In this section, we give certain theoretical analysis for our proposed multidividing ontology algorithm. Let
κ
=
sup
v
∈
V
K
(
v
,
v
)
and
Diam
(
V
)
=
sup
v
,
v
'
∈
V

v

v
'

. We divide this section into two parts: first, some useful lemmas are prepared; then, main results in our paper concerning approximation conclusions are presented. Our error analysis depends on integral operators and gradient learning, and more references on these tricks can be referred to in Mukherjee and Wu [18], Mukherjee et al. [19], Yao et al. [20], and Rosasco et al. [21].
4.1. Preliminary Results
Let sequence
{
f
→
t
}
t
∈
N
be the noisefree limit of the sequence (7) which is determined by
f
→
1
=
0
and
(10)
f
→
t
+
1
=
f
→
t

η
t
∑
a
=
1
k

1
∑
b
=
a
+
1
k
∫
Z
a
∫
Z
b
w
(
v
a

v
v
)
×
y
a

y
b
+
f
→
t
v
·
v
b

v
a
×
v
b

v
a
K
v
d
ρ
v
a
,
y
a
d
ρ
v
b
,
y
b

η
t
λ
t
f
→
t
.
Our error analysis for proving main result (Theorems 12 and 13 in the next subsection) consists of two parts: sample error and approximation error.
The main task in this subsection is to estimate the sample error
f
→
t
z

f
→
t
in terms of McDiarmidBernsteintype probability inequality and the multidividing sampling operator. For each
1
≤
a
≤
k
, the multidividing sampling operator
S
v
a
:
H
K
n
→
R
m
a
n
associated with a discrete subset
v
a
=
{
v
i
a
}
i
=
1
m
a
of
V
is defined by
(11)
S
v
a
(
f
→
)
=
f
→
v
i
a
i
=
1
m
a
=
f
→
v
1
a
,
f
→
v
2
a
,
…
,
f
→
v
m
a
a
T
.
The adjoint of the multidividing ontology sampling operator,
(
S
v
a
)
T
:
R
m
a
n
→
H
K
n
, is given by
(12)
S
v
a
T
(
c
)
=
∑
i
=
1
m
a
c
i
a
K
v
i
a
,
where
(13)
c
=
c
i
i
=
1
m
a
=
c
1
,
c
2
,
…
,
c
m
a
T
∈
R
m
a
n
.
Let us express (7) by virtue of the multidividing ontology sampling operator. Note that
(14)
f
→
t
z
v
i
a
·
v
j
b

v
i
a
v
j
b

v
i
a
=
(
v
j
b

v
i
a
)
v
j
b

v
i
a
T
f
→
t
z
(
v
i
a
)
=
(
v
j
b

v
i
a
)
v
j
b

v
i
a
T
S
v
f
→
t
z
i
a
.
For each pair of
(
a
,
b
)
with
1
≤
a
<
b
≤
k
, we single out one summation
∑
j
=
1
m
b
from (7) as
(15)
B
i
a
,
b
=
∑
j
=
1
m
b
w
i
a
,
j
b
(
v
j
b

v
i
a
)
v
j
b

v
i
a
T
∈
R
n
×
n
,
Y
i
a
,
b
=
∑
j
=
1
m
b
w
i
a
,
j
b
y
j
b

y
i
a
v
j
b

v
i
a
T
∈
R
n
.
We infer that
(16)
f
→
t
+
1
z
=
1

η
t
λ
t
f
→
t
z

η
t
∑
a
=
1
k

1
∑
b
=
a
+
1
k
m
a
m
b
×

∑
i
=
1
m
a
∑
j
=
1
m
b
Y
i
a
,
b
K
v
i
a
+
∑
i
=
1
m
a
∑
j
=
1
m
b
K
v
i
a
B
i
a
,
b
f
→
t
z
v
i
a
.
Denote
(17)
D
v
a
a
,
b
=
diag
{
B
1
a
,
b
,
B
2
a
,
b
,
…
,
B
m
a
a
,
b
}
∈
R
m
a
n
×
m
a
n
Y
→
a
a
,
b
=
Y
1
a
,
b
,
Y
2
a
,
b
,
…
,
Y
m
a
a
,
b
T
∈
R
m
a
n
.
Hence, we have
(18)
f
→
t
+
1
z
=
1

