This document outlines the plan for a seminar on kernel methods. The seminar will cover four main topics: kernel methods, random projections, deep learning, and more about kernels. The overall goal is to provide enough theoretical background to understand several related papers on convolutional kernel networks, fastfood kernel approximations, and deep fried convolutional neural networks. The document includes an agenda, definitions, theorems, and references to support understanding kernel methods and their applications.
8. The Plan
Main Goal
The main goal of these set of seminars is to have enough theoretical
background to understand the following papers
Julien Mairal et al., Convolutional Kernel Networks.
Quoc Viet Le et al., Fastfood: Approximate Kernel Expansions in
Loglinear Time.
Zichao Yang et al., Deep Fried Convnets.
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9. Greetings
Part of the content of these slides was done in collaboration with my
study group from the School of Mathematics at UNMSM (Universidad
Nacional Mayor de San Marcos, Lima - Perú). I want to thank the
members of the group for the great conversations and fun time studying
Support Vector Machines at the legendary office 308.
DSc. Jose R. Luyo Sanchez (UNMSM).
Lic. Diego A. Benavides Vidal (Currently a Master Student at
UnB).
Bach. Luis E. Quispe Paredes (UNMSM).
Also, I want to thank DSc. André M.S. Barreto, my supervisor, for give
me the freedom to choose my topic of research. As soon as I finish with
my obligatory courses at LNCC, I will start working in Reinforcement
Learning. :)
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11. Table of Contents I
Motivation
R to R2 Case
R2 to R3 Case
Cover’s Theorem
Definitions
Preliminaries to Cover’s Theorem
Cover’s Theorem
References for Cover’s Theorem
Mercer’s Theorem
Theory of Bounded Linear Operators
Integral Operators
Preliminaries to Mercer Theorem
Mercer’s Theorem
References for Mercer’s Theorem
Moore-Aronszajn Theorem
Reproducing Kernel Hilbert Spaces
11 / 125
12. Table of Contents II
Moore-Aronszajn Theorem
References for Moore-Aronszajn Theorem
Kernel Trick
Definitions
Feature Space based on Mercer’s Theorem
History
References
12 / 125
13. "Nothing is more practical than a good theory."
— From Vapnik’s preface to The Nature of Statistical Learning Theory
13 / 125
15. Motivation
Motivation
How we can split data that is not linear separable?
How we can utilize algorithms that works for linear separable data
that only depends on the inner product?
15 / 125
16. Motivation R to R2
Case
R to R2
Case
How to separate two classes?
0
R
R2
ϕ(x) = (x, x2)
ϕ
Figure: Separating the two classes of points by tranforming the points into a
higher dimensional space where the data is separable.
16 / 125
17. Motivation R2
to R3
Case
R2
to R3
Case
+
+
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Figure: Data which is not linear separable.
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21. Cover’s Theorem Definitions
During this section we will consider X a finite subset of Rd
X = {x1, x2, . . . , xN } (1)
where N a fixed natural number and xi in Rd for all 1 ≤ i ≤ N
21 / 125
22. Cover’s Theorem Definitions
Definition 2.1 (Homogenous Linear Threshold Function)
Consider a set of patterns represented by a set of vectors in a
d-dimensional Euclidean space Rd. A homogeneously linear threshold
function is defined in terms of a parameter vector w for every vector x
in Rd as
fw : Rd
→ {−1, 0, 1}
x → fw(x) =
1, If w, x > 0
0, If w, x = 0
−1, If w, x < 0
Note: The function fw can be written as fw(x) = sign( w, x ).
22 / 125
23. Cover’s Theorem Definitions
Thus every homogeneous linear threshold function naturally divides Rd
into two sets, the set of vectors x such that fw(x) = 1 and the set of
vectors x such that fw(x) = −1. These two sets are separated by the
hyperplane
H = {x ∈ Rd
| fw(x) = 0} = {x ∈ Rd
| w, x = 0} (2)
which is the (d − 1)-dimensional subspace orthogonal to the weight
vector w.
w w, x = 0
Figure: Some points of Rd
divided by an homogeneous linear threshold
function.
23 / 125
24. Cover’s Theorem Definitions
Definition 2.2 (Linearly Separable Dichotomies)
A dichotomy {X+, X−}, a binary partition1, of X is linearly separable
if and only if there exists a weight vector w in Rd and scalar b = 0 such
that
w, x > b, if x ∈ X+
w, x < b, if x ∈ X−
Definition 2.3 (Homogeneously Linearly Separable Dichotomies)
Let X be an arbitrary set of vectors in Rd. A dichotomy {X+, X−}, a
binary partition, of X is homogeneously linearly separable if and only if
there exists a weight vector w in Rd such that
w, x > 0, if x ∈ X+
w, x < 0, if x ∈ X−
1
X = X+
∪ X−
and X+
∩ X−
= ∅.
24 / 125
25. Cover’s Theorem Definitions
Definition 2.4 (Vectors in General Position)
Let X be an arbitrary set of vectors in Rd. A set of N vectors is in
general position in d-space if every subset of d or fewer vectors are
linearly independent.
Figure: Left: A set of vectors that are not in general position. Right: A set
of vectors that are in general position.
25 / 125
26. Cover’s Theorem Preliminaries to Cover’s Theorem
Lemma 2.5
Let X− and X+ subsets of Rd, and let y a point other than the origin
in Rd. Then the dichotomies {X+ ∪ {y}, X−} and {X+, X− ∪ {y}} are
both homogeneously linear separable if and only if {X+, X−} is
homogeneously linear separable by a (d − 1)-dimensional
subspace2containing y.
Proof.
Let W the set of separable vectors for {X+, X−} given by
W = w ∈ Rd
| w, x > 0, x ∈ X+
∧ w, x < 0, x ∈ X−
(3)
The set W can be rewritten as
W = w ∈ Rd
| w, x > 0, x ∈ X+
w ∈ Rd
| w, x < 0, x ∈ X−
(4)
2
(d − 1)−dimensional subspace is an hyperplane.
