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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 Rn is open and 0 < 1.
Definition 2.1.1 (i) If u : R is bounded and continuous, we
write
uC() := supx
u(x).
(ii) The thHolder seminorm of u : R is
[u]C0,() := supx,y, x 6=y
{ u(x) u(y)
x y
}
,
and the thHolder norm is
uC0, () := uC() + [u]C0, ().
(iii) The Holder space
Ck,()
11

12 CHAPTER 2. SOBOLEV SPACES
consists of all functions u Ck() for which the norm
uCk, () :=
k
DuC() +
=k
[Du]C0, () (2.1.1)
is finite.
Theorem 2.1.2 The space of functions Ck,() is a Banach space.
Proof. It is easy to verify that Ck,() is a norm (Exercise). It
suffices to check the completeness of Ck,().
1. Construction of limit functions.
Let {um}m=1 be a Cauchy sequence in C
k,(), i.e.,
um unCk,() 0, as m,n +.
Since any Cauchy sequence is a bounded sequence, so umCk, () is
uniformly bounded. Particularly, umC() is uniformly bounded, that
is, {um(x)} is a uniformly bounded sequence. Furthermore, the bound
edness of DumC() or [Dum]C0, () implies the equicontinuity of the
sequence {um(x)}. The ArzelaAscoli theorem implies there exists a
subsequence, still denoted by {um}m=1, which converges to a bounded
continuous function u(x) uniformly in . The same argument can be
repeated for any derivative of Dum of order up to and including k.
That is, there exist functions u C(), such that
Dum u, uniformly in ,  k.
2. Du = u.
Deduce by induction, first let = (1, 0, , 0), since the series
u1(x) ++
j=1
(uj+1 uj) = limm
um+1(x) = u(x),
uniformly in and the series
Du1(x) +
+
j=1
(Duj+1 Duj) = lim
mDum+1(x) = u(x),

2.1. HOLDER SPACES 13
uniformly in . The theory of series implies that u(x) is differentiable
in and Du = u. The induction method shows that Du = u is
valid for all multiindex such that  k.
3. u Ck,().
It suffices to show that for any multiindex such that  = k,
there holds [Du]C0, () < +.
For any x, y , x 6= y, since Dum Du uniformly in , there
exists a N large enough, such that:
DuN (z) Du(z)
1
2x y , z .
Calculate:
Du(x) Du(y) Du(x) DuN (x) + DuN (x) D
uN (y)
+DuN (y) Du(y)
x y + x yDuNC0,()
(1 + supm
DumC0, ())x y ,
Thus the boundedness of DumC0,() implies [Du]C0,() < +.
4. um u in Ck,().
It also suffices to show that for any multiindex such that  = k,
there holds
[Dum Du]C0, () 0, as m .
In fact, for any x, y , x 6= y, there holds
Dum(x) Du(x) (Dum(y) D
u(y))
lim supn
Dum(x) Dun(x) (D
um(y) Dun(y))
x y limn
[Dum Dun]C0, ().
Since {um}m=1 is a Cauchy sequence in C
k,(), [DumDu]C0,()
limn
[Dum Dun]C0,() 0 as m 0.

14 CHAPTER 2. SOBOLEV SPACES
2.2 Sobolev spaces
2.2.1 Lp spaces
The Lp() space consists of all the function u : R such that
up dx

2.2. SOBOLEV SPACES 15
Lemma 2.2.2 (Young inequality with ) Let 1 < p, q < , 1p +1q = 1, then
ab ap
p+
1
q/pbq
q, a, b > 0.
Lemma 2.2.3 (Holder inequality) Let 1 p, q , 1p +1q = 1,
then if u Lp(), v Lq(), we have
uv dx uLp()vLq().
Proof. 1. In the cases where p = or q = , it is easy, because there
exists a subset , with  = , such that sup
u = uL()
or sup
v = vL().
2. In the cases where 1 < p, q < . By the homogeneity of the
inequality, we may assume that uLp() = vLq() = 1. Then the
Young inequality implies that
uv dx
1
p
up dx+
1
q
uq dx = 1 = uLp()vLq().
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 + +1
pm= 1, then if uk L
pk() for k = 1, ,m, we have
u1 um dx
m
k=1ukLpk ().
Lemma 2.2.5 (Minkowski inequality) Assume 1 p , u, v
Lp(), then
u+ vLp() uLp() + vLp().