η
t
λ
t
f
→
t
z
+
η
t
∑
a
=
1
k

1
∑
b
=
a
+
1
k
m
a
m
b
×
∑
a
=
1
k

1
∑
b
=
a
+
1
k
S
v
a
T
Y
→
a
a
,
b
T

η
t
∑
a
=
1
k

1
∑
b
=
a
+
1
k
m
a
m
b
×
∑
a
=
1
k

1
∑
b
=
a
+
1
k
S
v
a
T
D
v
a
a
,
b
S
v
a
f
→
t
z
.
Thus, it confirms the following representation for the sequence
{
f
→
t
z
}
. For simplicity, let
∏
q
=
t
+
1
t
(
I

L
v
,
q
)
=
I
in the following contents.
Lemma 5.
Set
(19)
L
v
,
t
=
η
t
∑
a
=
1
k

1
∑
b
=
a
+
1
k
m
a
m
b
∑
a
=
1
k

1
∑
b
=
a
+
1
k
S
v
a
T
D
v
a
a
,
b
S
v
a
+
η
t
λ
t
I
.
If
{
f
→
t
z
}
is defined by (7), we deduce
(20)
f
→
t
z
=
Π
i
=
1
t

1
I

L
v
,
i
f
→
1
z
+
∑
i
=
1
t

1
∏
q
=
i
+
1
t

1
I

L
v
,
q
η
t
∑
a
=
1
k

1
∑
b
=
a
+
1
k
m
a
m
b
×
∑
a
=
1
k

1
∑
b
=
a
+
1
k
S
v
a
T
Y
→
a
a
,
b
T
.
We should discuss the convergence of the multidividing ontology operator
(21)
1
∑
a
=
1
k

1
∑
b
=
a
+
1
k
m
a
m
b
∑
a
=
1
k

1
∑
b
=
a
+
1
k
S
v
a
T
D
v
a
a
,
b
S
v
a
f
→
t
z
to the integral operator
L
K
,
s
:
H
K
n
→
H
K
n
determined by
(22)
L
K
,
s
f
→
=
∑
a
=
1
k

1
∑
b
=
a
+
1
k
∫
V
a
∫
V
b
w
(
v
a

v
b
)
(
v
b

v
a
)
v
b

v
a
T
·
f
→
v
a
K
v
a
d
ρ
V
a
v
a
d
ρ
V
b
v
b
∫
V
a
∫
V
b
,
where
f
→
∈
H
K
n
.
Lemma 6.
Let
z
=
{
z
1
,
z
2
,
…
,
z
k
}
be multidividing sample set independently drawn according to a probability distribution
ρ
on
Z
. Denote
(
H
,
·
)
as a Hilbert space and suppose that
F
:
Z
m
1
×
m
2
×
⋯
×
m
k
→
H
is measurable. If there is nonnegative
M
~
such that
F
(
z
)

E
z
(
F
(
v
)
)
≤
M
~
for each
v
∈
z
and almost every
z
∈
Z
m
1
×
m
2
×
⋯
×
m
k
, then for every
ɛ
>
0
,
(23)
P
z
∈
Z
m
1
×
m
2
×
⋯
×
m
k
F
z

E
z
F
z
≥
ɛ
≤
2
exp
{

ɛ
2
2
(
M
~
ɛ
+
σ
2
)
}
,
where
(24)
σ
2
=
∑
a
=
1
k
∑
i
=
1
m
a
sup
z
∖
{
v
i
a
}
∈
Z
m
1
×
m
2
×
⋯
×
(
m
a

1
)
×
⋯
×
m
k
E
v
i
{
F
(
z
)

E
v
i
(
F
(
z
)
)
2
}
.
For any
0
<
δ
<
1
, with confidence
1

δ
, one gets
(25)
F
z

E
z
F
z
≤
4
log
4
δ
M
~
+
σ
2
≤
4
(
1
+
m
Π
∑
i
=
1
k
m
Π
i
)
log
4
δ
M
~
.
By regarding
1
/
∑
a
=
1
k

