NOTES OF COURSE M2
”ANALYSE FONCTIONNELLE”
Jean-Yves CHEMIN
Laboratoire J.-L. Lions, Campus Pierre et Marie Curie
Sorbonne Université, Case 187
75 232 Paris Cedex 05, France
adresse électronique: [email protected]
September 13, 2020
2
Contents
1 Banach spaces, spaces of functions 7
1.1 Definition and basic properties and examples . . . . . . . . . . . . . . . . . . . 71.2 Spaces of functions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 91.3 Refined convolution inequalities . . . . . . . . . . . . . . . . . . . . . . . . . . . 181.4 Spaces of linear continuous maps . . . . . . . . . . . . . . . . . . . . . . . . . . 221.5 Around Cauchy-Lipschitz theorem . . . . . . . . . . . . . . . . . . . . . . . . . 25
2 Duality 29
2.1 The dual space of a normed space . . . . . . . . . . . . . . . . . . . . . . . . . 292.2 Examples of identification . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 33
3 More about Hilbert spaces 41
3.1 The orthogonality . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 413.2 Orthonormal basis . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 443.3 The adjoint of an operator, self-adjoint operators . . . . . . . . . . . . . . . . . 48
4 The Fourier transform 55
4.1 The Fourier transform on L1(Rd) . . . . . . . . . . . . . . . . . . . . . . . . . . 554.2 The inversion formula and the Fourier-Plancherel theorem . . . . . . . . . . . . 584.3 Sobolev spaces and Fourier transform . . . . . . . . . . . . . . . . . . . . . . . 604.4 The Rellich’s theorem . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 63
5 Dirichlet problem, weak derivatives, and tempered disribution 65
5.1 The Dirichlet problem . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 655.2 Basic ideas about distribution theory . . . . . . . . . . . . . . . . . . . . . . . . 695.3 The Schwartz space and tempered distributions . . . . . . . . . . . . . . . . . . 73
A Measure theory and definition of the Lp spaces 77
B Dirichlet problem, weak derivatives, and tempered disribution 81
B.1 Operations on tempered distributions . . . . . . . . . . . . . . . . . . . . . . . 83B.2 Two applications . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 95
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4
Introduction
This text consists in notes of a series of lectures given in the second year of the Master”Mathématiques of the modélisation” of Sorbonne Université.
5
6
Chapter 1
Banach spaces, spaces of functions
Introduction
1.1 Definition and basic properties and examples
Definition 1.1.1 Let E a vectorial space on K. A map N from E to R+ is a norm if andonly if the following three conditions are satisfied:
• N(x) = 0 if and only if x = 0,
• For any � in K, we have N(�x) = |�|N(x),
• N(x+ y) N(x) +N(y).
The couple (E,N) is called a normed space
Definition 1.1.2 Let N1 and N2 two norms on a vectorial space E. We say that the twonorms N1 and N2 are equivalent if and only if a constant C exists such that
8x 2 E , C�1N1(x) N2(x) CN1(x).
Notations We usally denote a norm by k · kE or by k · k.
Proposition 1.1.1 Let (E, k · k) a normed space, then the map defined by⇢
E ⇥ E �! R+(x, y) 7�! kx� yk
is a distance we E, called distance associated with the norm k · k.
The (easy) proof is an exercise left to the reader.
Agreement We always implicitely consider this metric structure on the space E.
Definition 1.1.3 Let (E, k · k) a normed space. We say that (E, k · k) is a Banach spaceif and only if the metric space (E, d) where d is the distance associated with the norm k · k(i.e. d(x, y) = kx� yk) is complete.
Let us exhibit examples of Banach spaces.
7
Proposition 1.1.2 For p in [1,1[, we consider the map defined by
k · kp
8><
>:
KN �! R+
(xj)1jN 7�!⇣ NX
j=1
|xj |p⌘ 1
p.
It is a norm on KN and (KN , k · kp) is a Banach space. Moreover, we have for any (p, q)of [1,1[2 such that p q,
kxk1def= sup
1jN|xj | kxk`q kxkp N1�
1q kxkq Nkxk1. (1.1)
Proof. Let us first prove Hölder Inequality which claims that for any (a, b) in KN ⇥KN ,
���NX
j=1
ajbj��� kakpkbkp0 with
1
p+
1
p0= 1. (1.2)
By homogeneity, we can assume that kakp = kbkp0 = 1. In order to prove it, we can invoquethe concavity of logarithm or simply Noticing that the function
x 7�! ax� ✓a1✓ � (1� ✓)x
11�✓
is negative because f 0(x) = a� x✓
1�✓ and f(a1�✓✓ ) = 0, we get
ab ✓a1✓ + (1� ✓)b
11�✓ . (1.3)
Applying this inequality gives
���NX
j=1
ajbj���
NX
j=1
|aj | |bj |
1p
NX
j=1
|aj |p +⇣1� 1
p
⌘|bj |p
0
1.
Inequality (1.2) implies that
NX
j=1
|xj + yj |p NX
j=1
|xj | |xj + yj |p�1 +NX
j=1
+|yj | |xj + yj |p�1
✓⇣ NX
j=1
|xj |p⌘ 1
p+⇣ NX
j=1
|yj |p⌘ 1
p
◆✓ NX
j=1
|xj + yj |(p�1)p
p�1
◆1� 1p
✓⇣ NX
j=1
|xj |p⌘ 1
p+⇣ NX
j=1
|yj |p⌘ 1
p
◆✓ NX
j=1
|xj + yj |p◆1� 1p
.
This gives the result for p in ]1,1[, the two cases p = 1 or p = 1 being obvious.Now let us prove (1.1). For any (p, q) in [1,1]2 such that p is less than q we have
xjkxk`q
1.
8
This implies that ���xj
kxk`q
���q
���
xjkxk`q
���p
·
By sommation, we get that
1 ⇣kxk`pkxk`q
⌘p
·
This gives the result. 2
Let us admit the following theorem which claims that when the vectorial space E is finitedimensionnal, all the norm are equivalent. This is due to the compactness of the unit ballin for finite dimensionnal space. Let us point out that this property is characteristic of finitedimensionnal normed space. We admit the following theorem.
Theorem 1.1.1 Let E be a normed space. If the dimension of E is finite then the closedunit ball is compact and all the norms on E are equivalentes. Conversely if the closed unitball is compact, then the dimension of normed space E is finite.
Exercise 1.1.1 For p in [1,1[, let us consider the space `p(N) of sequences x with value in Ksuch that X
n2N|x(n)|p < 1.
Let us define
kxk`pdef=⇣X
n2N|x(n)|p
⌘ 1p.
1) Prove that the space (`p(N), k · k`p) is a normed space.2) Moreover, if p q, then `p(N) is included in `q(N) and that
8x 2 `p(N) , kxk`p kxk`q .
3) Prove that (`p(N), k · k`p) is a Banach space.
Exercise 1.1.2 Prove that the set of sequences which have only a finite number of terms
which are not equal to 0 is dense in `p(N) for p in the intervalle [1,1[.
The following exercise demands an understanding of di↵erential calculus in infinite dimen-sionnal normed spaces.
Exercise 1.1.3 Prove that the function x 7�! kxkp`p
is di↵erentiable at any point x of `p(N)when p belongs to ]1,1[ and that it is twice di↵erentiable when p belongs to [2,1[.
1.2 Spaces of functions
Proposition 1.2.1 Let X a set and (E, k · k) a Banach space, we consider B(X,E) the setof functions f bounded from X to E, i.e. such that
kfkB(X,E)def= sup
x2X
kf(x)k < 1.
Then (B(X,E), k · kB(X,E)) is a Banach space. Moreover if X is equipped with a distance dwe define Cb(X,E) as the set of functions f of B(X,E) which are continuous. Then thespace (Cb(X,E), k · kB(X,E)) is a Banach space.
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Proof. Let (fn)n2N a Cauchy sequence of B(X,E). By definition, we have
8" , 9n0 / 8n � n0 , 8p , 8x 2 X , kfn(x)� fn+p(x)k < ". (1.4)
In particular, for any x, the sequence (fn(x))n2N is a Cauchy sequence of E. As this space iscomplete, this sequence is convergent. thus for any x of X, an element of E, denoted f(x)exists such that
limn!1
fn(x) = f(x).
Let us check that f belongs to B(X,E). Thanks to Inequality (1.4) applied with " = 1, wehave
8p , 8x 2 X , kfn0(x)� fn0+p(x)k < 1.
Passing to the limit when p tends to the infinity, we get
8x 2 X , kfn0(x)� f(x)k 1.
The function fn0 being bounded, so is the function f . Indeed we have
kf(x)� f(x0)k kf(x)� fn0(x)k+ kfn0(x)� fn0(x0)k+ kfn0(x0)� f(x0)k 2 + kfn0(x)� fn0(x0)k.
Now let us check that the sequence (fn)n2N converges to f in the space B(X,E). In order todo it, let us pass to the limit when p tends to infinity in the Inequality (1.4). This gives
8" > 0 , 9n0 2 N / 8n � n0 , 8x 2 X , kfn(x)� f(x)k ",
which proves that B(X,E) is complete.In order to prove that Cb(X,E) is complete, it is enough to prove that Cb(X,E) is a closed
subset of B(X,E) which means that a uniform limit of continuous functions is a continuousfunction. Let us consider a sequence (fn)n2N of elements of C(X,E) which converges to fin B(X,E), then f is continuous of X in E. By definition, for any " > 0, an integer n" existssuch that
8x 2 X , kfn0(x)� f(x)k <"
4· (1.5)
The repeated use of the triangular inequality and Inequality (1.5) imply that for any cou-ple (x0, x) of X ⇥X,
kf(x)� f(x0)k kf(x)� fn0(x)k+ kfn0(x)� fn0(x0)k+ kfn0(x0)� f(x0)k "
2+ kfn0(x)� fn0(x0)k.
The function fn0 being continuous, for any x0 of X, it exists ↵ > 0 such that
8x 2 X, d(x, x0) < ↵ =) kfn0(x)� fn0(x0)k <"
2·
We deduce that8x 2 X, d(x, x0) < ↵ =) kf(x)� f(x0)k < ",
which concludes the proof of the theorem. 2
10
Definition 1.2.1 Let p be in the interval [1,+1[. We denote Lp(X, dµ) the space of (equiv-alence classes modulo equality almost everywhere of) measurable functions f such that
Z
X
|f(x)|pdµ < 1.
If p = 1, we denote Lp(X, dµ) the space of (equivalence classes modulo equality almosteverywhere of) measurable functions such that
kfkL1def= sup
�� / µ{x / |f(x)| > �} > 0
= inf
�M / µ{x / |f(x)| > M}
= 0.
Let us prove that the two quantities in the definition of kfkL1 are equal. First, we note thatfor any pair (�,M) such that µ{x / |f(x)| > M} = 0 and µ{x / |f(x)| > �} > 0, we havethat � is strictly less than M . Hence,
sup�� / µ{x / |f(x)| > �} > 0
inf
�M / µ{x / |f(x)| > M} = 0
.
Now, let M1 be a real number strictly greater than sup�� / µ{x / |f(x)| > �} > 0
. By
definition of the upper bound, we have µ{x / |f(x)| > M1} = 0. Thus,
inf�M / µ{x / |f(x)| > M} sup
�� / µ{x / |f(x)| > �} > 0
.
The study of Lp spaces requires the introduction of the Hölder conjugate.
If p 2]1,1[ , p0 def= pp� 1 , if p = 1, p
0 déf= +1, and if p = +1 , p0 déf= 1.