26 / 125
27. Cover’s Theorem Preliminaries to Cover’s Theorem
y
w1
w2
w∗
Figure: We construct a hyperplane passing thought y which vector weight is
w∗
= − w2, y w1 + w1, y w2.
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28. Cover’s Theorem Preliminaries to Cover’s Theorem
The dichotomy {X+ ∪ {y}, X−} is homogeneously separable if and only
if there is a vector w in W such that w, y > 0 and the dichotomy
{X+, X− ∪ {y}} is homogeneously linearly separable if and only if there
is a w in W such that w, y < 0.
If {X+ ∪ {y}, X−} and {X+, X− ∪ {y}} are homogeneously separable
by w1 and w2 respectively, then we can construct a w∗ as
w∗
= − w2, y w1 + w1, y w2 (5)
such that separates {X+, X−} by the hyperplane
H = {x ∈ Rd | w∗, x = 0} passing thought y. We affirm that y
belongs to H. Indeed,
w∗
, y = − w2, y w1 + w1, y w2, y
= − w2, y w1, y + w1, y w2, y
= 0
28 / 125
29. Cover’s Theorem Preliminaries to Cover’s Theorem
We affirm that w∗, x > 0 if x in X+. In fact, let x in X+ then
w∗
, x = − w2, y w1 + w1, y w2, x
= − w2, y
>0
w1, x
>0
+ w1, y
>0
w2, x
>0
> 0
then w∗, x > 0 for all x in X+.
We affirm that w∗, x < 0 if x in X−. In fact, let x in X− then
w∗
, x = − w2, y w1 + w1, y w2, x
= − w2, y
<0
w1, x
>0
+ w1, y
>0
w2, x
<0
< 0
then w∗, x < 0 for all x in X−.
We conclude that {X+, X−} is homogeneously separable by the vector
w∗.
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30. Cover’s Theorem Preliminaries to Cover’s Theorem
Conversely, if {X+, X−} is homogeneously linear separable by an
hypeplane containing y then there exists w∗ in W such that w∗, y = 0.
We affirm that W is an open set. In fact, the set W can be rewritten as
W =
x∈X+
{w ∈ Rd
| w, x > 0}
x∈X−
{w ∈ Rd
| w, x < 0}
(6)
and the complement of this set is
Wc
=
x∈X+
{w ∈ Rd
| w, x ≤ 0}
x∈X−
{w ∈ Rd
| w, x ≥ 0}
(7)
The sets {w ∈ Rd | w, x ≤ 0}, x ∈ X+ and
{w ∈ Rd | w, x ≥ 0}, x ∈ X− are clearly closed due to the continuity
of the inner product then the finite union of closed sets is closed so we
can conclude that the set Wc is closed therefore W is an open set.
30 / 125
31. Cover’s Theorem Preliminaries to Cover’s Theorem
y
w∗
− ǫy
w∗
+ ǫy
w∗
Figure: {X+
∪ {y}, X−
} and {X+
, X−
∪ {y}} are homogeneously linearly
separable by the vectors w∗
+ y and w∗
− y respectively.
Since W is open, there exists an > 0 such that w∗ + y and w∗ − y
are in W. Hence, {X+ ∪ {y}, X−} and {X+, X− ∪ {y}} are
homogeneously linearly separable by the vectors w∗ + y and w∗ − y
respectively. Indeed, 31 / 125
32. Cover’s Theorem Preliminaries to Cover’s Theorem
We will prove that {X+ ∩ {y}, X−} is homegenously linear separable
by w∗ + y.
We affirm that w∗ + y, y > 0. In fact,
w∗
+ y, y = w∗
, y
=0
+ y, y (8)
= y 2
(9)
> 0 (10)
Therefore, w∗ + y, y > 0. Hence, {X+ ∪ {y}, X−} is homogeneously
linearly separable by w∗ + y.
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33. Cover’s Theorem Preliminaries to Cover’s Theorem
We will prove that {X+, X− ∩ {y}} is homegenously linear separable
by w∗ − y. We affirm that w∗ − y, y < 0. In fact,
w∗
+ y, y = w∗
, y
=0
+ y, y (11)
= − y 2
(12)
< 0 (13)
Therefore, w∗ + y, y < 0. Hence, {X+, X− ∪ {y}} is homogeneously
linearly separable by w∗ − y.
33 / 125
34. Cover’s Theorem Preliminaries to Cover’s Theorem
Lemma 2.6
A dichotomy of X separable by w if and only if the projection of the set
X onto the (d − 1)-dimensional orthogonal subspace to y is separable.
Proof.
Exercise :) (Intuitively it works but I don’t have an algebraic proof yet.)
y
w
X+
X−
Figure: Projecting the sets X+
and X−
to the hyperplane orthogonal to the
hyperplane passing thought y.
34 / 125
35. Cover’s Theorem Preliminaries to Cover’s Theorem
Theorem 2.7 (Function-Counting Theorem)
There are C(N, d) homogeneously linearly separable dichotomies of N
points in general position in Euclidean d-space, where
C(N, d) =
2
d−1
k=0
N−1
k , if N > d + 1
2N , if N ≤ d + 1
(14)
Proof.
To proof the theorem, we will use induction on N and d. Let C(N, d)
be the number of homogeneously linearly separable dichotomies of the
set X = {x1, x2, . . . , xN }. The base induction step is true because
C(1, d) = 2 if d ≥ 1 and C(N, 1) = 2 if N ≥ 1. Now, let’s prove that
the theorem is true for N + 1 points. Consider a new point xN+1 such
that X ∪ {xN+1} is in general position and consider the C(N, d)
homogeneously linearly separable dichotomies {X+, X−} of X.
