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Chapter 2
Sobolev Spaces
Sobolev spaces turn out often to be the proper setting in which to
apply ideas of functional analysis to get information concerning par-
tial differential equations. Here, we collect a few basic results about
Sobolev spaces. A general reference to this topic is Adams [1], Gilbarg-
Trudinger [29], or Evans [26].
2.1 Holder spaces
Assume Ω ⊂ Rn is open and 0 < γ ≤ 1.
Definition 2.1.1 (i) If u : Ω → R is bounded and continuous, we
write
‖u‖C(Ω) := supx∈Ω
|u(x)|.
(ii) The γth-Holder semi-norm of u : Ω → R is
[u]C0,γ(Ω) := supx,y∈Ω, x 6=y
|u(x) − u(y)|
|x− y|γ
,
and the γth-Holder norm is
‖u‖C0,γ (Ω) := ‖u‖C(Ω) + [u]C0,γ (Ω).
(iii) The Holder space
Ck,γ(Ω)
11
12 CHAPTER 2. SOBOLEV SPACES
consists of all functions u ∈ Ck(Ω) for which the norm
‖u‖Ck,γ (Ω) :=∑
|α|≤k
‖Dαu‖C(Ω) +∑
|α|=k
[Dαu]C0,γ (Ω) (2.1.1)
is finite.
Theorem 2.1.2 The space of functions Ck,γ(Ω) is a Banach space.
Proof. It is easy to verify that ‖ · ‖Ck,γ(Ω) is a norm (Exercise). It
suffices to check the completeness of Ck,γ(Ω).
1. Construction of limit functions.
Let um∞m=1 be a Cauchy sequence in Ck,γ(Ω), i.e.,
‖um − un‖Ck,γ(Ω) → 0, as m,n→ +∞.
Since any Cauchy sequence is a bounded sequence, so ‖um‖Ck,γ (Ω) is
uniformly bounded. Particularly, ‖um‖C(Ω) is uniformly bounded, that
is, um(x) is a uniformly bounded sequence. Furthermore, the bound-
edness of ‖Dum‖C(Ω) or [Dαum]C0,γ (Ω) implies the equicontinuity of the
sequence um(x). The Arzela-Ascoli theorem implies there exists a
subsequence, still denoted by um∞m=1, which converges to a bounded
continuous function u(x) uniformly in Ω. The same argument can be
repeated for any derivative of Dαum of order up to and including k.
That is, there exist functions uα ∈ C(Ω), such that
Dαum → uα, uniformly in Ω, ∀ |α| ≤ k.
2. Dαu = uα.
Deduce by induction, first let α = (1, 0, · · · , 0), since the series
u1(x) ++∞∑
j=1
(uj+1 − uj) = limm→∞
um+1(x) = u(x),
uniformly in Ω and the series
Dαu1(x) +
+∞∑
j=1
(Dαuj+1 −Dαuj) = limm→∞
Dαum+1(x) = uα(x),
2.1. HOLDER SPACES 13
uniformly in Ω. The theory of series implies that u(x) is differentiable
in Ω and Dαu = uα. The induction method shows that Dαu = uα is
valid for all multi-index α such that |α| ≤ k.
3. u ∈ Ck,γ(Ω).
It suffices to show that for any multi-index α such that |α| = k,
there holds [Dαu]C0,γ (Ω) < +∞.
For any x, y ∈ Ω, x 6= y, since Dαum → Dαu uniformly in Ω, there
exists a N large enough, such that:
|DαuN (z) −Dαu(z)| ≤1
2|x− y|γ , ∀ z ∈ Ω.
Calculate:
|Dαu(x) −Dαu(y)| ≤ |Dαu(x) −DαuN (x)| + |DαuN (x) −DαuN (y)|
+|DαuN (y) −Dαu(y)|
≤ |x− y|γ + |x− y|γ‖DαuN‖C0,γ(Ω)
≤ (1 + supm
‖Dαum‖C0,γ (Ω))|x− y|γ ,
Thus the boundedness of ‖Dαum‖C0,γ(Ω) implies [Dαu]C0,γ(Ω) < +∞.
4. um → u in Ck,γ(Ω).
It also suffices to show that for any multi-index α such that |α| = k,
there holds
[Dαum −Dαu]C0,γ (Ω) → 0, as m→ ∞.
In fact, for any x, y ∈ Ω, x 6= y, there holds
|Dαum(x) −Dαu(x) − (Dαum(y) −Dαu(y))|
≤ lim supn→∞
|Dαum(x) −Dαun(x) − (Dαum(y) −Dαun(y))|
≤ |x− y|γ limn→∞
[Dαum −Dαun]C0,γ (Ω).
Since um∞m=1 is a Cauchy sequence in Ck,γ(Ω), [Dαum−Dαu]C0,γ(Ω) ≤
limn→∞
[Dαum −Dαun]C0,γ(Ω) → 0 as m→ 0.
14 CHAPTER 2. SOBOLEV SPACES
2.2 Sobolev spaces
2.2.1 Lp spaces
The Lp(Ω) space consists of all the function u : Ω → R such that∫
Ω|u|p dx <∞, where 1 ≤ p ≤ ∞; with norm
‖u‖Lp(Ω) =
(
∫
Ω|u|p dx
)1/p, if 1 ≤ p <∞,
esssupΩ
|u|, if p = ∞.
In order to deduce some properties of Lp space, let us start from some
basic inequalities, which will be used latter.
Lemma 2.2.1 (Young inequality) Let 1 < p, q < ∞, 1p + 1
q = 1,
then
ab ≤ap
p+bq
q, a, b > 0.
Proof. Method 1. The mapping x 7→ ex is convex, and consequently,
ab = elog a+log b = e1p
log ap+ 1q
log bq
≤1
pelog ap
+1
qelog bq
=ap
p+bq
q.
