Compactness, Fredholm Alternative and Spectrum I. Compactness

Compactness, Fredholm Alternative and Spectrum
I. Compactness
Definition. We say that E ⊆ X is compact if from eavery open covering of E it
is possible to extract a finite subcovering of E.
Definition. A subset E of a normed space X is sequentially compact if for
every sequence {xk } ⊂ E there exists a subsequence {xk }, converging in X.
Theorem 1. Let X be a normed space and E ⊂ X. Then E is compact iff it is
sequentially compact.
• While a compact set is always closed and bounded, the following example exhibits
a closed and bounded set which is not compact in `2 .
Example. Consider the real Hilbert space
`2 = {x = {xk }∞
k=1 :
∞
X
x2k < ∞, xk ∈ R}
k=1
endowed with
hx, yi =
∞
X
xk yk ,
2
and kxk =
k=1
∞
X
x2k .
k=1
1
2
Let E = {ek }k≥1 , where e = {1, 0, 0, · · · }, e = {0, 1, 0, · · · }, etc.
and
Observe that E constitute an orthonormal basis in `2 . Thus, E is closed √
bounded in `2 . Howeve, E is not sequentially compact. Indeed, kei − ej k = 2,
if j 6= k, and therefore no subsequence of {ek }k≥1 can be convergent.
• In fact, only in finite-dimensional spaces each closed and bounded set is compact.
Theorem 1. Let B be a Banach space. B is finite-dimensional iff the unit ball
{x : kxk ≤ 1} is compact.
II. Weak Convergence and Compactness.
Definition. Let H be a Hilbert space with inner product h·, ·i and norm k · k. A
sequence {xk } ⊂ H converges weakly to x ∈ H and we write xk * x, if
hF, xk i∗ → hF, xi∗ ,
∀F ∈ H ∗ .
• The convergence in norm is called strong convergence.
• From Riesz’s representation theorem, it follows that {xk } ⊂ H converges
weakly to H iff
hxk , yi → hx, yi, ∀y ∈ H.
• The weak limit is unique, since xk → x and xk → z implv
hx − z, yi,
∀y ∈ H,
whence x = z.
Typeset by AMS-TEX
1
2
• Strong convergence implies weak convergence, since Schwarz’s inequality
gives
khxk , yik ≤ kxk − xkkyk.
– The two notions of convergence are equivalent in the finite-dimensiona spaces.
– It is not so in infinite dimensions, as the following example shows.
Example. Let H = L2 (0, 2π). The sequence vk = cos kx, k ≥ 1, is weakly
converhent to zero. In fact, ∀f ∈ L2 (0, 2π), the Riemann-Lbesgue theorem on
the Fourier coefficients of f implies that
Z 2π0
f (x) cos kx dx → 0, as k → ∞.
hf, vk i =
However
kuk kL2 (0,2π) =
√
π
and therefore {vk }k≥1 does not converge strongly.
Example. Let {ϕ}i=1,2,··· denote an orthonormal system in the Hilbert space H,
and we observe
√
kϕi − ϕj k = 2, ∀i, j ∈ N, i 6= j.
Consequently {ϕi }i=1,2,··· does not contains a strongly convergent subsequence.
According to Bessel’s inequality,
∞
X
|hϕ, f i|2 ≤ kf k2 < ∞, ∀f ∈ H,
i=1
and we infer
lim hϕi , f i = 0 = h0, f i ∀f ∈ H.
i→∞
We therefore obtain ϕ * 0 as i → ∞. Note that
k0k < 1 = lim inf kϕk.
k→∞
Principle of Uniform Boundedness. On the Hilbert space H let the sequence
of bounded functionals An : H → C with n ∈ N be given, such that each element
f ∈ H possesses a constant cf ∈ [0, +∞) with the property
(1)
|An f | ≤ cf ,
n = 1, 2, · · · .
Then we have a constant α ∈ [0, ∞) satisfying
kAn k ≤ α,
∀n ∈ N.
