CHAPTER ONE

FOUNDATION
1.1 Basic notions and results from Functional
Analysis
The purpose of this section is to refresh our minds on some basic fundamentals
required to facilitate a smooth understanding of the study of compact linear
operators and its applications.
Denition 1.1.1. A non-negative function jj jj on a vector space X is called
a norm on X if and only if
i) jjxjj 0 for every x 2 X (Positivity).
ii) jjxjj = 0 if and only if x = 0 (Nondegeneracy).
iii) jj(x)jj = jjjjxjj for every x 2 X (Homogeneity).
iv) jjx + yjj jjxjj + jjyjj for every x; y 2 X (subadditivity).
A vector space X with a norm jj jj is denoted by (X; jj jj) and is called a
normed linear space (or just a normed space).
A sequence (xn)n2N of elements in a normed linear space X is called Cauchy
if 8 > 0; 9N 2 N, such that for m; n 2 N, jjxm ? xnjj , m; n N
A Banach space is a normed linear space (X; jj jj) that is complete in the
canonical metric dened by (x; y) = jjx ? yjj for x; y 2 X i.e every Cauchy
sequence in X for the metric converges to some point in X.
1
Remark. Every normed linear space has a completion [3]
Denition 1.1.2. Let X and Y be K-linear spaces (K = R or K = C). A
map T : X ?! Y is called linear if
T(f + g) = T(f) + T(g) for all f; g 2 X; and all ; 2 K:
More generally, we can consider linear maps dened on a sub-space D(T)
of X and with values in Y . The Subspace D(T) X is called the domain of
T. We denote the image (or Range) of T by R(T) and is dened by
R(T) :=

y 2 Y : y = Tx for some x 2 D(T)