16 CHAPTER 2. SOBOLEV SPACES
Proof. From the Holder inequality, we have
u+ vpLp() =
u+ vp dx
u+ vp1(u + v) dx
(
u+ v
(p1) pp1 dx
)p1
p(
(
up dx)1/p + (
vp dx)1/p
)
= u+ vp1Lp()(uLp() + vLp()).
Lemma 2.2.6 (Interpolation inequality for Lp norms) Assume 1
s r t , and 1r =s +
(1)t . Suppose also u L
s() Lt().
Then u Lr(), and
uLr() uLs()u
(1)Lt().
Proof. From the Holder inequality, we have
ur dx =
uru(1)r dx
(
ur
sr dx
)rs(
u
(1)r t(1)r dx
)
(1)rt .
Theorem 2.2.7 (Completeness of Lp()) Lp() is a Banach space
if 1 p .
Proof. It is easy to check that Lp() is a norm (Exercise). It suffices
to verify the completeness of Lp()
1. Construction of the limit function.
Assume that 1 p < and let {um} be a Cauchy sequence in
Lp(). There is a subsequence {umj} of {um} such that
umj+1 umjp 1/2j , j = 1, 2 .

2.2. SOBOLEV SPACES 17
Let vm(x) =m
j=1umj+1(x) umj (x), then
vmp
m
j=1
umj+1 umjp
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
v(x)p dx lim inf
m
vm(x)
p dx 1.
Hence v(x) < a.e. in 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 . 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 .
2. u Lp() and um u in Lp().
For any > 0, there exists N such that if m,n N , then um
unp < . Hence by the Fatou Lemma again, we have
um(x) u(x)
p dx =
lim
jum(x) umj (x)
p dx
lim infj
um(x) umj (x)
p dx
p.
If m N , thus u = (u um) + um Lp(), and um up 0 as
m .
3. In the case where p = , let {um} be a Cauchy sequence in L(),
then there exists a subset A , with A = 0, such that if x \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 \ A, and um(x) is a Cauchy
sequence for any x \A, thus um(x) converges uniformly on \A to

18 CHAPTER 2. SOBOLEV SPACES
a bounded function u(x). Set u(x) = 0 if x A. We have u L()
and um u 0 as m .
Theorem 2.2.8 (Riesz Representation Theorem for Lp()) Let
1 < p

2.2. SOBOLEV SPACES 19
Motivation for definition of weak derivatives. Assume we are
given a function u C1(). Then if C0 (), we see from the
integration by parts that
uxi dx =
uxidx (i = 1, , n). (2.2.2)
More general now, if k is a positive integer, u Ck(), and =
(1, , n) is a multiindex of order  = 1 + + n = k, then
u(D) dx = (1)
(Du)dx. (2.2.3)
We next examine that formula (2.2.3) is valid for u Ck(), and
ask whether some variant of it might still be true even if u is not k times
continuously differentiable. Note that the lefthand 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 Du on the righthand 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 Du:
Definition 2.2.13 Suppose u, v L1loc(), and is a multiindex. We
say that v is the thweak partial derivative of u, written
Du = v,
provided that
uDdx = (1)
v dx (2.2.4)
for all test functions C0 ().
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, = (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, = (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()
consists of all locally summable functions u : R such that
for each multiindex with  k, Du exists in the weak sense
and belongs to Lp().
ii) If u W k,p(), we define its norm to be
uW k,p() :=
(
k
Dup dx
)1/p, (1 p