1
∑
b
=
a
+
1
k
m
a
m
b
{
∑
a
=
1
k

1
∑
b
=
a
+
1
k
(
S
v
a
)
T
(
D
v
a
)
a
,
b
S
v
a
(
f
→
t
z
)
}
and
L
K
,
s
as elements in
(
L
(
H
K
n
)
and
∥
·
∥
L
(
H
K
n
)
, the space of bounded linear multidividing ontology operators on
H
K
n
, Lemma 6 cannot be directly employed because
L
(
H
K
n
)
is not a Hilbert space, but a Banach space only. Therefore, we consider a subspace of
L
(
H
K
n
)
,
H
S
(
H
K
n
)
which is the space of HilbertSchmidt operators on
H
K
n
with inner product
A
,
B
H
S
(
H
K
n
)
=
T
r
(
B
T
A
)
. As
H
S
(
H
K
n
)
is a subspace of
L
(
H
K
n
)
, their norm relations are presented as
(26)
A
L
(
H
K
n
)
≤
A
H
S
(
H
K
n
)
,
A
B
H
S
(
H
K
n
)
≤
A
H
S
(
H
K
n
)
B
H
S
(
H
K
n
)
.
In addition,
H
S
(
H
K
n
)
is a Hilbert space and contains multidividing ontology operators
L
K
,
s
and
1
/
∑
a
=
1
k

1
∑
b
=
a
+
1
k
m
a
m
b
{
∑
a
=
1
k

1
∑
b
=
a
+
1
k
(
S
v
a
)
T
(
D
v
a
)
a
,
b
S
v
a
(
f
→
t
z
)
}
. By applying Lemma 6 to this Hilbert space, we obtain the following lemma.
Lemma 7.
Let
v
=
{
v
1
,
v
2
,
…
,
v
k
}
be multidividing sample set independently drawn from
(
V
,
ρ
V
)
. With confidence
1

δ
, one obtains
(27)
1
∑
a
=
1
k

1
∑
b
=
a
+
1
k
S
v
a
,
b
T
D
v
S
v
a
,
b

L
K
,
s
H
S
H
K
n
≤
34
n
κ
2
Diam
V
2
m
Π
/
∑
i
=
1
k
m
Π
i
s
n
+
2
log
4
δ
.
Proof.
Let
H
=
H
S
(
H
K
n
)
. Consider the multidividing ontology function
F
:
V
m
1
×
m
2
×
⋯
×
m
k
→
H
with values in
H
=
H
S
(
H
K
n
)
defined by
(28)
F
(
v
)
=
1
∑
a
=
1
k

1
∑
b
=
a
+
1
k
m
a
m
b
∑
a
=
1
k

1
∑
b
=
a
+
1
k
S
v
a
T
D
v
a
a
,
b
S
v
a
.
For
f
→
∈
H
K
n
, we confirm that
(29)
F
v
f
→
=
1
∑
a
=
1
k

1
∑
b
=
a
+
1
k
∑
i
=
1
m
a
∑
j
=
1
m
b
w
v
i
a

v
j
b
v
j
b

v
i
a
×
v
j
b

v
i
a
T
f
→
(
v
i
a
)
K
v
i
a
.
Recall that reproducing property of the RKHS
H
K
says that
(30)
f
(
v
)
=
f
,
K
v
K
,
∀
v
∈
V
,
f
∈
H
K
.
It implies that the rank of operator
A
v
:
H
K
→
H
K
determined by
A
v
(
f
)
=
f
(
v
)
K
v
=
f
,
K
v
K
K
v
is 1, and also in
H
S
(
H
K
)
. Furthermore,
∥
A
v
∥
H
S
(
H
K
)
=
K
(
v
,
v
)
. Let
A
→
v
be the operator on
H
K
n
which maps
f
→
to
f
→
(
v
)
K
v
. Then the above fact reveals that
A
→
v
H
S
(
H
K
n
)
≤
K
(
v
,
v
)
n
. Hence for any
v
∈
V
m
1
×
m
2
×
⋯
×
m
k
, we infer that
(31)
F
v
=
1
∑
a
=
1
k

