4 Fredholm's Alternative

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§4. Fredholm’s Alternative
Theorem 1 ( Riesz Representation )
Let ϕ be a bounded linear form defined from a Hilbert space H into
K = (R or C) then, there exists a unique element y in H such that, for all
x in H, we have
ϕ(x) = 〈x, y〉 .
In other words, all linear form in a Hilbert space H is a inner product
in H. Say
∀x ∈ H ∃!y ∈ H, ϕ(x) = 〈x, y〉 .
Proof
1. Existence
Suppose that, for all x ∈ H, ϕ(x) = 0, then the space H coincide with
the null space N(ϕ), so we may choose y = 0. That is to say
ϕ(x) = 0 = 〈x, 0〉 , for all x ∈ H.
Suppose that, ϕ(x) 6= 0 then the null space N(ϕ) is a closed proper
subspace of H. Say N(ϕ) ⊂ H, so we may find a non zero element z in the
orthogonal subspace N⊥(ϕ) of the null space N(ϕ). Say
H = N(ϕ)⊕N⊥(ϕ).
For all x ∈ H and z ∈ N⊥(ϕ) we get
x− ϕ(x)
ϕ(z)
z ∈ N(ϕ),
namely
ϕ
(
x− ϕ(x)
ϕ(z)
z
)
= ϕ(x)− ϕ(x)
ϕ(z)
ϕ(z) = 0.
Therefore, we write 〈
x− ϕ(x)
ϕ(z)
z, z
〉
= 0,
or still
〈x, z〉 = ϕ(x)
ϕ(z)
‖z‖2 ,
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hence, it comes
ϕ(x) =
ϕ(z)
‖z‖2
〈x, z〉
=
〈
x,
ϕ(z)
‖z‖2
z
〉
.
In other words, we obtain
y =
ϕ(z)
‖z‖2
z.
2. Uniqueness
Suppose that y1 and y2 be two elements of H such that
ϕ(x) = 〈x, y1〉 = 〈x, y2〉 , for all x ∈ H,
or still
〈x, y1 − y2〉 = 0, for all x ∈ H.
In particular for x = y1 − y2, we get
〈y1 − y2, y1 − y2〉 = ‖y1 − y2‖2 = 0.
Hence
y1 = y2
Theorem 2
Let A be an operator defined from a Hilbert space H1 into a Hilbert space
H2 then, there exists an adjoint operator of A denoted by A∗ defined from
H2 into H1 such that
〈Aϕ,ψ〉H2 = 〈ϕ,A
∗ψ〉H1 ,
for all ϕ ∈ H1 and ψ ∈ H2. Besides, we have
‖ A ‖=‖ A∗ ‖ .
Proof
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Define a linear functional U from H1 into K = (R or C), such that, for
all ϕ ∈ H1 and for all ψ ∈ H2, we get
H1 → K
ϕ 7→ U(ϕ) = 〈Aϕ,ψ〉
This functional is bounded, for
|U(ϕ)| = |〈Aϕ,ψ〉|
≤ ‖Aϕ‖ ‖ψ‖
≤ ‖A‖ ‖ϕ‖ ‖ψ‖
By the Riesz representation theorem there exists a unique element g ∈
H1 such that
U(ϕ) = 〈ϕ, g〉 ,
or still
U(ϕ) = 〈Aϕ,ψ〉 = 〈ϕ, g〉 ,
this equality defines an adjoint operator of A denoted by A∗ defined from
H2 into H1 by
A∗ψ = g.
In other words, we write
〈Aϕ,ψ〉 = 〈ϕ, g〉 = 〈ϕ,A∗ψ〉 .
It is easy to verify that, the operator A∗ is linear and unique, say
〈ϕ,A∗(α1ψ1 + α2ψ2)〉 = 〈Aϕ,α1ψ1 + α2ψ2〉
= α1 〈Aϕ,ψ1〉+ α2 〈Aϕ,ψ2〉
= α1 〈ϕ,A∗ψ1〉+ α2 〈ϕ,A∗ψ2〉
= 〈ϕ, α1A∗ψ1〉+ 〈ϕ, α2A∗ψ2〉
= 〈ϕ, α1A∗ψ1 + α2A∗ψ2〉 .
