 0} (B1 (0) × [−2, −1]) ≥ µ > 0
{u




 u ≤ 1, in Rn × [−2, 0].
There holds that u ≤ 1 − θ in Q1 .
Proof. For any 0 < µ ≤ αn , > 0, and function u on Rn × [−2, 0] satisfies:



 ut + (−∆)s u ≤ , in B2 (0) × [−2, 0]

 \
≤ 0} (B1 (0) × [−2, −1]) ≥ µ > 0
{u




 u ≤ 1, in Rn × [−2, 0].
For any 0 < R1 < 1 < R2 , let mR1 ,R2 : [−2, 0] → R be the solution of the ODE:

T
 m0
B1 (0)| − R2 mR1 ,R2 (t)
R1 ,R2 (t) = R1 |{x : u(x, t) ≤ 0}
 m
(−2) = 0.
R1 ,R2
That is, we have
Z
t
mR1 ,R2 (t) =
−2
\
R1 {x : u(x, s) ≤ 0} B1 (0) eR2 (s−t) ds ≥ 0,
for all −2 ≤ t ≤ 0.
For all t ∈ [−1, 0], then we know that
Z
t
mR1 ,R2 (t) ≥
−2
Z
t
≥
−2
Z
\
R1 {x : u(x, s) ≤ 0} B1 (0) eR2 (−2−t) ds
\
R1 {x : u(x, s) ≤ 0} B1 (0) eR2 (−2−0) ds
−1
≥
−2
\
R1 {x : u(x, s) ≤ 0} B1 (0) eR2 (−2−0) ds
52
Z
−1
\
{x : u(x, s) ≤ 0} B1 (0) ds
=
R1 e−2R1
=
\
R1 e−2R2 {u ≤ 0} (B1 (0) × [−2, −1])
≥
R1 e−2R2 µ.
−2
That is, we have
mR1 ,R2 (t) ≥ R1 e−2R2 µ,
(2.1.1)
for all t ∈ [−1, 0].
On the other hand, for all t ∈ [−2, 0], then
Z t
\
mR1 ,R2 (t) =
R1 {x : u(x, s) ≤ 0} B1 (0) eR2 (s−t) ds
−2
Z
0
≤
R1 nαn ds
−2
=
(2nαn )R1 .
1
1
, then mR1 ,R2 (t) ≤ . Let β be a smooth decreasing function on R such that
4nαn
2
β(t) = 1 if t ≤ 1, and β(t) = 0 if t ≥ 2. Consider the function
Now we restrict R1 ≤
η(x, t) = β(|x|) = β(|x|),
for all (x, t) ∈ Rn × [−2, 0].
It is easy to see that 0 ≤ η(x, t) ≤ 1 for all (x, t) ∈ Rn × [−2, 0], and β(·, t) is a radial function in Rn for
all t ∈ [−2, 0]. For any fixed t ∈ [−2, 0], then η(x, t) = 1 if |x| ≤ 1, and η(x, t) = 0 if |x| ≥ 2, that is, η(·, t)
looks like a radial bump function. Since β ∈ Cc∞ (R), then η(·, t) ∈ Cc∞ (Rn ) ⊂ S(Rn ), which implies that
(−∆)s η(·, t) ∈ C ∞ (Rn ), in particular, there exists some constant B > 0 such that |(−∆)s η(x, t)| ≤ B for all
(x, t) ∈ Rn × [−2, 0].
Now for any |x| ≥ 2, then
(−∆)s η(x, t)
Z
=
Cn,s P.V.
Rn
Z
=
Cn,s P.V.
Rn
η(x, t) − η(y, t)
dy
|x − y|n+2s
−η(y, t)
dy
|x − y|n+2s
Z
=
−Cn,s P.V.
Rn
η(y, t)
dy
|x − y|n+2s
Z
<
−Cn,s P.V.
|y|≤1
Z
=
−Cn,s P.V.
|y|≤1
Z
=
−Cn,s
|y|≤1
η(y, t)
dy
|x − y|n+2s
1
dy
|x − y|n+2s
1
dy
|x − y|n+2s
53
< 0.
By the continuity of (−∆)s η(·, ·), there exists some 0 < R3 < 2 such that (−∆)s η(x, t) ≤ 0 for all |x| ≥ R3 .
Let β1 = β(R3 ), since 0 < R2 < 2, then β1 > 0. Now if η(x, t) = β(|x|) ≤ β1 = η(R3 ), since β is decreasing,
then |x| ≥ R3 , which implies that
(−∆)s η(x, t) ≤ 0,
if η(x, t) ≤ β1 .
Claim I: There exists some 0 < R1∗ ≤ 1 < R2∗ < ∞ which just depend on s and n such that for all 0 < R1 ≤
R1∗ < 1 < R2∗ ≤ R2 , we have u(x, t) ≤ 1 − mR0 ,R1 (t) + 2 in Q1 .
1
Cn,s
∗
∗
∗
Let R1 = max
and R2∗ = 1+B
,
β1 . Assume for some 0 < R1 ≤ R1 and R2 ≤ R2 , we have
4nαn 2 · 3n+2s
u(x1 , t1 ) > 1 − mR1 ,R2 (t1 ) + 2 for some (x1 , t1 ) ∈ Q1 , in the following we will see what conditions R0 and R1
