ENTROPY-, APPROXIMATION- AND KOLMOGOROV NUMBERS
ON QUASI-BANACH SPACES
Bachelor Thesis
to achieve the Academic Degree
Bachelor of Science (B. Sc.)
in Mathematics
FRIEDRICH-SCHILLER-UNIVERSITÄT JENA
Fakultät für Mathematik und Informatik
submitted by Marcus Gerhold
born 30/09/1989 in Hohenmoelsen
Adviser: Prof. Dr. Dorothee D. Haroske
Jena, 25/08/2011
Abstract. In this bachelor's thesis we introduce three quantities for linear and
bounded operators on quasi-Banach spaces which are entropy numbers, approximation numbers and Kolmogorov numbers. At rst we establish the three quantities with
some basic properties and try to modify known content from the Banach space case.
We compare each one of them, with the corresponding other two and give estimates
concerning the mean values and limits. As an example, we analyze the identity op
erator between nite dimensional `p spaces id : `np → `nq for 0 < p, q ≤ ∞ and give
sharp estimates for entropy numbers. Furthermore we add some known estimates
for approximation numbers and Kolmogorov numbers. At last we examine some
renowned connections of these quantities to spectral theory on innite dimensional
Hilbert spaces, which are the inequality of Carl and the inequality of Weyl.
Contents
Part 1. Preparations
5
1.
Norms
2.
n
Sequence Spaces `p and
5
3.
Linear and Compact Operators
8
4.
Quotient Spaces and the Quotient Map
9
`p
8
Part 2. Entropy-, Approximation- and Kolmogorov Numbers
10
5.
Entropy Numbers on quasi-Banach Spaces
10
6.
Approximation Numbers on quasi-Banach Spaces
14
7.
Kolmogorov Numbers on quasi-Banach Spaces
16
Part 3. Relationship between the Numbers
23
8.
Relationship between Entropy and Approximation Numbers
23
9.
Relationship between Approximation and Kolmogorov Numbers
29
10.
Relationship between Entropy and Kolmogorov Numbers
11.
Axiomatic Theory of
s-Numbers
31
Part 4. Compact Embeddings
12.
13.
14.
n
id : `p
n
id : `p
n
id : `p
→
→
→
30
32
`nq and entropy numbers
`nq and approximation numbers
`nq and Kolmogorov numbers
Part 5. Relationship to Spectral Theory
32
41
46
48
15.
Preliminary Considerations
48
16.
The Inequality of Carl
49
17.
Hilbert Space Setting and the Inequality of Weyl
50
References
51
3
An expression of many thanks goes to my adviser Prof. Dr. Dorothee Haroske for
the constant support and always taking a lot of time for my concerns.
Part 1. Preparations
1. Norms
As the title suggests, this bachelor's thesis is about entropy-, approximation- and
Kolmogorov numbers of linear and bounded operators acting between quasi-Banach
spaces. One of the ideas of these quantities is to give a measure of how compact a compact operator is. To establish these numbers, we will rst need to clarify notations and
basic denitions concerning norms, Banach spaces,
`p
spaces, operators and quotient
maps, which is the main aim of this part. Therefore we start with the denitions of
norms, quasi-norms and
Denition 1.
Let
X
%-
norms.
be a linear space over a eld
K.
A mapping
k · k : X → [0, ∞)
is
called norm if it satises the following three conditions
kxk = 0 ⇐⇒ x = 0
kλxk = |λ| kxk for λ ∈ K, x ∈ X
kx + yk ≤ kxk + kyk for x, y ∈ X.
(i)
(ii)
(iii)
kx + yk ≤ C (kxk + kyk) with a constant C ≥ 1,
which is independent of x and y , we call k·k quasi-norm. If condition (iii) is substituted
%
%
%
by (iii) kx + yk ≤ kxk + kyk for % ∈ (0, 1], we call k · k % - norm.
If condition (iii) is substituted by (iii')
If we talk of Banach spaces as complete normed vector spaces, it is only natural that
the terms quasi-Banach space and
%
- Banach space arise, if a vector space is complete
with respect to the corresponding quasi- or
%
- norm.