2.2. SOBOLEV SPACES 21
ii) We write
um u in Wk,ploc ()
to mean
um u in Wk,p(U),
for each U .
Definition 2.2.19 We denote by
W k,p0 ()
the closure of C0 () in Wk,p().
Remark 2.2.20 (i) If p = 2, we usually write
Hk() = W k,2(), Hk0 () = Wk,20 (), (k = 0, 1, ).
The letter H is used, since  as we will see  Hk() is a Hilbert
space. Note that H0() = L2().
(ii) We henceforth identity functions in W k,p() which agree almost
everywhere.
Example 2.2.21 Let = B(0, 1), the open unit ball in Rn, and
u(x) = x (x , x 6= 0), > 0.
Then u W 1,p() if and only if 0 < ( + 1)p < n. In particular
u 6W 1,p() for each p n. (Exercise)
Example 2.2.22 Let {rk}k=1 be a countable, dense subset of =
B(0, 1) Rn. Set
u(x) =
k=1
1
2kx rk
(x ).
Then u W 1,p() if and only if ( + 1)p < n. If 0 < ( + 1)p < n,
we see that u belongs to W 1,p() and yet is unbounded on each open
subset of . (Exercise)
Theorem 2.2.23 (Properties of weak derivatives) Assume u, v
W k,p(),  k. Then

22 CHAPTER 2. SOBOLEV SPACES
(i) Du W k,p() and D(Du) = D(Du) = D+u for all
multiindices , with +  k;
(ii) For each , R, u + v W k,p() and D(u + v) =
Du+ Dv,  k;
(iii) If is an open subset of , then u W k,p();
(iv) If C0 (), then u Wk,p() and
D(u) =
(
)
DDu (Leibniz formula),
where
(
)
=!
!( )!.
Theorem 2.2.24 (Sobolev spaces as function spaces) For each k
= 0, 1, and 1 p , Sobolev space W k,p() is a Banach space;
W k,p() is reflexive if and only if 1 < p < ; For 1 p < , the
subspace W k,pC() is dense in W k,p(), thus W k,p() is separable;
Moreover, W k,2() is a Hilbert space with scalar product
(u, v)W k,2 =
k
DuDv dx.
Proof. 1. It is easy to check that W k,p() is a norm (Exercise).
Next to show that W k,p() is complete.
2. Construction of limit functions.
Assume {um}m=1 is a Cauchy sequence in W
k,p(). Then for each
 k, {Dum}m=1 is a Cauchy sequence in L
p(). Since Lp() is
complete, there exist functions u Lp(), such that
Dum u, in Lp(),
for each  k. In particular,
um u(0, ,0) =: u in Lp().
3. u W k,p() and Du = u,  k.

2.2. SOBOLEV SPACES 23
In fact, let C0 (), then
uD dx = lim
m
umD
dx
= limm
(1)
(Dum) dx
=
u dx.
4. um u in Wk,p().
Since Dum u = Du in Lp() for all  k, it follows that
um u in Wk,p().
5. Regard W k,p() as a closed subspace of a Cartesian product space
of spaces Lp(): Let N = N(n, k) =
0k
1 be the number of multi
indexes satisfying 0  k. For 1 p , let LpN =N
j=1Lp(),
the product norm of u = (u1, , uN ) in LpN being given by
uLpN=
(
N
j=1uj
pp
)1/p, if 1 p