should satisfy. Consider the function
wR1 ,R2 (x, t) = u(x, t) + mR1 ,R2 (t)η(x, t) − (2 + t),
for all (x, t) ∈ Rn × [−2, 0].
Since (x1 , t1 ) ∈ Q1 , then |x1 | ≤ 1 ≤ 1 + A|t1 |, and −1 ≤ t1 ≤ 0, which implies that η(x1 , t1 ) = 1. Hence
wR1 ,R2 (x1 , t1 )
= u(x1 , t1 ) + mR1 ,R2 (t1 )η(x1 , t1 ) − (2 + t1 )
= u(x1 , t1 ) + mR1 ,R2 (t1 ) − 2 − · t1
> 1 − t1
Since t1 ≤ 0.
> 1
On the other hand, for all (x, t) ∈ Rn × [−2, 0] such that |x| ≥ 2, then η(x, t) = 0, 2 + t ≥ 0, and
wR1 ,R2 (x, t)
=
u(x, t) + mR1 ,R2 (t) · η(x, t) − (2 + t)
=
u(x, t) − (2 + t)
≤
u(x, t)
≤
1.
Then there exists some (x0 , t0 ) ∈ B2 (0) × [−2, 0] such that
wR1 ,R2 (x0 , t0 )
= u(x0 , t0 ) + mR1 ,R2 (t0 )η(x0 , t0 ) − (2 + t0 )
=
sup
(x,t)∈Rn ×[−2,0]
≥ wR1 ,R2 (x1 , t1 )
>
1.
wR1 ,R2 (x, t)
54
By the way, for all x ∈ Rn , then
wR1 ,R2 (x, −2)
=
u(x, −2) + mR1 ,R2 (−2)η(x, −2) − (2 − 2)
= u(x, t)
≤ 1.
Hence, we have (x0 , t0 ) ∈ B2 (0) × (−2, 0]. Since (x0 , t0 ) is a global maximum point of wR1 ,R2 in Rn × [−2, 0],
then (wR1 ,R2 )t (x0 , t0 ) ≥ 0 (The inequality may happen when t0 = 0), and ∇wR1 ,R2 (x0 , t0 ) = 0, that is,
ut (x0 , t0 ) + m0R1 ,R2 (t0 )η(x0 , t0 ) + mR1 ,R2 (t0 )ηt (x0 , t0 ) − ≥ 0,
and ∇u(x0 , t0 ) + mR1 ,R2 (t0 )∇η(x0 , t0 ) = 0.
That is,
ut (x0 , t0 ) ≥ −m0R1 ,R2 (t0 ) · η(x0 , t0 ) − mR1 ,R2 (t0 )ηt (x0 , t0 ) + ,
and ∇u(x0 , t0 ) = −mR1 ,R2 (t0 ) · ∇η(x0 , t0 ).
On the other hand, we know that
ut (x0 , t0 ) ≥
−m0R1 ,R2 (t0 ) · η(x0 , t0 ) − mR1 ,R2 (t0 )ηt (x0 , t0 ) + = −m0R1 ,R2 (t0 )η(x0 , t0 ) + .
Sine m(t) ≥ 0 for all t ∈ [−2, 0], then |∇u(x0 , t0 )| = m(t0 )|∇η(x0 , t0 )|. Now consider the function
vR1 ,R2 (x) = u(x, t0 ) + mR1 ,R2 (t0 )η(x, t0 ) = wR1 ,R2 (x, t0 ) + (2 + t0 ),
for all x ∈ Rn .
Since (x0 , t0 ) is a global maximum point of wR1 ,R2 in Rn × [−2, 0], then for all x ∈ Rn , we have
u(x, t0 ) + mR1 ,R2 (t0 )η(x, t0 ) − (2 + t0 )
= wR1 ,R2 (x, t0 )
≤ wR1 ,R2 (x0 , t0 )
= u(x0 , t0 ) + mR1 ,R2 (t0 )η(x0 , t0 ) − (2 + t0 ).
Which implies that
vR1 ,R2 (x)
= u(x, t0 ) + mR1 ,R2 (t0 )η(x, t0 )
≤ u(x0 , t0 ) + mR1 ,R2 (t0 )η(x0 , t0 )
= vR1 ,R2 (x0 ),
for all x ∈ Rn .
That is, x0 is a global maximum point of vR1 ,R2 in Rn . Hence
u(x0 , t0 ) + mR1 ,R2 (t0 )η(x0 , t0 )
= vR1 ,R2 (x0 )
55
≥
vR1 ,R2 (x0 + y)
=
u(x0 + y, t0 ) + mR1 ,R2 (t0 )η(x0 + y, t0 ),
for all y ∈ Rn .
That is, u(x0 , t0 ) − u(x0 + y, t0 ) ≥ −m(t0 )[η(x0 , t0 ) − η(x0 + y, t0 )], for all y ∈ Rn , which implies that
\
(−∆)s u(x0 , t0 ) ≥ −m(t0 )(−∆)s η(x0 , t0 ). Define the set G(t0 ) = y : u(y, t0 ) ≤ 0
B1 (0). Now we estimate
the (−∆)s u(x0 , t0 ), then
(−∆)s u(x0 , t0 )
=
(−∆)s [u + mR1 ,R2 (t0 ) · η(·, t0 )](x0 , t0 ) − (−∆)s [mR1 ,R2 (t0 )η(·, t0 )](x0 , t0 )
−mR1 ,R2 (t0 )(−∆)s η(x0 , t0 ) + (−∆)s vR1 ,R2 (x0 , t0 )