As we will show in the following theorem, which can be found in [DL93, Ch. 2, Thm.
1.1.], it does not matter whether we are dealing with quasi-Banach spaces or
% - Banach
spaces, since both are equivalent.
Remark 2.
As the notation of the closed and open unit balls dier, we will use
BX = {x ∈ X : kxkX < 1}
and
B X = {x ∈ X : kxkX ≤ 1}
as the open and correspondingly the closed unit ball for a given normed space.
Theorem 3. Let X be a linear space. For each quasi-norm k · k with constant C ≥ 1
exists an equivalent % - norm k · k0 with % ∈ (0, 1].
Proof.
C as constant of the quasi-norm k · k we dene C0 := 2C . With C0 ≥ 2,
and f, g ∈ X we get the result kf + gk ≤ C (kf k + kgk) ≤ 2C max {kf k, kgk} =
C0 max {kf k, kgk}. By induction we derive that for f1 , . . . , fm ∈ X, m ∈ N
With
5
kf1 + . . . + fm k ≤ max C0j kfj k .
(1.1)
Since
1≤j≤m
C
is given, we dene
kf k0 :=
(1.2)
%
by
C0% = 2
inf
f =f1 +...+fm
and the mapping
k · k0 : X → [0, ∞)
by
{kf1 k% + . . . + kfm k% }1/% ,
where the inmum is taken over all decompositions of f . At rst we observe, that
%
C0 = 2 ⇐⇒ % = lnlnC20 , hence % ∈ (0, 1]. We now want to show that k · k0 is our desired
%
- norm. Therefore we need to investigate the three properties of Denition 1 of the
%
mapping. Properties (i) and (ii) of
- norms can be derived immediately. Property
(iii) is given through
kf + gk%0 =
≤
inf
f =f1 +...+fm
+
{kf1 k% + . . . + kfm k% }
{kg1 k% + . . . + kgm k% }
inf
g=g1 +...+gm
kf k%0 + kgk%0 .
=
k · k0 are equivalent
by showing, that there exist constants a, A > 0 in a way, that ak · k0 ≤ k · k ≤ Ak · k0 .
The rst inequality can easily be shown. Since the inmum in k · k0 is taken over all
decompositions, it is also taken over the trivial decomposition f = f , which yields
kf k0 = inf f =f1 +...+fm {kf1 k% + . . . + kfm k% }1/% ≤ inf f =f1 {kf1 k% }1/% = kf k. Thus a = 1.
Hence
k · k0
{kh1 k% | + . . . + khm k% }
inf
f +g=h1 +...+hm
is a
%
- norm. Now we need to show that
k·k
and
For the other inequality we dene
N (f ) =

0
C k
0
if
f =0
if
C0k−1 < kf k ≤ C0k .
Clearly we get
(1.3)
C0−1 N (f ) ≤ kf k ≤ N (f ).
At rst we show by induction
(1.4)
kf1 + . . . + fm k ≤ C0 (N (f1 )% + . . . + N (fm )% )1/% .
6
The case
m=1
follows immediately. We suppose that (1.4) has been established for
m = n − 1. Now for given f1 , . . . , fn ∈ X we can assume without loss of generality, that
kf1 k ≥ . . . ≥ kfn k (otherwise we renumber all fi ). If all N (fi ), i = 1, . . . , n are distinct,
we have
(1.3)
C0j ||fj || ≤ C0j N (fj ) ≤ C0 N (f1 ) ≤ C0 (N (f1 )% + . . . + N (fn )% )1/% .
(1.4) follows from (1.1).
j ∈ {1, . . . , n − 1} N (fj ) = N (fj+1 ) =
l ∈ Z. Using (1.1), we have kfj + fj+1 k ≤ C0 max {kfj k, kfj+1 k} = C0l+1 and
%(l+1)
%
furthermore N (fj + fj+1 ) ≤ C0
= 2(l+1) = 2l + 2l = N (fj )% + N (fj+1 )% . Combining
Let us now consider the case, where for certain
C0l ,
this result with our induction hypothesis, yields
kf1 + . . . + fm k
i.h.