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36. Cover’s Theorem Preliminaries to Cover’s Theorem
Since {X+, X−} is separable, either {X+ ∪ {xN+1}, X−} or
{X+, X− ∪ {xN+1}}. However, both dichotomies are separable, by
lemma (2.5), if and only if exists a separating vector w for {X+, X−}
lying in the (d − 1)-dimensional subspace orthogonal to xN+1. A
dichotomy of X is separable by such a w if and only if the projection of
the set X onto the (d − 1)-dimensional orthogonal subspace to xN+1 is
separable. By the induction hypothesis there are C(N, d − 1) such
separable dichotomies. Hence
C(N + 1, d) = C(N, d)
Number of Homogeneously Linearly
separable dichotomies of N points
in general position in Euclidean d-space
+ C(N, d − 1)
Number of Homogeneously Linearly
separable dichotomies of N points
in general position d − 1-subspace
36 / 125
37. Cover’s Theorem Preliminaries to Cover’s Theorem
C(N + 1, d) = C(N, d) + C(N, d − 1)
= 2
d−1
k=0
N − 1
k
+
d−2
k=0
N − 1
k
= 2
N − 1
0
+
d−1
k=1
N − 1
k
+
N − 1
k − 1
= 2
N
0
+
d−1
k=1
N
k
= 2
d−1
k=0
N
k
37 / 125
39. Cover’s Theorem Cover’s Theorem
Two kinds of randomness are considered in the pattern recognition
problem:
The pattern are fixed in position but are classified independently
with equal probability into one of two categories.
The patterns themselves are randomly distributed in space, and
the desired dichotomization maybe random or fixed.
Suppose that the dichotomy of X = {x1, x2, . . . , xN } is chosen are
random with equal probability from the 2N equiprobable possible
dichotomies of X.
Let P(N, d) be the probability that the random dichotomy is linear
separable.
P(N, d) =
C(N, d)
2N
=
1
2
N−1
d−1
k=0
N−1
k , if N > d + 1
1, if N ≤ d + 1
(16)
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40. Cover’s Theorem Cover’s Theorem
Figure: Behaviour of the probability P(N, d) vs N
d+1 [12, p.46].
If N
d+1 ≤ 1 then P(N, d + 1) = 1.
If 1 < N
d+1 < 2 and d → ∞ then P(N, d + 1) → 1.
If N
d+1 = 2 then P(N, d + 1) = 1
2.
40 / 125
41. Cover’s Theorem Cover’s Theorem
Theorem 2.8 (Cover’s Theorem)
A complex pattern classification problem cast in a high-dimensional
space nonlinearly, is more likely to be linearly separable than in a
low-dimensional space.
41 / 125
42. References for Cover’s Theorem
References for Cover’s Theorem
Main Source:
[7] Thomas Cover. “Geometrical and Statistical properties of systems
of linear inequalities with applications in pattern recognition”. In:
IEEE Transactions on Electronic Computer (), pp. 326–334.
Minor Sources:
[12] Ke-Lin Du and M. N. S. Swamy. Neural Networks and Statistical
Learning. Springer Science & Business Media, 2013.
[19] Simon Haykin. Neural Networks and Learning Machines. Third
Edition. Pearson Prentice Hall, 2009.
[39] Bernhard Schlköpf and Alexander Smola. Learning with Kernels:
Support Vector Machines, Regularization, Optimization, and
Beyond. The MIT Press, 2001.
[49] Sergios Theodoridis. Machine Learning: A Bayesian and
Optimization Perspective. Academic Press, 2015.
42 / 125
44. Mercer’s Theorem Integral Operators
Theorem 6.1 (Teorema de Mercer)
Let k a continous function in [a, b] × [a, b] such that
b
a
b
a
k(t, s)f(s)f(t) ds dt ≥ 0 (17)
for all f in L2([a, b]), then, for all t and s in [a, b] the series
k(t, s) =
∞
j=1
λjϕj(t)ϕj(s)
converges absolutely and uniformly in the set [a, b] × [a, b].
44 / 125
45. Mercer’s Theorem Integral Operators
Integral Operators
Definition 6.2 (Integral Operador)
Let k a measurable function in the set [a, b] × [a, b], then the integral
operator K associated to the function k is defined by
K : Γ → Ω
f → (Kf)(t) :=
b
a
k(t, s)f(s) ds
where Γ and Ω are space of functions. This operator is well defined
whenever the integral exists.
45 / 125
46. Mercer’s Theorem Integral Operators
Theorem 6.3
Let k a measurable complex Lebesgue function in L2([a, b] × [a, b]) and
let K the integral operator associated to the function k defined by
K : L2
([a, b]) → L2
([a, b])
f → (Kf)(t) =
b
a
k(t, s)f(s) ds
then the following affirmations are hold
1. The integral exists.
2. The integral operator associated to k is well defined.
3. The integral operator associated to k is linear.
4. The integral operator associated to k is a bounded operator.
Skip Proof
46 / 125
47. Mercer’s Theorem Integral Operators
Proof.
1. The integral exists because for almost every s in [a, b] the functions
k(t, .) and f(.) are Lebesgue measurable functions in [a, b].
2. To proof that the integral operator K is well defined we have to
show that the image of the operator is contained in L2([a, b]). Indeed,
because k is in L2([a, b] × [a, b]) then
k 2
L2([a,b]×[a,b]) =
b
a
b
a
|k(t, s)|2
ds dt < ∞ (18)
on the other hand,
47 / 125
48. Mercer’s Theorem Integral Operators
Proof.
Kf 2
L2([a,b]) = Kf, Kf L2([a,b])
=
b
a
(Kf)(t)(Kf)(t) dt
=
b
a
b
a
k(t, s)f(s) ds
b
a
k(t, s)f(s) ds dt
=
b
a
b
a
k(t, s)f(s) ds
2
dt
≤
b
a
b
a
|k(t, s)f(s)| ds
2
dt
≤
b
a
b
a
|k(t, s)|2
ds
b
a
|f(s)|2
ds dt (D. C-S)
48 / 125
49. Mercer’s Theorem Integral Operators
=
b
a
|f(s)|2
ds
b
a
b
a
|k(t, s)|2
ds dt
= f 2
L2([a,b])
b
a
b
a
|k(t, s)|2
ds dt
= f 2
L2([a,b]) k 2
L2([a,b]×[a,b])
then
Kf 2
L2([a,b]) ≤ f 2
L2([a,b]) k 2
L2([a,b]×[a,b]) (19)
using the previous inequality (19), by eq. (18) and due to f in L2([a, b])
we conclude
49 / 125
50. Mercer’s Theorem Integral Operators
Kf 2
L2([a,b]) ≤ f 2
L2([a,b]) k 2
L2([a,b]×[a,b]) < ∞ (20)
therefore, the functions Kf is in L2([a, b]) and we can conclude that the
integral operator K is well defined.