We say that p and p0 are conjugate exponents, and with the convention1
1 = 0, we have
1
p+
1
p0= 1.
An important result of this section is the following.
Theorem 1.2.1 For any p in [1,1], (Lp(X, dµ), k · kLp) is a Banach space.
Proof. We start with the case Now, we study the case where p is finite. The first step is toprove that (Lp(X, dµ), k · kLp) is a normed vector space, where
kfkLpdef=⇣Z
X
|f(x)|pdµ(x)⌘ 1
p.
This is clear for p = 1. When p belongs to ]1,1[, we first note that, since
|f(x) + g(x)|p 2p(|f(x)|p + |g(x)|p),
the set Lp(X, dµ) is a vectorial space. Let us show that it is a normed space. This relies onHölder’s inequality, which we state and prove.
Proposition 1.2.2 (Hölder’s inequality) Let (X,µ) be a measured space, f be a functionin Lp(X, dµ) and g be a function in Lp
0
(X, dµ). Then, the function fg is in L1(X, dµ), andwe have Z
X
|f(x)g(x)|dµ(x) kfkLpkgkLp0 .
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Proof. First, note that if p = 1 or p = 1, there is nothing to prove. Moreover, we canassume that kfkLp = kgkLp0 = 1, and that the functions f and g are non-negative. Using theconvexity inequality (1.3) page 8, we have
f(x)g(x) = (f(x)p)1p (g(x)p
0
)1p0 1
pf(x)p +
1
p0g(x)p
0
,
which proves Hölder’s inequality. Indeed, we integrating the above, and we getZ
X
|fg|dµ 1p
Z
X
|f |pdµ+ 1p0
Z
X
|g|p0dµ
1p+
1
p0= 1.
This ends the proof of the proposition. 2
Back to the proof of theorem 1.2.1. As f + g belongs to Lp, Hölder’s inequality implies thatZ
X
|f + g|pdµ Z
X
|f ||f + g|p�1dµ+Z
X
|g||f + g|p�1dµ
⇣Z
X
|f |pdµ⌘ 1
p⇣Z
X
|f + g|(p�1)p
p�1dµ⌘ 1
p0
+⇣Z
X
|g|pdµ⌘ 1
p⇣Z
X
|f + g|(p�1)p
p�1dµ⌘ 1
p0
✓⇣Z
X
|f |pdµ⌘ 1
p+⇣Z
X
|g|pdµ⌘ 1
p
◆⇣Z
X
|f + g|(p�1)p
p�1dµ⌘1� 1p
.
This implies that kf +gkLp kfkLp +kgkLp , and therefore that Lp(X, dµ) is a normed space.Now we will show that these spaces are complete. The proof relies on the following lemma.
Lemma 1.2.1 If E is a normed space such that, for any sequence (xn)n2N of elements of E,we have
X
n2NkxnkE < 1 =) SN
def=
NX
n=0
xn is convergent,
then E is a Banach space.
Proof. Let us recall that a Cauchy sequence that has a cluster point is convergent. Let (an)n2Nbe a Cauchy sequence in E: we will extract a convergent subsequence, and this will yield theresult. Define a map � from N to N by �(0) = 0 and
�(n+ 1)def= min
(m � �(n) + 1 , sup
p�0kam � am+pk
1
(2 + n)2
)·
Thus, for any integer n, we have
ka�(n+1) � a�(n)k 1
(1 + n)2·
Set xndef= a�(n+1) � a�(n). We have
X
n2Nkxnk
X
n2N
1
(1 + n)2< 1.
So, by hypothesis, the sequenceNX
n=0
xn = a�(N) � a0 is convergent; the lemma is proved. 2
12
Continuation of the proof of Theorem 1.2.1. Let (fn)n2N be a sequence of elements of Lp
such that X
n2NkfnkLp < 1.
Define the functions
SN (x) =NX
n=0
fn(x) and S+N(x) =
NX
n=0
|fn(x)|.
For any integer N , we have
kS+NkLp
X
n2NkfnkLp ,
which means that
8n 2 N ,Z
X
S+N(x)pdµ(x)
⇣X
n2NkfnkLp
⌘p
.
So the sequence (S+N(x))N2N is non-decreasing. The monotone convergence theorem A.0.1
yields that
Z
X
S+(x)pdµ(x) ⇣X
n2NkfnkLp
⌘p
with S+(x) =X
n2N|fn(x)|.
The function S+ takes finite values almost everywhere, and belongs to Lp. Thus, for almostany x, the series with general term fn(x) converges in C. We denote S its sum; S is an elementof Lp, and we have
���S �NX
n=0
fn���p
Lp=
Z
X
���S(x)�NX
n=0
fn(x)���p
dµ(x).
We know that���S(x)�
NX
n=0
fn(x)��� 2S+(x).
As S+ is in Lp, the dominated convergence theorem yields that
limN!1
���S �NX
n=0
fn���p
Lp= 0.
The theorem is proved in the case when p is finite.
Proof of Theorem 1.2.1 in the case of L1 First of all, we note that if kfkL1 = 0, there existsa sequence (�n)n2N which converges to 0 such that, for any n, the set {x / |f(x)| > �n} isnegligible. As the union of countably many negligible sets is also negligible, the set of points xsuch that f(x) is non-zero is negligible. Thus f is equal to 0 as an element of L1. The proofof the homogeneity of the L1 norm is left as an exercise, and uses the fact that
{x / |�f(x)| > M} =nx / |f(x)| > M|�|
o.
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Now, let us consider f and g two (equivalence classes of) functions in L1, and set two positivenumbersMf andMg such that the measures of the sets {x / |f(x)| > Mf} and { / |g(x)| > Mg}are equal to zero. Note that
�x / |f(x) + g(x)| > Mf +Mg
⇢ {x / |f(x)| > Mf} [ {x / |g(x)| > Mg},
so the space L1 is stable under addition. Moreover, as the infimum of a set is a lower boundof the set, we have kf + gkL1 Mf +Mg. As the infimum is the greatest lower bound, wehave kf + gkL1 kfkL1 + kgkL1 , thus we conclude that (L1(X, dµ), k · kL1) is a normedvector space.
In order to prove that (L1, k · kL1) is a Banach space, we are going to apply proposi-tion 1.2.1 page 9. Let us consider a Cauchy sequence (fn)n2N in (L1, k · kL1), and define theset Em,p,n as follows:
Em,p,ndef=nx 2 X / |fm(x)� fm+p(x)| >
1
n+ 1
o.
Here, the equivalence classes of the functions fn and representatives of these classes are de-noted the same way. As the sequence (fn)n2N is a Cauchy sequence, we can define a function �from N to N by
�(n) = minnm � �(n� 1) + 1 / 8m0 � m, 8p � 0 , kfm0 � fm0+pkL1 <
1
n+ 1
o·
By definition of the essential supremum, for any m � �(n), the measure of the set Em,p,n iszero. Thus, the set E defined as
Edef=
[
(n,p)2N2m��(n)
Em,p,n
has zero measure. We can therefore choose representatives of the (classes of) functions fn thatare equal to zero on the negligible set E (these representatives are still denoted fn). Thus,
8n , 8m � �(n) , 8p , 8x 2 X , |fm(x)� fm+p(x)| <1
n+ 1
and the sequence (fn)n2N is a Cauchy sequence in B(X,K). We can now use proposition 1.2.1page 9 to conclude that L1(X, dµ) is complete. 2
We are now going to make a few remarks on the Hölder inequality. We begin with thefollowing corollary.
Corollary 1.2.1 Let p and q be two real numbers strictly greater than 1, such that
1
p+
1
q 1.
Then the mapping ⇢Lp ⇥ Lq �! Lr(f, g) 7�! fg
is bilinear and continuous if1
r=
1
p+
1
q·
14
Proof. It su�ces to apply Hölder’s inequality with s = r/p. This yieldsZ
X
|f(x)g(x)|rdx kfkrLpkgkrLq ,
which means thatkfgkLr kfkLpkgkLq ,
and the corollary is proved. 2
Corollary 1.2.2 If X has finite measure, then Lp(X, dµ) ⇢ Lq(X, dµ) if p � q.
Proof. As the function 1 is clearly in the space L1(X, dµ)\L1(X, dµ), it belongs to Lp(X, dµ)for any p. We even have
8p 2 [1,1] , k1kLp = µ(X)1p .
Corollary 1.2.1 implies that
kfkLq µ(X)1s kfkLp with
1
s=
1
q� 1
p,
so the corollary is proved. 2
Now let us establish the classical convolution inequalities, known as Young’s inequalities.
Lemma 1.2.2 For all (p, q, r) in [1,1]3 such that1
p+
1
q= 1 +
1
r(1.6)
and for any couple (f, g) in Lp(Rd, dx)⇥ Lq(Rd, dx), we have
f ? g 2 Lr(Rd, dx) and kf ? gkLr(Rd,dx) kfkLp(Rd,dx)kgkLq(Rd,dx).
Proof. The case r = 1 reduces to the Hölder’s inequality which has been proved before.Letus now consider the case r < 1. Obviously, one can assume with no loss of generality that fand g are nonnegative and nonzero and that kfk
Lp(Rd,dx) = kgkLq(Rd,dx) = 1. Let us write that
(f ? g)(x) =
Z
Rdf
rr+1 (x� y) g
1r+1 (y)f
1r+1 (x� y) g
rr+1 (y) dy.
Once observed that (1.8) can be writtenr
r + 1
⇣1p+
1
q
⌘= 1, Hölder’s inequality implies that
(f ? g)(x) ✓Z
Rdfp(x� y)g
pr (y) dy
◆ r(r+1)p
✓Z
Rdf
qr (x� y)gq(y) dy
◆ r(r+1)q
.
Applying Hölder’s inequality with ↵ = rq/p (resp. � = rp/q) and the measure fp(x � y) dy)(resp. gq(y) dy) and using the invariance by translation of the Lebesgue, we get
(f ? g)(x) ✓Z
Rdfp(x� y)gq(y) dy
◆ 1r+1
⇣1p+
1q
⌘
Hence, raising the above inequality to the power r yields��(fp ? gq)(x)
��r (fp ? gq)(x).
Since obviously the left invariance of the measure µ combined with Fubini’s theorem impliesthat the convolution maps L1(Rd, dx) ⇥ L1(Rd, dx) into L1(Rd, dx) with norm 1, this yieldsthe desired result in the case r < 1. 2
15
We stress again that Hölder’s inequality is fundamental. It is also optimal in the followingsense.
Lemma 1.2.3 Let (X,µ) be a measured space. Let f be a measurable function and p be anelement of [1,1]. Assume that
supkgk
Lp01
g�0
Z
X
|f(x)|g(x)dµ(x) < +1. (1.7)
Then f belongs to Lp, and
kfkLp = supkgk
Lp01
����Z
X
f(x)g(x)dµ(x)
����.
Proof. We begin with the case p = 1. By choosing the constant function equal to 1 for g, wehave g bounded with an L1 norm equal to 1, and, using inequality (1.7), we get that
Z
X
|f(x)|g(x)dµ(x) < 1.
Now, consider g the function defined by
g(x) =f(x)
|f(x)| if f(x) 6= 0 and 0 otherwise.
The function g is bounded and its norm is equal to 1. Moreover,Z
X
f(x)g(x)dµ(x) =
Z
X
|f(x)|dµ(x).