Method 2. Let y = xp−1, x ≥ 0, then x = y1
p−1 = yq−1, Consider the
area of the region restricted by y = 0, x = a, and curve y = xp−1 is
I =
∫ a
0xp−1 dx =
ap
p;
and the area of the region restricted by x = 0, y = b, and curve
x = yq−1 is
II =
∫ b
0yq−1 dy =
bq
q;
The area of the rectangle [0, a]×[0, b] is ab, Thus ab ≤ I+II ≤ap
p+bq
q,
and the equality holds if and only if b = ap−1.
It is easy to obtain the following Young inequality with ε (Exercise):
2.2. SOBOLEV SPACES 15
Lemma 2.2.2 (Young inequality with ε) Let 1 < p, q < ∞, 1p +
1q = 1, then
ab ≤ εap
p+
1
εq/p
bq
q, a, b > 0.
Lemma 2.2.3 (Holder inequality) Let 1 ≤ p, q ≤ ∞, 1p + 1
q = 1,
then if u ∈ Lp(Ω), v ∈ Lq(Ω), we have
∫
Ω|uv| dx ≤ ‖u‖Lp(Ω)‖v‖Lq(Ω).
Proof. 1. In the cases where p = ∞ or q = ∞, it is easy, because there
exists a subset Ω′ ⊂ Ω, with |Ω′| = |Ω|, such that supΩ′
|u| = ‖u‖L∞(Ω)
or supΩ′
|v| = ‖v‖L∞(Ω).
2. In the cases where 1 < p, q < ∞. By the homogeneity of the
inequality, we may assume that ‖u‖Lp(Ω) = ‖v‖Lq(Ω) = 1. Then the
Young inequality implies that
∫
Ω|uv| dx ≤
1
p
∫
Ω|u|p dx+
1
q
∫
Ω|u|q dx = 1 = ‖u‖Lp(Ω)‖v‖Lq(Ω).
By induction argument, it is easy to show the following Generalized
Holder inequality (Exercise):
Lemma 2.2.4 (Generalized Holder inequality) Let 1 ≤ p1, · · · , pm ≤
∞, 1p1
+ · · · + 1pm
= 1, then if uk ∈ Lpk(Ω) for k = 1, · · · ,m, we have
∫
Ω|u1 · · · um| dx ≤
mΠ
k=1‖uk‖Lpk (Ω).
Lemma 2.2.5 (Minkowski inequality) Assume 1 ≤ p ≤ ∞, u, v ∈
Lp(Ω), then
‖u+ v‖Lp(Ω) ≤ ‖u‖Lp(Ω) + ‖v‖Lp(Ω).
16 CHAPTER 2. SOBOLEV SPACES
Proof. From the Holder inequality, we have
‖u+ v‖pLp(Ω) =
∫
Ω|u+ v|p dx
≤
∫
Ω|u+ v|p−1(|u| + |v|) dx
≤(
∫
Ω|u+ v|
(p−1) pp−1 dx
)p−1
p(
(
∫
Ω|u|p dx)1/p + (
∫
Ω|v|p dx)1/p
)
= ‖u+ v‖p−1Lp(Ω)(‖u‖Lp(Ω) + ‖v‖Lp(Ω)).
Lemma 2.2.6 (Interpolation inequality for Lp- norms) Assume 1 ≤
s ≤ r ≤ t ≤ ∞, and 1r = θ
s + (1−θ)t . Suppose also u ∈ Ls(Ω) ∩ Lt(Ω).
Then u ∈ Lr(Ω), and
‖u‖Lr(Ω) ≤ ‖u‖θLs(Ω)‖u‖
(1−θ)Lt(Ω).
Proof. From the Holder inequality, we have
∫
Ω|u|r dx =
∫
Ω|u|θr|u|(1−θ)r dx
≤(
∫
Ω|u|θr s
θr dx)
θrs(
∫
Ω|u|
(1−θ)r t(1−θ)r dx
)
(1−θ)rt .
Theorem 2.2.7 (Completeness of Lp(Ω)) Lp(Ω) is a Banach space
if 1 ≤ p ≤ ∞.
Proof. It is easy to check that ‖·‖Lp(Ω) is a norm (Exercise). It suffices
to verify the completeness of Lp(Ω)
1. Construction of the limit function.
Assume that 1 ≤ p < ∞ and let um be a Cauchy sequence in
Lp(Ω). There is a subsequence umj of um such that
‖umj+1 − umj‖p ≤ 1/2j , j = 1, 2 · · · .
2.2. SOBOLEV SPACES 17
Let vm(x) =m∑
j=1|umj+1(x) − umj (x)|, then
‖vm‖p ≤
m∑
j=1
‖umj+1 − umj‖p ≤
m∑
j=1
1/2j < 1.
Putting v(x) = limm→∞
vm(x), which may be infinite for some x, we obtain
by the Fatou Lemma that∫
Ω|v(x)|p dx ≤ lim inf
m→∞
∫
Ω|vm(x)|p dx ≤ 1.
Hence v(x) < ∞ a.e. in Ω and its absolutely convergence implies the
series
um1 +∞∑
j=1
(umj+1(x) − umj (x)) (2.2.1)
converges to a limit u(x) a.e. in Ω. Let u(x) = 0 whenever it is
undefined as the limit of series (2.2.1). Thus
limj→∞
umj (x) = u(x) a.e. in Ω.
2. u ∈ Lp(Ω) and um → u in Lp(Ω).
For any ε > 0, there exists N such that if m,n ≥ N , then ‖um −
un‖p < ε. Hence by the Fatou Lemma again, we have∫
Ω|um(x) − u(x)|p dx =
∫
Ωlim
j→∞|um(x) − umj (x)|
p dx
≤ lim infj→∞
∫
Ω|um(x) − umj (x)|
p dx
≤ εp.
If m ≥ N , thus u = (u − um) + um ∈ Lp(Ω), and ‖um − u‖p → 0 as
m→ ∞.
3. In the case where p = ∞, let um be a Cauchy sequence in L∞(Ω),
then there exists a subset A ⊂ Ω, with |A| = 0, such that if x ∈ Ω \A,
then for n,m = 1, · · · , there holds
|um(x)| ≤ ‖um‖∞, |um(x) − un(x)| ≤ ‖um − un‖∞.