:
We dene the Kernel (or Null space) of T denoted by N(T) to be the subspace
of X dened by
N(T) := fx 2 D(T) : Tx = 0g
T is said to be injective (or one to one ) if N(T) = f0g.
T is said to be surjective (or onto) if R(T) = Y .
Denition 1.1.3. A mapping T : X ?! Y is called a continuous linear
operator, if T is linear and is continuous at each point in a 2 X, that is
lim
x!a
Tx = Ta for all a 2 X :
The space of continuous linear operators from a Banach space X into a
Banach Space Y is denoted by B(X; Y ).
Proposition 1.1.4. [2]
Let X, Y be two normed spaces and BX be the closed unit ball of X. Let
T : X ?! Y be a linear map. Then the following are equivalent :
i) T is a bounded linear operator, i.e there exists a constant M > 0 such that
jjTxjjY MjjxjjX for all x 2 X.
ii) T is continuous.
iii) T is continuous at the origin (in the sense that if fxng is a sequence in
X. such that xn ! 0 as n ! 0 , then Txn ! 0 in Y as n ! 1).
iv) T is Lipschitz.
v) T(BX) is bounded (i.e there exists a constant C > 0 such that jjTxjj C
for all x 2 BX).
2
Furthermore, B(X; Y ) becomes naturally endowed with the operator norm,
jjTjj := sup
x2X;x6=0
jjTxjjY
jjxjjX
= sup
jjxjjX1
jjTxjjY ;
and if Y is a Banach space then so is B(X; Y ).
We also recall that every linear operator of a nite dimensional space is bounded.
Proposition 1.1.5. Every bounded linear map between normed linear space
has a unique extension between their completions [4].
Theorem 1.1.6. Given a continuous complex function G on [a; b] [a; b],
let T be dened on X = C[a; b] at each f by
(Tf)(x) =
Z b
a
G(x; y)f(y)dy for all x 2 [a; b]:
T is called an integral operator with the kernel function G. Then T 2 B(X; Y )
and
jjTjj = max
axb
Z b
a
jG(x; y)jdy
Proof.
Firstly, we show that T is well dened, i.e., for every f 2 C[a; b], Tf 2 C[a; b].
Let x1; x2 2 [a; b], then
jTf(x1) ? Tf(x2)j
Z b
a
jG(x1; t) ? G(x2; t)jjf(t)jdt
max
t2[a;b]
jG(x1; t) ? G(x2; t)jjjfjj1(b ? a)
Since G is continuous on the compact set C([a; b] [a; b]) by assumption, it
follows that G is uniformly continuous and so we deduce from the above inequality
that Tf is uniformly continuous. Thus Tf 2 C[a; b].
The linearity of T follows from the linearity of the integral.
Now we investigate the boundedness of T. We have,
j(Tf)(x)j
Z b
a
jG(x; y)jdy (1.1.1)
and so,
jjTfjj1 = max
axb
j(Tf)(x)j max
axb
Z b
a
jG(x; y)jdyjjfjj1
showing that T is bounded and
kTk max
axb
Z b
a
jG(x; y)jdy :
3
Hence
kTk M; where M :=
Z b
a
jG(x; y)jdy:
Now we dene,
S(x) :=
Z b
a
jG(x; y)jdy:
S is continuous on [a; b] and therefore it attains its maximum at some x0 2
[a; b]:
Let g(y) :=
8<
:
jG(x0;y)j
G(x0;y) if G(x0; y) 6= 0
0 otherwise
Then the function g is bounded and Lebesgue measurable and so belongs to
L1([a; b]). It follows from the fact that C[a; b] is dense in L1([a; b]) that there
exists a sequence (gn)n of elements of C[a; b] such that kgnk = max
y2[0;1]
jgn(y)j 1
and (gn) converges in L1([a; b]) to g. Hence we have ,
kTk = sup
jjfjj1
kTfk kTgnk = max
axb
j(Tgn)xj (Tgn)(x0)
) kTk (Tgn)(x0) =
Z b
a
K(x0; y)gn(y)dy ?!
Z b
a
K(x0; y)g(y)dy
therefore, kTk
Z b
a
K(x0; y)g(y)dy =
Z b
a
jK(x0; y)jdy = M
It follows that jjTjj M: Hence,
kTk = M
Corollary 1.1.7. Given a continuous complex valued function G on
[0; 1] [0; 1], let T be dened on X = C[0; 1] by
(Tf)(x) =
Z 1
0
G(x; y)f(y)dy ; 8f 2 C[a; b]:
Then T is a bounded linear map with
jjTjj = max
0x1
Z 1
0
jG(x; y)jdy :
Denition 1.1.8. A map T dened from a Banach space X into a Banach
space Y is called closed if its graph
G(T) = f(x; y) : x 2 D(T)g
is closed in X Y . In other words, T is closed if whenever xn ?! x and
Txn ?! y, we have x 2 D(T) and Tx = y
4
Remark. Every T 2 B(X; Y ) is closed
Theorem 1.1.9. Closed Graph Theorem
A closed linear map which maps a Banach space into a Banach space is con-
tinuous
Denition 1.1.10. A map T 2 B(X; Y ) is invertible if there is a bounded
linear map T?1 2 B(Y;X) such that T?1T = IX (the identity of X) and
TT?1 = IY (the identity operator of Y )
Corollary 1.1.11. Every continuous bijection between Banach spaces has a
continuous inverse.
Proof.
Let T 2 B(X; Y ). Then G(T) = f(x; Tx) : x 2 Xg is closed in X Y ,
G(T?1) = f(Tx; x) : x 2 Xg is closed in Y X
T?1 is a closed linear operator mapping Y into X.Therefore T?1 is bounded
by the closed graph theorem.
Denition 1.1.12. Let A be a subset of a Banach space X. A is said to
be precompact if any sequence of A has a Cauchy subsequence. A is totally
bounded if for every > 0, there exists a nite cover for A of open balls of
same radius
Denition 1.1.13. A is said to be compact if any sequence of A has a subse-
quence that converges to some point of A.
Remark. A set is said to be precompact in a Banach X if and only if its closure
is compact in X.
Proposition 1.1.14. Let A be a subset of a Banach space X. A is said to be
precompact if and only if it is totally bounded.
Proof.
Assume that A is a precompact subset of X, we show that A is totally
bounded.
Let > 0 and x 2 A
x 2 A ) B(x; )