3. Let α, β in R an f, g in L2([a, b]) then
K(αf + βg) =
b
a
[k(t, s)(αf(s) + βg(s))]ds
= α
b
a
k(t, s)f(s)ds + β
b
a
k(t, s)g(s)ds
= αK(f) + βK(g)
therefore the integral operator K is a linear operator.
50 / 125
51. Mercer’s Theorem Integral Operators
4. Due to (20) we have
Kf 2
L2([a,b]) ≤ f 2
L2([a,b])
b
a
b
a
|k(t, s)|2
ds dt
so that f L2([a,b]) = 0, then
Kf 2
L2([a,b])
f 2
L2([a,b])
≤
b
a
b
a
|k(t, s)|2
ds dt
then
Kf L2([a,b])
f L2([a,b])
≤
b
a
b
a
|k(t, s)|2
ds dt
1
2
51 / 125
52. Mercer’s Theorem Integral Operators
K = sup
f L2([a,b])=0
Kf L2([a,b])
f L2([a,b])
≤
b
a
b
a
|k(t, s)|2
ds dt
1
2
= k L2([a,b]×[a,b]) < ∞
in the last inequality using the equation (18) we can conclude that
K < ∞ so K is a bounded operator.
52 / 125
53. Mercer’s Theorem Integral Operators
Corollary 6.4
If k is a continuous measurable Lebesgue complex function in
[a, b] × [a, b] then the integral operator associated to k is in
L(L2([a, b]), L2([a, b])).
Proof.
As k is a continuous function then |k(t, s)| is a continuous function.
Moreover, every continuous function in a compact set [a, b] × [a, b] is
bounded then k en L2([a, b] × [a, b]).
53 / 125
54. Mercer’s Theorem Integral Operators
Lemma 6.5
Let ϕ1, ϕ2, . . . an orthonormal basis for L2([a, b]), the function defined
as Φij(s, t) = ϕi(s)ϕj(t), for all i, j in N is an orthonormal basis for
L2([a, b] × [a, b]).
Proof.
We affirm that the set B = { Φij | ∀i, j ∈ N } is orthonormal, in fact
Φjk, Φmn L2([a,b]×[a,b]) =
b
a
b
a
ϕj(s)ϕk(t)ϕm(s)ϕn(t) ds dt
=
b
a
b
a
ϕj(s)ϕk(t)ϕm(s) ϕn(t) ds dt
54 / 125
55. Mercer’s Theorem Integral Operators
Φjk, Φmn L2([a,b]×[a,b]) =
b
a
b
a
ϕj(s)ϕk(t)ϕm(s)ϕn(t) ds dt
=
b
a
b
a
ϕj(s)ϕk(t)ϕm(s) ϕn(t) ds dt
=
b
a
ϕj(s)ϕm(s) ds
b
a
ϕk(t)ϕn(t) dt
(T. Fubini)
= δjmδkn
where
δjmδkn =
1, if j = m ∧ k = n
0, in other case
(21)
therefore B is an orthonormal set.
55 / 125
56. Mercer’s Theorem Integral Operators
We affirm that B is a basis. To show that B is a basis we have to proof
if g is in L2 ([a, b] × [a, b]) and g, Φjk L2([a,b]×[a,b]) = 0, this implies that
g ≡ 0 almost everywhere this is because theorem ?? (2) then we can
conclude that B is an orthonormal basis for L2([a, b] × [a, b]). Indeed,
Let g in L2([a, b] × [a, b]), then
0 = g, Φjk L2([a,b]×[a,b]) =
b
a
b
a
g(s, t)ϕj(s) ϕk(t) ds dt
=
b
a
ϕj(s)
b
a
g(s, t)ϕk(t) dt
h
ds
=
b
a
ϕj(s) h ds
=
b
a
h ϕj(s) ds
= h, ϕj L2([a,b])
56 / 125
57. Mercer’s Theorem Integral Operators
then
h, ϕj L2([a,b]) = 0 (22)
where the function h is
h(s) =
b
a
g(s, t)ϕk(t) dt
the function h can be written in the following form
h(s) = g(s, .), ϕk L2([a,b]) , ∀k = 1, 2, . . . (23)
as the function h is orthonormal to every function ϕj this implies that
h ≡ 0 in almost every point s in [a, b] (theorem ?? (2)). By the
equation (23) and h ≡ 0 we can conclude that there is a set Ω which
measure is zero such that for all s which is not in Ω the function g(s, .)
is orthogonal to ϕk for all k = 1, 2, . . . therefore g(s, t) = 0 for all t and
each s which doesn’t belongs to Ω (theorem ?? (2)). Therefore
57 / 125
58. Mercer’s Theorem Integral Operators
b
a
b
a
|g(s, t)|2
dt ds = 0
so we conclude g ≡ 0 almost in everywhere point (t, s) in [a, b] × [a, b].
This proof that the set B is an orthonormal basis for L2([a, b]×[a, b]).
58 / 125
59. Mercer’s Theorem Integral Operators
Theorem 6.6
Let k a function defined in L2([a, b] × [a, b]) and let K the integral
operator associated to the function k defined as
K : L2
([a, b]) → L2
([a, b])
f → (Kf)(t) =
b
a
k(t, s)f(s) ds
then the adjoint opeator K∗ of the integral operator K is given by
(K∗
g)(t) =
b
a
k(s, t)g(s) ds
for all g in L2([a, b]).
59 / 125
60. Mercer’s Theorem Integral Operators
Proof.