That is, um(x) is uniformly bounded in Ω \ A, and um(x) is a Cauchy
sequence for any x ∈ Ω\A, thus um(x) converges uniformly on Ω\A to
18 CHAPTER 2. SOBOLEV SPACES
a bounded function u(x). Set u(x) = 0 if x ∈ A. We have u ∈ L∞(Ω)
and ‖um − u‖∞ → 0 as m→ ∞.
Theorem 2.2.8 (Riesz Representation Theorem for Lp(Ω)) Let
1 < p <∞ and L ∈ [Lp(Ω)]′. Then there exists v ∈ Lp′(Ω),1
p+
1
p′= 1
such that for all u ∈ Lp(Ω), there holds
L(u) =
∫
Ωu(x)v(x) dx.
Moreover, ‖v‖Lp′ (Ω) = ‖L‖[Lp(Ω)]′ , that is,
[Lp(Ω)]′ ≅ Lp′(Ω).
Theorem 2.2.9 (Riesz Representation Theorem for L1(Ω)) Let
L ∈ [L1(Ω)]′. Then there exists v ∈ L∞(Ω) such that for all u ∈ L1(Ω),
there holds
L(u) =
∫
Ωu(x)v(x) dx.
Moreover, ‖v‖L∞(Ω) = ‖L‖[L1(Ω)]′ , that is,
[L1(Ω)]′ ≅ L∞(Ω).
Theorem 2.2.10 (Reflexivity of Lp) Lp(Ω) is reflexive if and only
if 1 < p <∞.
Theorem 2.2.11 C∞0 (Ω) is dense in Lp(Ω) if 1 ≤ p <∞.
Theorem 2.2.12 (Separability of Lp(Ω)) Lp(Ω) is separable 1 ≤
p <∞.
2.2.2 Weak derivatives
Let C∞0 (Ω) denotes the space of infinitely differentiable functions φ :
Ω → R, with compact support in Ω. We will call a function φ belonging
to C∞0 (Ω) a test function.
2.2. SOBOLEV SPACES 19
Motivation for definition of weak derivatives. Assume we are
given a function u ∈ C1(Ω). Then if ϕ ∈ C∞0 (Ω), we see from the
integration by parts that
∫
Ωuφxi dx = −
∫
Ωuxiφdx (i = 1, · · · , n). (2.2.2)
More general now, if k is a positive integer, u ∈ Ck(Ω), and α =
(α1, · · · , αn) is a multi-index of order |α| = α1 + · · · + αn = k, then
∫
Ωu(Dαφ) dx = (−1)|α|
∫
Ω(Dαu)φdx. (2.2.3)
We next examine that formula (2.2.3) is valid for u ∈ Ck(Ω), and
ask whether some variant of it might still be true even if u is not k times
continuously differentiable. Note that the left-hand side of (2.2.3) make
sense if u is only locally summable: the problem is rather that if u is
not Ck, then the expression Dαu on the right-hand side of (2.2.3) has
no obvious meaning. We overcome this difficulty by asking if there
exists a locally summable function v for which formula (2.2.3) is valid,
with v replacing Dαu:
Definition 2.2.13 Suppose u, v ∈ L1loc(Ω), and α is a multi-index. We
say that v is the αth-weak partial derivative of u, written
Dαu = v,
provided that∫
ΩuDαφdx = (−1)|α|
∫
Ωvφ dx (2.2.4)
for all test functions ϕ ∈ C∞0 (Ω).
Lemma 2.2.14 (Uniqueness of weak derivative) The weak partial
derivative of u, if it exists, is uniquely defined up to a set of measure
zero.
Example 2.2.15 Let n = 1, Ω = (0, 2), and
u(x) =
x, if 0 < x ≤ 1,
1, if 1 ≤ x < 2.
20 CHAPTER 2. SOBOLEV SPACES
Define
v(x) =
1, if 0 < x ≤ 1,
0, if 1 ≤ x < 2.
From the definition 2.2.13, u′ = v. (Exercise)
Example 2.2.16 Let n = 1, Ω = (0, 2), and
u(x) =
x, if 0 < x ≤ 1,
2, if 1 ≤ x < 2.
We assert u′ does not exist in the weak sense. (Exercise)
2.2.3 Definition of Sobolev spaces
Fix 1 ≤ p ≤ ∞ and let k be a nonnegative integer. We define now cer-
tain function spaces, whose members have weak derivatives of various
orders lying in various Lp spaces.
Definition 2.2.17 i) The Sobolev space
W k,p(Ω)
consists of all locally summable functions u : Ω → R such that
for each multi-index α with |α| ≤ k, Dαu exists in the weak sense
and belongs to Lp(Ω).
ii) If u ∈W k,p(Ω), we define its norm to be
‖u‖W k,p(Ω) :=
(∑
|α|≤k
∫
Ω|Dαu|p dx
)1/p, (1 ≤ p <∞),
∑
|α|≤k
ess supΩ
|Dαu|, (p = ∞).
Definition 2.2.18 (Convergence in W k,p(Ω)) i) Let um∞m=1, u ∈
W k,p(Ω). We say um converges (strongly) to u in W k,p(Ω), writ-
ten
um → u in W k,p(Ω),
provided that
limm→∞
‖um − u‖W k,p(Ω) = 0.
2.2. SOBOLEV SPACES 21
ii) We write
um → u in W k,ploc (Ω)
to mean
um → u in W k,p(U),
for each U ⊂⊂ Ω.
Definition 2.2.19 We denote by
W k,p0 (Ω)
the closure of C∞0 (Ω) in W k,p(Ω).
Remark 2.2.20 (i) If p = 2, we usually write
Hk(Ω) = W k,2(Ω), Hk0 (Ω) = W k,2
0 (Ω), (k = 0, 1, · · · ).
The letter H is used, since - as we will see - Hk(Ω) is a Hilbert
space. Note that H0(Ω) = L2(Ω).
(ii) We henceforth identity functions in W k,p(Ω) which agree almost
everywhere.