A 6= ;
) 9 ax 2 A : ax 2 B(x; )
) 9 ax 2 A : d(ax; x) <
) 9 ax 2 A : x 2 B(ax; )
[
a2A
B(a; )
since x was arbitrarily chosen, we have that A
[
a2A
B(a; ), using the compactness
of A; 9 a1; a2:::an 2 A such that A
[n
i=1
B(ai; ) ) A
[n
i=1
B(ai; )
5
since A A.Hence A is totally bounded.
Conversely, suppose A is totally bounded, we show that A is precompact
(i.e A is compact).
Let fangn A, we show that fangn has a Cauchy subsequence
Case I. fan; n 1g is nite
Then there exists n1; ::; np elements of N such that
fan : n 1g = fan1 ; ::; anpg:
Dene for each i 2 f1; ::pg, Ei := fn 2 N : an = anig.
So we have N =
Sp
i=1 Ei.
Since N is innite, then one of the sets Ei is innite. Choose i0 2 f1:; ; :pg such
that Ei0 is innite.
Since Ei0 N, then it has a minimum element. Let m1 := minEi0
For k 1, mk+1 := minfEi0 n fm1;m2:::mkgg
Clearly, mk < mk+1, 8k 2 N, also amk = ani0 since mk 2 Ei0
famkgk1 is a constant subsequence of fang which is convergent.
Case II. fan; n 1g is innite.
For = 1
22 > 0; 9 x1; x2:::xm such that A
m[
i=1
B(xi;
1
22 ) since A is totally
bounded.
This implies that A
m[
i=1
B
(xi;
1
22 ) since B(xi; 1
22 ) B
(xi; 1
22 ).
So A
m[
i=1
B
(xi;
1
22 ) since
m[
i=1
B
(xi;
1
22 ) is closed.
It follows that A
m[
i=1
B(xi;
1
2
) which implies that
fangn
m[
i=1
B(xi;
1
2
), since fangn A
Therefore, 9 i0 2 f1; 2:::mg such that B(xio ; 1
2 ) contains innitely many
terms of the sequence.
I1 := fn 2 N : an 2 B(xio ;
1
2
)g:
I1 is innite and for any fangn2I1 , d(an; am) < 1 8m; n 2 I1.
For 1
23 > 0; 9×1; x2:::xr : A
[r
i=1
B(xi;
1
22 ) so, 9i1 2 f1; 2:::rg such that
B(xi1 ; 1
22 ) contains innitely many terms of the subsequence fangn2I1 .
Now dene I2 :=

n 2 I1 : an 2 B(xi1 ;
1
22 )