24 CHAPTER 2. SOBOLEV SPACES
p and n, such that
uLp (Rn) CDuLp(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, , xi1, yi, xi+1, , xn) dyi,
and so
u(x)
+
Du(x1, , xi1, yi, xi+1, , xn) dyi, i = 1, 2 , n.
Consequently
u(x)n
n1 n
i=1
(
+
Du(x1, , xi1, yi, xi+1, , xn) dyi
)1
n1.
Integrate this inequality with respect to x1: +
u(x)
nn1 dx1
+
n
i=1
(
+
Du(x1, , xi1, yi, xi+1, , xn) dyi
)1
n1dx1
=(
+
Du(y1, x2 , , xn) dy1
)1
n1
+
n
i=2
(
+
Du(x1, , xi1, yi, xi+1, , xn) dyi
)1
n1dx1
(
+
Du dy1
)1
n1( n
i=2
+
+
Du dx1dyi
)1
n1,
(2.2.6)
the last inequality follows from the Generalized Holder inequality.
Now integrate (2.2.6) with respect to x2:
+
+
u(x)
nn1 dx1 dx2
(
+
+
Du dx1dy2
)1
n1
+
n
i=1,i6=2I
1n1
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)
nn1 dx1 dx2
(
+
+
Du dx1dy2
)1
n1(
+
+
Du dy1dx2
)1
n1
n
i=3
(
+
+
+
Du dx1dx2dyi
)1
n1.
We continue by integrating with respect to x3, , xn, eventually to
find
Rnu(x)
nn1 dx
n
i=1
(
+
+
Du dx1 dxn
)1
n1
=(
RnDu dx
)n
n1.
(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 ,
where > 1 is to be selected. Then
(
Rnu(x)
nn1 dx
)n1
n
RnDu  dx =
Rnu1Du dx
(
Rnu(1)
pp1 dx
)p1
p(
RnDup
)1p .
(2.2.8)
We choose so thatn
n 1= ( 1)
p
p 1. That is, we get
=p(n 1)
n p> 1,
in which casen
n 1= ( 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:
uLq(Rn) CDuLp(Rn)
be true for all u C10(Rn).
Proof. For any u C10 (Rn) and > 0, define the scaled function
u(x) = u(x). Direct calculations show that
uLq(Rn) = n
q uLq(Rn)
and
DuLp(Rn) = 1n
p DuLp(Rn).
If the inequality in remark is true for all u C10 (Rn), replacing u by
u, there holds
uLq(Rn) C1n
p+ n
q DuLp(Rn).
If 1 np +nq 6= 0, upon sending to either 0 or , we will get C = .
Theorem 2.2.27 (Poincare inequality for W 1,p0 ()) Assume is
a bounded, open subset of Rn, Suppose u W 1,p0 () for some 1 p n implies (n 1) pp1 < n; so that
B(x,1)
dy
x y(n1) p
p1

2.2. SOBOLEV SPACES 29
Similarly, there holds
Wu(y) u(z) dz Cr1
np DuLp(Rn).
Substituting this estimate and (2.2.15) into (2.2.14) yields
u(x) u(y) Cr1n
p DuLp(Rn) = Cx y1n
p DuLp(Rn).
Thus
[u]C0, 1n/p(Rn) = supx 6=y
{ u(x) u(y)
x y1n/p}
CDuLp(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 CxX , 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 Rn with Lebesgue measure Ln()

30 CHAPTER 2. SOBOLEV SPACES
(ii) If 0 m < k np < m + 1, then Wk,p() Cm,(), for
0 km np ; the embedding is compact, if < kmnp .
Proof. 1. The embedding of the case where kp < n.
Assume that kp < n and u W k,p(). Then Du Lp() for all
 = k. The GagliardoNirenbergSobolev inequality implies
DuLp1 () CuW k,p(), for  = k 1, p1 =np
n p,
so u W k1,p1(). Similarly, we find u W k2,p2(), where p2 =np1n p1
=np
n 2p. Continuing, we eventually discover after k steps
that W 0,pk() = Lpk(), for pk =npn1n pn1
=np
n kp, and
uLpk () CuW k,p().
Since is bounded, thus for any 1 q p = npnkp , Wk,p() Lq().
2. The compactness of embedding.
That is to show that if {um}m=1 is a bounded sequence in W
k,p(),
there exists a subsequence {umj}j=1 which converges in L
q(). 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
= Rn and the functions {um}m=1 all have compact support in some
bounded open set U Rn, moreover,
supm
umW k,p(U) 0,m = 1, 2, ),
where :=1
n(x
) denotes the usual mollifier and satisfies
C0 (Rn), 0,
Rn(x) dx = 1.
A concrete example of is
(x) :=
Cexp(1
x2 1), if x < 1,
0, if x 1,

2.2. SOBOLEV SPACES 31
and constant C > 0 is selected so that
Rn(x) dx = 1. We may
suppose that the functions {um}m=1 all have support in U as well.
3. Claim
um um in Lq(U) as 0, uniformly in m. (2.2.17)
To prove this claim, we first note that if um is smooth, then
um(x) um(x) =
B(0,1)(y)(um(x y) um(x)) dy
=
B(0,1)(y)
1
0
d
dtum(x ty) dtdy
=
B(0,1)(y)
1
0Dum(x ty) y dtdy.
Thus
Uum(x) um(x) dx
B(0,1)(y)
1
0
UDum(x ty) dxdtdy
UDum(z) dz.
By approximation this estimate holds if um W1,p(U). Hence
um umL1(U) DumL1(U) DumLp(U),
the latter inequality holds since U is bounded. Owing to (2.2.16) we
thereby discover
um um in L1(U) as 0, uniformly in m. (2.2.18)
But then since 1 q < p, we using the Interpolation inequality for
Lpnorm that
um umLq(U) um um
L1(U)u
m um
1Lp(U)
,
where1
q= +
(1 )
p, 0 < 1. Consequently, (2.2.16) and the
GagliardoNirenbergSobolev inequality imply
um umLq(U) Cum um
L1(U);