≤
C0 (N (f1 )% + . . . + N (fj + fj+1 )% + . . . + N (fm )% )1/%
≤
C0 (N (f1 )% + . . . + N (fj )% + N (fj+1 )% + . . . + N (fm )% )1/% .
Thus we have proved (1.4) from which the wanted inequality can be derived from
kf1 + . . . + fm k
≤
(1.3)
≤
C02
N (f1 )
C0
%
+ ... +
N (fm )
C0
C02 (kf1 k% + . . . + kfm k% )1/% .
Now we can take the inmum over all decompositions of
denition of
k · k0 )
and get
% 1/%
A = C02 ,
f
(as it is done in the
and hence the equivalence of the norms
k·k
k · k0 .
and
This theorem will already be useful in the next statement, which is well known and
part of many lectures. Its proof, in the case for Banach spaces can be found in [DL93,
Ch. 3, Thm. 1.1.], but is done here for quasi-Banach spaces.
Theorem 4. Let [X, k · k] be a quasi-normed, linear vector space with constant CX and
U ⊂ X a linear subspace with dim U < n < ∞. Then for every element f ∈ X, there
exists an element g ∈ U with
kf − gk = inf kf − hk.
h∈U
Proof.
As we have shown in the preceding theorem, we nd an equivalent
every quasi-norm. So let
k · k0
be an equivalent
7
%
%
- Norm for
- norm for the quasi-norm
k · k.
By
denition of the inmum, there exists a sequence (hk )k∈N ∈ U for which kf
inf h∈U kf − hk0 . We also have khn k%0 ≤ kf k%0 + kf − hn k%0 . This means, that
bounded sequence on a nite dimensional subspace, hence
. Therefore we can nd a subsequence
j→∞
khnj − gk0 −→ 0.
hnj
j∈N
(hn )n∈N
⊂ (hn )n∈N
n→∞
− hn k0 −→
(hn )n∈N is a
is relatively compact
and an element
g∈X
with
Furthermore we get
j→∞
kf − gk%0 ≤ kf − hnj k%0 + khnj − gk%0 ≤ kf − gk%0 + kg − hnj k + khnj − gk%0 −→ kf − gk%0
j→∞
and by taking the % - th root kf − hnj k0 −→
n→∞
hn k0 −→ inf h∈U kf − hk0 and therefore kf − gk0
g ∈ U.
norm k · k.
gain
kf − gk0 . On the other hand kf −
= inf h∈U kf − hk0 . In particular we
With the established equivalence, this result is also valid for the quasi
2. Sequence Spaces
Denition 5.
For given
`np
and
`p
0 < p < ∞, n ∈ N
and a eld K (which
o
P
`np := a ∈ Kn : nj=1 |aj |p < ∞
1/p
P
n
p
|a
|
and k · kp :=
j
j=1
n
n
`∞ := a ∈ K : sup1≤j≤n aj < ∞
and k · k∞ := sup1≤j≤n |aj |
o
n
P∞
p
`p (N) := a = (aj )j∈N ⊂ K : j=1 |aj | < ∞
1/p
P
∞
p
|a
|
and k · kp :=
j
j=1
n
o
`∞ := a = (aj )j∈N ⊂ K : supj∈N |aj | < ∞
and k · k∞ := supj∈N |aj |
is
R
or
C)
we dene
n
(i)
(ii)
(iii)
(iv)
Remark 6.
Since the sequence spaces are well known, we shall use the following two
statements without proof in this bachelor's thesis. They can be found in [Tri92, Sect.
1.2.]
(1) For
(2) For
0 < p < 1 [`p (N) ; k · kp ] are quasi-Banach spaces.
1 ≤ p ≤ ∞ [lp (N) ; k · kp ] are Banach spaces.
3. Linear and Compact Operators
Denition 7.
(i)
Let
[X, k·kX ]
and
[Y, k·kY ]
be quasi-normed spaces.