:
6
I2 I1 and for n;m 2 I2,
d(an; am) d(an; ai1) + d(am; ai1)
1
22 +
1
22 =
1
2
:
Iteratively, for any Ik innite we can get Ik+1 innite with Ik+1 Ik and
8m; n 2 Ik+1
d(an; am) <
1
2k
Given nk choose nk+1 2 Ik+1 such that nk+1 > nk (this is possible because Ik+1
is innite). Now for j > k the index nj belongs toIk (becauseI1 I2 I3
:::is a nested sequence of sets)
Now for k < j
d(ank ; anj ) d(ank ; ank+1) + : : : + d(anj?1 ; anj )
1
2k + 1
2k+1 + : : : + 1
2j?1
= 1
2k (1 + 1
2 + : : : + 1
2j?k?1 )
1
2k?1 ?! 0; as k ?! 1
fankgk1 is a Cauchy subsequence of fang.Its convergence is guaranteed by the
completeness X. Hence A is precompact
Theorem 1.1.15. (Arzela-Ascoli)[6]
A subset A of the space of continuous functions C(K), where K is a non-empty
compact subset of RN, is relatively compact if and only if the two following con-
ditions are satised
i) A is uniformly bounded, i.e there exists M > 0 such that 8x 2 X;
jf(x)j M; 8f 2 K.
ii) A is equicontinuous i.e 8 > 0; 9 > 0:
8 x; y 2 X; kx ? yk ) jf(x) ? f(y)j ; 8f 2 K
.
Denition 1.1.16. An inner product (or scalar product) on a vector space X
is a scalar valued function, h ; i on X X such that
i) for each y 2 Xfixed; the functional x 7! hx; yi is linear..
ii) hx; yi = hy; xi; where the bar the complex conjugation.
iii) hx; xi 0; 8 x 2 X.
iv) 8 x 2 X, hx; xi = 0 if and only if x = 0.
7
The pair (X; h ; i) is called Pre-hilbertian (or inner product) space.
The function kkX =
p
hx; xi denes a canonical norm on X.
Denition 1.1.17. A Pre-hilbertian space (H; hi) is called a Hilbert space if
it is complete when equipped with the corresponding canonical norm.
Denition 1.1.18. Cauchy-Schwarz Inequality and parallelogram
law
Let hx; yi be an inner product on a vector space X.
Then
jhx; yij kxkkyk for all x; y 2 X: (1.1.2)
kx + yk2 + kx ? yk2 = 2(kxk2 + kyk2) for all x; y 2 X (1.1.3)
Proposition 1.1.19. The polarization identity.
Let X be an inner product space. Then for arbitrary x; y 2 X,
hx; yi =
1
4
fjjx + yjj2 ? jjx ? yjj2 + ijjx + iyjj2 ? ijjx ? iyjj2g (1.1.4)
where i2 = ?1.
Theorem 1.1.20. Jordan-Von Neumann.
The norm of a normed linear space X is given by an inner product if and only
if this norm satises the parallelogram law, i.e, if and only if,
kx + yk2 + kx ? yk2 = 2(kxk2 + kyk2); 8 x; y 2 X:
Denition 1.1.21. Let H be a Hilbert space and x; y 2 H. We say that x is
orthogonal to y denoted by x ? y if hx; yi = 0.
For M H, we say that x is orthogonal to M and write x ? M, if x is
orthogonal to every vector y in M.
The subset of vectors of H ortogonal to M is denoted by
M? = fx 2 H : x ? Mg
and is called the orthogonal complement of M in H.
Proposition 1.1.22. [1] Let M and N be arbitrary subspaces of a Hilbert
space H.Then the following holds
i) M? is a closed subspace of H
ii) M M??,
iii) if M N then N? M?
iv) (M?)? = M.
8
Denition 1.1.23. Given a closed subspace M of H, an operator P dened
on H is called the orthogonal projection onto M if
P(m + n) = m, for all m 2 M and n 2 M?
Theorem 1.1.24. The projection Theorem
Let H be a Hilbert space and M be a closed subspace of H. For an arbitrary
given vector x 2 H, there exists a unique vector m 2 M such that
jjx ? mjj jjx ? mjj for allm 2 M:
Furthermore, z 2 M is the unique vector m if and only if
(x ? z)?M:
Corollary 1.1.25. Direct Sum Decomposition
Let M be a closed subspace of a Hilbert Space, H. Then H = M M?:
Denition 1.1.26. An orthonormal system is a family f’igi2I of elements of
H such that h’i; ‘ji = i j , where, i j is the kronecker delta dened by
i j =

1 i = j;
0 i 6= j:
Example 1.1.27.

ei2nx ; n 2 Z

is an othonormal system for L2([0; 1])
It is easy to see that
hen; emi =
Z
[0;1]
ei2nxe?i2mxdx
=
Z
[0;1]
ei2(n?m)xdx =