1
∑
b
=
a
+
1
k
∑
i
=
1
m
a
∑
j
=
1
m
b
w
v
i
a

v
j
b
v
j
b

v
i
a
×
v
j
b

v
i
a
T
A
→
v
i
a
∈
H
S
(
H
K
n
)
.
Using the fact that
w
(
v
)
≤
1
/
s
n
+
2
and
A
→
v
H
S
(
H
K
n
)
≤
n
K
(
v
,
v
)
≤
n
κ
2
, we deduce that
(32)
F
v

E
v
i
F
v
H
S
H
K
n
≤
4
(
m
Π
/
∑
i
=
1
k
m
Π
i

1
)
κ
2
Diam
V
2
n
m
Π
/
∑
i
=
1
k
m
Π
i
2
s
n
+
2
.
Since
(33)
E
v
1
∑
a
=
1
k

1
∑
b
=
a
+
1
k
m
a
m
b
∑
a
=
1
k

1
∑
b
=
a
+
1
k
S
v
a
T
D
v
a
a
,
b
S
v
a
=
E
v
(
F
(
v
)
)
=
m
Π
/
∑
i
=
1
k
m
Π
i
m
Π
/
∑
i
=
1
k
m
Π
i

1
L
K
,
s
,
the stated result is held by combining Lemma 6 with
(34)
M
~
=
Diam
V
2
κ
2
n
8
(
m
Π
/
∑
i
=
1
k
m
Π
i

1
)
m
Π
/
∑
i
=
1
k
m
Π
i
2
s
n
+
2
and using the bound
L
K
,
s
H
S
(
H
K
n
)
≤
κ
2
n
(
Diam
(
V
)
)
2
/
s
n
+
2
.
In order to find the difference between
f
→
t
z
and
f
→
t
, the convergence of
(35)
1
∑
a
=
1
k

1
∑
b
=
a
+
1
k
m
a
m
b
∑
a
=
1
k

1
∑
b
=
a
+
1
k
S
v
a
T
Y
→
a
a
,
b
T
to the ontology function defined by (55) is studied.
Lemma 8.
Let
z
be a multidividing ontology sample independently drawn from
(
Z
,
ρ
)
. With confidence
1

δ
, one has
(36)
1
∑
a
=
1
k

1
∑
b
=
a
+
1
k
m
a
m
b
∑
a
=
1
k

1
∑
b
=
a
+
1
k
S
v
a
T
Y
→
a
a
,
b
T

f
→
ρ
,
s
H
K
n
≤
68
Diam
(
V
)
M
κ
m
Π
/
∑
i
=
1
k
m
Π
i
s
n
+
2
log
4
δ
.
Proof.
By applying Lemma 6 to the Hilbert space
H
=
H
K
n
and the ontology function
F
:
Z
m
1
×
m
2
×
⋯
×
m
k
→
H
K
n
given by
(37)
F
z
=
1
∑
a
=
1
k

1
∑
b
=
a
+
1
k
m
a
m
b
∑
a
=
1
k

1
∑
b
=
a
+
1
k
S
v
a
T
Y
→
a
a
,
b
T
=
1
∑
a
=
1
k

1
∑
b
=
a
+
1
k
∑
i
=
1
m
a
∑
j
=
1
m
b
w
v
i
a

v
j
b
v
j
b

v
i
a
v
j
b

v
i
a
K
v
i
a
,
we yield
E
z
(
F
(
z
)
)
=
(
m
Π
/
∑
i
=
1
k
m
Π
i

1
/
m
Π
/
∑
i
=
1
k
m
Π
i
)
f
→
ρ
,
s
. Hence, for almost every
z
∈
Z
m
1
×
m
2
×
⋯
×
m
k
, we get
(38)
F
z