The lemma is proved in this case.Now, let p be a real number strictly greater than 1, and consider an increasing sequence
of sets with finite measure (Kn)n2N, such that the union of these sets is equal to X. Set
f+n (x) = 1Kn\(|f |n)|f | and gn(x) =f+n (x)
p�1
kf+n kpp0
Lp
·
It is clear that the function f+n is non-negative, and belongs to L1 \ L1, so it belongs to Lq
for any q, and we have
kgnkp0
Lp0 =
1
kf+n kpLp
Z
X
f+n (x)(p�1) pp�1dµ(x) = 1.
By definition of functions fn and gn, we haveZ
X
|f(x)|1Kn\(|f |n)gn(x)dµ(x) =Z
X
f+n (x)gn(x)dµ(x)
=
✓Z
X
f+n (x)pdµ(x)
◆kfnk
�pp0
Lp
= kf+n kp�
pp0
Lp
= kf+n kLp .
16
Therefore, Z
X
f+n (x)pdµ(x)
✓sup
kgkLp
01g�0
Z
X
|f(x)|g(x)dµ(x)◆
p
.
The monotone convergence theorem applied to the non-decreasing sequence�(f+n )
p�n2N im-
mediately yields that
kfkLp supkgk
Lp01
g�0
Z
X
f(x)g(x)dµ(x).
Thus, if hypothesis (1.7) holds, the function f belongs to Lp. Now, we assume that f belongsto Lp. Then, if we set
g(x) =f(x)|f(x)|p�1
|f(x)|⇥ kfkpp0
Lp
,
we have
kgkp0
Lp0 =
1
kfkpLp
Z
X
|f(x)|(p�1)p
p�1dµ(x) = 1 and kfkLp =Z
X
f(x)g(x)dµ(x).
So the lemma is proved for every p in [1,1[. In the case where p = +1, consider a positivenumber � such that µ(|f | � �) is positive. Set E�
déf= (|f | � �), and let g0 be a non-negative
function in L1, with support in E�, and with integral equal to 1. Then, if
g(x) =f(x)
|f(x)|g0 ,
we get Z
X
|f(x)g(x)|dµ(x) =Z
X
|f(x)|g0(x)dµ(x) � �Z
X
g0dµ(x) � � .
The lemma is proved. 2A first application of this is the following proof of Minkowski inequality.
Proposition 1.2.3 Let (X1, µ1) and (X2, µ2) be two measure spaces and f, a nonnegativemeasurable function over X1 ⇥X2. For all 1 p q 1, we have
���kf(·, x2)kLp(X1,µ1)���Lq(X2,µ2)
���kf(x1, ·)kLq(X2,µ2)
���Lp(X1,µ1)
.
Proof. The result is obvious if q = 1. If q is finite, let us write that, using Fubini’s theorem,we have, with r
def= (q/p)0,
���kf(·, x2) kLp(X1,µ1)���Lq(X2,µ2)
=
Z
X2
✓Z
X1
fp(x1, x2) dµ1(x1)
◆ qp
dµ2(x2)
! 1q
=
sup
kgkLr(X2,µ2)=1
g�0
Z
X1⇥X2
fp(x1, x2)g(x2) dµ1(x1) dµ2(x2)
! 1p
Z
X1
✓sup
kgkLr(X2,µ2)=1
g�0
Z
X2
fp(x1, x2)g(x2) dµ2(x2)
◆dµ1(x1)
! 1p
.
17
Then, using Hölder’s inequality, we infer
���kf(·, x2)kLp(X1,µ1)���Lq(X2,µ2)
✓Z
X1
⇣Z
X2
f q(x1, x2) dµ2(x2)⌘ p
qdµ1(x1)
◆ 1p
,
whence the desired inequality. 2
1.3 Refined convolution inequalities
The purpose of this section is to improve the classical Young inequalities which clains that, if
1
p+
1
q= 1 +
1
r(1.8)
then kf ? gkLr kfkLpkgkLq . In order to state these refined inequalities, let us introduce theso caled weak Lp spaces.
Definition 1.3.1 For p in [1,1[, we denote by Lpw(X,µ)) space the set of measure functionssuch that
kfkpLpw(X,µ)
def= sup
�>0�pµ(|f | > �) < 1,
Let us notice that we have
µ(|f | > �) Z
(|f |>�)
⇣ |f(x)|�
⌘p
dµ(x) 1�p
kfkpLp.
Thus Lp is a subset of Lpw. Typical examples of weak Lp functions. The function | · |�↵ belongsfor L
d↵w (Rd, dx).
Theorem 1.3.1 Let (p, q, r) be in ]1,1[3 and satisfy (1.8). A constant C exists such that
kf ? gkLr CkfkLpkgkLqw
Let us notice that the above theorem implies the well-known Hardy-Littlewood-Sobolevinequalities on Rd .
Theorem 1.3.2 (Hardy-Littlewood-Sobolev inequality) Let ↵ be in ]0, d[ and (p, r)in ]1,1[2 satisfy
1
p+↵
d= 1 +
1
r· (1.9)
Then a constant C exists such that
k | · |�↵ ? fkLr(Rd) CkfkLp(Rd).
The proof of Theorem 1.3.1 relies on the atomic decomposition which is decribe in thefollowing proposition.
Proposition 1.3.1 Let (X,µ) be a measure space and p be in [1,1[. Let f be a nonnegativefunction in Lp. Then a sequence of positive real numbers (ck)k2Z and a sequence of nonnegativefunctions (fk)k2Z (the atoms) exist such that
f =X
k2Zckfk
18
where the support of the functions fk are pairwise disjoint and
µ(Supp fk) 2k+1, (1.10)
kfkkL1 2�kp , (1.11)
1
2kfkp
LpX
k2Zcpk 2kfkp
Lp. (1.12)
Remarks
• Let us notice that, because of Inequalities (1.10) and (1.11), the functions fk belongsto La or any a in [1,1] and satisfies
kfkkLa 21a 2
k
⇣1a�
1p
⌘
. (1.13)
• The fact that f belongs to Lp is given by the behavior of the sequence (ck)k2Z.
• As inferred by the definition given below, the sequence (ckfk)k2Z is independent of pand depends only on f .
Proof of Proposition 1.3.1. If µ(f > 0) is not finite, let us define, for any k in Z,
�kdef= inf
n� /µ(f > �) < 2k
o.
If µ(f > 0) is finite, let us define by k0 the smallest integer such that
µ(f > 0) 2k0 .
Then, for k k0, let us define
�kdef= inf
n� /µ(f > �) < 2k
o.
In all that follows in this proof, we shall implicitely consider that, if µ(f > 0) is finite, then,all the sequence define are 0 for k � k0. Then, let us define
ckdef= 2
kp�k and fk
def= c�1
k1(�k+1 �) is a decreasing function. Now let us consider a decreasingsequence (↵n,k)n2N which tends to �k. For any x in X, the sequence (1(f>↵n,k)(x))n2N is anon decreasing sequence. Thus the monotonic convergence theorem implies that
limn!1
µ(f > ↵n,k) = µ(f > �k) .
As ↵n,k is greaer than �k, then
8k 2 Z , µ(f > �k) 2k
19
Let us notice that (�k)k2Z is a decreasing sequence. By definition of the sequence (�k)k2Z,we get
8� < �k , µ(f > �) � 2k. (1.14)
The monotonic convergence theorem and the definition of (�k)k2Z, we have that
µ(f > �k) 2k (1.15)
which implies (1.10). It is obvious that kfkkL1 2�kp .
The point is now the proof of (1.12). As the support of the functions (fk)k2Z are pairwisedisjoint, one may write
kfkpLp
=X
k2ZcpkkfkkpLp .
Taking advantage of Inequalities (1.10) and (1.11), we claim that
8k 2 Z , kfkkpLp 2 .
which claims exactly that kfkpLp
2X
k2Zcpk.
Moreoverwhich, owing to the fact that f is a nonnegative function of Lp, converges to 0when k tends to +1.
Thanks to (1.15), we have µ(Supp fk) 2k+1. This givesX
k2Zcpk
=X
k2Z2k�p
k
= pX
k2Z
Z1
02k1]0,�k[(�)�
p�1 d�.
Using Fubini’s theorem, we get
X
k2Zcpk= p
Z1
0�p�1
✓ X
k /�k>�
2k◆d�.
By definition of sequence (�k)k2Z, having � < �k implies that µ(f > �) � 2k(See (1.14)).Thus we infer that
X
k2Zcpk
pZ
1
0�p�1
✓ X
k / 2kµ(f>�)
2k◆d�
2pZ
1
0�p�1µ(f > �) d�.
Now, the right inequality in (1.12) follows from the fact that, owing to Fubini’s theorem, wehave
kfkpLp
= p
Z1
0�p�1µ(|f | > �) d� (1.16)
which implies thatX
k2Zcpk 2kfkp
Lp. 2
20
Proof of Theorem 1.3.1 Let f and g be two nonnegative measurable functions on Rd. Let usconsider a nonnegative function h in Lr
0
and let us define
I(f, g, h)def=
Z
R2df(y)g(x� y)h(x) dµ(x) dµ(y).
Arguing by homogeneity, one can assume that kfkLp = kgkLqw = khkLr0 = 1. Defining
Cjdef=�y 2 Rd , 2�
j+1q < g(y) 2�
jq
and gjdef= 1Cjg,
we can write
I(f, g, h) X
j2ZI(f, gj , h).
Because kgkLqw= 1, we have kgjkLs 2
1s 2
�j
⇣1q�
1s
⌘
for all s in [1,1]. Thus if we directly applyYoung’s inequality with p, q and r, we find that I(f, gj , h) 2
1q so that series
�(I(f, gj , h)
�j2Z
has no reason to converge. In order to bypass this di�culty, one may introduce the atomicdecomposition for f and h given by Proposition 1.3.1 which leads to
I(f, g, h) =X
j,k,`
ckd`I(fk, gj , h`).
Using Young’s inequalities, we get for any (a, b) in [1,1]2 such that b a0 and for any ( ef,eh)in La ⇥ Lb,
I(fk, gj , h`) kfkkLakgjkLbkh`kLc0 with1
a+
1
b= 1 +
1
c·
Using Proposition 1.3.1, this gives
I(fk, gj , h`) 4⇥ 2j
⇣1b�
1q
⌘
2k
⇣1a�
1p
⌘
2`(1c0�
1r0 ) = 4⇥ 2�j
⇣1q�
1b
⌘
2k
⇣1a�
1p
⌘
2`(1r�
1c ).
Using the condition (1.8) on (p, q, r) and (a, b, c) implies
I(fk, gj , h`) 4⇥ 2�j
⇣1q�
1b
⌘
2k
⇣1a�
1p
⌘
2`
⇣1p+
1q�
1a�
1b
⌘
4⇥ 2(j�`)⇣
1b�
1q
⌘
2(k�`)
⇣1a�
1p
⌘
. (1.17)
As (p, q, r) is in ]1,1[2, a positive real number " exists, if1
adef=
1
p� " sg(k � `) and 1
bdef=
1
q� " sg(j � `)
then (a, b) is in [1,1]2 and 1a+
1
b� 1. With this choice of a and b, (1.17) becomes
I(fk, gj , h`) 4⇥ 2�"|j�`|�"|k�`|.
Now, using Young’s inequality for Z equipped with the counting measure, we deduce that
I(f, g, h) 4X
k,`
ckd`2�"|k�`|
X
j
2�"|j�`|
C"
X
k,`
ckd`2�"|k�`|
C"2
k(ck)k2Zk`pk(d`)`2Zk`p0 .
21
Condition (1.8) implies that r0 p0 and thus
I(f, g, h) C"2
k(ck)k`pk(d`)k`r0 .
