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36 II Basic Function Spaces and Related Inequalities
Finally, for λ (0, 1] and k N, by C
k
() we denote the closed subspace
of C
k
() consisting of all functions u whose derivatives up to the kth order
inclusive are older continuous (Lipschitz continuous if λ = 1) in , that is,
[u]
k
max
0≤|α|≤k
sup
x,y,x6=y
|D
α
u(x) D
α
u(y)|
|x y|
λ
< .
C
k
() is a Banach space with respect to the norm
kuk
C
k,λ kuk
C
k + [u]
k
, (II.1.12)
(Miranda 1978, §54).
Exercise II.1.1 Assuming bounded, use the Ascoli-Arzel`a theorem (see, e.g.,
Rudin 1987, p. 245) to show that from every sequence of functions uniformly
bounded in C
k+1
() it is always possible to select a subsequence converging in
the space C
k,λ
().
II.1.4 Classes of Domains and the ir Properties
We begin wi th a simple but useful result ho lding for arbitrary domains of R
n
.
Lemma II.1.1 Let be an arbitrary domain of R
n
. Then there exists an
open covering, O, of satisfying the following properties
(i) O is constituted by an at most countable number of open ball s {B
k
},
k I N, such that
B
k
, for al l k I ,
kI
B
k
= ;
(ii) For any fami ly F = {B
l
}, l I
0
with I
0
( I, there is B (O F) such
that [
lI
0
B
l
] B 6= ;
(iii) For any B, B
0
O, there exists a finite number of open balls B
i
O,
i = 1, . . ., N, such that
B B
1
6= , B
N
B
0
6= , B
j
B
j+1
6= , j = 1, . . . N 1 .
Proof. Since is open, for each x we may find an open ball B
r
x
(x)
. Clearly, the collection C {B
r
x
(x)}, x , satisfies
x
B
r
x
(x) =
. However, since is separable, we may determine an at most countable
subcovering, O, of C satisfying condition (i) in the lemma. Next, assume (ii)
is not true. Then, there would be at least one fa mily F
0
= {B
k
0
}, k
0
I
0
, with
I
0
( I such that
II.1.4 Classes of Domains and their Properties 37
"
[
k
0
I
0
B
k
0
#
\
B = , for al l B (O F
0
) .
Consequently, the sets
A
1
[
k
0
I
0
B
k
0
, A
2
[
k(II
0
)
B
k
are open, disjoint and satisfy A
1
A
2
= , contradicting the assumption that
is connected. Finally, let B, B
0
O and denote their centers by x and x
0
,
respectively. Since is open and connected, it is, i n particular, arc-connected.
Therefore, we may find a curve, γ, jo ining x and x
0
, that is homeomorphic
to the interval [0, 1]. Let O
0
O be a covering of γ. Since γ is compact, we
can extract from O
0
a finite covering that satisfies the property stated in the
lemma . ut
We next present certain classes of domains of R
n
, along with their relevant
properties. We begin with the following.
Definition II.1.1. Let be a domain with a bounded boundary, namely,
is either a bounded domain or it is a domain complement in R
n
of a compact
(not necessarily connected) set, namely, is an exterior domain.
4
Assume
that for each x
0
there is a ball B = B
r
(x
0
) and a real function ζ defined
on a domain D R
n1
such that in a system of coordinates {x
1
, . . . , x
n
} with
the origin at x
0
:
(i) The set B can be represented by an equation of the type x
n
=
ζ(x
1
, . . . , x
n1
);
(ii) Each x B satisfies x
n
< ζ(x
1
, . . . , x
n1
).
Then is said to be of class C
k
(or C
k
-smooth) [respectively, of cl ass C
k
(or C
k
-smooth), 0 < λ 1] if ζ C
k
(D) [respectively, ζ C
k
(D)]. If, in
particular, ζ C
0,1
(D), we say that is locally Lipschitz. L ikewise, we shall
say that σ is a boundary portion of class C
k
[respectively, of class C
k
]
if σ = B
r
(x
0
), for some r > 0, x
0
a nd σ admits a representation
of the form described in (i), (ii) with ζ of class C
k
[respectively of class C
k
].
If, in particular, ζ C
0,1
(D), we say tha t σ is a l ocally Lipschitz boundary
portion.