Proof. Step 1. Choose f0 ∈ H, ε > 0 and a constant c ≥ 0. Let A : H → C
denote a bounded linear functional such that
(2)
|Af | ≤ c ∀f ∈ H with kf − f0 k ≤ ε.
Then we have
2c
;
ε
indeed, if kxk ≤ 1, ∃f with kf − f0 k ≤ ε such thar x = 1ε (f − f0 ), and we infer
1
1
1
2c
|Ax| = Af − Af0 ≤ (|Af | + |Af0 |) ≤ .
ε
ε
ε
ε
kAk ≤
3
Step 2. If the statement (2) does not hold true, Step 1 together with the continuity
of the functional {An } enables us to construct a sequence of balls
Σn := {f ∈ H : kf − fn k ≤ εn },
n∈N
satisfying Σ1 ⊃ Σ2 ⊃ · · · with εn & 0 as n → ∞ and an index-sequence 1 < n1 <
n2 < · · · such that
|Anj x| ≥ j, ∀x ∈ Σj , j ∈ N,
which yields a contradiction to (1).
• We have observed that the norm in a Hilbert space is strongly continuous.
With respect to weak convergence, the norm is only lower semincontinuous,
as poperty (b) in the following theorem shows.
Theorem 2. Let {xk } ⊂ H such that xk weakly converges to x. Then
(a) {xk } is bounded,
(b) kxk ≤ lim inf k→∞ kxk k.
Proof. (a) follows from the principle of uniform boundedness. For (b), it is enough
to observe that
kxk2 ≤ lim hxk , xi ≤ kxk lim inf kxk k.
k→∞
k→∞
• The usefulness of weak convegence is revealed by the following compactness
result.
Theorem 3. Every bounded sequence in a Hilbert space H contains a subsequence
which is weakly convergent to an element x ∈ H.
Proof. Step 1. The sequence {xj } is bounded and thus we have a constant c ∈
[0, +∞) such that kxj k ≤ c, j = 1, 2, · · · . On account of
∀j ∈ N,
|hx1 , xj i| ≤ ckx1 k,
(1)
(1)
we find a subsequence {xj } ⊂ {xj } such that limj→∞ hx1 , xj i exists.
- Noting that
(1)
|hx2 , xj i| ≤ ckx2 k, ∀j ∈ N,
(2)
(1)
(2)
we select a further subsequence {xj } ⊂ {xj } such that limj→∞ hx2 , xj i exists.
- The continuation of this process gives a chain of subsequence
(1)
(2)
(k)
{xj } ⊃ {xj } ⊃ {xj } ⊃ · · · ⊃ {xj }
such that
(k)
lim hxi , xj i
j→∞
exists for i = 1, · · · k.
(k)
- The Cantor’s diagonal procedure then gives us the sequence xk
(k)
lim hxi , xk i
k→∞
exists for all i ∈ N.
such that
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Step 2. Denote the linear subspace of all fnite linear combinations of xj by M,
namely
N (x)
X
M = {x : x =
αi xi , αi ∈ C, N (x) ∈ N}.
i=1
Then
(k)
lim hx, xk i
k→∞
exists for all x ∈ M.
Making the transition to the linear closed subspace M ⊂ M ⊂ H,
(k)
lim hy, xk i
k→∞
exists for all y ∈ M.
Step 3. Due the projection theorem, each element y ∈ H can be represented in
the form
y = y1 + y2 , , with y1 ∈ M, y1 ∈ M⊥ .
This implies the existence of
(k)
(k)
(k)
lim hy, xk i = lim hy1 + y2 , xk i = lim hy1 , xk i,
k→∞
k→∞
k→∞
∀y ∈ H.
Step 4. From Step 3, we may define a linear functional A in H by setting
(k)
Ay = lim hy, xk i,
k→∞
∀y ∈ H.
Consider the bounded linear functional
(i)
Ai (y) = hxi , yi,
y ∈ H.