The theorem is proved. 2
1.4 Spaces of linear continuous maps
Proposition 1.4.1 Let E and F two normed spaces; we denote by L(E,F ) the set of con-tinuous linear maps from E to F . The map defined by
kLkL(E,F )def= sup
kxkE1kL(x)kF
is a norm on L(E,F ). If (F, k ·kF ) is a Banach space, then (L(E,F ), k ·kL(E,F )) also a Banachspace
Proof. We admit the proof of the fact that L(E,F ) is a vectorial space. Let us check thethree properties of a norm. Let us assume that kLkL(E,F ) = 0. This implies that L(x) = 0for any x of in the unit ball. Then for any x 6= 0,
L
✓x
2kxkE
◆=
1
2kxkEL(x) = 0 .
From this we deduce that L(x) = 0 and donc that L = 0. Let L1 and L2 be two elementsof L(E,F ). We have
kL1(x) + L2(x)kF kL1(x)kF + kL2(x)kF .
Thus for any x of E with norm less than or equal to 1, we have
kL1(x) + L2(x)kF kL1kL(E,F ) + kL2kL(E,F ).
By definition of the upper bound, we get
kL1 + L2kL(E,F ) kL1kL(E,F ) + kL2kL(E,F ).
And let L be in L(E,F ) and � in K, we have
k�L(x)kF = |�| kL(x)kF .
Thussup
kxkE1k�L(x)kF = |�| sup
kxkE1kL(x)kF .
And thus k · kL(E;F ) is a norm on L(E;F ).
Now let us assume that (F, ·kF ) is complete ; let us consider a Cauchy sequence impliesthat for any x of E, the sequence (Ln(x))n2N is a Cauchy sequence of F . The space F beingcomplete, for any x of E, an element of F , denoted L(x) exusts such that
limn!1
Ln(x) = L(x).
22
Now we have to prove that the map defined by x 7! L(x) belongs to L(E,F ) and que
limn!1
Ln = L in L(E,F ).
The uniqueness of the limit ensures that L is a linear map. Comme the sequence (Ln)n2N isa Cauchy on, it is bounded. Thus a constant C exists such that
supn2N
kxkE1
kLn(x)kF C.
Passing to the limit when n tends to infinityty, we deduce that
supkxkE1
kL(x)kF C.
Thus the linear map L is continuous. Let us proof the convergence of the sequence (Ln)n2Nto L in L(E,F ). The sequence (Ln)n2N being a Cauchy one, for any " positive, il existe aninteger n0 exists such that, for any integer n � n0,
supkxkE1p2N
kLn(x)� Ln+p(x)kF < ".
Passing to the limit when p tends to infinity, we get
supkxkE1
kLn(x)� L(x)kF " .
This concludes the proof of the proposition. 2
Notation When E = F , we denote L(E,F ) by L(E).
Exercise 1.4.1 Let (E, k · kE) and (F, k · kF ) be two normed spaces. For any L in L(E,F )we have
supkxkE1
kL(x)kF = supkxkE=1
kL(x)kF .
Remark The particular case when F = K will be the purpose of Chapter 2.
Proposition 1.4.2 Let E, F and G be three vectorial spaces and (L1, L2) an elementof L(E,F )⇥ L(F,G). The map L2 � L1 belongs to L(E,G) and
kL2 � L1kL(E,G) kL1kL(E,F )kL2kL(F,G).
Proof. We have for any x in the unit ball of E,
kL2 � L1(x)kG kL2kL(F,G)kL1(x)kF kL2kL(F,G)kL1kL(E,F ).
This gives the result. 2
Remark Even if F = K (it is the case of linear forms), the upper bound is not alwaysreached as shown by the following exercise.
23
Exercise 1.4.2 Let (an)n2N a sequence of elements of ]0, 1[. Let us assume that the (an)n2Ntends to 0. Let us define the linear form `a on `1(N) by
h`a, xidef=X
n2N(1� an)x(n).
Prove that k`ak(`1(N))0 = 1 and that for any x of the unit ball of `1(N), |h`a, xi| is less than 1.
Let us study the set U(E) of invertible elements of L(E). One of the basic results we theelements invertibles of L(E) is the following.
Theorem 1.4.1 Let (E, k · k) a Banach space; the open ball of center Id and radius 1 isincluded in U(E) the set of invertible elements of L(E) which means that
8A 2 BL(E)(Id, 1) , 9!A�1 2 L(E) / A �A�1 = A�1 �A = Id .
Proof. It relies on the following proposition.
Proposition 1.4.3 Let E is a Banach space and (xn)n2N a sequence of elements of E suchthat X
n2NkxnkE < 1.
Then the sequence SNdef=
NX
n=0
xn converges.
Proof. Let us prove that the sequence (SN )N2N is a Cauchy one. Indeed we have
kSN+P � SNk N+PX
n=N+1
kxnk.
The fact that the series (kxnkE)n2N is convergent implies that the sequence (SN )N2N is aCauchy one. The proposition is proved. 2
Continuation of the proof of Theorem 1.4.1. An element A of BL(E)(Id, 1) writes A = Id�Lwith kLkL(E) less than 1. Let us define
SN =NX
n=0
Ln.
Using Proposition 1.4.3, we claims that the sequence (SN )N2N converges to an element A�1
of L(E). Moreover we haveSNA = SNA = Id�LN+1.
Passing to the limit when N tends to infinity, we conclude the proof of this theorem. 2
Corollary 1.4.1 The set U(E) is an open subset of L(E) and the map Inv defined by⇢
U(E) �! U(E)L 7�! L�1
is continuous.
24
Proof. Let L0 an element of U(E). We can write
L = L0 � (L0 � L) =�Id�(L0 � L)L�10
�L0.
If twe assume that
kL� L0kL(E) <1
kL�10 kL(E),
we infer fromTheorem 1.4.1 that Id�(L0 � L)L�10 is invertible and also L aussi. Thus theopen ball of center L0 and radius kL�10 k
�1L(E) is included in U(E). Theorem 1.4.1 also implies
that
L�1 = L�10
1X
n=0
�(L0 � L)L�10
�n.
From this we infer that, if L belongs to B(L0, ⇢0) with ⇢0 less than kL�10 k�1L(E), we have
��L�1 � L�10 kL(E) kL�10 kL(E)
1X
n=1
kL� L0knL(E)kL�10 k
n
L(E)
kL� L0kL(E)kL�10 k2L(E)
1X
n=0
kL� L0knL(E)kL�10 k
n
L(E)
kL� L0kL(E)kL�10 k2L(E)
1
1� ⇢kL�10 kL(E)·
This conclut the proof of the corollary. 2
Remark In fact, twe proved that if ⇢0 is less than kL�10 k�1L(E), the map L 7! L
�1 is k
lipschitzian on BL(E)(L0, ⇢0) with
k =kL�10 k2L(E)
1� ⇢0kL�10 kL(E)·
The following exercise demands an understanding of di↵erential calculus in infinie dimen-sionnal normed spaces.
Exercise 1.4.3 Prove that the map Inv is C1 map on U(E).
1.5 Around Cauchy-Lipschitz theorem
In this section we present some very classical applications of functionnal analysis to the reso-lution of ordinary di↵erential equations. Let us start with the linear ones.
Theorem 1.5.1 Let E be a Banach space1, I an open interval of R and A a continuous mapfrom I to L(E), the set of continuous linear maps from E into E. Let t0 be in I, a unique C1function x from I to E exists such that
(ODE)
8><
>:
du
dt= A(t)u(t)
u(0) = u0.1The reader who is not familar with the integration of continuous functions with value in a Banach space
can assume that E is finite dimensionnal vectorial
25
Proof. Uniqueness is obvious because as it is a linear equation, it is enough to prove that if xis a solution of (ODE) which initial data 0, then it is identically zero. The set of t in I suchthat x(t) = 0 is a closed subset of I. It is open because if t1 is sucht at x(t1) = 0, then wehave by integration
supt2[t1�↵,t1+↵1]
kx(t)kE ↵ supt2[t1�↵,t1+↵1]
kA(t)kL(E) supt2[t1�↵,t1+↵1]
kx(t)kE
Taking ↵ small enough implies that x is identically 0 near t1. Let us prove the existence ofsolutions of this equation. Let � be a positive real number, let us introduce the space E�definited by
E� =nx 2 C(I, E) / kxk�
def= sup
t2I
kx(t)k exp⇣��Z
t
0A(t0)dt0
⌘< 1
o
with A(t0) def= kA(t0)kL(E) . It is an exercise left to the reader to prove that E� is a Banachspace. The solution of (ODE) are the same as the solutions of
(Id�PA)u = u0 with PAu(t)def=
Zt
0A(t0)u(t0)dt0.
Let us prove that
kPAkL(E�) 1
�(1.18)
which will imply the theorem because the constant function with value x0 obviously belongsto E⇤ for any positive real number �. By definition of A we have
kPAx(t)k exp⇣��Z
t
0A(t0)dt0
⌘
Z
t
0exp⇣��Z
t
t0
A(t00)dt00⌘A(t0) exp
⇣��Z
t0
0A(t00)dt00
⌘kx(t0)kdt0.
By definition of k · k�, we infer that
kPAx(t)k exp⇣��Z
t
0A(t0)dt0
⌘ kxk�
Zt
0exp⇣��Z
t
t0
A(t00)dt00⌘A(t0)dt0.
An obvious computation of integral implies that, for any t in I,
kPAx(t)k exp⇣��Z
t
0A(t0)dt0
⌘ 1�kxk�
which is exactly (1.18) ; the theorem is proved. 2
Let us state and prove the classical theorem about existence and uniqueness of solution of anordinary di↵erential equation associated with a Lipschitz function. This uses Picard’s fixed pointtheorem.
Theorem 1.5.2 (Cauchy-Lipschitz) Let E be a Banach space and f a continuous function wehaven open subset U of R⇥E with value in E. Let us consider a point (t0, x0) of U such that an open
26
interval J0 containing t0 and a ball B0 centered in x0 and of radius R0 such that J0 ⇥ B0 is includedin U and such that the two functions
M0(t)def= sup
x2B0kf(t, x)kE and K0(t)
def= sup
(x1,x2)2B20
kf(t, x1)� f(t, x2)kEkx1 � x2kE
are integrable on J0. Then, for any subinterval J of J0 containing t0 which satisfiesZ
JM0(t)dt R0, (1.19)
a unique continuous function x on J with values in B0 exists such that, for any t in J
x(t) = x0 + F (x)(t) with F (x)(t)def=
Z t
t0
f(t0, x(t0))dt0.
Proof. Let us denote by X the set of continuous functions on J with values in B0. Let us consider xin X. By definition of M0 and using Condition (1.19), we get, for any t in J
���Z t
t0
f(t0, x(t0))dt0���E
���Z t
t0
kf(t0, x(t0))kEdt0���
Z
JM0(t)dt R0.
Thus the function F maps X into itself. Then for any positive �, let us consider the metric space Xequipped with the distance
d�(x1, x2)def= sup
t2J
✓exp��
���Z t
t0
K0(t0)dt0
���◆kx1(t)� x2(t)kE .
Let us estimate d�(F (x1), F (x2)). In order to do it, let us write that, for t � t0,
kF (x1)(t)� F (x2)(t)kE Z t
t0
kf(t0, x1(t0))� f(t0, x2(t0))kE dt0
Z t
t0
K0(t0)kx1(t0)� x2(t0)kE dt0.