Example 2.2.21 Let Ω = B(0, 1), the open unit ball in Rn, and
u(x) = |x|−α (x ∈ Ω, x 6= 0), α > 0.
Then u ∈ W 1,p(Ω) if and only if 0 < (α + 1)p < n. In particular
u 6∈W 1,p(Ω) for each p ≥ n. (Exercise)
Example 2.2.22 Let rk∞k=1 be a countable, dense subset of Ω =
B(0, 1) ∈ Rn. Set
u(x) =
∞∑
k=1
1
2k|x− rk|
−α (x ∈ Ω).
Then u ∈ W 1,p(Ω) if and only if (α + 1)p < n. If 0 < (α + 1)p < n,
we see that u belongs to W 1,p(Ω) and yet is unbounded on each open
subset of Ω. (Exercise)
Theorem 2.2.23 (Properties of weak derivatives) Assume u, v ∈
W k,p(Ω), |α| ≤ k. Then
22 CHAPTER 2. SOBOLEV SPACES
(i) Dαu ∈ W k−|α|,p(Ω) and Dβ(Dαu) = Dα(Dβu) = Dα+βu for all
multi-indices α, β with |α+ β| ≤ k;
(ii) For each λ, µ ∈ R, λu + µv ∈ W k,p(Ω) and Dα(λu + µv) =
λDαu+ µDαv, |α| ≤ k;
(iii) If Ω′ is an open subset of Ω, then u ∈W k,p(Ω′);
(iv) If ξ ∈ C∞0 (Ω), then ξu ∈W k,p(Ω) and
Dα(ξu) =∑
β≤α
(
α
β
)
DβξDα−βu (Leibniz′ formula),
where
(
α
β
)
=α!
β!(α − β)!.
Theorem 2.2.24 (Sobolev spaces as function spaces) For each k
= 0, 1, · · · and 1 ≤ p ≤ ∞, Sobolev space W k,p(Ω) is a Banach space;
W k,p(Ω) is reflexive if and only if 1 < p < ∞; For 1 ≤ p < ∞, the
subspace W k,p∩C∞(Ω) is dense in W k,p(Ω), thus W k,p(Ω) is separable;
Moreover, W k,2(Ω) is a Hilbert space with scalar product
(u, v)W k,2 =∑
α≤k
∫
ΩDαuDαv dx.
Proof. 1. It is easy to check that ‖ · ‖W k,p(Ω) is a norm (Exercise).
Next to show that W k,p(Ω) is complete.
2. Construction of limit functions.
Assume um∞m=1 is a Cauchy sequence in W k,p(Ω). Then for each
|α| ≤ k, Dαum∞m=1 is a Cauchy sequence in Lp(Ω). Since Lp(Ω) is
complete, there exist functions uα ∈ Lp(Ω), such that
Dαum → uα, in Lp(Ω),
for each |α| ≤ k. In particular,
um → u(0,··· ,0) =: u in Lp(Ω).
3. u ∈W k,p(Ω) and Dαu = uα, |α| ≤ k.
2.2. SOBOLEV SPACES 23
In fact, let ψ ∈ C∞0 (Ω), then
∫
ΩuDαψ dx = lim
m→∞
∫
ΩumD
αψ dx
= limm→∞
(−1)|α|∫
Ω(Dαum)ψ dx
=
∫
Ωuαψ dx.
4. um → u in W k,p(Ω).
Since Dαum → uα = Dαu in Lp(Ω) for all |α| ≤ k, it follows that
um → u in W k,p(Ω).
5. Regard W k,p(Ω) as a closed subspace of a Cartesian product space
of spaces Lp(Ω): Let N = N(n, k) =∑
0≤|α|≤k
1 be the number of multi-
indexes α satisfying 0 ≤ |α| ≤ k. For 1 ≤ p ≤ ∞, let LpN =
NΠ
j=1Lp(Ω),
the product norm of u = (u1, · · · , uN ) in LpN being given by
‖u‖LpN
=
(
N∑
j=1‖uj‖
pp
)1/p, if 1 ≤ p <∞,
max1≤j≤N
‖uj‖∞, if p = ∞.
Since a closed subspace of a reflexive (Resp. separable, uniformly con-
vex) Banach space is also reflexive (Resp. separable, uniformly convex),
thus W k,p(Ω) is reflexive if and only if 1 < p <∞; and W k,p(Ω) is sep-
arable if 1 ≤ p <∞. Furthermore,(
C∞(Ω))N
is dense in LpN and thus
in W k,p(Ω) if 1 ≤ p <∞.
6. If p = 2, define the scalar product as
(u, v)W k,2 =∑
|α|≤k
∫
ΩDαuDαv dx.
It is easy to verify W k,2 = Hk is a Hilbert space.
2.2.4 Inequalities
Theorem 2.2.25 (Gagliardo-Nirenberg-Sobolev inequality) Let
1 ≤ p < n, p∗ =np
n− p. There exists a constant C, depending only on
24 CHAPTER 2. SOBOLEV SPACES
p and n, such that
‖u‖Lp∗ (Rn) ≤ C‖Du‖Lp(Rn), (2.2.5)
for all u ∈ C10 (Rn).
Proof. 1. First assume p = 1.
Since u has compact support, for each i = 1, 2 · · · , n and x ∈ Rn
we have
u(x) =
∫ xi
−∞uxi(x1, · · · , xi−1, yi, xi+1, · · · , xn) dyi,
and so
|u(x)| ≤
∫ +∞
−∞|Du(x1, · · · , xi−1, yi, xi+1, · · · , xn)| dyi, i = 1, 2 · · · , n.
Consequently
|u(x)|n
n−1 ≤nΠ
i=1
(
∫ +∞
−∞|Du(x1, · · · , xi−1, yi, xi+1, · · · , xn)| dyi
)1
n−1.