The principle of uniform boundedness gives us a constant C ∈ [0, ∞) such that
kx(n)
n k = kAn k ≤ C,
∀n ∈ N.
This and Theorem 1(b) imply kAk ≤ C. Thus A is a bounded linear functional
and Riesz representtion theorem then gives us an element x ∈ H satisfying
A(y) = hx, yi,
∀y ∈ H.
We then obtain
(k)
lim hy, xk i = lim Ak (y) = A(y) = hx, yi,
k→∞
k→∞
∀y ∈ H,
(k)
which means that the sequence {xk } converges weakly. III. Compact Operators
• Let V1 and V2 be normed linear spaces. Every operator in L(V1 , V2 ) transforms
bounded sets in V1 into bounded sets in V2 .
The subclass of operators that transform bounded sets into pre-compact sets
is particularly important.
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Definition. A subset Σ of a normed linear space V is called precompact, if each
sequence {ym } ⊂ Σ contains a stronly convegent subsequence {ymk }, which means
lim kymk − yn` k = 0.
k,`→∞
Definition. Let V1 and V2 be normed linear spaces and L ∈ L(V1 , V2 ). L is said
to be compact if for every bounded subset E of V1 , the image L(E) is pre-compact
in V2 . This means that each sequence {xm } in V1 with kxm kV1 ≤ r, for some r ∈ R
and all n ∈ N, contains a subsequence {xmk } such that {Lxmk } ⊂ V2 converges
strongly.
Proposition 4. A compact operator K : V1 → V2 is bounded.
Proof. If K : V1 → V2 were not bounded, there would exist a sequence {xn } ⊂ V1
with kxm k = 1, for all n ∈ N and kKxm k2 → ∞ as n → ∞. Therefore, we cannot
select a convergent subsequence from {Kxn } in V2 . • An equivalent characterization of compact operators between Hilbert spaces may
be given in terms of weak convergence.
Indeed, an operator is compact iff it converts weak convergence to strong
convergence. Precisely, we have the following.
Proposition 5. Let H1 and H2 be Hilbert spaces and L ∈ L(H1 , H2 ). L is
compact iff for every sequence {xk } ⊂ H1 ,
xk * x in H1 weakly implies
Lxk → Lx in H2 strongy.
Proof. (⇒) Let {xm } ⊂ H1 be a sequence with kxm k ≤ 1, m ∈ N. According to
Theorem 3, we have a subsequence {xmk } ⊂ {xm } which converges weakly to x in
H1 as k → ∞.
Then, by assumption
Kxmk → Kx strongly,
k → ∞.
Consequently, the operator K : H1 → H2 is compact.
(⇐) Now let K be compact and {xm } denote a sequence weakly converges to x. We
then have to prove Kxm → Kx strongly as m → ∞ in H2 .
If the latter statement were false, there would exist a number d > 0 and a subsequence {xmk } of {xm } satisfying
(3)
kKxmk − Kxk ≥ d > 0,
∀n ∈ N.
– On the other hand, by Theorem 2(b) the sequence {xm } is bounded in H1 . Since
the operator is compact, we have a further subsequence
{e
xmk } ⊂ {xmk } with K x
emk → y strongly, as n → ∞.
– Moreover, {xm } converges weakly to Kx, which implies y = Kx. Hence kKxmk −
Kxk → 0 as mk → ∞, in contradiction with (3). Example. From Theorem 3, the identity operator I : H → H is compact iff
dim H < ∞.
Also, any bounded operator with finite dimensional range is compact.
• The following proposition is useful.
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Proposition 6. Let L : H1 → H2 be compact. If G ∈ L(H2 , H3 ) or G ∈
L(H0 , H1 ), then the operator G ◦ L or L ◦ G is compact.
Proof. Since L is compact, the sequence xk * x weakly in H1 as k → ∞ is
transformed into the sequence Lxk which converges strongly to Lx in H2 as k → ∞.