A : X −→ Y is called
∀x ∈ X ∃c > 0 : kAxkY ≤ ckxkX .
A linear mapping
if
8
bounded operator,
(vii)
L (X, Y) := {A : X −→ Y, A linear and bounded} (If X = Y we will write
L (X) := L (X, X) .)
If T ∈ L (X, Y) we dene R (T ) = {y ∈ Y : ∃x ∈ X : T x = y} as range of
the operator T .
An operator A ∈ L (X, Y) is called compact, if the range of every bounded
set in X is relatively compact in Y.
K (X, Y) := {A ∈ L (X, Y) , A compact}.
rank T := dim R (T )
A is called nite rank operator if R (A) ⊆ Y and rank A < ∞.
Remark 8.
Again, these denitions as well as many conclusions of them are well known
(ii)
(iii)
(iv)
(v)
(vi)
and we will take them for granted. For completeness we shall list a few, which will be
used in this bachelor's thesis. Their proofs however can be found in [Har11, Folg. 1.18.;
Bem. in 2.1.3.; Folg. 2.12.].
(1) If
[X, d]
compact if and only if there exists a nite
(2)
(3)
A ⊂ X,
for A for
is an arbitrary complete metric space and
ε
- net
A is relatively
every ε > 0.
in Y.
then
K ∈ K (X, Y) if and only if K B X is relatively compact
Let [X, k · kX ] be a quasi-normed space, [Y, k · kY ] a quasi-Banach space and
A ∈ L (X, Y). If there exists a sequence of operators (An )n ⊂ L (X, Y) and
n→∞
rank Ak = nk < ∞ ∀k ∈ N for which kA − An k −→ 0. Then A ∈ K (X, Y).
(As mentioned above, the proof can be found in [Har11, Folg. 2.12.] for Banach
spaces. It is certied quickly that the fact stays correct for quasi-Banach spaces.)
4. Quotient Spaces and the Quotient Map
Denition 9.
subspace of
[X, k · kX ]
a normed vector space over a eld
K
and
U ⊂X
linear
.
X/U := {x + U ; x ∈ X} = {[x]U : x ∈ X} the quotient space of X
with respect to U .
X
The mapping QU : X −→ X/U is called canonical quotient map of X with
respect to U .
We dene addition as [x]U + [y]U = x + y + U = [x + y]U and multiplication
with a scalar λ ∈ K as λ [x]U = λx + U = [λx]U .
For x ∈ X we dene k [x]U kX/U := inf y∈U kx + ykX
(i)
We call
(ii)
(iii)
(iv)
Remark
that
X
Let
10
.
Since we only need one result, following from these denitions, which is
X/U, k · kX/U
is a normed vector space, we won't prove it in this bachelor's thesis.
The proof can be found in [Har11, Satz 3.13.].
9
The following lemma is taken from [CS90, p. 49] and is slightly modied here.
Lemma 11. Let [X, k · kX ] be a quasi-normed vector space and U ⊂ X a linear subspace
of X. Then
QXU (BX ) = BX/U .
Proof.
[x]U ∈ QXU (BX ),
Let
then
[x]U = { x − u; u ∈ U }
where
x ∈ BX
hence
kxkX < 1.
By Denition 9 as inmum, we get
k [x]U kX/U = inf kx − ukX ≤ kxkX < 1.
u∈U
0∈U
[x]U ∈ X/U with k [x]u kX/U < 1 we know that
= [x]U and kxkX < 1. This yields [x]U ∈ QXU (BX ).
On the other hand, if we take
exists
x∈
X, such that QXU x
there
Part 2. Entropy-, Approximation- and Kolmogorov Numbers
5. Entropy Numbers on quasi-Banach Spaces
In this part we will now introduce the three quantities, beginning with entropy numbers. There is more than one way to introduce them, but in the proceeding denition
we follow the notation of [ET96, Subsect.
1.3.1., Def.
1.], which is based on dyadic
entropy numbers.
Denition 12.
Let
and
X
Y
be quasi-Banach spaces,
n∈N
and further
T ∈ L (X, Y).