Integrate this inequality with respect to x1:∫ +∞
−∞|u(x)|
nn−1 dx1
≤
∫ +∞
−∞
nΠ
i=1
(
∫ +∞
−∞|Du(x1, · · · , xi−1, yi, xi+1, · · · , xn)| dyi
)1
n−1dx1
=(
∫ +∞
−∞|Du(y1, x2 · · · , · · · , xn)| dy1
)1
n−1
·
∫ +∞
−∞
nΠ
i=2
(
∫ +∞
−∞|Du(x1, · · · , xi−1, yi, xi+1, · · · , xn)| dyi
)1
n−1dx1
≤(
∫ +∞
−∞|Du| dy1
)1
n−1( n
Πi=2
∫ +∞
−∞
∫ +∞
−∞|Du| dx1dyi
)1
n−1,
(2.2.6)
the last inequality follows from the Generalized Holder inequality.
Now integrate (2.2.6) with respect to x2:
∫ +∞
−∞
∫ +∞
−∞|u(x)|
nn−1 dx1 dx2
≤(
∫ +∞
−∞
∫ +∞
−∞|Du| dx1dy2
)1
n−1
∫ +∞
−∞
nΠ
i=1,i6=2I
1n−1
i dx2,
2.2. SOBOLEV SPACES 25
where
I1 =
∫ +∞
−∞|Du| dy1, Ii =
∫ +∞
−∞
∫ +∞
−∞|Du| dx1dyi, i = 3, · · · , n.
Applying once more the Generalized Holder inequality, we find
∫ +∞
−∞
∫ +∞
−∞|u(x)|
nn−1 dx1 dx2
≤(
∫ +∞
−∞
∫ +∞
−∞|Du| dx1dy2
)1
n−1(
∫ +∞
−∞
∫ +∞
−∞|Du| dy1dx2
)1
n−1
·nΠ
i=3
(
∫ +∞
−∞
∫ +∞
−∞
∫ +∞
−∞|Du| dx1dx2dyi
)1
n−1.
We continue by integrating with respect to x3, · · · , xn, eventually to
find
∫
Rn
|u(x)|n
n−1 dx ≤nΠ
i=1
(
∫ +∞
−∞· · ·
∫ +∞
−∞|Du| dx1 · · · dxn
)1
n−1
=(
∫
Rn
|Du| dx)
nn−1
.
(2.2.7)
This is conclusion for p = 1.
2. Consider now the case that 1 < p < n. We apply (2.2.7) to v := |u|γ ,
where γ > 1 is to be selected. Then
(
∫
Rn
|u(x)|γn
n−1 dx)
n−1n
≤
∫
Rn
|D|u|γ | dx = γ
∫
Rn
|u|γ−1|Du| dx
≤ γ(
∫
Rn
|u|(γ−1) pp−1 dx
)p−1
p(
∫
Rn
|Du|p)
1p .
(2.2.8)
We choose γ so thatγn
n− 1= (γ − 1)
p
p − 1. That is, we get
γ =p(n− 1)
n− p> 1,
in which caseγn
n− 1= (γ − 1)
p
p − 1=
np
n− p= p∗. Thus, (2.2.8)
becomes (2.2.5).
26 CHAPTER 2. SOBOLEV SPACES
Remark 2.2.26 The exponent p∗ =np
n− pis called critical Sobolev
exponent, because it is the unique exponent q = p∗ to make the following
inequality:
‖u‖Lq(Rn) ≤ C‖Du‖Lp(Rn)
be true for all u ∈ C10(Rn).
Proof. For any u ∈ C10 (Rn) and λ > 0, define the scaled function
uλ(x) = u(λx). Direct calculations show that
‖uλ‖Lq(Rn) = λ−n
q ‖u‖Lq(Rn)
and
‖Duλ‖Lp(Rn) = λ1−np ‖Du‖Lp(Rn).
If the inequality in remark is true for all u ∈ C10 (Rn), replacing u by
uλ, there holds
‖u‖Lq(Rn) ≤ Cλ1−np+ n
q ‖Du‖Lp(Rn).
If 1− np + n
q 6= 0, upon sending λ to either 0 or ∞, we will get C = ∞.
Theorem 2.2.27 (Poincare inequality for W 1,p0 (Ω)) Assume Ω is
a bounded, open subset of Rn, Suppose u ∈ W 1,p0 (Ω) for some 1 ≤ p <
n. Then there exists a constant C, depending only on p, n and Ω, such
that
‖u‖Lq(Ω) ≤ C‖Du‖Lp(Ω), (2.2.9)
for each q ∈ [1, p∗], the constant C depending only on p, q, n and Ω.
Proof. Since u ∈W 1,p0 (Ω), then exists a sequence of functions um∞m=1 ⊂
C∞0 (Ω) converging to u in W 1,p
0 (Ω). We extend each function um(x)
to be 0 on Rn \ Ω and apply Theorem 2.2.25 to discover ‖u‖Lp∗ (Ω) ≤
C‖Du‖Lp(Ω). Since |Ω| <∞, the Holder inequality implies ‖u‖Lq(Ω) ≤
C‖u‖Lp∗(Ω) if 1 ≤ q ≤ p∗.
Remark 2.2.28 In view of Theorem 2.2.27, on W 1,p0 (Ω), the norm
‖Du‖Lp(Ω) is equivalent to ‖u‖W 1,p0 (Ω) if Ω is bounded.
2.2. SOBOLEV SPACES 27
Theorem 2.2.29 (Morrey inequality) Assume n ≤ p < ∞, γ :=
1 −n
p. Then there exists a constant C, depending only on p and n,
such that
‖u‖C0, γ(Rn) ≤ C‖u‖W 1,p(Rn), (2.2.10)
for all u ∈ C1(Rn).
Proof. 1. First choose any ball B(x, r) ⊂ Rn. We claim there exists
a constant C, depending only on n, such that
−
∫
B(x,r)|u(x) − u(y)| dy ≤ C
∫
B(x,r)
|Du(y)|
|y − x|n−1dy. (2.2.11)
To prove this, fix any point w ∈ ∂B(0, 1). Then if 0 < s < r, there
holds
|u(x+ sw) − u(x)| =∣
∣
∫ s
0
d
dtu(x+ tw) dt
∣
∣
=∣
∣
∫ s
0Du(x+ tw) · w dt
∣
∣
≤
∫ s
0|Du(x+ tw)| dt.