(i) If G : H2 → H3 is continuous, we infer
G ◦ Lxk → G ◦ Lx as n → ∞ in H3 .
(ii) If G : H0 → H1 is continuous, the sequence yk * y weakly in H0 as k → ∞ is
transformed into the sequence Gyk which converges weakly to Gy in H1 as n → ∞.
Since L is compact, the sequence L ◦ Gyk converges strongly to L ◦ Gy in H2 as
k → ∞. IV. The Fredholm Alternative
The Fredholm alternative concerns compact linear operators from a space V into
itself and is an extension of the theory of linear mappings in finite dimensional
spaces.
• The result we are going to present are extensions of well known facts concerning
the solvability of linear algebraic systems of the form
(4)
Ax = b,
where A is an n × n matrix and b ∈ Rn .
– The following dichotomy holds: either (4) has a unique solution for every b or
the homogeneous equation Ax = 0 has nontrivial solutions.
• More precisely, system (4) is solvable iff b belong to the column space of A,
which is the orthogonal complement of ker(A> ).
– If w1 , ws span ker (A> ), this amounts to asking the s compatibility conditions,
0 ≤ s ≤ n,
b · wj = 0, j = 1, · · · , s.
Finally, ker(A) and ker(A> ) have the same dimension and if v1 , · · · , vn span
ker(A), the general solution of (4) is given by
x=x+
s
X
cj vj
j=1
where x is a particular solution of (4) and c1 , · · · , cs are arbitrary constants.
• The extension to infinite-dimensional spaces requires some care.
Theorem 7. Let K be a compact linear mapping of a normed linear space V into
itself. Then either
(i) the homogeneous equation x − Kx = 0 has a nontrivial solution x ∈ V or
(ii) for each y ∈ V the equation x − Kx = y has a unique solution x ∈ V .
Furthermore, in case (ii), the operator (I − K)−1 whose existence is asserted there
is also bounded.
To prove Theorem 7, we first establish the following.
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Proposition 8. Let K be a compact linear mapping of a normed linear space V
into itself. Let Φ = I − K. Then ∃ a constant C such that
dist (x, N (Φ)) ≤ CkΦxk,
∀v ∈ V.
Proof. Indeed, suppose that the result is not true. Then there exists a sequence
{xn } ⊂ V satisfying kΦxn k = 1 and dn = dist(xn , N (Φ)) → ∞.
- Choose a sequence {yn } ⊂ N (Φ) such that dn ≤ kxn − yn k ≤ 2dn . Then if
zn =
xn − yn
kxn − yn k
we have kzn k = 1 and kΦzn k ≤ d−1
n → 0, so that the sequence {Φzn } converges
to 0.
- But since K is compact, by passing to a subsequence if necessary, we may
assume that the sequence {Kzn } converges to an element y0 ∈ V .
- Since zn = (Φ + K)zn , we then also have {zn } converging to y0 and hence
y0 ∈ N (Φ).
dist (zn , N (Φ)) =
inf
y∈N (Φ)
kzn − yk
=kxn − yk−1
inf
y∈N (Φ)
xn − yn − kxn − yn ky =kxn − yn k−1 dist (xn , N (Φ)) ≥
1
.
2
From Proposition 8, we obtain the following result.
Proposition 9. Let K be a compact linear mapping of a normed linear space V
into itself. Let Φ = I − K. Then R(Φ) is a closed subspace of V .
Proof of Proposition 9. Let {xm } be a sequence in V whose image Φxm converges
to an element y ∈ V . To show tht R(Φ) is closed, we must show that
y = Φx
for some x ∈ V .
In fact, by Proposition 8, the sequence {dn } where dn = dist (xn , N (Φ)) is bounded.
Choose as before a sequence {yn } ⊂ N (Φ) such that dn ≤ kxn − yn k ≤ 2dn . Write
wn = xn − yn ,
we consequently have that the sequence {wn } is bounded while the sequence {Φwn }
conveges to y.