Then we dene
(
en (T ) := inf
ε > 0 : ∃y1 , . . . , y2n−1 ∈ Y : T B X ⊆
n−1
2[
)
yi + εB Y
i=1
as the
Remark
n
- th (dyadic) entropy number of the operator
13
.
T.
This is a denition of entropy numbers based on an operator, but it is also
possible to introduce them rst on an arbitrary set :
(
en (A, X) := inf
ε > 0 : ∃x1 , . . . , x2n−1 ∈ X : A ⊆
n−1
2[
xi + εB X
)
i=1
from which the above denition is established through
en (T ) = en T B X , Y .
The following theorem, though altered in notation to t in the context of quasiBanach spaces, can be found in [CS90, Sect. 1.3.]. The last part of the theorem, which
is (Ce ) was slightly modied taken from [Har10, Satz 3.30.].
10
Theorem 14. (Properties of en (T ))
Let X,Y and W be quasi-Banach spaces with constants CX ,CY and CW . Further let
T ∈ L (X, Y).
CY e1 (T ) ≥ kT k ≥ e1 (T ) ≥ e2 (T ) ≥ . . . ≥ 0
Let S ∈ L (X, Y), n, m ∈ N, then we have
em+n−1 (S + T ) ≤ CY (em (S) + en (T )) which is equivalent to
em+n−1 (S + T )% ≤ em (S)% + en (T )% for an equivalent % - norm with
% ∈ (0, 1].
Let S ∈ L (Y, W), n, m ∈ N , then we have em+n−1 (ST ) ≤ em (S) en (T ).
T ∈ K (X, Y) ⇐⇒ limn→∞ en (T ) = 0
(Me )
(Ae )
(Pe )
(Ce )
Proof.
ε > 0 , the monotonicinf x∈A kxk ≤ inf x∈B kxk if B ⊆ A. To prove
By denition of the entropy numbers as inmum over all
ity (Me ) is immediately derived, since
CY e1 (T ) ≥ kT k ≥ e1 (T ),
we show two inequalities. The rst is obtained through
T B X ⊆ kT kB Y =⇒ ∃y1 ∈ Y : T B X ⊆ y1 + kT kB Y
y1 =0
ε > 0 where for some y1 ∈ Y, T B X ⊆
y1 + εB Y is valid and we get e1 (T ) ≤ kT k. Now we prove the opposite inequality
and let ε > e1 (T ) ≥ 0. This yields that there exists y1 ∈ Y for which T B X ⊆
y1 + εB Y . Furthermore for an arbitrary x ∈ B X there exist η1 , η2 ∈ B Y for which
T x = y1 + εη1 and T (−x) = −T (x) = y1 + εη2 . Subtracting the second from the rst
and taking the inmum over all such
equality yields,
2T x = ε (η1 − η2 ) ⇐⇒ T x =
ε
(η1 − η2 ) .
2
ε
ε
=⇒ kT xkY ≤ k (η1 − η2 ) kY ≤ CY (kη1 kY + kη2 kY ) ≤ CY ε
2
2 | {z } | {z }
≤1
Because the right-hand side is independent of
x,
≤1
the inequality is maintained, if we
x ∈ B X . Hence kT k ≤ CY ε.
get kT k ≤ CY e1 (T ).
take the supremum over all those
inmum over all
ε > e1 (T )
and
Let us now have a look at the additivity (Ae ).
arbitrary
n−1
N ≤2
λ > en (T ) and µ > em (S).
m−1
and M ≤ 2
, such that
11
S and T , we choose
y1 , . . . , yN , z1 , . . . , zM , where
With given
For these exist
Now we can take the
(5.1)
T BX
N
[
⊆
yi + λB Y
S BX ⊆
and
i=1
i ∈ {1, . . . , N }, j ∈ {1, . . . , M }
x ∈ BX
(5.2)
to choose one of the
yi
and
zj
for
such that
yi + λB Y
Therefore it follows, that for
λy + µz .
zi + µB Y .