32 CHAPTER 2. SOBOLEV SPACES
whence assertion (2.2.17) follows from (2.2.18).
4. Next we claim
for each fixed > 0, the sequence {um}m=1
is uniformly bounded and equicontinuous.(2.2.19)
Indeed, if x Rn, then
um(x)
B(x,)(x y)um(y)dy
L(Rn)umL1(U) C
n 0 so
small that
um umLq(U) /2, (2.2.21)
for m = 1, 2, .
We now observe that since functions {um}m=1, and thus functions
{um}m=1, have support in some fixed bounded set U R
n, we may
use (2.2.19) and the ArzelaAscoli theorem, to obtain a subsequence
{umj}j=1 {u
m}
m=1 which converges uniformly on U . In particular
therefore
lim supj,k
umj umk
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 = 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 umkLq(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 kl,r(), (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 Du C0,1nr () for all  k l0 1.
Observe also that 1n
r= 1
n
p+ l0 = [
n
p]+1
n
p. Thus u Cm, ()
with m = k l0 1 = k [n
p] 1, = 1
n
r= [
n
p] + 1
n
p, and the
stated estimate follows easily.
8. The ArzelaAscoli theorem implies the compactness of embedding
W k,p() Cm,() provided that < = k m np .
Denote (u) =
u dx = average of u over .
Theorem 2.2.33 (Poincare inequality for W 1,p()) Let Rn
be bounded and connected, with C1 boundary . Assume 1 p .
Then there exists a constant C, depending only on p, n and , such
that
u (u)Lp() CDuLp(), (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 W1,p()
such that
uk (uk)Lp() > kDukLp(). (2.2.25)

34 CHAPTER 2. SOBOLEV SPACES
Normalize by defining
vk :=uk (uk)
uk (uk)Lp(), k = 1, 2, .
Then, there holds
(vk) = 0, vkLp() = 1, (2.2.26)
and from (2.2.25), we have
DvkLp() 0 there
exists a C = C(, n) such that for u H1(B1) with
{x B1;u = 0}
B1
,
there holds
B1
u2 C
B1
Du2.
Proof. Suppose not. Then there exists a sequence {um} H1(B1)
such that
{x B1;um = 0}
B1
,
and
B1
u2m = 1,
B1
Dum2 0 as m .
Hence we may assume um u0 H1(B1) strongly in L
2(B1) and
weakly in H1(B1). Then limm
(
Dum2 D(um u0)
2)
= Du02 and
Du02 lim
mDum
2 = 0. Thus u0 is a nonzero constant. So
0 = limm
B1
um u02 lim
m
{um=0}um u0
2
u02 inf
m
{um = 0}
> 0.
This contradiction completes the proof.
Definition 2.2.36 For a domain with Ckboundary = , k
N, 1 < p

36 CHAPTER 2. SOBOLEV SPACES
Theorem 2.2.37 (Trace Theorem) Assume is a domain with Ck
boundary = , k N, 1 < p < . Then W k1p,p() is a Banach
space.
Theorem 2.2.38 (Extension Theorem) For any domain with Ck
boundary = , k N, 1 < p < , there exists a continuous linear
extension operator ext : W k1p,p() W k,p() such that
(
ext(u))
=
u, for all u W k1p,p().
Theorem 2.2.39 Suppose is a domain with Ckboundary =
, k N, 1 < p < . Then W k,p() Wk 1
p,p() W k1,p()
and both embeddings are compact.
EXERCISES
1. Prove The Young inequality with  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,(), Lp() and W k,p() are norms.
8. Prove Theorem 2.2.31.