Kf, g L2([a,b]) =
b
a
(Kf(t)) g(t) dt
=
b
a
b
a
k(t, s)f(s) ds g(t) dt
=
b
a
b
a
k(t, s)f(s)g(t) ds dt
=
b
a
b
a
k(t, s)f(s)g(t) dt ds (T. Fubini)
=
b
a
f(s)
b
a
k(t, s)g(t) dt ds
=
b
a
f(s)
b
a
k(t, s)g(t) dt ds
= f, K∗
g L2([a,b])
60 / 125
61. Mercer’s Theorem Integral Operators
where K∗g is defined by
K∗
g(s) :=
b
a
k(t, s)g(t) dt
is the auto-adjoint operator of K.
61 / 125
62. Mercer’s Theorem Integral Operators
Theorem 6.7
Let k a function in L2([a, b] × [a, b]) and let K the integral operator
associated to k defined as
K : L2
([a, b] × [a, b]) → L2
([a, b])
f → (Kf)(t) =
b
a
k(t, s)f(s) ds
then the integral operator K is a compact operator.
Skip Proof
62 / 125
63. Mercer’s Theorem Integral Operators
Proof.
During this proof we will write k, Φij instead of k, Φij L2([a,b]×[a,b]).
First of all, we will build a sequence of operator with finite range which
converges in norm to the integral operator K as follows:
Let ϕ1, ϕ2, . . . an orthonormal basis for L2 ([a, b]). Then, the functions
defined by
Φij(t, s) = ϕi(t)ϕj(s) ∀i, j = 1, 2, . . .
, by lemma 6.5 this functions form an orthonormal basis for
L2 ([a, b] × [a, b]).
The function k by the lemma ?? (2) can be written as
k(t, s) =
∞
i=1
∞
j=1
k, Φij Φij(t, s)
and we defined a sequence of functions {kn}∞
n=1, where the n-th
function is defined as
63 / 125
64. Mercer’s Theorem Integral Operators
kn(t, s) :=
n
i=1
n
j=1
k, Φij Φij(t, s)
then the sequence {k − kn}∞
n=1 converge to 0 in norm in
L2([a, b] × [a, b]) i.e.
lim
n→∞
k − kn L2([a,b]×[a,b]) = 0
which is equivalent in notation to
k − kn L2([a,b]×[a,b]) → 0 (24)
on the other hand, let Kn the integral operador associated to the
function kn defined in L2 ([a, b]) as
(Knf)(t) :=
b
a
kn(t, s)f(s) ds
Kn is a bounded operator
64 / 125
65. Mercer’s Theorem Integral Operators
(due to kn is a linear combination of functions in L2([a, b]), a vector
space, and by theorem (6.3) we can conclude that the operador is
linear and bounded) with finite range because Kn is in
span{ϕ1, . . . , ϕn}, in fact
(Knf)(t) =
b
a
kn(t, s)f(s) ds
=
b
a
n
i=1
n
j=1
k, Φij Φij(t, s)
f(s) ds
=
b
a
n
i=1
n
j=1
k, Φij Φij(t, s)f(s) ds
=
n
i=1
n
j=1
b
a
k, Φij Φij(t, s)f(s) ds
65 / 125
66. Mercer’s Theorem Integral Operators
=
n
i=1
n
j=1
b
a
k, Φij ϕi(t)ϕj(s)f(s) ds
=
n
i=1
n
j=1
ϕi(t)
b
a
k, Φij ϕj(s)f(s) ds
=
n
i=1
ϕi(t)
n
j=1
b
a
k, Φij ϕj(s)f(s) ds
=
n
i=1
ϕi(t)
b
a
n
j=1
k, Φij ϕj(s)f(s) ds
66 / 125
67. Mercer’s Theorem Integral Operators
=
n
i=1
b
a
n
j=1
k, Φij ϕj(s)f(s)
ds
αi
ϕi(t)
=
n
i=1
αiϕi(t)
where
αi =
b
a
n
j=1
k, Φij ϕj(s)f(s)
ds ∀1 ≤ i ≤ n
so Kn in span{ϕ1, . . . , ϕn} hence the operator Kn is an operator with
finite range.
67 / 125
68. Mercer’s Theorem Integral Operators
On the other hand, because the operador K is linear and bounded then
K ≤
b
a
b
a
|k(t, s)|2
ds dt
1
2
= k L2([a,b]×[a,b]) (25)
By the equation (25) applied to the operator K − Kn we have
K − Kn ≤ k − kn L2([a,b]×[a,b])
and by the equation (24) we have
K − Kn ≤ k − kn L2([a,b]×[a,b]) → 0
so we can conclude that
K − Kn → 0
and applying the theorem ?? (puesto Kn es un operador de rango
finito) to the last equation we can conclude that the operator K is a
compact operator.
68 / 125
69. Mercer’s Theorem Preliminaries to Mercer Theorem
Lemma 6.8
Let k a continuous complex function defined in [a, b] × [a, b] which holds
b
a
b
a
k(t, s)f(s)f(t) ds dt ≥ 0 (26)
for all f in L2([a, b]) then the following statements are hold
1. The integral operator associated to k is a positive operator.
2. The integral operator associated to k is an auto-adjoint operator.
3. The number k(t, t) is real for all t in [a, b].
4. The number k(t, t) holds k(t, t) ≥ 0, for all t in [a, b].
69 / 125
70. Mercer’s Theorem Preliminaries to Mercer Theorem
Lemma 6.9
If k is a continuous complex function in [a, b] × [a, b] then the function
h defined as follows
h(t) =
b
a
k(t, s)ϕ(s) ds (27)
is continuous in [a, b] for all ϕ in L2([a, b]).
70 / 125
71. Mercer’s Theorem Preliminaries to Mercer Theorem
Lemma 6.10
Let {fn}∞
n=1 a sequence of continous real functions in [a, b] such that
satisfies the next conditions:
1. f1(t) ≤ f2(t) ≤ f3(t) ≤ ... for all t in [a, b] ({fn}∞
n=1 is a
monotonous increasing sequence of functions).
2. f(t) = lim
n→∞
fn(t) is a continous function in [a, b].
and we define the set Fn as
Fn := { t | f(t) − fn(t) ≥ } , ∀n ∈ N
then
1. Fn+1 ⊂ Fn for all n in N.
2. The set Fn is closed.
3.
∞
n=1
Fn = ∅ .