- Since K is compact, by passing to a subsequence if necessary, we may assume
that the sequence {Kwn } conveges to an element w0 ∈ V .
- Thus, since wn = Φwn + Kwn , the sequence {wn } converges to y + w0 .
By the continuity of Φ, we have Φ(y + w0 ) = y. To proceeed further, we need the following simple result.
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Lemma 10 (Riesz). Let V be a normed linear space and M a proper closed
subspace of V . Then, for any θ < 1, there exists an element xθ ∈ V satisfying
|xθ k = 1 and dist (xθ , M ) ≥ θ.
Proof of Lemma 10. Let x ∈ V \ M . Since M is closed, we have
dist(x, M ) = inf kx − yk = d > 0.
y∈M
Consequently there exists an element yθ ∈ M such that
kx − yθ k ≤
d
,
θ
so that, defining
xθ =
x − yθ
,
kx − yθ k
we have kxθ k = 1 and for any y ∈ M ,
x − yθ − kyθ − xky
kyθ − xk
d
≥
≥ θ. kyθ − xk
kxθ − yk =
From Proposition 9 and Lemma 10, we obtain the following.
Proposition 11. Let K be a compact linear mapping of a normed linear space V
into itself.
(i) Let Rj = Φj (V ), j ∈ N. Then ∃k ∈ N such that Rj = Rk for all j ≥ k.
(ii) Let Nj = Φ−j (0), j ∈ N. Then ∃` ∈ N such that Rj = R` for all j ≥ `.
Proof. (i) By Proposition 9, the sets Rj form a non-increasing sequence of closed
subspaces of V . Suppose no two of these spaces coincide. Then
Rj $ Rj−1 , j ≥ 1.
Hence, by Lemma 10, ∃a sequence {yj } ⊂ V such that yj ∈ Rj , kyj k = 1 and
dist (yj , Rj+1 ) ≥ 12 . Thus, if j > k,
Kyk − Kyj =yk + (−yj − Φyk − Φyj )
=yk − y for some y ∈ Rk+1 .
Hence kKyk − Ky` k ≥ 12 , contrary to the compactness of K.
(ii) Let Nj = Φ−j (0), j ∈ N, which is closed by the continuity of Φ.
- Then the sets Nj form a non-decreasing sequence of closed subspaces of V .
- By applying an anologous argument based on Lemma 10 to that used in (i), we
obtain N` = Nk for all ` ≥ some integer `. 9
Proof of Theorem 6. It is convenient to split the proof into four stages.
Step 1. Claim: If N (Φ) = ∅, then R(Φ) = V .
By Proposition 11 (i), for some k ∈ N and for all y ∈ V , Φk y ∈ Rk = Rk+1 and so
Φk y = Φk+1 x
for some x ∈ V .
Therefore
Φk (y − Φx) = 0,
and so y = Φx, since Φ−k (0) = Φ−1 (0) = 0. Consequently, R(Φ) = Rk = V .
Step 2. Claim: If R(Φ) = V , then R(Φ) = ∅.
Let ` ∈ N be as indicated in Proposition 11 (ii). If R(Φ) = V , then
∀y ∈ N` , ∃x ∈ V such that y = Φ` x.
Consequently, Φ2` x = 0, so that x ∈ N2` = N` , whence y = Φ` x = 0.
Step 3. In case (ii), the boundedness of the operator Φ−1 = (I − K)−1 follows
from Proposition 7. V.. The Adjoint of a Bounded Operator
The concept of adjoint operator extends the notion of transpose of an m × n
matrix A and plays a crucial role in determining compatibility conditions for the
solvability of several problems.
The transpose A⊥ is characterized by the identity
hAx, yiRm = hx, A⊥ yiRn .
We extend precisely this relation to define the adjoint of a bounded linear operator.