1 if n = m
0 if n 6= m:
Denition 1.1.28. [1] A Hilbert space H is separable if H contains a countable
dense subset. Equivalently, a Hilbert space H is said to be separable if there
exists a sequence of vectors v1; v2; :::; vk; ::: which span a dense subspace of H.
Theorem 1.1.29.
A Hilbert space admits a countable orthonormal basis if and only if it is sepa-
rable.
Theorem 1.1.30. Riesz representation theorem
Let f be a continuous linear form on Hilbert space i.e f 2 H. Then there
exist a unique uf 2 H such that hf; vi = hv; uf i for all v 2 H.
Furthermore, we have kfkH = kufkH
9
Denition 1.1.31. Let X and Y be Banach spaces and T 2 B(X; Y ), dene
the dual (also called adjoint) operator as a map T : Y ?! X dened by
Tf = f T
that is,
(Tf)(x) = f(T(x)) for all x 2 X:
T is called the (topological) dual or adjoint operator of T.
Remark. T 2 B(Y ;X)
Denition 1.1.32. Adjoint operators on Hilbert spaces
Let T 2 B(H1;H2), the adjoint of T is the unique map T : H2 ?! H1 such
that
hTx; yi = hx; Tyi for all x 2 H1 and all y 2 H2
Example 1.1.33. Let be a bounded complex valued Lebesque measurable
function on [a; b]. Let
T : L2([a; b]) ?! L2([a; b])
be the bounded linear operator dened by T(f) = f, that is,
(Tf)(t) = (t) f(t) for a:e: t 2 [a; b]:
For all f; g 2 L2([a; b]) we have
hTf; gi =
Z b
a
(Tf)(t)g(t)dt =
Z b
a
(t)f(t)g(t)dt = hf; gi:
Thus T(g) = g :
Theorem 1.1.34. Let T : L2([a; b]) ?! L2([a; b]) be the bounded operator
dened by
(Tf)(t) =
Z b
a
G(t; s)f(s)ds
where G is in L2([a; b] [a; b]).
For all g 2 L2([a; b]),
(Tg)(t) =
Z b
a
G(s; t)g(s)ds :
Proof.
10
hTf; gi =
Z b
a
Z b
a
G(t; s)f(s)ds

g(t)dt
=
Z b
a
f(s)
Z b
a
G(t; s)g(t)dt

ds by Fubini’s thorem
= hf; gi
where
g(s) =
Z b
a
G(t; s)g(t)dt
1.2 Complexication of real Banach spaces
Many of the classical Banach functions spaces exist in real or complex-valued
versions. Examples are the Lp()-spaces and C(K)-spaces. Usually one is in-
terested in knowing whether a theory carried for real Banach spaces also holds
for complex Banach spaces (or vice-versa). An approach of solution is given by
the Complexication theory of real Banach spaces. Complexication preserves
norm and allows us to extend all basic notions on any arbitrary real Banach
space to Complex Banach space.
Denition 1.2.1. A complex vector space EC is a complexication of a real
vector space E if the two following conditions holds.
a) There is a one-to-one real linear map j : E ?! EC
b) complex-span (j(E)) = EC.
There are, however various alternative concrete descriptions, some of which
include Ordered pair, Tensor and Linear operator descriptions of complexi-
cation.
Hence T is well dened ,infact T 2 B(`2)
Ordered pair description of a complexication.
If E is a real vector space, we can make E E a vector space by dening
(x; y) + (u; v) := (x + u; y + v) 8x; y; u; v 2 E
( + i)(x; y) := (x ? y; x + y) 8x; y 2 E; 8; 2 R:
11
Consider the map
j : E ?! E E
x 7! (x; 0):
Clearly,
j(x + y) = j(x) + j(y) for any x; y 2 E and 2 R;
Ker(j) = fx 2 E : j(x) = (0; 0)g = f0g;
and
Complex ? span(j(E)) = E E:
The map j satises the conditions a) F(T(B)) ) F(T(B))andb)above; andsothiscomplexvectorspaceisacomplexificationofE:ItisconvenienttodenoteitbyEEiE:andalsosuppressrefrencetojbywrittingz = x+iy for the element z =
(x; y) = j(x) + ij(y):Itisnaturaltowritex=For other two descriptions, we refer to[4].
Denition 1.2.2. Let E be a real Banach space and EC := EiE.
?
EC; jj jj