71 / 125
72. Mercer’s Theorem Preliminaries to Mercer Theorem
Theorem 6.11 (Dini’s Theorem)
Let {fn}∞
n=1 a sequence of continous real functions in [a, b] such that
satisfies the next conditions:
1. f1(t) ≤ f2(t) ≤ f3(t) ≤ ... for all t ∈ [a, b] ({fn}∞
n=1 is a
monotonous increasing sequence of functions).
2. f(t) = lim
n→∞
fn(t) es continua en [a, b].
Then the sequence of functions {fn}∞
n=1 converges uniformently to the
function f in [a, b].
72 / 125
73. Mercer’s Theorem Mercer’s Theorem
Theorem 6.12 (Teorema de Mercer)
Let k a continous function in [a, b] × [a, b] such that
b
a
b
a
k(t, s)f(s)f(t) ds dt ≥ 0 (28)
for all f in L2([a, b]), then, for all t and s in [a, b] the series
k(t, s) =
∞
j=1
λjϕj(t)ϕj(s)
converges absolutely and uniformly in the set [a, b] × [a, b].
Skip Proof
73 / 125
74. Mercer’s Theorem Mercer’s Theorem
Proof.
Applying Cauchy-Schwarz inequality to the set of functions
λmϕm(t), λmϕm(t), . . . , λnϕn(t)
and
λmϕm(s), λmϕm(s), . . . , λnϕn(s)
we have
n
j=m
|λjϕj(t)ϕj(s)| ≤
n
j=m
λj|ϕj(t)|2
1
2
n
j=m
λj|ϕj(s)|2
1
2
(29)
Fixing t = t0 and by lemma ?? (5) applied to the series
n
j=m
λj|ϕj(t0)|2
given 2 > 0 implies the existence of an integer N such that for all n, m
n > m ≥ N satifies
74 / 125
75. Mercer’s Theorem Mercer’s Theorem
n
j=m
λj|ϕj(t0)ϕj(s)| ≤
n
j=m
λj|ϕj(t0)|2
1
2
<
n
j=m
λj|ϕj(s)|2
1
2
≤C
< C
, ∀s ∈ [a, b] where C2 = max
t∈[a,b]
k(t, t) and by Cauchy’s criteria for
uniform series we conclude that
∞
j=1
λjϕj(t)ϕj(s) converges absolutely
and uniformently in s for each t (t0 was arbitrary).
The next step is to prove that the series
∞
j=1
λjϕj(t)ϕj(s) converges to
k(t, s). Indeed, let ˜k(t, s) the function defined by
˜k(t, s) :=
∞
j=1
λjϕj(t)ϕj(s)
75 / 125
76. Mercer’s Theorem Mercer’s Theorem
and let the function f defined in L2 ([a, b]) and t = t0 fixed, the uniform
convergence of the series in s and the continuity of each function ϕj
(because ϕj is a continous function) implies that ˜k(t0, s) is continous as
a function of s. Moreover, Let
LHS =
b
a
k(t0, s) − ˜k(t0, s) f(s) ds
then
LHS =
b
a
k(t0, s)f(s) ds −
b
a
˜k(t0, s)f(s) ds
= (Kf)(t0) −
b
a
∞
j=1
λjϕj(t0)ϕj(s)
f(s) ds
= (Kf)(t0) −
b
a
∞
j=1
λjϕj(t0)ϕj(s)f(s)
ds
= (Kf)(t0) −
∞
j=1
b
a
λjϕj(t0)ϕj(s)f(s) ds
76 / 125
78. Mercer’s Theorem Mercer’s Theorem
Therefore, ˜k(t0, s) = k(t0, s) almost everywhere for s in [a, b]. As
˜k(t0, s) and k(t0, s) are continous then ˜k(t0, s) = k(t0, s) for all s in [a, b]
therefore ˜k(t0, .) = k(t0, .) and as t0 was arbitrary then ˜k ≡ k so that
k(t, s) = ˜k(t, s) =
∞
j=1
λjϕj(t)ϕj(s)
In particular, k(t, t) =
∞
j=1
λj|ϕj(t)|2 for all t in [a, b] and applying
Dini’s Theorem 6.11 to the functions
fn(t) =
n
j=1
λj|ϕj(t)|2
78 / 125
79. Mercer’s Theorem Mercer’s Theorem
({fn}∞
n=1 is a sequence of increasing monotone functions and {fn}∞
n=1
converges to the continous function k(t, t) pointwise) we can conclude
that the sequence of functions {fn}∞
n=1 converge uniformently in [a, b].
By definition of uniformently series there is a 2 > 0 which doesn’t
depends on t, there is an integer N such that for all n, m ≥ N we have
n
j=m
λj|ϕj(t)|2
< 2
, ∀t ∈ [a, b]
utilizing the relationship (29) and the lemma ?? (3) implies that
79 / 125
80. Mercer’s Theorem Mercer’s Theorem
n
j=m
λj|ϕj(t)ϕj(s)| ≤
n
j=m
λj|ϕj(t)|2
1
2
<
n
j=m
λj|ϕj(s)|2
1
2
≤C
< C
,∀(t, s) ∈ [a, b] × [a, b] where C2 = max
s∈[a,b]
k(s, s). Using Cauchy’s criteria
for series to the series
∞
j=1
λjϕj(t)ϕj(s) we conclude that this series
converges absolutely and uniformently in [a, b] × [a, b].
80 / 125
81. References for Mercer’s Theorem
References for Mercer’s Theorem
Main Sources:
[17] Israel Gohberg, Seymour Goldberg, and Marinus A. Kaashoek.
Basic Classes of Linear Operators. Birkhäuser, 2003.
[22] Harry Hochstadt. Integral Equations. Wiley, 1989.
Minor Sources:
[13] Nelson Dunford and Jacob T. Schwartz. Linear Opertors Part II:
Spectral Theory Self Adjoint Operators in Hilbert Space.
Interscience Publishers, 1963.
[30] James Mercer. “Functions of positive and negative type and their
connection with the theory of integral equations”. In:
Philosophical Transactions of the Royal Society (1909),
pp. 415–446.