i=1
These inclusions allow us, for any given
Tx ∈
M
[
x ∈ BX
and
exist
Sx ∈ zj + µB Y
y ,z ∈ B Y
such that
(S + T ) x = yi + zj +
However we nd that
kλy + µzkY ≤ CY (λ kyk + µ kzk ) ≤ CY (λ + µ) .
|{z}
|{z}
≤1
(S + T ) x ∈
=⇒
(5.1)
=⇒ (S + T ) B X ∈
≤1
yi + zj + CY (λ + µ) B Y
N [
M
[
yi + zj + CY (λ + µ) B Y
i=1 j=1
To obtain the wanted inequality, we need to have a look at the number of elements
in the following set, which is
(5.3)
# {yi + zj , i = 1, . . . , N, j = 1, . . . , M } ≤ N M ≤ 2n−1+m−1 = 2(n+m−1)−1 .
=⇒ en+m−1 (S + T ) ≤ CY (λ + µ)
And taking the inmum over all those
λ
and
µ,
we get
en+m−1 (S + T ) ≤ CY (en (T ) + em (S)) .
As we have shown in Theorem 3 we can nd an equivalent % - norm, such that
em+n−1 (S + T )% ≤ em (S)% + en (T )% . In particular, we would have in (5.2)
kλy + µzk%Y ≤ λ% kyk%Y + µ% kzk%Y ≤ λ% + µ%
| {z }
| {z }
≤1
if we considered a
%
≤1
- norm. The next step would be
12
=⇒
(S + T ) x ∈
(5.1)
=⇒ (S + T ) B X ∈
n
o
yi + zj + (λ% + µ% )1/% B Y
N [
M n
[
o
yi + zj + (λ% + µ% )1/% B Y .
i=1 j=1
By the same arguments as above, we would get
en+m−1 (S + T ) ≤ (λ% + µ% )1/%
and by taking the inmum over all
λ
and
µ,
we had
en+m−1 (S + T )% ≤ en (S)% + em (T )% .
The multiplicativity (Pe ) can be shown by similar arguments. That is for another
W and operators T ∈ L (X, Y)
λ > en (T ) and µ > em (S), such that
given quasi-Banach space
can choose
(5.4)
T BX ⊆
N
[
yi + λB Y
and
S BY ⊆
i=1
as well as
M
[
S ∈ L (Y, W)
zj + µB W
we
j=1
y1 , . . . , yN , z1 , . . . , zM and N < 2n−1 , M < 2m−1 . The right-hand inclusion
S is equivalent to S λB Y
= λS B Y ⊆ M
λz
+
λµB
, so that applying the
W
j
j=1
operator S to the left-hand side of (5.4) amounts to
for
ST B X ⊆
N [
M
[
Syi + λzj + λµB W .
i=1 j=1
ε > 0, yields
λ > en (T ) and
Counting the elements as in (5.3) and taking the inmum over all such
en+m−1 (ST ) ≤ λµ. The last step is taking the inmum
µ > em (S). Therefore en+m−1 (ST ) ≤ en (T ) em (S).
over all these
The last property, which is compactness (Ce ) is immediately established through
limn→∞ en (T ) = 0 means in
particular, that for ε > 0 exists n0 ∈ N such that for all n ≥ n0 we nd that en (T ) < ε.
Choosing those y1 , . . . , y2n−1 , we have found a nite ε- net for T B X . This is possible
for all ε > 0, hence T B X is relatively compact. On the other hand, if T B X is
relatively compact, there exists a nite ε - net for every ε > 0. Since this is only a
question of denition, we can choose only these ε - nets, which have a dyadic number
of elements. Hence limn→∞ en (T ) = 0.
the denition of relatively compactness and Remark 8.
13
6. Approximation Numbers on quasi-Banach Spaces
Let us now dene approximation numbers. By doing so, we follow the notation of
[ET96, Subsect. 1.3.1., Def. 2.].
Denition 15.
Let
X
and
Y
T ∈ L (X, Y).
be quasi-Banach spaces and
For
n∈N
we
dene
an (T ) := inf {kT − Sk : S ∈ L (X, Y) ,
(6.1)
as
n
- th approximation number of the operator
rank
S < n}
T.