• Let L ∈ L(H1 , H2 ). If y ∈ H2 is fixed, the real-valued map
Ty : x 7→ hLx, yiH2
defines an element of H1∗ . In fact
|Ty x| = |hLx, yi| ≤ kLxkH2 kykH2 ≤ kLkL(H1 ,H2 ) kykH2 kxkH1
so that kTy k ≤ kLkL(H1 ,H2 ) kykH2 .
– From Riesz’s theorem, there exists a unique w ∈ H1 depending on y, which we
denote by w = L∗ y, such that
Ty x = hx, L∗ yiH1 ,
∀x ∈ H1 , ∀y ∈ H2 .
This defines L∗ as an operator from H2 into H1 , which is called the adjoint of
L. Precisely:
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Definition. The operator L∗ : H2 → H1 defined by the identity
(5)
hLx, yiH2 = hx, L∗ yiH1 ,
∀x ∈ H1 , ∀y ∈ H2
is called the adjoint of L.
• The following properties are immediate consequences of the definition of adjoint.
Proposition 12. Let L, L1 ∈ L(H1 , H2 ) and L2 ∈ L(H2 , H3 ). Then
(a) L∗ ∈ L(H2 , H1 ). Moreover, L∗∗ = L and
kL∗ kL(H2 ,H1 ) = kLkL(H1 ,H2 ) .
(b) (L2 L1 )∗ = L∗1 L∗2 . In particulat, if L is an isomorphism, then
(L−1 )∗ = (L∗ )−1 .
• The following proposition is useful.
Proposition 13. Let L : H1 → H2 be compact. Then L∗ : H2 → H1 is compact.
Proof. We use Proposition 5. Let {xk } ⊂ H2 and xk converges to 0 weakly.
Claim: kL∗ xk |k → 0 strongly in H2 . Indeed, since L ∈ L(H2 , H1 ), we have
L∗ xk * 0 weakly in H1 .
Thus, the compactness of L and Proposition 5 entails LL∗ xk → 0 strongly in H2 .
Then, since the weak convergence of xk yields kxk k ≤ M for some constant M ,
kL∗ xk k2H2 =hL∗ xk , L∗ xk iH1 = hxk , LL∗ xk iH2
≤kxk kkLL∗xk kH2 ≤ M kLL∗xk kH2 → 0. • The next theorem extends relations well known in the finite-dimensional case.
Theorem 14. Let L ∈ L(H2 , H1 ). Then
(a) R(L) = N (L∗ )⊥ .
(b) N (L) = R(L∗ )⊥ .
Proof. (a) (i) Let z ∈ R(L). Then, ∃x ∈ H1 such that z = Lx and, if y ∈ N (L∗ ),
we have
hz, yiH2 = hLx, yiH2 = hx, L∗ yiH1 = 0.
Thus, R(L) ⊆ N (L∗ )⊥ . Since N (L∗ )⊥ is closed, it follows that
R(L) ⊆ N (L∗ )⊥ .
(ii) On the other hand, if z ∈ R(L)⊥ , for every x ∈ H1 , we have
0 = hLx, ziH2 = hx, L∗ ziH1
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whence L∗ z = 0. Therefore
R(L)⊥ ⊆ N (L∗ ),
which is equivalent to
N (L∗ )⊥ ⊆ R(L).
(b) Substituting L∗ for L in (a), we deduce
R(L∗ ) = N (L)⊥
which is equivalent to N (L) = R(L∗ )⊥ . VI. Fredholm’s Alternative in Hilbert Spaces.
We introduce some terminology.
Definition. Let V1 , V2 be Hilbert spaces and Φ : V1 → V2 . Then Φ is said to be
a Fredholm operator if N (Φ) and R(Φ)⊥ have finite dimensions.
The index of Φ is the integer
ind Φ = dim N (Φ) − dim R(Φ)⊥ = dim N (Φ) − dim N (Φ∗ ).
We have:
Theorem 15 (Fredholm’s Alternative). Let V be a Hilbert space and K ∈
L(V ) be a compact operator. Then Φ : I − K is a Fredholm operator with zero
index. Moreover, Φ∗ = I − K ∗ ,
(5)
R(Φ) = N (Φ∗ )⊥
and
(6)
N (Φ) = {0} ⇐⇒ R(Φ) = V.