[55] Stephen M. Zemyan. The Classical Theory of Integral Equations:
A Concise Treatment. Birkhauser, 2010.
81 / 125
83. Moore-Aronszajn Theorem Reproducing Kernel Hilbert Spaces
Reproducing Kernel
Definition 10.1 (Reproducing Kernel)
A function k defined by
k: E × E → C
(s, t) → k(s, t)
is a Reproducing Kernel of a Hilbert Space H if and only if
1. For all t in E, k(., t) is an element of H.
2. For all t in E and for all ϕ in H,
ϕ, k(., t) H = ϕ(t) (30)
The condition (30) is called Reproducing Property because the value of
the function ϕ in the point t is reproduced by the inner product of ϕ
with k(., t).
83 / 125
84. Moore-Aronszajn Theorem Reproducing Kernel Hilbert Spaces
Definition 10.2 (Reproducing Kernel Hilbert Space)
A Hilbert Space of complex functions which has a Reproducing Kernel
is called Reproducing Kernel Hilbert Space (RKHS).
Hilbert Space
Banach Space
Reproducing Kernel Hilbert Space (RKHS)
84 / 125
85. Moore-Aronszajn Theorem Reproducing Kernel Hilbert Spaces
Theorem 10.3
For all t and s in E the following property is hold
k(s, t) = k(., t), k(., s) H
Proof.
Let g a function defined by g(.) = k(., t). Due to k(., t) is a reproducing
kernel of H this implies that g(.) is an element of the Hilbert Space H.
Moreover, due to the reproducing property we have
g(s) = k(s, t) = g, k(., s) H = k(., t), k(., s) H
this shows that k(s, t) = k(., t), k(., s) .
85 / 125
86. Moore-Aronszajn Theorem Reproducing Kernel Hilbert Spaces
Examples of Reproducing Kernel Hilbert Spaces
A Finite Dimensional Example
Theorem 10.4
Let β = {e1, e2, . . . , en} an orthonormal basis of H and let define the
function k as follows
k: E × E → C
(s, t) → k(s, t) =
n
i=1
ei(s)ei(t)
then k is a reproducing kernel.
86 / 125
87. Moore-Aronszajn Theorem Reproducing Kernel Hilbert Spaces
Proof.
For all t in E, we have
k(., t) =
n
i=1
ei(t)ei(.)
belongs to H (this is due to k(., t) is a linear combination of elements of
the basis β). On the other hand, for all function ϕ of H we have
ϕ(.) =
n
i=1
λiei(.)
then
87 / 125
88. Moore-Aronszajn Theorem Reproducing Kernel Hilbert Spaces
Proof.
ϕ, k(., t) H =
n
i=1
λiei(.),
n
i=1
ei(t)ei(.)
H
=
n
i=1
λi ei(.),
n
i=1
ei(t)ei(.)
H
=
n
i=1
n
j=1
λiei(t) ei, ej H
=1
=
n
i=1
λiei(t) = ϕ(t), ∀t ∈ E
88 / 125
89. Moore-Aronszajn Theorem Reproducing Kernel Hilbert Spaces
Corollary 10.5
Every finite dimensional Hilbert Space H has a reproducing Kernel.
Proof.
Let β = {v1, . . . , vn} a basis for the Hilbert Space H. Using the
Gram-Schmidt process on the set β we can build an orthonormal basis
ˆβ = { ˆv1, . . . , ˆvn}. Using the previous theorem we on this new basis ˆβ
conclude that
k: E × E → C
(s, t) → k(s, t) =
n
i=1
vi(s)vi(t)
is a Reproducing Kernel for H.
89 / 125
90. Moore-Aronszajn Theorem Reproducing Kernel Hilbert Spaces
For every t in E, we define the functional evalutation operator et of g in
the point t as the application
et : H → C
g → et(g) = g(t)
g(t)
t
g
Figure: The functional evaluation et associated to any function g is the value
g(t) in the point t. 90 / 125
91. Moore-Aronszajn Theorem Reproducing Kernel Hilbert Spaces
Theorem 10.6
A Hilbert spaces of complex function in E has a reproducing kernel if
and only if all the functional evaluations et, t in E, are continous in H.
91 / 125
92. Moore-Aronszajn Theorem Reproducing Kernel Hilbert Spaces
Corollary 10.7
Let H an RKHS then all sequence which converges in norm converge
pointwise to the same limit.
92 / 125
93. Moore-Aronszajn Theorem Reproducing Kernel Hilbert Spaces
Definition 10.8 (Semidefinite positive function)
A function k : E × E → C is called semidefinite positive or positive
type function if
∀n ≥ 1, ∀(a1, . . . , an) ∈ Cn
, ∀(x1, . . . , xn) ∈ En
,
n
i=1
n
j=1
aiajk(xi, xj) ≥ 0
(31)
93 / 125
94. Moore-Aronszajn Theorem Reproducing Kernel Hilbert Spaces
Lemma 10.9
Let H a Hilbert Space with inner product ., H (Not necesary an
RKHS) and let ϕ : E → H, then, the function k defined as
k : E × E → C
(x, y) → k(x, y) = ϕ(x), ϕ(y) H
is a semidefinite positive function.
94 / 125
96. Moore-Aronszajn Theorem Reproducing Kernel Hilbert Spaces
Lemma 10.11
Let L a semdefinite positive function in E × E, then,
1. For all x in E
L(x, x) ≥ 0
2. For all (x, y) in E × E holds
L(x, y) = L(y, x)
3. The function L is semidefinite positive.
4. |L(x, y)|2 ≤ L(x, x)L(y, y).
96 / 125
97. Moore-Aronszajn Theorem Reproducing Kernel Hilbert Spaces
Lemma 10.12
A real function L defined on E × E is a semidefinite positive function if
and only if
1. The L the function is symetric.
2. ∀n ≥ 1, ∀(a1, a2, . . . , an) ∈ Rn, ∀(x1, x2, . . . , xn) ∈ En,
n
i=1
n
j=1
aiajk(xi, xj) ≥ 0
97 / 125
98. Moore-Aronszajn Theorem Moore-Aronszajn Theorem
Definition 10.13 (pre-RKHS)
A space which satifies the following properties.