Z
vR1 ,R2 (x0 ) − vR1 ,R2 (x0 + y)
dy
= −mR1 ,R2 (t0 )(−∆)s η(x0 , t0 ) + Cn,s P.V.
|y|n+2s
Rn
Z
vR1 ,R2 (x0 ) − vR1 ,R2 (x0 + y)
0
s
0 0
dy
≥ −mR1 ,R2 (t )(−∆) η(x , t ) + Cn,s P.V.
|y|n+2s
x0 +y∈G(t0 )
=
Since vR1 ,R2 (x0 ) ≥ vR1 ,R2 (x0 + y) for all y ∈ Rn .
For any x0 + y ∈ G(t0 ), then
vR1 ,R2 (x0 ) − vR1 ,R2 (x0 + y)
=
wR1 ,R2 (x0 , t0 ) + (2 + t0 ) − u(x0 + y, t0 ) − mR1 ,R2 (t0 ) · η(x0 + y, t0 )
≥
wR1 ,R2 (x0 , t0 ) − mR1 ,R2 (t0 )η(x0 + y, t0 )
≥
wR1 ,R2 (x0 , t0 ) −
1
2
Since 0 ≤ η(x0 + y, t0 ) ≤ 1, and 0 ≤ mR1 ,R2 (t0 ) ≤
|y|
1
2
1
2
>
1−
=
1
2
≤
|x0 | + |y + x0 |
<
3.
Hence, we get
(−∆)s u(x0 , t0 ) ≥
−mR1 ,R2 (t0 )(−∆)s η(x0 , t0 ) + Cn,s
Z
x0 +y∈G(t0 )
Cn,s
|G(t0 )|
2 · 3n+2s
≥
−mR1 ,R2 (t0 )(−∆)s η(x0 , t0 ) +
≥
−mR1 ,R2 (t0 )(−∆)s η(x0 , t0 ) + R1 |G(t0 )|.
1
2
[3 + A]n+2s
dy
56
If η(x0 , t0 ) ≤ η1 , then (−∆)s β(x0 , t0 ) ≤ 0. Since (x0 , t0 ) ∈ B2+A (0) × [−2, 0], then
≥ ut (x0 , t0 ) + (−∆)s u(x0 , t0 )
≥
−m0R1 ,R2 (t0 )η(x0 , t0 ) + −mR1 ,R2 (t0 )(−∆)s η(x0 , t0 ) + R1 |G(t0 )|
−m0R1 ,R2 (t0 )η(x0 , t0 ) + R1 |G(t0 )| + Since (−∆)s η(x0 , t0 ) ≤ 0 and mR1 ,R2 (t0 ) ≥ 0
h i
\
= − R1 {x : u(x, t0 ) ≤ 0} B1 (0) − R2 mR1 ,R2 (t0 ) η(x0 , t0 ) + R1 |G(t0 )| + ≥
= − R1 |G(t0 )| − R2 mR1 ,R2 (t0 ) η(x0 , t0 ) + R1 |G(t0 )| + = R1 |G(t0 )|[1 − η(x0 , t0 )] + R2 m(t0 )η(x0 , t0 ) + >
Since 0 < η(x0 , t0 ) ≤ 1.
Which is a contradiction. If η(x0 , t0 ) > β1 , since |(−∆)s η(x0 , t0 )| ≤ B and (x0 , t0 ) ∈ B2 (0) × [−2, 0], then
≥
≥
ut (x0 , t0 ) + (−∆)s u(x0 , t0 )
−m0R1 ,R2 (t0 )η(x0 , t0 ) + −mR1 ,R2 (t0 )(−∆)s η(x0 , t0 ) + R1 |G(t0 )|
= −m0R1 ,R2 (t0 )η(x0 , t0 ) + R1 |G(t0 )| − mR1 ,R2 (t0 )(−∆)s η(x0 , t0 ) + ≥
−m0R1 ,R2 (t0 )η(x0 , t0 ) + R1 |G(t0 )| − BmR1 ,R2 (t0 ) + = −m0R1 ,R2 (t0 )η(x0 , t0 ) + R1 |G(t0 )| − BmR1 ,R2 (t0 ) + h i
\
= − R1 {x : u(x, t0 ) ≤ 0} B1 (0) − R2 mR1 ,R2 (t0 ) η(x0 , t0 ) + R1 |G(t0 )| − BmR1 ,R2 (t0 ) + = − R1 |G(t0 )| − R2 mR1 ,R2 (t0 ) η(x0 , t0 ) + R1 |G(t0 )| − BmR1 ,R2 (t0 ) + = R1 |G(t0 )|[1 − η(x0 , t0 )] + [R2 · η(x0 , t0 ) − B]mR1 ,R2 (t0 ) + ≥ [R2 η(x0 , t0 ) − B]mR1 ,R2 (t0 ) + Since 0 < η(x0 , t0 ) ≤ 1
≥ [R2 β1 − B]mR1 ,R2 (t0 ) + ≥ mR1 ,R2 (t0 ) + Since R2 ≥
>
1+B
β1
.
Which is a contradiction. In a summary, therefore, we can conclude that for all 0 < R1 ≤ R1∗ < 1 < R2∗ ≤ R2 ,
R1 · e−2R2 µ
θ
R1 · e−2R2 µ
we have u(x, t) ≤ 1 − mR0 ,R1 (t) + 2 in Q1 . In this case, let θ =
and 0 = =
, by
2
2
4
57
(2.1.1), then
m(t) − 20
≥ R1 e−2R2 µ − 2
=
R1 e−2R2 µ
4
R1 e−2R2 µ
2
= θ,
for all t ∈ [−1, 0].
Which implies that u(x, t) ≤ 1 − θ for all (x, t) ∈ B1 (0) × [−1, 0].
Lemma 2.1.3 (The Diminish of Oscillation). There exists some constant θ, α, 0 , r > 0 which just depend on s
and n such that for all function f and u satisfy:


 ut + (−∆)s u = f,
in Q1


α



M
sup
|u(x1 , t1 ) − u(x2 , t2 )| u ≤ , for all M > 1
r

 BM (0)×[−1,0]



 sup |u(x1 , t1 ) − u(x2 , t2 )| ≤ 1,
and
kf kL∞ (Q1 ) ≤ 0 .

Q1
There holds that sup |u(x1 , t1 ) − u(x2 , t2 )| ≤ 1 − θ.
Qr
Proof. Define the constant R ≥ 2 such that


 2,
1
2
R=
1
1

 2 + 2 2s
, if s < .
2
if s ≥
αn
> 0, by Lemma 2.1.2, then there exists some 0 > 0 and θ > 0 which just depend on s, µ and n
2
such that for function w on Rn × [−2, 0] satisfies:


 wt + (−∆)s w ≤ 30 , in B2 (0) × [−2, 0]


 \
≤ 0} (B1 (0) × [−2, −1]) ≥ µ > 0
{w




 w ≤ 1, in Rn × [−2, 0].
Let µ =
(2R)2α
1
There holds that w ≤ 1 − θ in Q1 . Since lim
=
, then we can find some small α0 > 0 such that
α&0 2s − α
2s
2α
(2R)
1
1
whenever α0 ≥ α > 0, we have Cn,s nαn R−2s
−
≤ 0 . Let r =
, where 0 < l < 1 such that
2s − α
2s
2Rl
2lα = 1, these above θ, α, 0 , r > 0 are what we need. Now for any function f and u satisfy:



ut + (−∆)s u = f,
in Q1


α



M
sup
|u(x1 , t1 ) − u(x2 , t2 )| u ≤ , for all M > 1
r
BM (0)×[−1,0]