E
z
i
F
z
H
K
n
≤
16
M
κ
Diam
(
V
)
(
m
Π
/
∑
i
=
1
k
m
Π
i

1
)
m
Π
/
∑
i
=
1
k
m
Π
i
2
s
n
+
2
.
Lemma 6 implies that for any
0
<
δ
<
1
, with confidence
1

δ
, we obtain
(39)
1
∑
a
=
1
k

1
∑
b
=
a
+
1
k
m
a
m
b
∑
a
=
1
k

1
∑
b
=
a
+
1
k
S
v
a
T
Y
→
a
a
,
b
T

m
Π
/
∑
i
=
1
k
m
Π
i

1
m
Π
/
∑
i
=
1
k
m
Π
i
f
→
ρ
,
s
1
∑
a
=
1
k

1
∑
b
=
a
+
1
k
m
a
m
b
∑
a
=
1
k

1
∑
b
=
a
+
1
k
S
v
a
T
Y
→
a
a
,
b
T
H
K
n
≤
32
1
+
1
/
m
Π
/
∑
i
=
1
k
m
Π
i
M
κ
Diam
(
V
)
m
Π
/
∑
i
=
1
k
m
Π
i
s
n
+
2
log
4
δ
.
Finally, conclusion follows from the fact that
f
→
ρ
,
s
H
K
n
≤
4
Diam
(
V
)
M
κ
/
s
n
+
2
.
Obviously, for
{
f
→
t
z
}
, the sequence
{
f
→
t
}
has a similar expression as (20).
Lemma 9.
Let
L
K
,
λ
i
,
η
i
=
η
i
L
K
,
s
+
η
i
λ
i
I
be an ontology operator on
H
K
n
and suppose that
∏
q
=
i
+
1
t

1
(
I

L
K
,
λ
a
k
,
η
k
)
=
I
. For the ontology operator
L
K
,
s
determined by (22) and
{
f
→
t
}
by (10), one obtains
(40)
f
→
t
=
∏
i
=
1
t

1
(
I

L
K
,
λ
i
,
η
i
)
f
→
1
+
∑
i
=
1
t

1
∏
q
=
i
+
1
t

1
(
I

L
K
,
λ
k
,
η
k
)
η
i
f
→
ρ
,
s
.
The sample error
f
→
t
z

f
→
t
H
K
n
is stated in the following conclusion.
Theorem 10.
Let
{
f
→
t
z
}
be obtained by (5) and
{
f
→
t
}
by (10). Suppose that
η
i
≤
1
and
λ
i
+
1
≤
λ
i
≤
1
for all
i
∈
N
. Then for any
0
<
δ
<
1
, with confidence
1

δ
, one infers that
(41)
f
→
t
z

f
→
t
H
K
n
≤
34
Diam
V
κ
m
Π
/
∑
i
=
1
k
m
Π
i
λ
t

1
2
s
n
+
2
×
κ
n
Diam
V
+
4
λ
t

1
M
log
8
δ
.
Proof.
Let
(42)
f
→
ρ
,
t
z
=
∑
i
=
1
t

1
∏
q
=
i
+
1
t

1
(
I

L
v
,
k
)
η
i
f
→
ρ
,
s
+
∏
i
=
1
t

1
(
I

L
v
,
i
)
f
→
1
z
.
Let
Z
1
⊆
Z
m
1
×
m
2
×
⋯
×
m
k
with measure at least
1

δ
such that (36) establishes for any
z
∈
Z
1
. Thus, from the positivity of the multidividing ontology operator
(
S
v
a
)
T
(
D
v
a
)
a
,
b
S
v
a
(for each pair of
(
a
,
b
)
) on
H
K
n
and the assumption
∏
q
=
t
+
1
t