1. Every the evaluation functionals et are continous in H0.
2. Toda sucesión de Cauchy {fn}∞
n=1 en H0 que converge
puntualmente a 0 también converge en norma a 0 en H0.
is called a pre-RKHS with reproducing kernel.
98 / 125
99. Moore-Aronszajn Theorem Moore-Aronszajn Theorem
Theorem 10.14
Let H0 a subset of CE, the space of complex functions in E, with inner
product ., . H0 and associated norm . H0 then the hilbert space H with
the following properties
1. H0 ⊂ H ⊂ CE and the topology induced by ., . H0 en H0 concide
with the topology induced H0 by H.
2. H has a reproducing kernel k.
exists if and only if
1. All the functional evaluatins et are continous in H0.
2. Every Cauchy sequence {fn}∞
n=1 in H0 which converge pointwise to
0 converge to 0 in norm.
99 / 125
101. Moore-Aronszajn Theorem Moore-Aronszajn Theorem
Theorem 10.15 (Moore-Aronszajn Theorem)
Let k a semidefinite positive function in E × E then exits a unique
Hilbert Space H of functions in E with reproducing kernel k such that
the subspace H0 of H defined as
H0 = span{k(., t) | t ∈ E}
is dense in H and H is a set of functions in E which is the limit of
inner products of Cauchy sequences in H0.
101 / 125
102. References for Moore-Aronszajn Theorem
References for Moore-Aronszajn Theorem
Main Sources:
[4] Alain Berlinet and Christine Thomas. Reproducing kernel Hilbert
spaces in Probability and Statistics. Kluwer Academic Publishers,
2004.
[42] D. Sejdinovic and A. Gretton. Foundations of Reproducing Kernel
Hilbert Space I. url:
http://www.stats.ox.ac.uk/~sejdinov/RKHS_Slides1.pdf
(visited on 03/11/2012).
[43] D. Sejdinovic and A. Gretton. Foundations of Reproducing Kernel
Hilbert Space II. url: http://www.gatsby.ucl.ac.uk/
~gretton/coursefiles/RKHS_Slides2.pdf (visited on
03/11/2012).
[44] D. Sejdinovic and A. Gretton. What is an RKHS? url:
http://www.gatsby.ucl.ac.uk/~gretton/coursefiles/RKHS_
Notes1.pdf (visited on 03/11/2012).
102 / 125
104. Kernel Trick Definitions
Definition 13.1 (Kernel)
Let X a non-empty set. A function k : X × X → K is called kernel in
X if and only if there is Hilbert Space H and a mapping Φ : X → H
such that for all s, t it holds
k(t, s) := Φ(t), Φ(s) H (32)
The function Φ is called feature mapping and H feature space of k.
104 / 125
105. Kernel Trick Definitions
Example 13.2
Consider X = R and the function k defined by
k(s, t) = st = s√
2
s√
2
,
t√
2
t√
2
where the feature mappings are Φ(s) = s and ˜Φ(s) = s√
2
s√
2
and the
features spaces are H = R and ˜H = R2 respectly.
105 / 125
106. Kernel Trick Feature Space based on Mercer’s Theorem
Feature Space based on Mercer’s Theorem
The Mercer’s theorem allows to define a feature mapping for the kernel
k as follows
k(t, s) =
∞
j=1
λjϕj(t)ϕj(s)
= λjϕj(t)
∞
j=1
, λjϕj(s)
∞
j=1 2([a,b])
we can take 2([a, b]) as the feature space.
106 / 125
107. Kernel Trick Feature Space based on Mercer’s Theorem
Theorem 13.3
The application Φ defined as
Φ : [a, b] → 2
([a, b])
t → λjϕj(t)
∞
j=1
es well defined and satifies
k(t, s) = Φ(t), Φ(s) 2([a,b]) (33)
107 / 125
108. Kernel Trick Feature Space based on Mercer’s Theorem
Theorem 13.4 (Mercer Representation of RKHS)
Let X a compact metric space and k : X × X → R a continous kernel.
We defined the set H as
H =
f ∈ L2
(X) f =
∞
j=1
ajϕj where
aj
λj
∞
j=1
∈ 2
(34)
with inner product
∞
j=1
ajϕj,
∞
j=1
bjϕj
H
=
∞
j=1
ajbj
λj
(35)
then H is a RKHS with reproducing kernel k.
108 / 125
110. History
Timeline
Table: Timeline of Support Vector Machines Algorithm Development
1909 • Mercer Theorem — James Mercer.
"Functions of Positive and Negative Type, and their Connection with the
Theory of Integral Equations".
1950 • "Moore-Aronzajn Theorem" — Nachman Aronszajn.
"Reproducing Kernel Hilbert Spaces".
1964 • Introduced the geometrical interpretation of the kernels as
inner products in a feature space — Aizerman, Braverman
and Rozonoer.
"Theoretical Foundations of the Potential Function Method in Pattern
Recognition Learning".
1964 • Original SVM algorithm — Vladimir Vapnik and Alexey
Chervonenkis.
"A Note on One Class of Perceptrons"
110 / 125
111. History
Timeline
Table: Timeline of Support Vector Machines Algorithm Development
1965 • Cover’s Theorem — Thomas Cover.
"Geometrical and Statistical Properties of Systems of Linear Inequalities
with Applications in Pattern Recognition".
1992 • Support Vector Machines — Bernhard Boser, Isabelle
Guyon and Vladimir Vapnik.
"A Training Algorithm for Optimal Margin Classifiers".
1995 • Soft Support Vector Machines — Corinna Cortes and
Vladimir Vapnik.
"Support Vector Networks".
111 / 125
113. References
References I
[1] Yaser S. Abu-Mostafa, Malik Magdon-Ismail, and
Hsuan-Tien Lin. Learning From Data: A short course. AML
Book, 2012.
[2] Nachman Aronszajn. “Theory of Reproducing Kernels”. In:
Transactions of the American Mathematical Society 68 (1950),
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