As before, the following theorem and its proof in case of Banach spaces can be found
in [CS90, Sect.
2.1.]
and is adopted in this bachelor's thesis to t in the context of
quasi Banach spaces.
Theorem 16. (Properties of an (T ))
Let X,Y and W be quasi-Banach spaces with constants CX ,CY and CW . Further let
T ∈ L (X, Y).
kT k = a1 (T ) ≥ a2 (T ) ≥ . . . ≥ 0
Let S ∈ L (X, Y), n, m ∈ N, then we have
am+n−1 (S + T ) ≤ CY (am (S) + an (T )) which is equivalent to
am+n−1 (S + T )% ≤ am (S)% + an (T )% for an equivalent % - norm with
% ∈ (0, 1].
Let S ∈ L (Y, W), n, m ∈ N , then we have am+n−1 (ST ) ≤ am (S) an (T ).
rank T < n =⇒ an (T ) = 0
dim X ≥ n =⇒ an (idX→X ) = an (idX ) = 1
limn→∞ an (T ) = 0 =⇒ T ∈ K (X, Y)
(Ma )
(Aa )
(Pa )
(Ra )
(Na )
(Ca )
Proof.
The monotonicity (Ma ) is obviously derived, since
B ⊆ A.
Having a closer look at
a1 (T ) = inf



a1
kT − Sk : S ∈ L (X, Y) , rank
{z
|
S≡0


S < 1 = kT k.
}
λ > an (T )
and
µ > am (S),
where
That means nothing else, than
∃L, R ∈ L (X, Y) ,
rank
L < n,
if
we get
To prove the additivity (Aa ) we start with
n, m ∈ N.
inf x∈A ||x|| ≤ inf x∈B ||x||
rank
R < m : kT − Lk < λ
14
and
kS − Rk < µ.
Now we dene
M := L + R.
Clearly
M ∈ L (X, Y)
and rank
M < n + m − 1.
Therefore
k (S + T ) − M k =
sup k [(S + T ) − M ] xkY
kxkX =1
!
≤ CY
sup k (S − R) xk + sup k (T − L) xk
kxkX =1
kxkX =1
= CY (kS − Rk + kT − Lk) ≤ CY (λ + µ) .
If we now take the inmum over all such operators
M ∈ L (X, Y)
M <
λ and
with rank
n + m − 1, we get an+m−1 (S + T ) < CY (λ + µ). Taking the inmum over
µ amounts to an+m−1 (S + T ) ≤ CY (an (T ) + am (S)) , which is again equivalent to
an+m−1 (S + T )% ≤ an (S)% + am (T )% . We will not prove the equivalence, since the idea
is the same, as we have seen in (Ae ) of Theorem 14.
The multiplicativity (Pa ) is established through a similar proof, in which we let
λ > an (T )
and
µ > am (S)
for given
S ∈ L (Y, W)
∃L ∈ L (X, Y) ,
and
rank
∃R ∈ L (Y, W) ,
We go on by dening
rank
and
n, m ∈ N.
As above we get
L < n : kT − Lk < λ
R < m : kS − Rk < µ.
M := RT + SL − RL ∈ L (X, W).
Hence
kST − M k = kST − RT − SL + RLk = k (S − R) (T − L) k
≤ kS − RkkT − Lk < λµ.
Furthermore rank
M ≤ rank SL + rank (R (T − L)) ≤ rank L + rank R < n + m − 1.
Knowing, that there exists such an operator, we can take the inmum over these, which
amounts to
an+m−1 (ST ) ≤ λµ.
Taking the inmum over
λ
and
µ
yields
an+m−1 (ST ) ≤ an (S) am (T )
The rank property (Ra ) is quickly derived, because of rank
T < n,
it is a fair com-
petitor for the inmum, which results to
0 ≤ an (T ) = inf {kT − Sk : S ∈ L (X, Y) , rank S < n} ≤ kT − T k = 0.
Next we will prove (Na ) for which the monotonicity is needed in
kidX k = 1.
