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1.1 Basic notions and results from Functional
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.
Remark. Every normed linear space has a completion 
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
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
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. 
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).
Furthermore, B(X; Y ) becomes naturally endowed with the operator norm,
jjTjj := sup
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 .
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
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 )
jjTjj = max
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
jG(x1; t) G(x2; t)jjf(t)jdt
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,
jG(x; y)jdy (1.1.1)
jjTfjj1 = max
showing that T is bounded and
jG(x; y)jdy :
kTk M; where M :=
Now we dene,
S is continuous on [a; b] and therefore it attains its maximum at some x0 2
Let g(y) :=
G(x0;y) if G(x0; y) 6= 0
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
and (gn) converges in L1([a; b]) to g. Hence we have ,
kTk = sup
kTfk kTgnk = max
) kTk (Tgn)(x0) =
K(x0; y)gn(y)dy !
K(x0; y)g(y)dy =
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
G(x; y)f(y)dy ; 8f 2 C[a; b]:
Then T is a bounded linear map with
jjTjj = max
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
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-
Denition 1.1.10. A map T 2 B(X; Y ) is invertible if there is a bounded
linear map T1 2 B(Y;X) such that T1T = IX (the identity of X) and
TT1 = IY (the identity operator of Y )
Corollary 1.1.11. Every continuous bijection between Banach spaces has a
Let T 2 B(X; Y ). Then G(T) = f(x; Tx) : x 2 Xg is closed in X Y ,
G(T1) = f(Tx; x) : x 2 Xg is closed in Y X
T1 is a closed linear operator mapping Y into X.Therefore T1 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
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.
Assume that A is a precompact subset of X, we show that A is totally
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; )
since x was arbitrarily chosen, we have that A
B(a; ), using the compactness
of A; 9 a1; a2:::an 2 A such that A
B(ai; ) ) A
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 =
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
22 ) since A is totally
This implies that A
22 ) since B(xi; 1
22 ) B
22 ) since
22 ) is closed.
It follows that A
) which implies that
), 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 ;
I1 is innite and for any fangn2I1 , d(an; am) < 1 8m; n 2 I1.
23 > 0; 9×1; x2:::xr : A
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 ;
I2 I1 and for n;m 2 I2,
d(an; am) d(an; ai1) + d(am; ai1)
Iteratively, for any Ik innite we can get Ik+1 innite with Ik+1 Ik and
8m; n 2 Ik+1
d(an; am) <
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(anj1 ; anj )
2k + 1
2k+1 + : : : + 1
2k (1 + 1
2 + : : : + 1
2k1 ! 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)
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.
The pair (X; h ; i) is called Pre-hilbertian (or inner product) space.
The function kkX =
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
Let hx; yi be an inner product on a vector space X.
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 =
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.  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.
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
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:
ei2nx ; n 2 Z
is an othonormal system for L2([0; 1])
It is easy to see that
hen; emi =
1 if n = m
0 if n 6= m:
Denition 1.1.28.  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.
A Hilbert space admits a countable orthonormal basis if and only if it is sepa-
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
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
(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
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 =
(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
where G is in L2([a; b] [a; b]).
For all g 2 L2([a; b]),
G(s; t)g(s)ds :
hTf; gi =
ds by Fubini’s thorem
= hf; gi
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-
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:
Consider the map
j : E ! E E
x 7! (x; 0):
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;
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=<e z and y = =mz.
For other two descriptions, we refer to.
Denition 1.2.2. Let E be a real Banach space and EC := EiE.
EC; jj jj
is called a complexication of E if
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..
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
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
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,
jj(cos t)x (sin t)yjjE jjx + iyjjEC
jjx + iyjjEC inf
(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,
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
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
jjxcos(t) + ysin(t)jj
jjxcos(t) + ysin(t)jj
jjxcost + ysintjj
jjx + iyjjT = jjx iyjjT
We have shown that property c) and d) are satised. Hence jjjjT is a reasonable
Let us also verify the inequality in (1.2.1)
From proposition (1.2.6)
jjx + iyjjT jjx + iyjj inf
(jjxcost ysintjjE + jjxsint + ycostjjE)
jjxcost ysintjjE + sup
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
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
(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
(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
jj~L(x + iy)jjT jj~ Ljjjjx + iyjjT =) jj~Ljj jjLjj
jj~Ljj = jjLjj
Taylor’s procedure is just one of innitely many procedures with similar
1.3 Some function spaces (Lp, Sobolev spaces)
Denition 1.3.1. Let 1 p < 1;
be an open bounded subset of Rn. We
) as the set of measurable functions f :
! R such that
jf(x)jpdx < +1
) as the set of measurable functions f :
! R such that esupjfj < 1
esupjfj = inffk > 0; jf(x)j k a:e x 2
For f 2 Lp(
), we dene,
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
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
@1×1 : : :@nxn
A locally integrable function v i.e element of L10
) is called the th weak
derivative of u 2 L10
), if it satises
u(x)D(x)dx = (1)jj
v(x)(x)dx; 8 2 D(
) denotes the set of C1-functions on
with compact support in
Let x 2 Rn, we write x = (x0; xn) with x0 2 Rn1, x0 = (x1; x2; :::xn1). We
consider the following notations.
+ = 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(
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
iii) (B+) =
\ U; (B0) = @
Let 1 p +1, m 2 N,. The Sobolev space Wm;p(
) is dened by
) = 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(
denoted by Hm(
) 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
; p) such that
) 8u 2 W1;p
is connected and satises a C1 boundary condition, then there exists a
; p) such that
); 8u 2 W1;p(
, is the mean value of u on
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.