1 1
2 2


and
kf kL∞ (Q1 ) ≤ 0 .
 sup |u(x , t ) − u(x , t )| ≤ 1,
Q1
58
Consider the function v defined by:
x
t
,
v(x, t) = u
,
2R (2R)2s
for all (x, t) ∈ Rn × [−(2R)2s , 0].
By the Lemma 2.1.1, then v satisfies
vt + (−∆)s v = (2R)−2s f˜,
Where f˜(x, t) = f
in Q2R .
x
t
for all (x, t) ∈ Q2R . Let
,
2R (2R)2s
1
1
m=
sup u(x, t) + inf u(x, t) =
sup v(x, t) + inf v(x, t) .
Q1
Q2R
2 Q1
2 Q2R
Then either
|B (0)|
\
nαn
1
=
> 0,
{v ≤ m} (B1 (0) × [−2, −1]) ≥
2
2
or
|B (0)|
\
nαn
1
=
> 0.
{v ≥ m} (B1 (0) × [−2, −1]) ≥
2
2
\
If {v ≤ m} (B1 (0) × [−2, −1]) ≥ µ > 0, since sup |u(x1 , t1 ) − u(x2 , t2 )| ≤ 1, then for all (x, t) ∈ Q2R , we
Q1
have
v(x, t) − m
1
sup u(x, t) + inf u(x, t)
≤ sup v(x, t) −
Q1
2 Q1
Q1
1
= sup u(x, t) −
sup u(x, t) + inf u(x, t)
Q1
2 Q1
Q1
1
sup u(x, t) − inf u(x, t)
=
Q1
2 Q1
≤
1
.
2
Let w(x, t) = min{1, 2[v(x, t)−m]} for all (x, t) ∈ Rn ×[−2R, 0], then w(x, t) ≤ 1 for all (x, t) ∈ Rn ×[−2R, 0],
and w(x, t) = 2[v(x, t) − m] for all (x, t) ∈ Q2R , which implies that wt (x, t) = 2vt (x, t), and ∇w(x, t) = 2∇v(x, t)
for all (x, t) ∈ Q2R . Let us compute (−∆)s w(x, t) − 2(−∆)s v(x, t) in QR , in fact for all (x, t) ∈ QR , then
w(x, t) = 2[v(x, t) − m], and
=
=
=
(−∆)s w(x, t) − 2(−∆)s v(x, t)
Z
Z
w(x, t) − w(x + y, t)
v(x, t) − v(x + y, t)
Cn,s P.V.
dy − 2Cn,s P.V.
dy
n+2s
|y|
|y|n+2s
Rn
Rn
Z
w(x, t) − 2v(x, t) − w(x + y, t) + 2v(x + y, t)
Cn,s P.V.
dy
|y|n+2s
Rn
Z
−2m − w(x + y, t) − 2v(x + y, t)
Cn,s P.V.
dy
|y|n+2s
n
R
59
Z
= Cn,s P.V.
x+y∈B2R
−2m − w(x + y, t) + 2v(x + y, t)
dy
|y|n+2s
−2m − w(x + y, t) + 2v(x + y, t)
dy
|y|n+2s
x+y6∈B2R
Z
Z
−2m + 2m
−2m − w(x + y, t) + 2v(x + y, t)
= Cn,s P.V.
dy
+
C
P.V.
dy
n,s
n+2s
|y|
|y|n+2s
x+y∈B2R
x+y6∈B2R
Z
−2m − w(x + y, t) + 2v(x + y, t)
= Cn,s P.V.
dy
|y|n+2s
x+y6∈B2R
Z
+Cn,s P.V.
If w(x+y, t) = 2[v(x+y, t)−m], then −2m−w(x+y, t)+2v(x+y, t) = −2m−2v(x+y, t)+2m+2v(x+y, t) = 0.
So only possible is that w(x + y, t) = 1. Hence,
(−∆)s w(x, t) − 2(−∆)s v(x, t)
Z
= Cn,s P.V.
x+y6∈B2R , 2v(x+y,t)−2m>1
−2m − 1 + 2v(x + y, t)
dy.
|y|n+2s
Then
2[v(x + y, t) − m] ≤ 2
|u(x2 , t2 ) − u(x3 , t3 )|
sup
B |x+y| (0)×[−1,0]
2R
≤ 2
|x + y|
α
1
2Rl
α
≤ 2(2Rl) [|x| + |y|]α
=
2lα (2R)α [|x| + |y|]α
=
(2R)α [|x| + |y|]α .
On the other hand, we know that |y| ≥ |y +x|−|x| ≥ 2R−R = R ≥ |x|, which implies that 2[v(x+y, t)−m] ≤
(2R)α · [2|y|]α = (4R)α · |y|α . Hence, we can obtain
Z
s
s
−(∆) w(x, t) − 2(−∆) v(x, t) ≤ Cn,s
(4R)α |y|α − 1
dy
|y|n+2s
|y|≥R
"
Z
Z
= Cn,s (4R)α
|y|−n−2s+α dy −
|y|≥R
4α · Rα α−2s
1
= Cn,s nαn
R
− R−2s
2s − α
2s
(2R)2α
1
≤ Cn,s nαn R−2s
−
2s − α
2s
|y|≥R
≤ 0 .
Which implies that for all (x, t) ∈ QR , we have
wt + (−∆)s w
=
2vt + 2(−∆)s v + (−∆)s w − 2(−∆)s v
≤
2vt + 2(−∆)s v + 0
#
−n−2s
|y|
dy
60
=
2(2R)−2s · f˜ + 0
≤
2kf˜kL∞ (QR ) + 0
=
2kf kL∞ (Q1 ) + 0
≤
20 + 0
=
30 .
Since R ≥ 2
1
1
1
1
, then R2s ≥ R ≥ 2. If s < , since R = 2 + +2 2s , then R2s ≥ 2 2s ·2s = 2. Since R ≥ A + 2, then
2
2
B2+A (0) × [−2, 0] ⊂ QR , and
If s ≥
wt + (−∆)s w ≤ 30 ,
in B2+A (0) × [−2, 0].
\
Since {v ≤ m} (B1 (0) × [−2, −1]) ≥ µ > 0, and w(x, t) = v(x, t) − m for all (x, t) ∈ QR , which implies
\
that {w ≤ 0} (B1 (0) × [−2, −1]) ≥ µ > 0. In a summary, we know that w satisfies:



wt + (−∆)s w ≤ 20 , in B2 (0) × [−2, 0]


 \
≤ 0} (B1 (0) × [−2, −1]) ≥ µ > 0
{w




 w ≤ 1, in Rn × [−2, 0].
Therefore, we get w ≤ 1 − θ in B1 (0) × [−1, 0], that is, v − m ≤ 1 − θ in B1 (0) × [−1, 0] = Q1 . Hence we get
2[u(x, t) − m] ≤ 1 − θ,
1 = Qr .
in Q 2R
Hence, we get
sup |u(x1 , t1 ) − u(x2 , t2 )|
≤ sup |u(x1 , t1 ) − m| + sup |u(x1 , t1 ) − m|
Qr
Qr
Qr
≤
1−θ 1−θ
+
2
2
=
1 − θ.
|B (0)|
\
nαn
1
If {v ≥ m} (B1 (0) × [−2, −1]) ≥
=
> 0, let w̃(x, t) = min {1, 2[m − v(x, t)]} for all (x, t) ∈
2
2
Rn × [−(2R)2s , 0]. By the same argument, we know that sup |u(x1 , t1 ) − u(x2 , t2 )| ≤ 1 − θ. Therefore, we can
Qr
conclude that sup |u(x1 , t1 ) − u(x2 , t2 )| ≤ 1 − θ.
Qr
Theorem 2.1.1. Let 0 < s < 1, f ∈ L∞ (Q1 ) and u ∈ L∞ (Rn × [−1, 0]) such that
ut + (−∆)s u = f,
in Q1 .
61
Then there exists some 0 < α < 1 and C > 0 which just depend on s and n such that
α |u(x, t) − u(0, 0)| ≤ C |x|α + |t| 2s kukL∞ (Rn ×[−1,0]) + kf kL∞ (Q1 ) ,
for all (x, t) ∈ Rn × [−1, 0].
Proof. Since b ∈ L∞ (Q1 ), let θ, α0 , 0 > 0 and 0 < r < 1 be the ones in the Lemma 2.1.3 which just depend
on s and n. Since f ∈ L∞ (Q1 ) and u ∈ L∞ (Rn ) , then we can consider the function:
1
v(x, t) =
2kukL∞ (Rn ×[−1,0]) +
kf kL∞ (Q1 )
0
u(x, t),
for all (x, t) ∈ Rn × [−1, 0].
Notice that
1
|v(x, t)| =
2kukL∞ (Rn ×[−1,0]) +
1
≤
2kukL∞ (Rn ×[−1,0]) +
≤
kf kL∞ (Q1 )
0
1
,
2
kf kL∞ (Q1 )
0
· |u(x, t)|
· kukL∞ (Rn ×[−1,0])
for all (x, t) ∈ Rn × [−1, 0].
Which implies that kvkL∞ (Rn ×[−1,0]) and
|v(x1 , t1 ) − v(x2 , t2 )| ≤ 1. Also we know that v satisfies
sup
Rn ×[−1,0]
the PDE:
f
vt + (−∆)s v =
2kukL∞ (Rn ×[−1,0]) +
kf kL∞ (Q1 )
0
= g,
in Q1 .
Notice that
|f (x, t)|
|g(x, t)| =
2kukL∞ (Rn ×[−1,0]) +
kf kL∞ (Q1 )
0
kf kL∞ (Q1 )
≤
2kukL∞ (Rn ×[−1,0]) +
≤ 0 ,
kf kL∞ (Q1 )
0
for all (x, t) ∈ Q1 .
That is, kgkL∞ (Q1 ) ≤ 0 . Since rα → 1, as α & 0, then there exists some 0 < α ≤ min{2s, α0 } such the
1 − θ ≤ rα .
Claim I: sup |v(x1 , t1 ) − v(x2 , t2 )| ≤ rαk for all k = 0, 1, 2, · · · .
Qrk
We prove the Claim I by the induction. When k = 0, then sup |v(x1 , t1 ) − v(x2 , t2 )| ≤
Q1
sup
|v(x1 , t1 ) −
Rn ×[−1,0]
v(x2 , t2 )| ≤ 1, that is, the Claim I is true for k = 0. Now assume the Claim I is true for 0 ≤ k ≤ m, we need to
prove the Claim I is also true for k = m + 1. In fact, we consider the function