Hence∫
∂B(0,1)|u(x+ sw) − u(x)| dS ≤
∫ s
0
∫
∂B(0,1)|Du(x+ tw)| dSdt
=
∫ s
0
∫
∂B(0,1)|Du(x+ tw)|
tn−1
tn−1dSdt.
Let y = x + tw, so that t = |x − y|. Then converting from polar
coordinates, we have
∫
∂B(0,1)|u(x+ sw) − u(x)| dS ≤
∫
B(x,s)
|Du(y)|
|x− y|n−1dy
≤
∫
B(x,r)
|Du(y)|
|x− y|n−1dy.
Multiply by sn−1 and integrate from 0 to r with respect to s:
∫
B(x,r)|u(x) − u(y)| dy ≤
rn
n
∫
B(x,r)
|Du(y)|
|y − x|n−1dy.
28 CHAPTER 2. SOBOLEV SPACES
It follows (2.2.11).
2. supRn
|u| ≤ C‖u‖W 1,p(Rn).
Now fix x ∈ Rn. We apply (2.2.11) as follows:
|u(x)| ≤ −
∫
B(x,1)|u(x) − u(y)| dy + −
∫
B(x,1)|u(y)| dy
≤ C
∫
B(x,1)
|Du(y)|
|y − x|n−1dy + C‖u‖Lp(B(x,1))
≤ C‖Du‖Lp(Rn)
(
∫
B(x,1)
dy
|x− y|(n−1) p
p−1
)p−1
p+ C‖u‖Lp(Rn)
≤ C‖u‖W 1,p(Rn).(2.2.12)
The last estimate holds since p > n implies (n− 1) pp−1 < n; so that
∫
B(x,1)
dy
|x− y|(n−1) p
p−1
<∞.
As x ∈ Rn is arbitrary, (2.2.12) implies
supRn
|u| ≤ C‖u‖W 1,p(Rn). (2.2.13)
3. [u]C0, 1−n/p(Rn) ≤ C‖Du‖Lp(Rn).
Choose any two points x, y ∈ Rn and write r := |x − y|. Let
W := B(x, r) ∩B(y, r). Then
|u(x) − u(y)| ≤ −
∫
W|u(x) − u(z)| dz + −
∫
W|u(y) − u(z)| dz. (2.2.14)
On the other hand, (2.2.11) allows us to estimate
−
∫
W|u(x) − u(z)| dz ≤ C −
∫
B(x,r)|u(x) − u(z)| dz
≤ C‖Du‖Lp(Rn)
(
∫
B(x,1)
dz
|x− z|(n−1) p
p−1
)p−1
p
≤ C(
rn−(n−1) pp−1
)p−1
p ‖Du‖Lp(Rn)
= Cr1−n
p ‖Du‖Lp(Rn).(2.2.15)
2.2. SOBOLEV SPACES 29
Similarly, there holds
−
∫
W|u(y) − u(z)| dz ≤ Cr1−
np ‖Du‖Lp(Rn).
Substituting this estimate and (2.2.15) into (2.2.14) yields
|u(x) − u(y)| ≤ Cr1−n
p ‖Du‖Lp(Rn) = C|x− y|1−n
p ‖Du‖Lp(Rn).
Thus
[u]C0, 1−n/p(Rn) = supx 6=y
|u(x) − u(y)|
|x− y|1−n/p
≤ C‖Du‖Lp(Rn).
This inequality and (2.2.13) complete the proof.
2.2.5 Embedding theorems and Trace theorems
Definition 2.2.30 Let (X, ‖ · ‖X), (Y, ‖ · ‖Y ) be Banach spaces. X is
(continuously) embedded into Y , denoted X → Y , if there exists an
injective linear map i : X → Y and a constant C such that
‖i(x)‖Y ≤ C‖x‖X , ∀x ∈ X.
X is compactly embedded into Y , denoted X →→ Y , if i maps
bounded subsets of X into precompact subsets of Y .
Theorem 2.2.31 (Exercise)
(i) For Ω ⊂ Rn with Lebesgue measure Ln(Ω) <∞, 1 ≤ p < q ≤ ∞,
we have Lq(Ω) → Lp(Ω). This ceases to be true if Ln(Ω) = ∞;
(ii) Suppose Ω is a precompact domain in Rn, let m = 0, 1, · · · , and
0 ≤ α < β ≤ 1. Then Cm,β(Ω) →→ Cm,α(Ω) compactly.
Theorem 2.2.32 (Sobolev embedding theorem) Let Ω ⊂ Rn be
a bounded domain with Lipschitz boundary, k ∈ N, 1 ≤ p ≤ ∞. Then
the following hold:
(i) If kp < n, then W k,p(Ω) → Lq(Ω) for 1 ≤ q ≤ p∗ = npn−kp ; the
embedding is compact, if q < npn−kp ;
30 CHAPTER 2. SOBOLEV SPACES
(ii) If 0 ≤ m < k − np < m + 1, then W k,p(Ω) → Cm,α(Ω), for
0 ≤ α ≤ k−m− np ; the embedding is compact, if α < k−m− n
p .
Proof. 1. The embedding of the case where kp < n.
Assume that kp < n and u ∈ W k,p(Ω). Then Dαu ∈ Lp(Ω) for all
|α| = k. The Gagliardo-Nirenberg-Sobolev inequality implies
‖Dβu‖Lp1 (Ω) ≤ C‖u‖W k,p(Ω), for |β| = k − 1, p1 =np
n− p,
so u ∈ W k−1,p1(Ω). Similarly, we find u ∈ W k−2,p2(Ω), where p2 =np1
n− p1=
np
n− 2p. Continuing, we eventually discover after k steps
that W 0,pk(Ω) = Lpk(Ω), for pk =npn−1
n− pn−1=
np
n− kp, and
‖u‖Lpk (Ω) ≤ C‖u‖W k,p(Ω).
Since Ω is bounded, thus for any 1 ≤ q ≤ p∗ = npn−kp , W k,p(Ω) → Lq(Ω).