From this we deduce that, by definition of d�,
K�(t)def= exp
⇣��Z t
t0
K0(t0)dt0
⌘kF (x1)(t)� F (x2)(t)kE
Z t
t0
K0(t0) exp
⇣��Z t
t0K0(t
00)dt00⌘exp⇣��Z t0
t0
K0(t00)dt00
⌘kx1(t0)� x2(t0)kE dt0
d�(x1, x2)Z t
t0
K0(t0) exp
⇣��Z t
t0K0(t
00)dt00⌘dt0
1�d�(x1, x2).
The map F satisfies
8(x1, x2) 2 X2 , d�(F (x1), F (x2)) 1
�d�(x1, x2) .
Picard’s fixed point theorem allows to conclude the proof. 2
27
28
Chapter 2
Duality
2.1 The dual space of a normed space
Let us recall briefly the notion of duality in finite dimensionnal spaces. If E is a vectorialspace, the set E? denotes the space of linear forms defined on E. Moreover, when ` is a linearform on E and x a vector of E, we denote
h`, xi déf= `(x).
The first key concept is the concept of transposition for linear maps between vectorial spaces.Let us consider a linear map L from E to F . We define the transposed linear map from F ?
into E? by
htL(`), xi = h`, L(x)i.
Sait with other words,tL(`) = ` � L.
Let us point out that if L1 and L2 are two linear maps, then we have
t(L1 � L2) = tL2 � tL1.
Indeed, by definition of the transposition, we have
ht(L1 � L2)(`),�!h i = h`, L1 � L2(
�!h )i
= h`, L1(L2(�!h ))i
= htL1(`), L2(�!h )i
= htL2(tL1(`)),�!h i.
In the case when the space vectoriel E is of finite dimension, we choose a basis (�!e j)1jNof E. It is easy to prove that that the family (e?
j)1jN of linear forms defined par
he?j ,�!e ki = �j,k
is a basis of the space vectoriel E? called the dual basis. Thus in the case of the finitedimension, the set of continuous linear forms (all of them are) is a vectorial space of the samedimension.
29
In this chapter, we study the case when E is a infinite dimensionnal space.
Definition 2.1.1 We called dual topological space (ou simply dual space) of a K-vectorialspace (E, k · k) and denote it by E0 the space of linear continuous forms on E. It is equippedwith the norm
k`kE0def= sup
kxk1|h`, xi|.
Now we prove the Hahn-Banach theorem about extension of continuous linear form onseparable normed spaces. The precise theorem is the following.
Theorem 2.1.1 (Hahn-Banach) Let (E, k · kE) be a separable real normed space and Fany subspace of E. Let us consider a continuous linear form on F , i.e. a linear form on Fsuch that
8x 2 F , hL, xi CkxkE .
Then an element eL of E0 exists such that
8x 2 E , heL, xi CkxkE and eL|F = L.
As a first corollary of this theorem, we can observe that it proves that for a separablenormed space, the dual space E0 is non reduced to the null linear form. Let us establish thefollowng corollary.
Corollary 2.1.1 Let (E, k · kE) be a separable Banach space. For any x0 in E, a linear formof E0 exists such that
hL, x0i = kx0kE and kLkE0 = 1 .
Proof. Let us assume that x0 6= 0. Let us consider F = Rx0 and let us define
hL0,�x0)def= �kx0kE
Applying Hahn-Banach theorem, we get that some element L of E0 exists such that
hL, xi kxkE and hL,�x0i = �kx0kE .
This proves the corollary. 2
Proof of Theorem 2.1.1 It mainly relies on this lemma which is a convexity lemma.
Lemma 2.1.1 Let F be a subspace of a real vectorial space E. We consider a convex func-tion f on E and a linear form L on F such that
8x 2 F , hL, xi f(x).
Let e be any vector in the complement of F . Then a linear form eL exists on F +Re such that
8y 2 F + Re , heL, yi f(y) and eL|F = L . (2.1)
Proof. A linear form eL on F + Re extending L is determined by the choice of ↵ def= heL, ei.Condition (2.1) writes
8� 2 R , 8x 2 F , hL, xi+ �↵ f(x+ �e).
30
This can be written
8� > 0 , 8x 2 F , hL, xi � f(x� �e)�
↵ f(x+ �e)� hL, xi�
·
If we prove that
mdef= sup
x2F
�>0
hL, xi � f(x� �e)�
M def= infx2F
�>0
f(x+ �e)� hL, xi�
, (2.2)
then the lemma is proved by considering some ↵ in the interval [m,M ]. Condition (2.2) isequivalent to prove that
8(x, x0) 2 F 2 , 8(�,�0) 2]0,1[2 , 8e 2 E , hL, xi � f(x� �e)�
f(x0 + �0e)� hL, x0i
�0· (2.3)
Because of the linearity of L, this condition writes
hL,�0x+ �x0i �f(x0 + �0e) + �0f(x� �e).
Dividing the above inequality by � + �0 and using again the linearity of L we claim thatCondition (2.3) is equivalent to
hL, �0x+ �x0
�+ �0i �f(x
0 + �0e) + �0f(x� �e)�+ �0
·
The convexity of f implies that
f⇣�x0 + �0x
�+ �0
⌘= f
⇣�(x0 + �0e) + �0(x� �e)�+ �0
⌘
�f(x0 + �0e) + �0f(x� �e)
�+ �0·
The hypothesis on L and f ensures (2.3) and thus (2.2). This proves the lemma. 2
Continuation of the proof of Theorem 2.1.1 Because the space E is separable, it is possibleto construct a countable family of linearly independant vectors (en)n2N of E \F such that thevectorial space F
def= F +Vect{en} is dense in E. Applying step by step the above lemma, we
construct a linear form L on F such that
8x 2 F , hL, xi CkxkE .
The map of the extension theorem for dense subspaces concludes the proof of Theorem 2.1.12
Comparing with the case when the vectorial space E is finite dimensionnal, a totally newphenomena appears. As we shall see late on, the dual space of a Banach space can be verydi↵erent form the space itsself.
Now we are going to exhibit a natural bilinear form on the space E0 ⇥ E.
Proposition 2.1.1 Let E a normed space. Then the map h·, ·i defined par
h·, ·i⇢
E0 ⇥ E �! K(`, x) 7�! h`, xi.
is a continuous bilineare form.
Proof. By definition of the norm on E0, we have |h`, xi| k`kE0kxk. This means exactly thatthe bilinear form is continuous. 2
31
For reason that will appear later on, it is relevant to define a weaker way on convergencefor sequences of E0.
Definition 2.1.2 Let (Ln)n2N be a sequence in E0 where E is a Banach space. We say thatthe sequence (Ln)n2N weakly-? converges to some L in E0 if and only if
8x 2 E , limn!1
hLn, xi = hL, xi .
Let us first notice that, if a sequence (`n)n2N of E0 converges ro somme ` in the sense ofnorm, i.e. lim
n!1k`n � `kE0 = 0, then, for any x in E,
��h`n, xi � `, xi�� k`n � `kE0kxkE
and thus8x 2 E , lim
n!1h`n, xi = h`, xi .
Exercise 2.1.1 Let (xn)n2N and (`n)n2N be two sequences in E and E0 respectively, where Eis a Banach space. Then,
⇣limf?n!1
`n = ` and limn!1
kxn � xk = 0⌘=) lim
n!1h`n, xni = h`, xi.
Let us point out that the Banach-Steinhaus theorem (the proof of which is based on Bairetheorem) implies that when a sequence (`n)n2N is weakly-? convergent, it is bounded. In fact,up to an extraction, this property has the following reciproque, which is an important resultabout bounded sequences in the dual space E0 The weak-? convergence for sequences of linearforms is exactly the point wise convergence. In this context, this type of convergence has thefollowing very important compactness result.
Theorem 2.1.2 (weak-? compactness) Let E be a separable space. Let us consider asequence (`n)n2N bounded in E0. An element ` of E0 and an extraction function existsuch that
limf?n!1
` (n) = `.
The above theorem means that for any bounded sequence of E0 it is possible to extract aweakly-? convergent sequence. So it is a question of importance to know that if a Banachspace F can be identify (or not) to the dual space of a separable Banach space E. Let us givea precise mathematical meaning to this question.
Proposition 2.1.2 Let E and F be two Banach spaces, and let B be a bilinear continuousfunctional on F ⇥ E. The mapping �B defined by
�B
(F �! E0
y 7�! �B(y) : h�B(y), xidef= B(y, x)
(2.4)
is in L(F,E0).
Proof. As the bilinear form B is continuous, we can write that
|h�B(y), xi| CkykF kxkE .
By definition of the norm on E0, we have
k�B(y)kE0 CkykF ;
hence the proposition. 2
32
Definition 2.1.3 Let E and F be two Banach spaces, and let B be a bilinear continuousfunctional on F ⇥ E. We say that B identifies E0 with F if and only if the map �B definedby (2.4) is an isomorphism from F to E0.
In other words, this means that there exists a constant C such that, for any continuouslinear functional ` on E, there exists a unique element y in F such that
8x 2 E , h`, xi = B(x, y) and C�1kykF k`kE0 CkykF .
Let us make the following key remark: if a space F is identified to the dual space of aseparable Banach space E by a bilinear form B, then for any bounded sequence (yn)n2N of E,an extraction function ' and a vector y of F exists such that
8x 2 E , limn!1
B(y'(n), x) = B(y, x) . (2.5)
2.2 Examples of identification
We first treat the case of Hilbert spaces. Let us give the definition of an inner product.
Definition 2.2.1 Let E and F be two K-vector spaces. An map ` from E to F is said to beantilinear if, for any pair (x, y) in E ⇥ F , and any scalar � in K, we have
`(�x+ y) = �`(x) + `(y).
A map f from E⇥E to K is said to be sesquilinear1 if it is linear with respect to the firstvariable, and antilinear with respect to the second one. In other words,
8� 2 K , 8(x, y, z) 2 E3 f(�x+ y, z) = �f(x, y) + f(y, z)f(z,�x+ y) = �f(z, x) + f(z, y).
A map f from E ⇥ E to K is an inner product if and only if it is sesquilinear, Hermitianand positive definite, i.e. it satisfies
8(x, y) 2 E2 , f(y, x) = f(x, y) (Hermitian)8x 2 E , f(x, x) 2 R+, (positive)f(x, x) = 0 () x = 0 (definite).
Remark. An inner product is often denoted (·|·), and we write k · k2 = (·|·).
Proposition 2.2.1 Let E be a K-vector space, and let (·|·) be an inner product on E. Themap x 7! (x|x) 12 is a norm on E, and we have the following properties.
8(x, y) 2 E2 , |(x|y)| kxkkyk (Cauchy-Schwarz inequality).
8(x, y) 2 E2 , kxk2 + kyk2 = 2����x+ y
2
����2
+ 2
����x� y2
����2
(Appolonius’s theorem).
1from the Latin “sesqui”, meaning “one and a half”
33
Proof. Obviously, we have
0 ����(x|y)
x
kxk � kxky����2
.
By expanding the right-hand side, we get
0 |(x|y)|2 � 2
Let us definem
def= inf
�F (u) , u 2 H
.
By definition of the infimum, a sequence (un)n2N of elements of H exists such that
limn!1
F (un) = m. 2
Let us proved that such a sequence is a Cauchy sequence. Because of Appolonius’s theorem,we have, for any couple of ontegers (n, p),
1
2kun+pk2H +
1
2kunk2H =
����un+p + un
2
����2
H
+
����un+p � un
2
����2
H
.