w(x, t) = r−αm v(rm x, r2sm t),
for all (x, t) ∈ Rn × [−r−2s , 0].
62
By the Lemma 2.1.1, then w satisfies:
wt (x, t) + (−∆)s w(x, t) = rm(2s−α) g(rm x, r2sm t) = h(x, t),
for all (x, t) ∈ Qr−m .
Since 0 < r < 1, and m ≥ 0, then r−m ≥ 1, in particular, we have
wt (x, t) + (−∆)s w(x, t) = rm(2s−α) g(rm x, r2sm t) = h(x, t),
for all (x, t) ∈ Q1 .
Since 0 < α ≤ 2s, and 0 < r < 1, then
|h(x, t)|
≤
rm(2s−α) |g(rm x, r2sm t)|
≤
kgkL∞ (Q1 )
≤
0 ,
for all (x, t) ∈ Qr−m .
By induction, we know that sup |v(x1 , t1 ) − v(x2 , t2 )| ≤ rαm , then
Qrm
sup |w(x1 , t1 ) − w(x2 , t2 )| = r−αm sup |v(x1 , t1 ) − v(x2 , t2 )|
Q1
Qrm
≤ r
=
−αm αm
r
1.
For all 1 ≤ l ≤ m, by induction, we know that sup |v(x1 , t1 ) − v(x2 , t2 )| ≤ rα(m−l) , then
Qrm−l
sup |w(x1 , t1 ) − w(x2 , t2 )| = r−αm sup |v(x1 , t1 ) − v(x2 , t2 )|
Qr−l
Qrm−l
≤ r−αm rα(m−l)
≤ r−αl .
For m < l, since 0 < r < 1, then r−αm ≤ r−αl . Hence
sup |w(x1 , t1 ) − w(x2 , t2 )| = r−αm sup |v(x1 , t1 ) − v(x2 , t2 )|
Qr−l
Qrm−l
≤ r−αm
sup
|v(x1 , t1 ) − v(x2 , t2 )|
Rn ×[−1,0]
≤ r−αm
≤ r−αl .
In a summary, we have
sup |w(x1 , t1 ) − w(x2 , t2 )| ≤ r−αl ,
Qr−l
for all l = 0, 1, 2, · · · ,.
63
For any M > 1, since 0 < r < 1, then there exists some l ≥ 1 such that r−l ≥ M > r−l+1 , which implies
M
. Hence
that r−l ≤
r
|w(x1 , t1 ) − w(x2 , t2 )|
sup
≤
sup |w(x1 , t1 ) − w(x2 , t2 )|
Qr−l
BM ×[−1,0]
≤
≤
r−αl
α
M
.
r
Now by the Lemma 2.1.3, then sup |w(x1 , t1 ) − w(x2 , t2 )| ≤ 1 − θ ≤ rα . Since w(x, t) = r−αm v(rm x, r2sm t)
Qr
for all (x, t) ∈ Rn × [−r−m , 0], then v(x, t) = rαm w(r−m x, r−2sm t) for all (x, t) ∈ Rn × [−1, 0], which implies
that
sup
|v(x1 , t1 ) − v(x2 , t2 )| =
rαm sup |w(x1 , t1 ) − w(x2 , t2 )|
Qrm+1
Qr
αm α
≤
r
=
rα(m+1) .
r
That is, the Claim I is true for k = m + 1. By the induction, we can conclude that the Claim I is true for
all k = 0, 1, 2, · · · , that is, sup |v(x1 , t1 ) − v(x2 , t2 )| ≤ rαk , and Qrk = Brk (0) × [−r2sk , 0] for all k = 0, 1, 2, · · · .
Qr k
1
1
Now for any (x, t) ∈ Q1 such that |x| + |t| 2s < 1, then there exists some k ≥ 0 such that rk+1 ≤ |x| + |t| 2s ≤ rk ,
in particular, x ∈ Qrk . By the Claim I, then
|v(x, t) − v(0, 0)|
≤
sup |v(x1 , t1 ) − v(x2 , t2 )|
Qrk
≤
rαk
=
rα(k+1) r−α
iα
h
1
r−α |x| + |t| 2s
≤
α ≤ r−α |x|α + |t| 2s ,
Since 0 < α < 1.