Besides, we get
‖A∗ψ‖2 = 〈A∗ψ,A∗ψ〉
= 〈AA∗ψ,ψ〉
≤ ‖AA∗ψ‖ ‖ψ‖
≤ ‖A‖ ‖A∗ψ‖ ‖ψ‖ ,
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after simplification, we obtain
‖A∗ψ‖ ≤ ‖A‖ ‖ψ‖
Hence, we get the following inequality
‖A∗‖ ≤ ‖A‖ , (1)
Inversely,
‖Aϕ‖2 = 〈Aϕ,Aϕ〉
= 〈ϕ,A∗Aϕ〉
≤ ‖ϕ‖ ‖A∗Aϕ‖
≤ ‖ϕ‖ ‖A∗‖ ‖Aϕ‖ ,
after simplification, we obtain
‖Aϕ‖ ≤ ‖A∗‖ ‖ϕ‖
Hence, we get the following inequality
‖A‖ ≤ ‖A∗‖ . (2)
The relations (1) and (2), give the equality of the norms
‖A‖ = ‖A∗‖ .
Theorem 3
Let A be a compact operator defined from a Hilbert space H1 into a
Hilbert space H2 then, the adjoint operator A∗ defined from H2 into H1 is
also a compact operator.
Proof
Let ψn be a bounded sequence ofH2, that is to say, there exists a constant
M > 0, such that ‖ψn‖ n
Let B∗ be an operator defined on H2 given by
B∗ = A∗ + V ∗,
where V ∗ is an operator of finite dimensional defined as follows
V ∗(ψ) =
n∑
k=1
〈ψ,ψ∗k〉ϕ∗k. (8)
Interchanging the roles, let ψ0 be an element of N(I−B∗) then, we write
the equation
(I −B∗)ψ0 = 0
ψ0 −A∗ψ0 − V ∗ψ0 = 0,
or explicitly,
ψ0 −A∗ψ0 −
n∑
k=1
〈ψ0, ψ∗k〉ϕ∗k = 0,
in order to obtain m= 0 then, for all ψ in H2, we
have
〈(I −A)ϕ,ψ〉 = 0
⇒ 〈ϕ, (I −A∗)ψ〉 = 0, ∀ψ ∈ H2
it follows that
ϕ ∈ R⊥(I −A∗),
or still
N(I −A) ⊂ R⊥(I −A∗).
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Inversely, suppose that, ϕ ∈ R⊥(I −A∗) then, for all ψ in H2, we have
〈ϕ, (I −A∗)ψ〉 = 0
⇒ 〈(I −A)ϕ,ψ〉 = 0, ∀ψ ∈ H2,
it follows that
(I −A)ϕ ∈ H⊥2 = {0} ⇔ ϕ ∈ N(I −A),
or still
R⊥(I −A∗) ⊂ N(I −A).
Whence
R⊥(I −A∗) = N(I −A) ⇔ R(I −A∗) = N⊥(I −A).
Theorem 6 (Fredholm Alternative)
Let A be a compact operator defined from a Hilbert space H into itself,
then the equations
ϕ−Aϕ = f , (9)
and its adjoint
ψ −A∗ψ = g, (10)
admit a unique solutions for all second member if the homogeneous equations
ϕ−Aϕ = 0,
and
ψ −A∗ψ = 0,
admit uniquely the trivial solutions ϕ = 0 and ψ = 0.
Contrarily, if the homogeneous equations have the same number of solu-
tions ϕ1, ϕ2, ..., ϕn and ψ1, ψ2, ..., ψn respectively then, the nonhomogeneous
equations (9) and (10) have solutions if and only if, for all k = 1, 2, ..., n,
we have
〈f, ψk〉 = 0,
and
〈g, ϕk〉 = 0.
The general solution of the equation (9) is given by
ϕ = ϕ0 +
n∑
k=1
αkϕk,
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and the one of the equation (10) is given by
ψ = ψ0 +
n∑
k=1
αkψk,
where ϕ0 and ψ0 are any particular solutions of the equations (9) and (10)
respectively, a1, a2, ..., an are arbitrary constants.
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Bibliography
[1] R. KRESS. Linear integral equations. Applied Mathematical Sci-
ences 82, Springer-Verlag, Heidelberg (1989).
[2] M. NADIR. Cours d’analyse fonctionnelle, université de Msila
2004.
Address. Prof. Dr. Mostefa NADIR
Department of Mathematics
Faculty of Mathematics and Informatics
University of Msila
28000 ALGERIA
E-mail. mostefanadir@yahoo.fr
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