If is sufficiently smoo th, of class C
1
, for example, then the unit outer
normal , n, to is well defined and continuous. However, in several inter-
esting cases, we need less regularity on , but still would like to have n
well-defined. In this regard, we have the following result, for whose proof we
refer to Neˇcas (1967, Chapitre II, Lemme 4.2).
Lemma II.1.2 Let be locally Lipschitz. Then the unit outer normal n
exists almost everywhere on .
4
Hereafter, the whole space R
n
will be considered a particular exterior domain.
38 II Basic Function Spaces and Related Inequalities
We shall now consider a special class of bounded domains called star-
shaped (or star-like) with respect to a point. For such domains, there exist
x (which we may, occasionally, assume to be the origin of coo rdinates)
and a continuous, positi ve function h on the unit sphere such that
=
x R
n
: |x x| < h
x x
|x x|

. (II.1.13)
Some elementary properties of star-shaped domains are collected in the fol-
lowing exercises.
Exercise II.1.2 Show that is star-shaped with respect to x if and only if every
ray starting from x intersects at one and only one p oint.
Exercise II.1.3 Assume star-shaped with respect to the ori gin and set
(ρ)
= {x R
n
: x = ρy, for some y }. (II.1.14)
Show that
(ρ)
if ρ (0, 1) and
(ρ)
if ρ > 1.
The following useful result holds.
Lemma II.1.3 Let be locally Lipschitz. Then, there exist m locally Lip-
schitz bounded domai ns G
1
, . . . , G
m
such that
(i)
m
i=1
G
i
;
(ii) The domains
i
= G
i
, i = 1, . . . , m, are (locally Lipschitz and)
star-shaped with respect to every po int of a ball B
i
with B
i
i
.
Proof. Let x
0
. By assumption, there is B
r
(x
0
) and a function ζ = ζ(x
0
),
x
0
= (x
1
, . . . , x
n1
) D R
n1
such that
|ζ(ξ
0
) ζ(η
0
)| < κ|ξ
0
η
0
|, ξ
0
, η
0
D,
for som e κ > 0 and, moreover, points x = (x
0
, x
n
) B
r
(x
0
) satisfy
x
n
= ζ(x
0
), x
0
D,
while points x B
r
(x
0
) satisfy
x
n
< ζ(x
0
), x
0
D.
We may (and will) take x
0
to be the origin of coordi nates. Denote next, by
y
0
(0, . . ., 0, y
n
) the point of intersection of the x
n
-axis with B
r
(x
0
)
and consider the cone Γ (y
0
, α) with vertex at y
0
, axis x
n
, and semiaperture
α < π/2. It is easy to see that, taking α sufficiently small, every ray ρ starting
from y
0
and lying in Γ (y
0
, α) intersects B
r
(x
0
) at (one and) only one
point. In fact, assume ρ cuts B
r
(x
0
) at two points z
(1)
and z
(2)
and
II.1.4 Classes of Domains and their Properties 39
denote by α
0
< α the angle formed by ρ with the x
n
-axis. Possibly rotating
the coordinate system around the x
n
-axis we may assume without loss
5
z
(1)
= (z
(1)
1
, 0 , . . ., 0, ζ(z
(1)
1
, 0 , . . ., 0)), z
(1)
1
> 0
z
(2)
= (z
(2)
1
, 0 , . . ., 0, ζ(z
(2)
1
, 0 , . . ., 0)), z
(2)
1
> 0
and so, at the same time,
tan α
0
=
z
(1)
1
ζ(z
(1)
1
, 0 , . . ., 0) y
n
tan α
0
=
z
(2)
1
ζ(z
(2)
1
, 0 , . . ., 0) y
n
implying
|ζ(z
(1)
1
, 0 , . . ., 0) ζ(z
(2)
1
, 0 , . . ., 0)|
|z
(1)
1
z
(2)
1
|
=
1
tan α
0
1
tan α
.
Thus, i f (say)
tan α
1
2κ
,
ρ will cut B
r
(x
0
) at only one point. Next, denote by σ = σ(z) the
intersection of Γ (y
0
, α/2) with a plane orthogonal to x
n
-axis at a point z =
(0, . . ., z
n
) with z
n
> y
n
, and set
R = R(z) dist (σ, z).