2. The compactness of embedding.
That is to show that if um∞m=1 is a bounded sequence in W k,p(Ω),
there exists a subsequence umj∞j=1 which converges in Lq(Ω). For
simplicity, we just prove the case k = 1. By induction argument, one
get the proof of general case.
By extension argument, without loss of generality assume that
Ω = Rn and the functions um∞m=1 all have compact support in some
bounded open set U ⊂ Rn, moreover,
supm
‖um‖W k,p(U) <∞. (2.2.16)
Let us first study the smooth functions:
uεm := ηε ∗ um (ε > 0,m = 1, 2, · · · ),
where ηε :=1
εnη(x
ε) denotes the usual mollifier and η satisfies
η ∈ C∞0 (Rn), η ≥ 0,
∫
Rn
η(x) dx = 1.
A concrete example of η is
η(x) :=
Cexp(1
|x|2 − 1), if |x| < 1,
0, if |x| ≥ 1,
2.2. SOBOLEV SPACES 31
and constant C > 0 is selected so that
∫
Rn
η(x) dx = 1. We may
suppose that the functions uεm∞m=1 all have support in U as well.
3. Claim
uεm → um in Lq(U) as ε→ 0, uniformly in m. (2.2.17)
To prove this claim, we first note that if um is smooth, then
uεm(x) − um(x) =
∫
B(0,1)η(y)(um(x− εy) − um(x)) dy
=
∫
B(0,1)η(y)
∫ 1
0
d
dtum(x− εty) dtdy
= −ε
∫
B(0,1)η(y)
∫ 1
0Dum(x− εty) · y dtdy.
Thus∫
U|uε
m(x) − um(x)| dx ≤ ε
∫
B(0,1)η(y)
∫ 1
0
∫
U|Dum(x− εty)| dxdtdy
≤ ε
∫
U|Dum(z)| dz.
By approximation this estimate holds if um ∈W 1,p(U). Hence
‖uεm − um‖L1(U) ≤ ε‖Dum‖L1(U) ≤ ε‖Dum‖Lp(U),
the latter inequality holds since U is bounded. Owing to (2.2.16) we
thereby discover
uεm → um in L1(U) as ε→ 0, uniformly in m. (2.2.18)
But then since 1 ≤ q < p∗, we using the Interpolation inequality for
Lp-norm that
‖uεm − um‖Lq(U) ≤ ‖uε
m − um‖θL1(U)‖u
εm − um‖1−θ
Lp∗(U),
where1
q= θ +
(1 − θ)
p∗, 0 < θ ≤ 1. Consequently, (2.2.16) and the
Gagliardo-Nirenberg-Sobolev inequality imply
‖uεm − um‖Lq(U) ≤ C‖uε
m − um‖θL1(U);
32 CHAPTER 2. SOBOLEV SPACES
whence assertion (2.2.17) follows from (2.2.18).
4. Next we claim
for each fixed ε > 0, the sequence uεm∞m=1
is uniformly bounded and equi-continuous.(2.2.19)
Indeed, if x ∈ Rn, then
|uεm(x)| ≤
∫
B(x,ε)ηε(x− y)|um(y)|dy
≤ ‖ηε‖L∞(Rn)‖um‖L1(U) ≤C
εn<∞,
for m = 1, 2, · · · . Similarly
|Duεm(x)| ≤
∫
B(x,ε)|Dηε(x− y)| |um(y)|dy
≤ ‖Dηε‖L∞(Rn)‖um‖L1(U) ≤C
εn+1<∞,
for m = 1, 2, · · · . Thus the claim (2.2.19) follows these two estimates.
5. Now fix δ > 0. We will show that there exists a subsequence
umj∞j=1 of um∞m=1 such that
lim supj,k→∞
‖umj − umk‖Lq(U) ≤ δ. (2.2.20)
To see this, let us first employ assertion (2.2.17) to select ε > 0 so
small that
‖uεm − um‖Lq(U) ≤ δ/2, (2.2.21)
for m = 1, 2, · · · .
We now observe that since functions um∞m=1, and thus functions
uεm∞m=1, have support in some fixed bounded set U ⊂ Rn, we may
use (2.2.19) and the Arzela-Ascoli theorem, to obtain a subsequence
uεmj
∞j=1 ⊂ uεm∞m=1 which converges uniformly on U . In particular
therefore
lim supj,k→∞
‖uεmj
− uεmk
‖Lq(U) = 0. (2.2.22)
Then (2.2.21) and (2.2.22) imply (2.2.20).
2.2. SOBOLEV SPACES 33
6. We next employ assertion (2.2.20) with δ = 1,1
2,1
3, · · · and use
the standard diagonal argument to extract a subsequence uml∞l=1 of
um∞m=1 such that
lim supl,k→∞
‖uml− umk
‖Lq(U) = 0.
Conclusion (i) of theorem is proved by steps 1 - 6. Next to prove
conclusion (ii) of theorem.
7. Assume that 0 ≤ m < k − np < m+ 1. Then as above we see
u ∈W k−l,r(Ω), (2.2.23)
for r =np
n− lpprovided that lp < n. We choose the integer l0 = [
n
p].
Consequently, (2.2.23) holds for r =np
n− pl0> n. Hence (2.2.23) and
Morrey inequality imply that Dαu ∈ C0,1−nr (Ω) for all |α| ≤ k− l0 − 1.
Observe also that 1−n
r= 1−
n
p+ l0 = [
n
p]+1−
n
p. Thus u ∈ Cm, γ(Ω)
with m = k − l0 − 1 = k − [n
p] − 1, γ = 1 −
n
r= [
n
p] + 1 −
n
p, and the
stated estimate follows easily.
8. The Arzela-Ascoli theorem implies the compactness of embedding
W k,p(Ω) → Cm,α(Ω) provided that α < γ = k −m− np .
Denote (u)Ω = −∫
Ω u dx = average of u over Ω.
Theorem 2.2.33 (Poincare inequality for W 1,p(Ω)) Let Ω ⊂ Rn
be bounded and connected, with C1 boundary ∂Ω. Assume 1 ≤ p ≤ ∞.