Substracting h`, un+p + uni to this equality gives, by definition of F ,����un+p � un
2
����2
H
= F (un+p) + F (un)� 2F✓un+p + un
2
◆·
By definition of m, we have
����un+p � un
2
����2
H
F (un+p) + F (un)� 2m.
Using that the sequence (F (un))n2N converges to m, we infer that
8" > 0 , 9n" / 8n � n" , 8p 2 N , F (un+p) + F (un)� 2m "
4·
Thus (un)n2N is a Cauchy sequence and thus converges to some u in H. As F is continuous,we infer that F (u) = m. Then, for any h in H and any t in R, we have
F (u) F (u+ th) = kuk2H � h`, ui+ t�(h|u)� h`, hi
�+ t2khk2H .
Thus for all t in R and all h in H, we have
t�(h|u)� h`, hi
�+ t2khk2H � 0
which implies that8h 2 H , (h|u)� h`, hi = 0 .
The theorem id proved. 2
Remark In the case when the Hilbert space is complex, The same method proves that if `is an element of H0, then v exists such that
8h 2 H , 2
Now let us investigate the case of Lp space. The main result of this section is the followingtheorem, which explains how continuous linear functionals on Lp spaces, when p is real, canbe represented.
Theorem 2.2.2 Let p be a real number (strictly) greater than 1, and let B be the bilinearform defined by
B
8<
:
Lp0 ⇥ Lp �! K(g, f) 7�!
Z
X
f(x)g(x)dµ(x).
Then the bilinear form B identifies the dual of Lp(X, dµ) with Lp0
(X, dµ).
Before proving this theorem, we will show an important consequence of it, and make afew comments about the result.
Corollary 2.2.1 Let p be a number in [1,1[. Consider a bounded sequence (gn)n2N in Lp0
.
There exists an extraction function � and a function g in Lp0
such that
8f 2 Lp , limn!1
Z
X
g�(n)(x)f(x)dµ(x) =
Z
X
g(x)f(x)dµ(x). (2.6)
Proof. We apply theorem 2.1.2 page 32 to the sequence (Ln)n2N of linear functionals on Lp
defined by
hLn, fidef=
Z
X
gn(x)f(x)dµ(x).
This yields the existence of an extraction function � and a linear form L such that
8f 2 Lp , limn!1
Z
X
g�(n)(x)f(x)dµ(x) = hL, fi.
Theorem 2.2.2 above then ensures the existence of a function g in Lp0
such that
8f 2 Lp , limn!1
Z
X
g(x)f(x)dµ(x) = hL, fi.
The corollary is proved. 2
Let us consider the case where X is an open set in Rd, and µ is the Lebesgue measure.Note that, as p is finite, the space Cc(⌦) is dense in Lp(⌦). We let the reader establish, as anexercise, that assertion (2.6) is equivalent to the following:
8' 2 Cc(⌦) , limn!1
Z
X
g�(n)(x)'(x)dµ(x) =
Z
X
g(x)'(x)dµ(x). (2.7)
We are going to observe that this property is false when p0 is equal to 1, in other wordswhen p is infinity. Indeed, let � be a function in D(Rd) with integral equal to 1. We kwonthat if ("n)n2N is a sequence of positive numbers converging to 0, we have
8' 2 Cc(⌦), limn!1
Z
Rd
1
"dn�⇣ x"n
⌘'(x)dx = '(0).
But the linear functional �0 defined on Cc(⌦) by h�0,'i = '(0) is the Dirac measure at 0. Nofunction g in L1 can represent this via the bilinear form B because, if such was the case,
Z
⌦g(x)'(x)dx = '(0)
implies that g is zero except at x = 0, and therefore g is zero almost everywhere.
36
Proof of theorem 2.2.2 when p belongs to ]1, 2]. The theorem was proved in the case p = 2, thisis exactly Theorem 3.2.3. The setX is assumed to be the union of countably many subsetsKn,each with finite measure. Without loss of generality, we can assume that the sequence (Kn)n2Nis non-decreasing, in the sense that Kn ⇢ Kn+1. Let � be a linear functional on Lp(X, dµ),and let K be any subset of X with finite measure. We denote �K the restriction of � to K,that is the linear form defined by
h�K , fidef= h�,1Kfi.
We are now going to work in the space Lp(K, dµ). Let � be a linear form on Lp(K, dµ).Assuming for now that there exists a unique function gK on K such that
8f 2 Lp(K, dµ) , h�, fi = B(gK , f) =Z
K
gK(x)f(x)dµ(x), (2.8)
and such thatkgKkLp0 = k�Kk(Lp)0 , (2.9)
we consider the non-decreasing sequence (Kn)n2N of subsets ofX with finite measure such thatthe union of the Kn is the whole space X. Then there exists a sequence of functions (gn)n2Nsatisfying
8f 2 Lp(Kn, dµ) , h�Kn , fi = B(gn, f).
By lemma 1.2.3, we know that
gn 2 Lp0
and that kgnkLp0 = k�Knk(Lp)0 k�k(Lp)0 .
Moreover, the uniqueness of the function gn satisfying (2.8) and (2.9) ensures that, if m � n,
1Kngm = gn,
because the function 1Kngm also satisfies relations (2.8) and (2.9) for the linear form �Kn .The sequence (|gn|p
0
)n2N is therefore an non-decreasing sequence of functions such that
supn
Z
⌦|gn(x)|p
0
dµ(x) k�kp0
(Lp)0 .
The monotone convergence theorem ensures that the function
g(x) = limn!1
gn(x) 2 Lp0
and that kgkLp
0 k�k(Lp)0 .
Lemma 1.2.3 then ensures that kgkLp
0 = k�k(Lp)0 . Moreover, for any function f in Lp and forany integer n, we have
h�,1Knfi =Z
X
g(x)1Knf(x)dµ(x).
The dominated convergence theorem then finishes the proof o↵. But we are yet to proverelations (2.8) and (2.9). By Hölder’s inequality, we can say that
kfkLp =✓Z
K
|f(x)|pdµ(x)◆ 1
p
✓Z
K
|f(x)|p2pdµ(x)
◆ p2⇥
1p
µ(K)1p(1�
p2 )
kfkL2µ(K)1p�
12 .
37
Thus, the linear form � appears as a continuous linear form on the space L2, so there existsa function g in L2 such that
8f 2 L2 , h�, fi =Z
X
f(x)g(x)dµ(x).
We deduce that, for any function f in L2, we haveZ
K
|f(x)g(x)|dµ(x) k�k(Lp)0kfkLp .
The space L1 \Lp is dense in Lp. As p is between 1 and 2 here, and since the measure of Kis finite, L2 is dense in Lp. By lemma 1.2.3, we deduce that g belongs to Lp
0
, and kgkLp
0 =k�k(Lp)0 . Thus, (2.8) and (2.9) are proved.
Proof of theorem 2.2.2 when p belongs to ]2,1[. We start by proving that, for p in theinterval ]2,1[, the following inequality holds:
8(a, b) 2 R2 , |a|p + |b|p2
����a+ b
2
���p
+p(p� 1)2p�1
���a� b2
���p
. (2.10)
As the function t 7! |t|p is twice continuously di↵erentiable on R, the second-order Taylorformula with integral remainder between a and (a+ b)/2, then between b and (a+ b)/2, yields
|a|p =���a+ b
2
���p
+ pa� b2
a+ b
2
���a+ b
2
���p�2
+ p(p� 1)⇣a� b
2
⌘2 Z 1
0(1� ✓)
���a+ b
2+ ✓
a� b2
���p�2
d✓
and |b|p =���a+ b
2
���p
+ pb� a2
a+ b
2
���a� b2
���p�2
+ p(p� 1)⇣a� b
2
⌘2 Z 1
0(1� ✓)
���a+ b
2+ ✓
b� a2
���p�2
d✓ .
The mean of these two equalities gives
|a|p + |b|p2
=���a+ b
2
���p
+p(p� 1)
2
⇣a� b2
⌘2 Z 1
0(1� ✓)
✓���a+ b
2+ ✓
a� b2
���p�2
+���b� a2
+ ✓a+ b
2
���p�2◆d✓ .
If we have���a+ b
2+ ✓
a� b2
��� 1
4|b� a| and
���a+ b
2+ ✓
b� a2
��� 1
4|b� a|, (2.11)
then
✓|b� a| =���✓a� b2
� ✓ b� a2
���
=���a+ b
2+ ✓
a� b2
� a+ b2
� ✓ b� a2
���
���a+ b
2+ ✓
a� b2
���+���a+ b
2� ✓ b� a
2
���
12|b� a|.
38
Assuming that a and b are distinct (there is nothing worth seeing otherwise), we get thatassertion (2.11) can only be satisfied if ✓ is less than or equal to 1/2. Hence,
|a|p + |b|p2
����a+ b
2
���p
+p(p� 1)
2
⇣b� a2
⌘2���b� a4
���p�2
Z 112
(1� ✓)d✓
����a+ b
2
���p
+p(p� 1)2p�1
���b� a2
���p
,
which proves inequality (2.10). The proof of the theorem will now continue, following the linesof that of theorem 2.2.1 page 34. Let ` be a continuous linear form on Lp(X, dµ). Considerthe function
Fp
8<
:
Lp(X, dµ) �! Rf 7�! 1
pkfkp
Lp� h`, fi
The function Fp is bounded from below, because
Fp(f) �1
pkfkp
Lp� k`k(Lp)0kfkLp .
Set mpdef= inf
f2LpFp(f). We consider a minimising sequence (fn)n2N, that is a sequence such
thatFp(fn) = mp + "n with lim
n!1"n = 0.
We are going to prove that this is a Cauchy sequence. First, we apply inequality (2.10)with a = f(x) and b = g(x), then we integrate with respect to the measure dµ; we get that
8(f, g) 2�Lp(X, dµ)
�2, kfkp
Lp+ kgkp
Lp� 2���f + g
2
���p
Lp+
p(p� 1)2p�1
���f � g2
���p
Lp. (2.12)
Applying this inequality with f = fn and g = fm, we can write
1
pkfnkpLp +
1
pkfmkpLp �
2
p
���fn + fm
2
���p
Lp+
(p� 1)2p�1
���fn � fm
2
���p
Lp.
After subtracting h`, fn + fmi, the inequality becomes
1
pkfnkpLp � h`, fni+ kfmk
p
Lp� h`, fmi
� 2✓���
fn + fm2
���p
Lp� h`, fn + fm
2i◆+
(p� 1)2p�1
���fn � fm
2
���p
Lp.
By definition of Fp, this can be written as
(p� 1)2p�1
���fn � fm
2
���p
Lp Fp(fn) + Fp(fm)� 2Fp
⇣fn + fm2
⌘
As the sequence (fn)n2N minimises Fp, we deduce that
(p� 1)2p�1
���fn � fm
2
���p
Lp "n + "m,
which ensures that (fn)n2N is a Cauchy sequence. Let f be its limit. As the function Fp iscontinuous, we deduce that Fp(f) = mp; in other words, the infimum of Fp is a minimum.
39
Now, for a given function h in Lp, consider the function gh from R to R defined by
gh(t)def= Fp(f + th).
Using the di↵erentiability theorem for functions defined by integrals, we see that the func-tion gh is di↵erentiable, and that
g0h(t) =
Z
X
(f(x) + th(x))|f(x) + th(x)|p�2h(x)dµ(x)� h`, hi.