The last formula shows that Φ is one-to-one iff it is onto.
In other words, uniqueness for the equation
(7)
x − Kx = f
is equivalent to existence for every f ∈ V .
• The same thing hlds for the adjoint Φ∗ = I − K ∗ and the associated equation
y − K ∗ y = g.
• Let d = dim R(Φ)⊥ = dim N (Φ∗ ) > 0.
Then (5) says that the equation (7) is solvable iff f ⊥ N (Φ∗ ), that is, iff hf, yi =
0, for every y of
y − K ∗ y = 0.
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If y1 , · · · , yn span N (Φ∗ ), this amounts to asking that the d compatibility
relations
hf, yj i = 0,
j = 1, · · · , n
are necessary and sufficient conditions for the solvability of (7).
To prove Theorem 15, it remains to show that (i) dim N (Φ) < ∞ and (ii)
dim N (Φ) = dim N (φ∗ ).
(i) If (i) were not correct, there would exist an orthonormal system {ϕi } satisfying
0 = Φϕi = ϕi − Kϕi ,
i ∈ N.
Since the operator K is compact, we can select a strongly convergent√subsequence
{ϕij } of {ϕi } in V . This contradics the statement kϕi − ϕj k = 2, ∀i, j ∈ N
with i 6= j.
(ii) To prove (ii), we assume w.l.o.g. d = dim N (Φ) ≤ dim N (Φ∗ ) = d∗ < ∞, for
otherwise we could replace Φ by Φ∗ and Φ∗ by Φ∗∗ .
Consider the orthonormal basis {ϕ1 , · · · , ϕd } of N (Φ) and the orthonormal basis
{ψ1 , · · · , ψd∗ } of N (Φ∗ ). We consider the Fredholm operator
Sx = Φx −
d
X
hϕi , xiψi ,
x ∈ V.
i=1
On account of Theorem 14 (a), the null space of the operator S satisfies
N (S) = {x ∈ V : Sx = 0} = {0}.
Theorem 6 then implies that the mapping S : H → H is surjective. Consequently
d = d∗ , i.e. dim N (Φ) = N (φ∗ ).
II.3. Solvability for Abstract Variational Problem
• Let us go back to the variational problem
a(u, v) = hF, vi∗ ,
∀v ∈ V,
and suppose that Lax-Milgram cannot be applied, since, for instance, a in not
V -coercive.
• The problem involves two Hilbert spaces:
V , the space where we seek the solution, and
V ∗ , which the data F belongs to.
Let us introduce a third space H, intermediate between V and V ∗ .
• In boundary value problem, usually H = L2 (Ω), with Ω bounded domain in Rn ,
while V is a Sobolev space.
• In practice, we often meet a pair of Hilbert spaces V , H with the following
properties:
(i) V ,→ H, i.e. V is continuously embedded in H. Recall that this simply
means that the identity operator IV →H , from V to H, is continuous or, equivalently that there exists C such that
(8)
kukH ≤ CkukV ,
∀u ∈ V.
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(ii) V is dense in H.
Using Riesz’s theorem, we may identify H with H ∗ .
Also, we may continuously embed H into V ∗ , so that any element in H can
be regarded as an element of V ∗ . Indeed, for any fixed u ∈ H, the functional Tu
defined by
(9)
hTu , vi∗ = hu, viH ,
∀v ∈ V
is continuous in V , since the Schwartz inequality and (8) give
|hu, viH | ≤ kukH kvkH ≤ CkukH kvkV .
Then we have a continuous map u 7→ Tu from H into V ∗ with kT ukV ∗ ≤ CkukH .
- Moreover, the map u 7→ Tu is one-to-one; indeed, if Tu = 0, then
hu, viH = 0,
∀v ∈ V,
which forces u = 0, using the fact that V is dense in H.