Then there exists a constant C, depending only on p, n and Ω, such
that
‖u− (u)Ω‖Lp(Ω) ≤ C‖Du‖Lp(Ω), (2.2.24)
for all u ∈W 1,p(Rn).
Proof. Argue by contradiction. If the stated estimate is false, that is,
for each integer k = 1, 2, · · · , there would exist a function uk ∈W 1,p(Ω)
such that
‖uk − (uk)Ω‖Lp(Ω) > k‖Duk‖Lp(Ω). (2.2.25)
34 CHAPTER 2. SOBOLEV SPACES
Normalize by defining
vk :=uk − (uk)Ω
‖uk − (uk)Ω‖Lp(Ω), k = 1, 2, · · · .
Then, there holds
(vk)Ω = 0, ‖vk‖Lp(Ω) = 1, (2.2.26)
and from (2.2.25), we have
‖Dvk‖Lp(Ω) <1
k, k = 1, 2, · · · . (2.2.27)
Particularly, vk∞k=1 is a bounded in W 1,p(Ω). From the compact em-
bedding, there exists a subsequence vkj∞j=1 ⊂ vk
∞k=1, and a function
v ∈ Lp(Ω) such that
vkj→ v in Lp(Ω).
From (2.2.26), it follows that
(v)Ω = 0, ‖v‖Lp(Ω) = 1.
On the other hand, (2.2.27) implies that for every ϕ ∈ C∞0 (Ω), there
holds∫
Ωvϕxi dx = lim
j→∞
∫
Ωvkjϕxi dx = − lim
j→∞
∫
Ω(vkj
)xiϕdx = 0,
consequently, v ∈ W 1,p(Ω) and Dv = 0 a.e. in Ω. Thus v is constant,
since Ω is connected, furthermore, (v)Ω = 0, hence v ≡ c = 0, which
contradicts the fact that ‖v‖Lp(Ω) = 1.
Theorem 2.2.34 (Poincare inequality for a ball) Suppose 1 ≤ p
< n. Then there exists a constant C, depending only on p and n, such
that
‖u− (u)x,r‖Lp(B(x,r)) ≤ Cr‖Du‖Lp(B(x,r)), (2.2.28)
holds for each ball B(x, r) ⊂ Rn and each function u ∈W 1,p0 (B(x, r)).
Proof. The case Ω = B(0, 1) follows from Theorem 2.2.33. In general,
if u ∈W 1,p0 (B(x, r)), set
v(y) := u(x+ ry), (y ∈ B(0, 1)).
2.2. SOBOLEV SPACES 35
Then v ∈W 1,p0 (B(0, 1)), and there holds
‖v − (v)0,1‖Lp(B(0,1)) ≤ C‖Dv‖Lp(B(0,1)),
Changing variables, we recover estimate (2.2.28).
Lemma 2.2.35 (Poincare-Sobolev inequality) For any ε > 0 there
exists a C = C(ε, n) such that for u ∈ H1(B1) with∣
∣x ∈ B1;u = 0∣
∣ ≥ ε∣
∣B1
∣
∣,
there holds∫
B1
u2 ≤ C
∫
B1
|Du|2.
Proof. Suppose not. Then there exists a sequence um ⊂ H1(B1)
such that∣
∣x ∈ B1;um = 0∣
∣ ≥ ε∣
∣B1
∣
∣,
and∫
B1
u2m = 1,
∫
B1
|Dum|2 → 0 as m→ ∞.
Hence we may assume um → u0 ∈ H1(B1) strongly in L2(B1) and
weakly in H1(B1). Then limm→∞
(
|Dum|2 −|D(um −u0)|2)
= |Du0|2 and
|Du0|2 ≤ lim
m→∞|Dum|2 = 0. Thus u0 is a nonzero constant. So
0 = limm→∞
∫
B1
|um − u0|2 ≥ lim
m→∞
∫
um=0|um − u0|
2
≥ |u0|2 inf
m
∣
∣um = 0∣
∣ > 0.
This contradiction completes the proof.
Definition 2.2.36 For a domain Ω with Ck-boundary ∂Ω = Γ, k ∈
N, 1 < p <∞, denote W k− 1p,p(Γ) as the set of equivalent classes
u+
W k,p0 (Ω); u ∈W k,p(Ω)
, endowed with norm
‖u|Γ‖W
k− 1p ,p
(Γ)= inf
‖v‖W k,p(Ω) : u− v ∈W k,p0 (Ω)
.
The elements of W k− 1p,p(Γ) is call the traces u|Γ of functions u ∈
W k,p(Ω).
36 CHAPTER 2. SOBOLEV SPACES
Theorem 2.2.37 (Trace Theorem) Assume Ω is a domain with Ck-
boundary ∂Ω = Γ, k ∈ N, 1 < p < ∞. Then W k− 1p,p(Γ) is a Banach
space.
Theorem 2.2.38 (Extension Theorem) For any domain Ω with Ck-
boundary ∂Ω = Γ, k ∈ N, 1 < p < ∞, there exists a continuous linear
extension operator ext : W k− 1p,p(Γ) →W k,p(Ω) such that
(
ext(u))∣
∣
Γ=
u, for all u ∈W k− 1p,p(Γ).
Theorem 2.2.39 Suppose Ω is a domain with Ck-boundary ∂Ω =
Γ, k ∈ N, 1 < p < ∞. Then W k,p(Γ) → Wk− 1
p,p(Γ) → W k−1,p(Γ)
and both embeddings are compact.
EXERCISES
1. Prove The Young inequality with ε - Lemma 2.2.2.
2. Prove the generalized Holder inequality - Lemma 2.2.4.
3. Complete Example 2.2.15.
4. Complete Example 2.2.16.
5. Complete Example 2.2.21.
6. Complete Example 2.2.22.
7. Verify ‖ · ‖Ck,γ(Ω), ‖ · ‖Lp(Ω) and ‖ · ‖W k,p(Ω) are norms.
8. Prove Theorem 2.2.31.