Clearly, taking z sufficiently close to y
0
(z = z, say), σ(z) will be entirely
contained in and, further, every ray starting from a point of σ(z) and lying
within Γ (y
0
, α/2) will f orm with the x
n
-axis an angle less than α and so, by
what we have shown, it will cut B
r
(x
0
) at only one point. Let C be a
cylinder with axis coincident with the x
n
-axis and such tha t
C = Γ (y
0
, α/2) .
Then, setting
G = C B
r
(x
0
),
we have that G is locally Lipschitz and that G is star-shaped with respect
to all points of the ball B
R(z)
(z). Since x
0
is arbi trary, we may form an
open covering G of constituted by domains of the type G. However,
is compact and, therefore, we may select from G a finite subset {G
1
, . . . , G
m
}
satisfying all conditions in the lemma, which is thus completely proved. ut
5
Clearly, the Lipschitz constant κ is invariant by this transformation.
40 II Basic Function Spaces and Related Inequalities
Other relevant properties related to star-shaped domains are described in
the following exercises.
Exercise II.1.4 Assume that the function h in (II.1.13) i s Lipschitz continuous,
so that, by Lemma II.1.2, the outer unit normal n = n(x) on exists for a.a. x.
Then, setting F (x) n(x) · (x x), show that ess inf
x
F (x) > 0.
Exercise II.1.5 Assume bounded and locally Lipschitz. Prove that
=
m
[
i=1
i
,
where each
i
is a locally Lipschitz and star-shaped domain with respect to every
point of a ball B
i
with B
i
i
. Hint: Use Lemma II.1.3.
We end this section by recalling the following classical result, whose proof
can b e found, e.g., in Neˇcas (1967, Chapitre 1, Proposition 2.3).
Lemma II.1.4 Let K be a compa ct subset of R
n
, and let O = {O
1
, ··· , O
N
}
be an o pen covering of K. Then, there exist functions ψ
i
, i = 1, . . . , N satis-
fying the following properties
(i) 0 ψ
i
1 , i = 1, . . ., N ;
(ii) ψ
i
C
0
(O
i
) , i = 1, . . ., N ;
(iii)
P
N
i=1
ψ
i
(x) = 1 , for all x K .
The fa mily {ψ
i
} is referred to as partition of unity in K subordinate to the
covering O.
II.2 The Lebesgue Spaces L
q
For q [1, ), let L
q
= L
q
() denote the linear space of all (equivalence
classes of) real Leb esgue-measurable functions u defined in such that
kuk
q
Z
|u|
q
1/q
< . (II.2.1)
The functional (II.2.1) defines a no rm in L
q
, wi th respect to which L
q
becomes
a Banach space. Likewise, denoting by L
= L
() the linear space of all
(equivalence classes of) Lebesgue-measurable real-valued functions u defined
in with
kuk
ess sup
|u| < (II.2.2)
one shows that (II.2 .2) is a norm and that L
endowed with this norm is a
Banach space. For a proo f of the ab ove properties see, e.g., Miranda (1978,
§47). For q = 2, L
q
is a Hilbert space under the scalar product
II.2 The Lebesgue Spaces L
q
41
(u, v)
Z
uv, u, v L
2
.
Whenever confusion of domains might occur, we shall use the notation
k · k
q,Ω
, k · k
,Ω
, and (· , ·)
.
Given a sequence {u
m
} L
q
() and u L
q
(), 1 q , we thus have
that u
m
u, namely, {u
m
} converges (strongly) to u, if and only if
lim
k→∞
ku
k
uk
q
= 0 .
The following two basic properties, collected in as many lemmas, will be
frequently used throughout. The first one is the classical Lebesgue dominated
convergence theorem (Jones 2001, Chapter 6 §C), while the other one relates
convergence in L
q
with pointwise convergence; see Jones (2001, Corollary at
p. 234)
Lemma II.2.1 Let {u
m
} be a sequence of measurable functions on , and
assume that
u(x) lim
m→∞
u
m
(x) exists for a.a. x ,
and that there is U L
1
() such that
|u
m
(x)| |U(x)| for a.a x .
Then u L
1
() and
lim
m→∞
Z
u
m
=
Z
u .
Lemma II.2.2 Let {u
m
} L
q
() and u L
q
(), 1 q , with u
m
u.
Then, there exists {u
m
0
} {u
m
} such tha t
lim
m
0
→∞
u
m
0
(x) = u(x) , for a.a. x .