is called a complexication of E if
?
EC; jjjj

is a complex Banach space, jjjjjE
is the original norm of E (i.e jjx + i0jj = jjxjj; 8x 2 E) and
jjx + iyjj = jjx ? iyjj 8x; y 2 E:
Now we might ask the Question: Is there a norm on EC which makes EC
a complex Banach space and induces the original norm onE ?
The answer is armative and there are innitely many ways to do so.[4].
Proposition 1.2.3. Let EC be a complexication of the real space E endowed
with a norm jj jj such that (E; jj jj) is Banach. Then jj jjT as dened below
denes a norm on EC.
jjx + iyjjT := sup
0t2
jj(cos t)x ? (sin t)yjj
All other complexication norms jj jj on EC are equivalent to jj jjT . Indeed
jjx + iyjjT jjx + iyjj 2jjx + iyjjT 8 x; y 2 E: (1.2.1)
Denition 1.2.4. Let E be a real Banach space. We say that a norm on the
complexication EC is reasonable if
c) jjj(x)jj = jjxjj 8x 2 E
d) jjx + iyjj = jjx ? iyjj x; y 2 E
When EC is equipped with such a norm, we call it a reasonable complexication
of E
12
Proposition 1.2.5. Let EC be a reasonable complexication of the real Banach
space E. for any x; y 2 E we have jjxjjE jjx+iyjjEC and jjyjjE jjx+iyjjEC
Proof. By property (c),
2jjxjjE = jj(x + iy) + (x ? iy)jjEC jjx + iyjjEC + jjx ? iyjjEC
An application of property d) gives jjxjjE jjx + iyjjEC
Similarly we have the other inequality.
Proposition 1.2.6. Let EC be a complexication of the real Banach space E
For any x; y 2 E we have,
sup
0t2
jj(cos t)x ? (sin t)yjjE jjx + iyjjEC
and
jjx + iyjjEC inf
0t2
(jj(cos t)x ? (sin t)yjjE + jj(sin t)x + (cos t)yjjE):
Proof. For each 0 t 2
jjx+iyjjEC = jjeit(x+iy)jjEC = jj((cos t)x?(sin t)y)+i((sin t)x+(cos t)y)jjEC.
Using proposition (1.2.5) on the left and the triangle inequality on the right,
we have
jj(cos t)x ? (sin t)yjjE jjx + iyjjEC and jjx + iyjjEC jj(cos t)x ? (sin t)yjjE +
jj(sin t)x + (cos t)yjjE.
Hence, the result follows immediately.
Let us check verify that jj jjT is a reasonable complexication norm.
i)For any x 2 E, jjxjjT = jjx + i0jjT = sup
0t2
jjxcost ? 0sintjj = jjxjj
ii) jjxcost ? ysintjj = jjxcos(?t) + ysin(?t)jj and the function
t 7! xcost ? ysint is periodic with period of 2 for all x; y 2 E.Therefore,
jjx + iyjjT = sup
0t2
jjxcost ? ysintjj
= sup
t2R
jjxcos(?t) + ysin(?t)jj
= sup
t2R
jjxcos(t) + ysin(t)jj
= sup
0t2
jjxcost + ysintjj
jjx + iyjjT = jjx ? iyjjT
We have shown that property c) and d) are satised. Hence jjjjT is a reasonable
complexication norm.
Let us also verify the inequality in (1.2.1)
From proposition (1.2.6)
13
jjx + iyjjT jjx + iyjj inf
0t2
(jjxcost ? ysintjjE + jjxsint + ycostjjE)
sup
0t2
jjxcost ? ysintjjE + sup
0t2
jjxsint + ycostjjE
= 2jjx + iyjjT
jjx + iyjjT jjx + iyjj 2jjx + iyjjT
The norm jj jjT was rst considered by A.Y Taylor [ad]. (EC; jj jjT ) is known
as Taylor complexication of E.
There is a useful alternative description of jjx + iyjjT :
jjx + iyjjT = sup
0t2
jjxcost ? ysintjj
= sup
0t2
sup
jjfjjE1
jf(x)cost ? f(y)sintj
= sup
jjfjjE1
p
f(x)2 + f(y)2 ; 8x; y 2 E
Another feature of the Taylor complexication, is that it is a general complex-
ication whose denition is not tied to any specic characteristic of the real
Banach space E which is being complexied. Moreover, this procedure allows
us to extend continuous linear maps between real Banach space to complex
linear maps between their complexications without increasing the norm. If
L : E ?! F is a linear map between real vector spaces E and F, there is a
unique complex-linear extension ~L : EC ?! FC given by
~L
(x + iy) = L(x) + iL(y)
Proposition 1.2.7. Let E and F be real Banach spaces. If L 2 L(E; F), then
~L
2 L