1
(
1

η
q
λ
q
)
=
1
, we have that for any
z
∈
Z
1
,
(43)
f
→
t
z

f
→
ρ
,
t
z
H
K
n
=
∑
a
=
1
k

1
∑
b
=
a
+
1
k
S
v
a
T
Y
→
a
a
,
b
T

f
→
ρ
,
s
∑
i
=
1
t

1
∏
q
=
i
+
1
t

1
I

L
v
,
q
η
i
×
∑
a
=
1
k

1
∑
b
=
a
+
1
k
S
v
a
T
Y
→
a
a
,
b
T
1
∑
a
=
1
k

1
∑
b
=
a
+
1
k
m
a
m
b
×
∑
a
=
1
k

1
∑
b
=
a
+
1
k
S
v
a
T
Y
→
a
a
,
b
T

f
→
ρ
,
s
L
H
K
n
≤
∑
i
=
1
t

1
∏
q
=
i
+
1
t

1
I

L
v
,
k
L
H
K
n
68
Diam
V
M
κ
m
Π
/
∑
i
=
1
k
m
Π
i
s
n
+
2
log
4
δ
≤
68
Diam
(
V
)
M
κ
m
Π
/
∑
i
=
1
k
m
Π
i
s
n
+
2
log
4
δ
∑
i
=
1
t

1
∏
q
=
i
+
1
t

1
(
1

η
q
λ
q
)
η
i
.
In terms of
η
i
λ
i
=
1

(
1

η
i
λ
i
)
and
1
≤
λ
i
λ
t

1

1
, we get
(44)
∑
i
=
1
t

1
∏
q
=
i
+
1
t

1
1

η
q
λ
q
η
i
≤
1
λ
t

1
∑
i
=
1
t

1
∏
q
=
i
+
1
t

1
1

η
q
λ
q

∑
i
=
1
t

1
∏
q
=
i
t

1
1

η
q
λ
q
=
1
λ
t

1
{
1

∏
q
=
1
t

1
(
1

η
q
λ
q
)
}
.
By virtue of the assumptions on
η
i
,
λ
i
, we infer that
(45)
∑
i
=
1
t

1
∏
q
=
i
+
1
t

1
(
1

η
q
λ
q
)
η
i
≤
1
λ
t

1
,
which implies that
(46)
f
→
t
z

f
→
ρ
,
t
z
H
K
n
≤
log
4
δ
68
Diam
(
V
)
M
κ
s
n
+
2
m
Π
/
∑
i
=
1
k
m
Π
i
λ
t

1
for any
z
∈
Z
1
.
Now, we consider the estimate of
f
→
t
z

f
→
ρ
,
t
z
H
K
n
. Let
Z
2
⊆
Z
m
1
×
m
2
×
⋯
×
m
k
with measure at least
1

δ
such that (27) is established for any
z
∈
Z
2
. In view of (26), for each
z
∈
Z
2
we yield
(47)
1
∑
a
=
1
k

1
∑
b
=
a
+
1
k
m
a
m
b
∑
a
=
1
k

1
∑
b
=
a
+
1
k
S
v
a
T
D
v
a
a
,
b
S
v
a

L
K
,
s
L
H
K
n
≤
log
2
δ
34
n
κ
2
Diam
V
2
s
n
+
2
m
Π
/
∑
i
=
1
k
m
Π
i
.
Using the fact that
L
K
,
λ
j
,
n
j

L
v
,
j
=
η
j
(
L
K
,
s

(
1
/
∑
a
=
1
k

1
∑
b
=
a
+
1
k
m
a
m
b
)
{
∑
a
=
1
k

1
∑
b
=
a
+
1
k
(
S
v
a
)
T
(
D
v
a
)
a
,
b
S
v
a
}
)
, we obtain that for any
z
∈
Z
2
,
(48)
f
→
t

f
→
ρ
,
t
z
H
K
n
=
∑
i
=
1
t

1
∏
q
=
i
+
1
t

1
I

L
v
,
q

∏
l
=
i
+
1
t

1
I

L
K
,
λ
l
,
n
j
η
i
f
→
ρ
,
s
H
K
n
=
∑
i
=
1
t

1
∑
j
=
i
+
1
t

1
∏
q
=
j
+
1
t

1
I

L
v
,
q
L
K
,
λ
j
,
n
j

L
v
,
q
×
∏
l
=
i
+
1
t

1
I

L
K
,
λ
l
,
n
j
η
i
f
→
ρ
,
s
∑
i
=
1
t

1
∑
j
=
i
+
1
t

1
∏
q
=
j
+
1
t

1
H
K
n
≤
∑
i
=
1
t

1
∑
j
=
i
+
1
t

1
∏
q
=
j
+
1
t

1
1

η
q
λ
q
η
j
×
17
κ
2
Diam
V
2
n
m
Π
/
∑
i
=
1
m
i
'
s
n
+
2
log
2
δ
∏
l
=
i
+
1
j