The point t = 0 is a minimum of the function gh, so its derivative vanishes at this point,which means that
8h 2 Lp(X, dµ) ,Z
X
f(x)|f(x)|p�2h(x)dµ(x) = h`, hi,
which ends the proof of Theorem 2.2.2. 2
40
Chapter 3
More about Hilbert spaces
3.1 The orthogonality
Definition 3.1.1 Let E be a vector space endowed with an inner product, and let x and ybe two elements of E. The vectors x and y are said to be orthogonal (denoted x ? y) if andonly if (x|y) = 0.
We recall the highly famous Pythagoras equality:
x ? y =) kx+ yk2 = kxk2 + kyk2.
Proposition 3.1.1 Let E be a K-vector space endowed with an inner product, and let A bea subset of E. Define
A?def= {x 2 E/ 8a 2 A , x ? a}.
The set A? is a closed vector subspace of E.
Proof. The linear map La from E to K defined by x 7! (x|a) is continuous, so its kernel is aclosed vector subspace of E. It is clear that
A? =\
a2A
kerLa,
so A? is a closed vector subspace of E, as it is an intersection of the same. This proves theproposition. 2
Exercise 3.1.1 Let E be a K vector space endowed with a Hermitian inner product, andlet A be a subset of E. Prove that A
?= A? = Span(A)?.
Definition 3.1.2 Let H be a K-vector space endowed with an inner product (·|·). Thespace H is called a Hilbert space if and only if the space H is complete for the norm associatedwith the inner product.
The Hilbert space structure is absolutely fundamental. Many simple methods and conceptsfrom geometry in Euclidean spaces with finite dimension will reappear here.
Pythagoras’s equality has a following ”infinite dimensionnal” extension.
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Theorem 3.1.1 Let (xn)n2N be a sequence of elements of a Hilbert space H such that,for n 6= m, xn ? xm. Then the series
X
n
xn is convergent if and only if the seriesX
n
kxnk2
is convergent, and in that case,
���X
n2Nxn���2=X
n2Nkxnk2
Proof. Let us set Sq =X
pq
xp. Pythagoras’s theorem yields
X
pq
kxpk2 = kSqk2. (3.1)
If the right-hand side of the above equality is assumed to be convergent, then it is bounded
independently of q, and the seriesX
n
kxnk2 is convergent. Conversely, if the seriesX
n
kxnk2
is convergent, then
kSq+q0 � Sqk2 =q+q0X
q+1
kxpk2
1X
q+1
kxpk2.
By taking the limit in equality (3.1), we finish the proof of the theorem. 2
Exercise 3.1.2 Let (xn)n2N be a sequence of elements of a Hilbert space H such that1X
0
kxnk2 < 1.
Assume that there exists an integer N0 such that, if |n�m| � N0, the vectors xn and xm areorthogonal. Prove that the series
Pnxn is convergent, and that there exists a constant C,
which only depends on N0, such that
���X
n2Nxn���2 C
X
n2Nkxnk2.
Many remarkable properties of Hilbert spaces are based on the following projection theo-rem.
Theorem 3.1.2 (projection onto a closed convex) Let � be a closed convex subset ofa Hilbert space H. Then, for any point x in H, there exists a unique point in �, which wedenote p�(x) and call the projection of x on �, such that
kx� p�(x)k = inf�2�
kx� �k.
Proof. Recall Appolonius’s equality: if � and �0 are two points in �, then we have
1
2k� � �0k2 = kx� �k2 + kx� �0k2 � 2
����x�� + �0
2
����2
· (3.2)
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Note that, since the set � is convex, the point� + �0
2is also in �. We first prove uniqueness.
Assume that there are two points �1 and �2 in � which both minimise kx � gk for g 2 �.Then, Appolonius’s equality yields
1
2k�1 � �2k2 = kx� �1k2 + kx� �2k2 � 2
����x��1 + �2
2
����2
= 2d2 � 2����x�
�1 + �22
����2
2d2 � 2d2 = 0.
This provides uniqueness. Now we prove existence. By definition of the infimum, there existsa minimising sequence, that is a sequence (�n)n2N of elements of � such that
limn!1
kx� �nk = ddef= inf
�2�kx� �k.
We use relation (3.2) again to get that, for any pair of integers (n,m),
1
2k�n � �mk2 = kx� �nk2 + kx� �mk2 � 2
����x��n + �m
2
����2
·
As � is convex, the midpoint between �n and �m is also in �, thus,
1
2k�n � �mk2 kx� �nk2 + kx� �mk2 � 2d2.
Given the definition of the sequence (�n)n2N, the above yields that it is a Cauchy sequence.As H is a Hilbert space, it is complete, and, � being a closed subset of H, it is also complete,so the sequence (�n)n2N converges to a point in �. This ends the proof of the theorem. 2
Proposition 3.1.2 Let � be a closed convex subset of a Hilbert space H, and let x and x0be two points in H. We denote p�(x) the unique point in � such that d(x,�) = kx� p�(x)k.Then, for any � in �,
Nearly all properties in Hilbert spaces can be seen as corollaries of the above theorem 3.1.2.
Corollary 3.1.1 Let F be a closed vector subspace of a Hilbert space H. Then,
H = F � F? and x = pF (x) + pF?(x).
Proof. Let x be an element of H. For any point f in F , and for any real number �, we have
kx� pF (x) + �fk2 = �2kfk2 + 2�
Theorem 3.2.1 In every separable Hilbert space H, there exists an orthonormal basis.
Proof. Let (an)n2N be a total countable subset of H. We can assume that the family (an)n2Nconsists of linearly independent vectors. Indeed, we define the following extraction function
�(0) = min{n / an 6= 0} and �(n) = min�m/ am 62 Span{a0, . . . , a�(n)i
.
We let the reader check, as an exercise, that the vector subspace generated by the a�(n) is thesame as the one spanned by the an. We will now apply the Gram-Schmidt orthonormalisationmethod to this family. Let us recall the principle. First, we set
e1 =a1ka1k
·
We assume that the terms (ej)1jn satisfy property (3.3), so that
Span{a1, · · · , an} = Span{e1, · · · , en}.
Set
en+1 =fn+1
kfn+1kwith fn+1 = an+1 �
nX
j=1
(an+1|ej)ej .
It is quick to check the equalities in (3.3) are satisfied by en+1. The theorem is proved. 2
Theorem 3.2.2 Let H be a separable Hilbert space, and let (en)n2N be an orthonormal basisof H. The map I defined by ⇢
H �! `2(N)x 7�! ((x|en))n2N
is a linear isometric bijection. In particular, the Bessel and Parseval equalities hold in H:
x =X
n2N(x|en)en and kxk2 =
X
n2N|(x|en)|2.
Proof. The main point to prove is that I is indeed a map from H to `2(N). We set
xqdef=X
pq
(x|ep)ep.
It is clear that
(x|xq) =X
pq
(x|ep)(x|ep)
=X
pq
|(x|ep)|2
= (xq|xq).
The Cauchy-Schwarz inequality states that kxqk2 kxk ⇥ kxqk. We deduce that, for anyinteger q, we have
qX
p=0
|(x|ep)|2 kxk2.
So I(x) does belong to `2(N).The fact that the map I is one to one is deduced from corollary 3.1.2, which states that
the orthogonal set of a total subset is the singleton {0}. That the map I is onto and anisometry, is deduced from theorem 3.1.1. Thus, theorem 3.2.1 is proved. 2
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The duality in Hilbert spaces is described by the following theorem.
Theorem 3.2.3 (Riesz representation theorem) Let H be a Hilbert space. We considerthe map � defined by
�
⇢H �! H0x 7�! �(x) : h 7! (h|x)
This is an antilinear isometric bijection.
Proof. The fact that k�(x)kH0 kxk (and therefore �(x) is a continuous linear form) isobtained by using the Cauchy-Schwarz inequality. It is obvious that � is antilinear. Moreover,
h�(x), xkxki = kxk ;
so � is an isometry. An isometry is always one to one, so it remains to prove that � is onto.To do this, let us consider an element ` in H0 \ {0}. Its kernel is a closed subspace of H, notequal to H. Let x be a non-zero vector in (ker `)?, such that h`, xi = kxk2. Both ` and �(x)have the same kernel, and there exists a non-zero h0 such that h`, h0i = h�(x), h0i, so �(x) = `.Thus, the theorem is proved. 2
We can deduce the following two consequences.
Definition 3.2.2 Let (xn)n2N be a sequence of elements of a Hilbert space H, and let x be anelement of H. The sequence (xn)n2N is said to be weakly convergent to x, denoted limf
n!1xn = x
or (xn)n2N * x, if and only if
8h 2 H , limn!1
(h|xn) = (h|x).
Observing this definition, we note that
limfn!1
xn = x () limfn!1
? �(xn) = �(x).
Among other things, the following theorem explains the terminology, why the convergenceis called weak.
Example Let us consider ((en)n2N a Hilbertian basis of a separable Hilbert space H. Then
limfn!1
en = 0 and kenk = 1 .
Theorem 3.2.4 Let (xn)n2N be a sequence of elements of a Hilbert space H, and let x bean element of H. Then, we have:
if limn!1
kxn � xk = 0, then limfn!1
xn = x ; (3.4)
if limfn!1
xn = x and limn!1
kxnk = kxk, then limn!1
kxn � xk = 0. (3.5)
Moreover, if (xn)n2N * x, then the sequence (xn)n2N is bounded.
Proof. The first point of the theorem is a direct consequence of the fact that
|(h|xn)� (h|x)| khk ⇥ kxn � xk.
For the second point, it su�ces to write that
kxn � xk2 = kxnk2 � 2
As the sequence (xn)n2N is weakly convergent to x, we have
�2 limn!1
3.3 The adjoint of an operator, self-adjoint operators
The adjoint of a linear map is a well-known notion in finite dimension. Given an inner producton a vector space H with finite dimension d, the adjoint of A is denoted A?, and defined bythe relation
(Ax|y) = (x|A?y).It is common knowledge that, if (Ai,j)1i,jd is the matrix of the map A in an orthonormalbasis, then the matrix representing A? in the same basis is such that A?
i,j= Aj,i. Moreover,
there is a classic linear algebra theorem which states that if A is self-adjoint (i.e. A = A?),then it can be diagonalised in an orthonormal basis.
In this section, we generalise these concepts and results in Hilbert spaces with infinitedimension.
Theorem 3.3.1 Let A be a continuous linear operator on a Hilbert space H. There exists aunique linear operator A?, which is continuous on H, such that
8(x, y) 2 H⇥H , (Ax|y) = (x|A?y).
Moreover, the map from L(H) to itself, defined by A 7! A?, is an antilinear isometry.
We extend from this theorem the following definition.
Definition 3.3.1 The map A? given above is called the adjoint of the operator A. Anelement A of L(H) is self-adjoint if and only if A = A?.
Proof of theorem 3.3.1. Uniqueness of the operator A? is a consequence of the fact that theorthogonal of H is {0}. Let LA be the map defined by
LA⇢
H �! H0y 7�! LA(y) : x 7�! (Ax|y).
It is clear that the map LA is a continuous antilinear map from H to H0. Set A?def= ��1 �LA.
By definition of A?, we have
(x|A?y) = (LAy|x) = (Ax|y).
So, the operator A? satisfies the desired relation with respect to A. We also have
kAk = supkxk1,kyk1
|(Ax|y)|
= supkxk1,kyk1
|(x|A?y)|
= kA?k.
So, theorem 3.3.1 is proved.
Exercise 3.3.1 Let H be a real Hilbert space, and let ⌦ be an open subset of H. We definethe “gradient” operator by
grad
⇢C1(⌦;R) �! C(⌦,H)
g 7�! ��1Dg
Let g be a C1 function from ⌦ to ⌦0, another open set of H, and let f be a C1 functionfrom ⌦0 to R. Compute grad(f � g).