We want now to collect som e inequalities in L
q
spaces that will be fre-
quently used throughout. Fo r 1 q , we set
q
0
= q/(q 1);
one then shows the older inequality
Z
|uv| kuk
q
kvk
q
0
(II.2.3)
for all u L
q
(), v L
q
0
() (Miranda 1978, Teorema 47.I). The number
q
0
is called the older conjugate of q. In particular, (II.2.3) shows that the
42 II Basic Function Spaces and Related Inequalities
bilinear fo rm (u, v) is meaningful whenever u L
q
() and v L
q
0
(). In
case q = 2, inequality (II.2.3) is referred to as the Schwarz inequality. More
generally, one has the generalized older inequalit y
Z
|u
1
u
2
. . .u
m
| ku
1
k
q
1
ku
2
k
q
2
· . . . · ku
m
k
q
m
, (II.2.4)
where
u
i
L
q
i
() , 1 q
i
, i = 1, . . ., m ,
m
X
i=1
q
1
i
= 1 .
Both inequalities (II.2.3) and (II.2.4) are an easy consequence of the Young
inequali ty:
ab
εa
q
q
+ ε
q
0
/q
b
q
0
q
0
(a, b, ε > 0) (II.2.5)
holdi ng for all q (1, ). When q = 2, relation (II.2.5) is known as the
Cauchy inequality.
Two noteworthy consequences of inequality (II.2.3) are the Minkowski in-
equality:
ku + vk
q
kuk
q
+ kvk
q
, u, v L
q
(), (II.2.6)
and the interpolation (o r convexity) inequality:
kuk
q
kuk
θ
s
kuk
1θ
r
(II.2.7)
valid for all u L
s
() L
r
() with 1 s q r , and
q
1
= θs
1
+ (1 θ)r
1
, θ [0, 1].
Another important inequality is the generalized Minkowski inequality re-
ported in the following lemma, and for whose proof we refer to Jones (2001,
Chapter 11, §E).
6
Lemma II.2.3 Let
1
, and
2
be domains of R
n
and R
m
, respectively, with
m, n 1. Suppose that u :
1
×
2
R is a Lebesgue measurable function
such that, for some q [1, ],
Z
2
Z
1
|u(x, y)|
q
dx
1/q
dy < .
Then,
Z
1
Z
2
u(x, y) dy
q
dx
1/q
< ,
and the following inequality holds
6
Actually, it can be proved that (II.2.6) is just a particular case of (II.2.8), hence
the adjective “generalized”; see Jones (2001, p. 272).
II.2 The Lebesgue Spaces L
q
43
Z
1
Z
2
u(x, y) dy
q
dx
1/q
Z
2
Z
1
|u(x, y)|
q
dx
1/q
dy . (II.2.8)
Exercise II.2.1 Assume bounded. Show that if u L
(), then
lim
q→∞
kuk
q
= kuk
.
Exercise II.2.2 Prove inequality (II.2.5). Hint: Minimize the function
t
q
/q t + 1/q
0
.
Exercise II.2.3 Prove inequalities (II.2.6) and (II.2.7).
We shall now list some of the basic properties of the spaces L
q
. We begi n
with the following (see, e.g. Miranda 1978, §51).
Theorem II.2.1 For 1 q < , L
q
is separable, C
0
() being, in pa rticular,
a dense subset
Note tha t the above property is not true if q = , since C() is a closed
subspace of L
()); see Miranda, loc. cit..
Concerning the density of smooth functions in L
q
, one can prove something
more than what stated in Theorem II.2.1, namely, that every function in L
q
,
1 q < , can be approximated by functions from C
0
(). This fact follows
as a particular case of a general smoothing procedure that we are going to
describe. To this end, given a real (measurable) function u in , we shall
write
u L
q
loc
()
to mean
u L
q
(
0
), for any bounded domain
0
with
0
.
Likewise, we write
u L
q
loc
()
to mean
u L
q
(
0
), for any bounded domain
0
.
Clearly, for bounded we have L
q
loc
() = L
q
(). Now, let j C
0
() be a
non-negative function such tha t
(i) j(x) = 0, for |x| 1,
(ii)
Z
R
n
j = 1.