1
(
1

η
l
λ
l
)
η
i
f
→
ρ
,
s
H
K
n
.
By changing the order of summation, we determine that
(49)
f
→
t

f
→
ρ
,
t
z
H
K
n
≤
34
κ
2
Diam
V
2
n
m
Π
/
∑
i
=
1
k
m
Π
i
s
n
+
2
log
4
δ
·
∑
j
=
2
t

1
∏
q
=
j
+
1
t

1
1

η
q
λ
q
η
j
×
∑
i
=
1
j

1
∏
l
=
i
+
1
j

1
(
1

η
l
λ
l
)
η
i
f
→
ρ
,
s
H
K
n
.
According to (45), we can verify that
f
→
t

f
→
ρ
,
t
z
H
K
n
is bounded by
(50)
log
4
δ
34
κ
2
Diam
V
2
n
m
Π
/
∑
i
=
1
k
m
Π
i
s
n
+
2
×
∑
j
=
2
t

1
∏
q
=
j
+
1
t

1
1

η
q
λ
q
η
j
1
λ
j

1
f
→
ρ
,
s
H
K
n
≤
log
(
4
δ
)
34
κ
2
Diam
V
2
n
m
Π
/
∑
i
=
1
k
m
Π
i
s
n
+
2
1
λ
j

1
2
f
→
ρ
,
s
H
K
n
.
In view of the above fact and (46), we obtain that for any
z
∈
Z
1
∩
Z
2
,
(51)
f
→
t

f
→
ρ
,
t
z
H
K
n
≤
log
2
δ
68
κ
2
Diam
V
2
n
m
Π
/
∑
i
=
1
k
m
Π
i
s
n
+
2
+
34
κ
2
Diam
V
2
n
m
Π
/
∑
i
=
1
k
m
Π
i
s
n
+
2
f
→
ρ
,
s
H
K
n
.
However, the measure of the subset
Z
1
∩
Z
2
of
Z
m
1
×
m
2
×
⋯
×
m
k
is at least
1

2
δ
. The desired conclusion follows after substituting
δ
for
δ
/
2
.
The following result is Theorem 4 in Dong and Zhou [23]; it also holds in multidividing setting and we skip the detailed proof.
Theorem 11.
Let
{
λ
t
,
η
t
}
t
∈
N
be determined by (53). Then, we deduce that
(52)
f
→
t

f
→
λ
t
*
H
K
n
≤
t
2
γ
+
α

1
4
γ
C
λ
1
η
1
,
γ
+
α
,
1

γ
+
exp
λ
1
η
1

log
e
λ
1
η
1
1

γ

α
×
f
→
ρ
,
s
H
K
n
λ
1
.
4.2. Main Results
The first main result in our paper implies that
{
f
→
t
z
}
is a good approximation of a noisefree limit for the ontology function (6) as a solution of (8) which we refer as multidividing ontology function
f
→
λ
*
.
Theorem 12.
Let
0
<
γ
,
α
<
1
, and
λ
1
and
η
1
>
0
satisfy
2
γ
+
α
<
1
and
λ
1
η
1
<
1
. For any
t
∈
N
, take
(53)
λ
t
=
λ
1
t

α
.
Define
{
f
→
t
z
}
by (7) and
{
f
→
λ
*
}
by (8). If

y

≤
M
is almost established, then for any
0
<
δ
<
1
, with confidence
1

δ
, one has
(54)
f
→
t
z

f
→
λ
t
*
H
K
n
≤
C
~
log
8
δ
t
2
γ
m
Π
/
∑
i
=
1
k
m
Π
i
s
n
+
2
+
t
2
γ
+
α

1
×
1
+
f
→
ρ
,
s
H
K
n
,
where constant
C
~
independent of
m
1
,
m
2
,
…
,
m
k
,
t
,
s
or
δ
and
f
→
ρ
,
s
is the multidividing ontology function determined by
(55)
f
→
ρ
,
s
=
∑
a
=
1
k