- Thus the map u 7→ Tu is a continuous embedding.
This allows the identification of u with an element of V ∗ , which means that,
instead of (9), we can write
hu, vi∗ = hu, viH ,
∀v ∈ V,
regarding u on the left as an element of V ∗ and on the right as an element of H.
Finally, it can be shown that V and H are dense in V ∗ .
Thus, we have
V ,→ H ,→ V ∗
with dense embeddigs. We call (V, H, V ∗ ) a Hilbert triplet.
• To state the main result we introduce weakly coercive forms and their adjoints.
Definition. The bilinear form a(u, v) is said to be weakly coercive with respect
to the pair (V, H) if there exist λ0 ∈ R and α > 0 such that
a(u, v) + λ0 kvk2 ≥ αkuk2V ,
∀v ∈ V.
Definition. The adjoint forms a∗ of a is given by
a∗ (u, v) = a(v, u).
Set
N (a) = {u : a(u, v) = 0 ∀v ∈ V },
N (a∗ ) = {w : a∗ (w, v) = 0 ∀v ∈ V }.
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Theorem 16. Let (V, H, V ∗ ) be a Hilbert triplet, with V compactly embedded in
H. Let F ∈ V ∗ and a be a bilinear fom in V , continuous and weakly coercive with
respect to (V, H). Then
(a) Either the equation
(10)
a(u, v) = hF, vi∗ ,
∀v ∈ V
has a unique solution u and kuk ≤ CkF kV ∗ ,
(b) or
dim N (a) = dim N (a∗ ) = d < ∞
and (10) is solvable iff hF, vi∗ = 0 for every w ∈ N (a∗ ).
Some comments are in order. The following dichotomy holds either (10) has a
unique solution for every F ∈ V ∗ , or the homogeneous equation a(u, v) = 0 has
nontrivial solutins.
The same conclusion holds for the adjoint equation
a∗ (u, v) = hF, vi∗ ,
∀v ∈ V.
If w1 , · · · , wd span N (a∗ ), (10) is solved iff the d compatibility conditions
hF, wj i∗ ,
j = 1, · · · , d,
holds. In this case, equation (10) has infinitely many solutions given by
u=u+
d
X
cj zj ,
j=1
where u is a particular solution of (10), z1 , · · · , zd span N (a) and c1 , · · · , cd are
arbitrary constants.
Proof of Theorem 16. The strategy is to write equation
(11)
a(u, x) = hF, vi∗
in the form
(IV − K)u = g,
where IV is the identity operator in V and K : V → V is compact.
Let J : V → V ∗ , the embedding of V into V ∗ .
– Recall that J is the composition of the embedding IV →H and IH→V ∗ .
– Since IV →H is compact and IH→V ∗ is continuous, we infer from Proposition 6
that J is compact.
– We write (11) in the form
aλ0 (u, v) = a(u, v) + λ0 hu, viH = hλ0 Ju + F, vi∗ ,
where λ0 > 0 is such that aλ0 is coercive.
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Since, for each fixed u ∈ V , the linear map
v 7→ aλ0 (u, v)
is continuous in V , there exists L ∈ L(V, V ∗ ) such that
hLu, vi∗ = aλ0 (u, v),
∀u, v ∈ V.
Thus, the equation (11) is equivalent to
hLu, vi∗ = hλ0 Ju + F, vi∗ ,
∀v ∈ V,
and therefore to
(12)
Lu = λ0 Ju + F.
Since aλ0 is V -coercive, from the Lax-MIlgram theorem, the operator L is an
isomorphism between V and V ∗ and (12) can be written in the form
u − λ0 L−1 Ju = L−1 F.
Letting g = L−1 F ∈ V and K = λ0 L−1 J, (12) becomes
(IV − K)u = g,
where K : V → V .
Since J is compact and L−1 is continuous, K is compact.
Applying the Fredholm Alternative Theorem and rephrasing the conclusion in
terms of bilinear forms we conclude the proof.