(EC; jj jjT ); (FC; jj jjT )

and jj~ Ljj = jjLjj
Proof. Since ~L extends L, we have jj~Ljj jjLjj.
On the other hand, if x; y 2 E then
jj~L(x + iy)jjT = jjL(x) + iL(y)jjT = sup
0t2
jjL(x)cost ? L(y)sintjjF
= sup
0t2
jjL(xcost ? ysint)jjF
jjLjj sup
0t2
jjxcost ? ysintjjE
jj~L(x + iy)jjT jj~ Ljjjjx + iyjjT =) jj~Ljj jjLjj
Hence,
jj~Ljj = jjLjj
14
Taylor’s procedure is just one of innitely many procedures with similar
properties.
1.3 Some function spaces (Lp, Sobolev spaces)
Denition 1.3.1. Let 1 p < 1;
be an open bounded subset of Rn. We
dene
Lp(
) as the set of measurable functions f :
?! R such that
Z

jf(x)jpdx < +1
L1(
) as the set of measurable functions f :
?! R such that esupjfj < 1
where,
esupjfj = inffk > 0; jf(x)j k a:e x 2
g
For f 2 Lp(
), we dene,
jjfjjp =
Z

jf(x)jp
1
p ; 1 p < 1:
jjfjj1 = esupjfj; if p = 1:
Theorem 1.3.2. The following properties holds for Lp space
i) Lp-space is Banach for 1 p 1
ii) Lp-space is Re exive for 1 < p < 1
iii) Lp-space is Separable for 1 p < 1
F(T(B)) ) F(T(B))Itisalsoexpedienttorecallsomenotationsandbasicresultsfromdistributiontheory:
Denition 1.3.3. The Space L10
(
) is the space of all Lebesgue measurable
functions in
having absolute value integrable on each compact subset of

A multi-index is a vector (1; 2:::n) 2 Nn.
The length of is given by jj = 1 + ::: + n
We also dene the generalized derivative
D =
@jj
@1×1 : : :@nxn
Denition 1.3.4.
A locally integrable function v i.e element of L10
(
) is called the ? th weak
derivative of u 2 L10
(
), if it satises
Z

u(x)D(x)dx = (?1)jj
Z

v(x)(x)dx; 8 2 D(
)
Where D(
) denotes the set of C1-functions on
with compact support in

15
Let x 2 Rn, we write x = (x0; xn) with x0 2 Rn?1, x0 = (x1; x2; :::xn?1). We
consider the following notations[5].
Rn
+ = fx = (x0; xn) 2 Rn : xn > 0g
B = fx = (x0; xn) 2 Rn : jjx0jj < 1; jxnj < 1g
B+ = fx = (x0; xn) 2 B : xn > 0g
B0 = fx = (x0; xn) 2 B : xn = 0g
Denition 1.3.5. We say that an open subset
Rn is of class Cm(
)(m; integer)
if for every x 2 @
, there exist an open neighbourhood U of x in Rn and a
map : B ?! U such that,
i) is a bijection
ii) 2 Cm( B; U);?1 2 Cm(U
; B)
iii) (B+) =
U; (B0) = @
U
Denition 1.3.6.
Let 1 p +1, m 2 N,. The Sobolev space Wm;p(
) is dened by
Wm;p(
) = fu 2 Lp(
) j Du 2 LP (
) for all jj mg
We shall be working with the case p = 2 .The Sobolev space Wm;p(
) are
denoted by Hm(
). H1
0(
) is the closure of D(
) in H1(
)
. Finally, we shall consider two important results which are very instrumental
to the application of spectral theorem of compact self adjoint operators to elliptic
partial dierential equations.
Proposition 1.3.7. Poincare Inequality
Let 1 p < 1 and
a bounded open subset of RN. Then there exist a
constant C(
; p) such that
jjujjLp(
) CjjrujjLp(
) 8u 2 W1;p
0 (
):
If
is connected and satises a C1 boundary condition, then there exists a
constant C(
; p) such that
jju ? ujjLp(
) CjjrujjLp(
); 8u 2 W1;p(
)
u =
1
j
j
Z

u(x)dx
, is the mean value of u on

16
Denition 1.3.8.
Let E and F be two normed vector spaces such that E F.We say Ec ,! F is
compact embeddings if any bounded subset of E is precompact in F ,or equiv-
alently any bounded sequence of E has a subsequence that converges in F.
17
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