1
∑
b
=
a
+
1
k
∫
V
a
∫
V
b
w
a
,
b
s
v
a

v
b
f
ρ
v
b

f
ρ
v
a
×
(
v
b

v
a
)
K
v
d
ρ
V
(
v
a
)
d
ρ
V
(
v
b
)
.
The proof of Theorem 12 follows from Theorems 10 and 11 and an exact expression for the constant
C
~
relying on
α
,
η
1
,
λ
1
,
κ
,
n
,
γ
,
M
and
Diam
(
V
)
can be easily determined.
The second main result in our paper follows from Theorem 10 and the technologies raised in [23].
Theorem 13.
Assume that for certain
0
<
τ
≤
2
/
3
,
c
ρ
>
0
and for any
s
>
0
, the marginal distribution
ρ
V
satisfies
(56)
ρ
V
v
∈
V
:
inf
u
∈
R
n
∖
V
u

v
≤
s
≤
c
ρ
2
s
4
s
,
and the density
p
(
v
)
of
d
ρ
V
(
v
)
exists and for any, any
u
,
v
∈
V
satisfies
(57)
sup
v
∈
V
p
(
v
)
≤
c
ρ
,
p
v

p
u
≤
c
ρ
u

v
τ
.
Suppose that the kernel
K
∈
C
3
and
∇
f
ρ
∈
H
K
n
. Let
0
<
β
<
1
/
(
4
+
2
n
+
4
γ
/
τ
)
and
0
<
γ
<
2
/
5
. Take
λ
t
=
t

γ
,
η
t
=
t
(
5
/
2
)
γ

1
, and
s
=
s
(
m
1
,
m
2
,
…
,
m
k
)
=
(
κ
c
ρ
)
2
/
τ
m
Π
/
∑
i
=
1
k
m
Π
i

β
γ
/
τ
and suppose that
(
m
Π
/
∑
i
=
1
k
m
Π
i
)
β
≤
t
≤
2
(
m
Π
/
∑
i
=
1
k
m
Π
i
)
β
; then for any
0
<
δ
<
1
, with confidence
1

δ
, one infers that
(58)
f
→
t
z

∇
f
ρ
(
L
ρ
V
2
)
n
≤
C
~
ρ
,
K
1
m
Π
/
∑
i
=
1
k
m
Π
i
θ
log
(
4
δ
)
,
where
(59)
θ
=
min
1
2

2
β

n
+
2
β
γ
τ
,
β
γ
2
and constant
C
~
ρ
,
K
is independent of
m
1
,
m
2
,
…
,
m
k
,
t
or
δ
.
Proof.
Obviously, under the assumptions
K
∈
C
3
, (56) and (57), we get
(60)
f
→
ρ
,
s
H
K
n
≤
C
ρ
,
K
(
c
ρ
n
2
π
n
/
2
κ
2
∇
f
ρ
H
K
n
+
s
)
.
Furthermore, by virtue of Proposition 15 in Mukherjee and Zhou [22], we have
(61)
f
→
t
*

∇
f
ρ
(
L
ρ
V
2
)
n
≤
C
ρ
,
K
{
∇
f
ρ
H
K
n
λ
+
s
λ
}
,
where constant
C
ρ
,
K
relies on
ρ
and
K
. Theorem 10 and these estimates reveal that with confidence
1

δ
, we yield
(62)
f
→
t
z

∇
f
ρ
(
L
ρ
V
2
)
n
≤
C
~
'
1
+
∇
f
ρ
H
K
n
×
log
4
δ
t
2
γ
m
Π
/
∑
i
=
1
k
m
Π
i
s
n
+
2
+
t
2
γ
+
α

1
+
s
t
γ
+
t

γ
/
2
.
The learning rate (58) is determined according to the selection of the parameters.