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Proposition 3.3.1 Let A be a continuous linear map on a Hilbert space H. The followingproperties hold.
The?operation is involutory, i.e. A = A??. We also have
(kerA)? = ImA? and kerA = (ImA?)?. (3.6)
Finally, if a sequence (xn)n2N is weakly convergent to x, then the sequence (Axn)n2N is weaklyconvergent to Ax.
Proof. The first point is simple: since (Ax|y) = (x|A?y), we have (A?y|x) = (y|Ax). To provethe equalities in (3.6), write that
8y 2 H , (Ax|y) = 0 () 8y 2 H , (x|A?y) = 0.
Restating the above equivalence in terms of sets yields that kerA and (ImA?)? are equal.The other relation is deduced from this by taking the orthogonal.
Let (xn)n2N be a sequence that converges weakly to x. For any y in H, we have
(y|Axn) = (A?y|xn).
So, the weak convergence of the sequence (xn)n2N implies the weak convergence of the se-quence (Axn)n2N. 2
Proposition 3.3.2 Let A be a self-adjoint operator in L(H). Then,
kAkL(H) = supkuk=1
|(Au|u)|.
Proof. First, we note thatkAkL(H) = sup
kuk=1kvk=1
Example. If u(E) is a vector space with finite dimension, then the operator u is compact.Indeed, if A is a bounded subset of E, the image of A is a bounded subset of the space u(E),which has finite dimension, and the closure of such a set is compact.
Exercise 3.3.2 Let E be the vector space of functions f from [0, 1] to R such that
supx 6=x0
|f(x)� f(x0)||x� x0| 12
< 1
1) Prove that
kfkEdef= |f(0)|+ sup
x 6=x0
|f(x)� f(x0)||x� x0| 12
is a norm on E, and that (E, k · kE) is a Banach space.2) Prove that the map
◆
⇢E �! C([0, 1],R)f 7�! f
is compact.
Exercise 3.3.3 Let (A(n,m))(n,m)2N2 be a sequence of real numbers such thatX
(n,m)2N2A2n,m
is finite.
1) Prove that the relation
(Ax)(n) def=X
m2NA(n,m)x(m)
defines a continuous operator A from `2(N) to itself, and that kAkL(`2) ⇣ X
(n,m)2N2A2n,m
⌘ 12.
2) Let AN be an operator defined by (ANx)(n) = 1{0,··· ,N}(n)(Ax)(n). Prove that
limN!1
kAN �AkL(`2) = 0.
3) Deduce that the operator A is compact.
We now state the diagonalisation theorem for compact, self-adjoint operators on a Hilbertspace.
Theorem 3.3.2 Let A be a compact, self-adjoint operator on a separable Hilbert space Hwith infinite dimension. There exists a sequence of real numbers (�j)j2N, such that (|�j |)j2Nis nonincreasing and converges to 0, such that:
• for any j such that �j is non-zero, �j is an eigenvalue of A, and Ejdef= kerA � �j Id
is a subspace of H with finite dimension; moreover, if �j and �j0 are distinct, theircorresponding eigenspaces are orthogonal;
• if E def= Span[
j2N�j 6=0
Ej , then kerA = E?;
• we have kAkL(H) = maxj2N
|�j |.
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Before we prove this theorem, we recall and re-prove the diagonalisation theorem for self-adjoint operators on Euclidean (or Hermitian) spaces with finite dimension.
Theorem 3.3.3 Let A be a self-adjoint operator on a Hilbert space with finite dimension N .There exists a finite sequence (�j)1jN and an orthonormal basis (ej)1jN such that, forany j, ej is an eigenvector corresponding to the eigenvalue �j .
Proof. Consider the quadratic (or Hermitian) form associated with A, that is
QA
⇢H �! Rx 7�! (Ax|x)
This is a continuous function on H. We denote SH the unit sphere of H. As the space hasfinite dimension, a point xM exists in SH such that
|(AxM |xM )| = M0def= sup
x2SH|(Ax|x)| (3.7)
We aim to show that xM is an eigenvector of A, corresponding to the eigenvalue ±M0. Thisproof is general, it does not depend on whether the dimension of the space is finite or not.Let us state a lemma.
Lemma 3.3.1 Let H be a Hilbert space (with finite or infinite dimension). Let x0 be a vectorin SH such that
|(Ax0|x0)| = M0def= sup
x2SH|(Ax|x)|
Then x0 is an eigenvector of A, corresponding to the eigenvalue M0 or �M0.
Proof. By switching A for �A if needed, we can assume that (Ax0|x0) = M0. Observe that,for any non-zero vector y in H,
✓A⇣ ykyk
⌘���y
kyk
◆ M0,
which implies that
8y 2 H , F (y) def= M0kyk2 � (Ay|y) � 0. (3.8)
We can observe that the point x0 satisfies F (x0) = 0, and is a minimum of the function F .Therefore, the di↵erential of F at the point x0 is equal to zero. Alternatively, we can give adirect argument seeing that
8t 2 R , 8h 2 H , M0kx0 + thk2 ��A(x0 + t)|x0 + th
�� 0
As F (x0) = 0, by expanding the above, we get
8t 2 R , 8h 2 H , 2t
Back to the proof of theorem 3.3.3. We are now going to prove a very simple linear algebraresult, which also holds regardless of whether the dimension of the space is finite or not.
Lemma 3.3.2 Let A be a self-adjoint operator on a Hilbert space H. When two eigenvectorsof A are assoicated with distinct eigenvalues, they are orthogonal.
Proof. Let u and v be two eigenvalues corresponding to distinct eigenvalues � and �0. Wehave
�(u|v) = (Au|v) = (Av|u) = �0(u|v).
This yields (�� �0)(u|v) = 0, so the lemma is proved. 2
Back to the proof of theorem 3.3.3. We define
E±0def= ker(A±M0 Id) and H1
def= (E+0 � E
�
0 )? = (E+0 )
? \ (E�0 )?. (3.9)
Lemma 3.6 states, in particular, that
(A±M0 Id)(H1) ⇢ (E±0 )? and therefore A(H1) ⇢ H1 (3.10)
We reiterate the process on H1, and find either that there exist one or two non-zero eigenval-ues ±M1, or that 0 is an eigenvalue. As the dimension of H is finite, the decreasing sequenceof subspaces Hj is finite, and theorem 3.3.3 is proved. 2
Proof of theorem 3.3.2. We follow the same lines as above, but some complications arise dueto the space having infinite dimension. Here are the three main tricky points to prove, orderedby increasing di�culty:
• the fact that the eigenspaces corresponding to non-zero eigenvalues have finite dimen-sion;
• the fact that the sequence of eigenvalues converges to 0;
• the existence of a point xM that attains the supremum M0.
For the first point, we need to prove that the unit ball of the eigenspace Ej is compact,which, by theorem 1.1.1 page 9, ensures that the dimension of Ej is finite.
Let (xn)n2N be a bounded sequence of Ej . As the operator A is compact, we can extract asubsequence (x'(n))n2N such that the sequence (Ax'(n))n2N converges. But, we have Ax'(n) =�jx'(n), and �j is non-zero, hence the unit ball of Ej is compact.
In order to prove the second point, let us consider an infinite sequence (�j)j2N of distincteigenvalues, and let (ej)j2N be a sequence of vectors with norm 1, such that ej is an eigenvectorcorresponding to the eigenvalue �j . The sequence (ej)j2N is weakly convergent to 0 (theproof is left as an exercise), and we can extract a strongly convergent subsequence from this.As Aej = �jej , we have
limj!1
|�j | kejk = 0.
As the norm of ej is equal to 1 for every j, it is the sequence (�j)j2N that converges to 0.
Now we need to construct the sequence (�j)j2N. The crucial element is the followinglemma.
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Lemma 3.3.3 Let H be a separable Hilbert space with infinite dimension, and let A be acompact, self-adjoint operator on H. There exists a vector xM on the unit sphere of H suchthat
|(AxM |xM )| = M0def= sup
kxk=1|(Ax|x)|.
Moreover, AxM = ±M0xM .
Proof. If M0 = 0, then by proposition 3.3.2, we have A = 0, and there is nothing to prove. Sonow let M0 be positive. As (Ax|x) is real, and switching A for �A if needed, we assume that
M0 = supkxk=1
(Ax|x).
By definition, there exists a sequence (xn)n2N of elements of the unit ball B such that
limn!1
(Axn|xn) = M0
As the operator A is compact, there exist a subsequence, still denoted (xn)n2N, and an ele-ment y in H such that
limn!1
Axn = y.
By using theorem 3.2.5, we can extract another subsequence and assume that there exists x1in H such that
limfn!1
xn = x1.
We aim to prove that Ax1 = y. As the operator A is self-adjoint, we have, for any z in H,
limn!1
(xn|Az) = (x1|Az) = (Ax1|z).
It is also clear thatlimn!1
(Axn|z) = (y|z).
As a result, for any z in H, we have (y|z) = (Ax1|z), which implies that y = Ax1. So, byproposition 3.2.1,
limn!1
(Axn|xn) = (Ax1|x1).
The supremum M0 is therefore attained at a point x which is, of course, non-zero, since M0is strictly positive. Moreover, we would have, for any x in H \ {0},
(Ax|x) = kxk2✓A
x
kxk
����x
kxk
◆
= M0kxk2 .
This implies that kx1k = 1. And thus the sequence (xn)n2N converges to x1 in norm. 2We can apply lemma 3.3.3 and work in the space
H1def=⇣ker(A�M0 Id) + ker(A+M0 Id)
⌘?
= ker(A�M0 Id)? \ ker(A+M0 Id)?.
We prove that H1 is invariant under A. It is a simple lemma.
Lemma 3.3.4 Let B be an element of L(H). Then (kerB)? is invariant under B?.
Proof. Let x 2 (kerB)? and y 2 kerB. We have
(y|B?x) = (By|x) = 0,
hence B?x is in the orthogonal space of kerB. The lemma is proved. 2
53
Back to the proof of theorem 3.3.2. If we apply the above lemma with B = A±M0 Id, we getthat
(A±M0 Id)�ker(A±M0 Id)
�? ⇢
�ker(A±M0 Id)
�?,
because A is self-adjoint. This immediately implies that
A�ker(A±M0 Id)
�? ⇢
�ker(A±M0 Id)
�?,
hence AH1 ⇢ H1. We now study the operator A restricted to the Hilbert space H1. Theoperator A|H1 is is a compact, self-adjoint operator on H1 (exercise). Set
M1def= sup
kxk=1x2H1
(Ax|x).
By the preceding study, if M1 6= 0, this supremum is a maximum attained for at least onevector with norm 1 in (ker(A�M1 Id) + ker(A+M1 Id))?. Thus M1 < M0. This process isreiterated as follows. Set
Hjdef=�E0 + · · ·+ Ej�1
�?
and supkuk=1u2Hj+1
|(Au|u)| = Mj+1.
If Mj+1 = 0, proposition 3.3.2 implies that Hj+1 = kerA, and the procedure ends. If Mj+1 ispositive, the procedure continues.
If the sequence (Mj)j2N is a sequence of positive real numbers, set
E? =\
j2NHj .
This means that8j 2 N , sup
kuk=1u2E
?
|(Au|u)| |�j |,
hencesupkuk=1u2E
?
|(Au|u)| = 0,
which implies, using Proposition 3.3.2, that A|E? = 0. The proof of Theorem 3.3.2 is complete.2
Exercise