1
1
For all (x, t) ∈ Rn × [−1, 0] such that |x| + |t| 2s ≥ 1, since 0 < α < 1, then [|x| + |t| 2s ]α > 1. Hence we know
that
|v(x, t) − v(0, 0)|
≤
2kvkL∞ (Rn ×[−1,0])
≤
1
≤
[|x| + |t| 2s ]α
≤
|x|α + |t| 2s .
1
α
64
In a summary, we can conclude that there exists some constant A > 0 which just depends on s and n such
that for all (x, t) ∈ Rn × [−1, 0], we have
α |v(x, t) − v(0, 0)| ≤ A |x|α + |t| 2s .
By the definition of v, that is,
1
v(x, t) =
2kukL∞ (Rn ×[−1,0]) +
kf kL∞ (Q1 )
0
u(x, t),
for all (x, t) ∈ Rn × [−1, 0].
Then we know that
|u(x, t) − u(0, 0)| =
≤
kf kL∞ (Q1 )
α 2kukL∞ (Rn ×[−1,0]) +
A · A |x|α + |t| 2s
0
α
α C |x| + |t| 2s [kukL∞ (Rn ×[−1,0]) + kf kL∞ (Q1 ) ], for all (x, t) ∈ Rn × [−1, 0].
Remark 2.1.1. References: Caffarelli and Silvestre[7], Silvestre[34], Silvestre[35], Silvestre[35], etc..
2.2. Nonlocal ABP Estimates and Harnack Inequality.
Theorem 2.2.1. Let u ≤ 0 in Rn \B1 and Γ be its concave envelop in B3 . Assume M + u(x) ≥ −f (x). Then
there is a finite family of open cubes Qj , j = 1, · · · , m with diameters dj and constants C(n, λ, Λ) > 0 and
µ(n, λ, Λ) > 0 such that the followings hold:
T
Qj = ∅ for all 1 ≤ i < j ≤ m.
m
[
b. {x : u(x) = Γ(x)} ⊂
Qj .
a. Qi
c. {x : u(x) = Γ(x)}
1
− 2−σ
d. dj ≤ ρ0
\ j=1
Qj 6= ∅.
, where ρ0 =
"
#n
1
√
.
8 n
e. |∇Γ(Qj )| ≤ C sup f |Qj |.
Qj
(
"
#n )
√
f. y ∈ 8 nQj : u(y) > Γ(y) − C sup f d2j ≥ µ|Qj |.
Qj
Theorem 2.2.2. Let u ≥ 0 in Rn , M − u ≤ C0 and M + u ≥ −C0 in B2 . Assume σ ≥ σ0 for some σ0 > 0.
Then there exists some constant C(n, λ, Λ) > 0 such that
sup u ≤ C[u(0) + C0 ].
B1
2
65
Theorem 2.2.3. Let σ > σ0 for some σ0 > 0 and u ∈ L∞ (Rn ) such that M + u ≥ −C0 and M − u ≤ C0 in B1 .
Then there exists some α(n, λ, Λ, σ0 ) > 0 and C(n, λ, Λ, σ0 ) > 0 such that u ∈ C α (B 21 ), and
kukC α (B 1 ) ≤ C kukL∞ (Rn ) + C0 . .
2
Theorem 2.2.4. Let σ > σ0 and I be a nonlocal elliptic operator with respect to L1 . Assume u is a bounded
function such that Iu = 0 in B1 . Then there exists some α(n, λ, Λ, σ0 ) > 0 and C(n, λ, Λ, σ0 ) > 0such that
u ∈ C 1+α (B 21 ) and
kukC 1+α (B 1 ) ≤ C kukL∞ (Rn ) + |I0| .
2
Remark 2.2.1. References: Caffarelli and Silvestre[7], Dong and Kim[10], Gilbarg and Trudinger[16], Kim and
Lee[20], Kim and Lee[21].
66
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