44 II Basic Function Spaces and Related Inequalities
A typical exampl e is
j(x) =
(
c exp[1/(1 |x|
2
)] if |x| < 1
0 if |x| 1,
with c chosen i n such a way that property (ii) is satisfied. The regular izer (or
mollifier) in the sense o f Friedrichs u
ε
of u L
1
loc
() is then defined by the
integral
u
ε
(x) = ε
n
Z
R
n
j
x y
ε
u(y)dy, ε < dist (x, ) .
This function has several interesting properties, some of which will be recalled
now here. First o f all, we observe that u
ε
is infinitely differentiabl e at each
x with dist (x, ) > ε. Moreover, if u L
q
loc
() we may extend it by
zero outside , so that u
ε
becomes defined f or all ε > 0 and all x R
n
. Thus,
in particular, if u L
q
(), 1 q < , one can show (Miranda 1978, §51; see
also Exercise II.2.10 for a generalization)
ku
ε
k
q
kuk
q
for all ε > 0 ,
lim
ε0
+
ku
ε
uk
q
= 0.
(II.2.9)
Exercise II.2.4 Show that for u C
0
(),
lim
ε0
+
u
ε
(x) = u(x) holds uniformly in x .
Exercise II.2.5 For u L
q
(), 1 q < , show the inequality
sup
R
n
|D
α
u
ε
(x)| ε
n/q−|α|
kD
α
jk
q
0
,R
n
kuk
q,Ω
, |α| 0.
We next observe that, by writing u
ε
(x) as follows:
u
ε
(x) = ε
n
Z
|ξ|<ε
j
ξ
ε
u(x + ξ),
it becomes apparent that, if u is of compact support i n and ε is chosen less
than the distance of the support of u from , then u
ε
C
0
(). The latter,
together with (II.2 .9)
2
and the density of C
0
in L
q
, yields that C
0
() is a
dense subspace of L
q
(), 1 q < . The proof of this property, along with
some of its consequences, is left to the reader in the following exercises.
Exercise II.2.6 Prove that C
0
() is dense in L
q
(), 1 q < . Hint. Use the
density of C
0
() in L
q
() (Miranda 1978, §51) along with the properties of the
mollifier.
II.2 The Lebesgue Spaces L
q
45
Exercise II.2.7 Prove the existence of a basis in L
2
() constituted by functions
from C
0
(). Hint: Use the separability of L
2
along with the density of C
0
into L
2
.
Exercise II.2.8 Let u L
q
(), 1 q < . Extend u to zero in R
n
and
continue to denote by u the extension. Show the following continuity in the mean
property: Given ε > 0 there is δ > 0 such that for every h R
n
with |h| < δ the
following inequality holds
Z
|u(x + h) u(x)|
q
dx < ε
q
.
Hint: Show the property f or u C
0
(), then use the density of C
0
in L
q
.
Exercise II.2.9 Assume u L
1
loc
(). Prove that
Z
= 0, for all ψ C
0
(), implies u 0, a.e. in .
Hint: Consider the function
sign u =
8
<
:
1 if u > 0
1 if u 0.
For a fixed bounded
0
with
0
,
sign u L
1
(
0
)
and so sign u can be approximated by functions from C
0
(
0
).
Exercise II.2.10 Let u L
q
(R
n
), 1 q < , and for z R
n
and k n set
z
(k)
= (z
1
, . . . , z
k
) , z
(k)
= (z
k+1
, . . . , z
n
) .
Moreover, define
u
(k)
(x) = ε
k
Z
R
k
j
x
(k)
y
(k)
ε
«
u(y
(k)
, y
(k)
) dy
(k)
.
Show the following properties, for each y
(k)
R
nk
:
ku
(k)
k
q,R
k
ku(·, y
(k)
)k
q,R
k
for all ε > 0 ,
lim
ε0
+
ku
(k)
u(·, y
(k)
)k
q,R
k
= 0.
Hint: Use the generalized Minkowski inequality, the result in Exercise II.2.8 and
Lebesgue dominated convergence theorem (Lemma II.2.1).
Let v L
q
0
(), with q
0
the older conjugate of q. Then, by (II.2 .3), the
integral
`(u) =
Z
vu, u L
q
() (II.2.10)
defines a linear functional on L
q
. However, for q [1, ), every li near func-
tional must be of the form (II.2.10). Actually, we have the following Riesz
representation theorem for whose proof we refer to Miranda (1978, §48).