If we can show, that
an (idX ) ≥ 1
15
an (idX ) ≤ a1 (idX ) =
the equality is established. For that, let
dim X ≥ n and L ∈ L (X, X) and rank L < n. Hence there exists x0 ∈ X, x0 6= 0 for
which Lx0 = 0. Without loss of generality, we can say, that kx0 kX = 1 (otherwise, we
scale it).
=⇒ 1 = kx0 kX = kx0 − Lx0 kX ≤ sup k (idX − L) xkX = kidX − Lk
|{z}
kxkX =1
=0
If we take the inmum over all such
of the
n
L,
we end up on (6.1), which is the denition
- th approximation number. Hence
an (idX ) ≥ 1
and with our rst step, the
equality is proven.
The last property, which is compactness (Ca ), is immediately given, because
T ∈
L (X, Y) and limn→∞ an (T ) = 0. This means, that there exists a sequence of nite rank
operators, that converges to T . Hence T ∈ K (X, Y) (See Remark 8)
Remark
17
.
We have seen that property (Ce ) of Theorem 14 is an equivalence, whereas
(Ca ) of Theorem 16 is only an implication. We should point out, that we have no loss of
information when we switch from Banach spaces to quasi-Banach spaces, which means
that the opposite implication is not even valid in the Banach space case.
We do however have a loss of information when considering the quasi-Banach space
case concerning (Ra ), since we have an equivalence in the Banach space case.
(See
[CS90, Sect. 2.4., A4].)
7. Kolmogorov Numbers on quasi-Banach Spaces
At last we introduce Kolmogorov numbers. We will follow the notation of [Har10,
Abschn. 3.3.] in this part.
Denition 18.
Let
X
and
Y
be quasi-Banach spaces and
T ∈ L (X, Y).
For
n∈N
we
call
dn (T ) =
the
n-th
Remark
19
sup inf ||T x − y||Y
inf
Un ⊂Y, dim Un <n ||x|| ≤1 y∈Un
X
Kolmogorov number of the operator
.
T.
This is again a denition, which is based on an operator
Banach space and
A ⊂ X, n ∈ N
If
X is a quasi-
then Kolmogorov numbers can also be introduced
as
dn (A, X) :=
T.
sup inf ||x − y||X
inf
Un ⊂X, dim Un <n x∈A y∈Un
16
from which the above number is established through
dn (T ) = dn (T (B X ), Y).
Accord-
ing to this denition of Kolmogorov numbers based on sets, we can make the following
statement:
Let
X
be a quasi-Banach space with
dim X ≥ n
dk B X , X = 1,
Proof.
that
for
for
n ∈ N.
Then
k = 1, . . . , n.
d1 B X , X = supx∈B X kxk = 1. Furthermore we can easily see,
B X , X if k ≥ m. This is because
At rst we notice
dk B X , X ≤ dm
dm (B X , X) =
≥
and of course
inf
sup
inf kx − ykX
Um ⊂X, dim Um <m ||x|| ≤1 y∈Um
X
inf
inf kx − ykX = dm+1 (T )
sup
Um+1 ⊂X, dim Um+1 <m+1 ||x|| ≤1 y∈Um+1
X
inf x∈A ||x|| ≤ inf x∈B ||x||
if
B ⊆ A.
Hence
dn B X , X ≤ d1 B X , X = 1.
Now we need to show that dn B X , X
≥ 1. At rst we will clarify that for all
such subspaces Un ⊂ X, with dim Un < n there exists xn ∈ X, xn 6= 0 such that
inf y∈U ky − xn kX = kxn kX . The case xn ∈ Un is obvious, so for a given subspace
Un ⊂ X, we choose an arbitrary ξ ∈ X\Un . With Theorem 4 we know, that there exists
a best approximation. This means that there exists yn ∈ Un , such that 0 < kξ − yn kX =
inf u∈Un kξ − ukX . Hence by dening xn := ξ − yn ∈ X, we get xn 6= 0 and
kxn kX = kξ − yn kX = inf kξ − yn − (u − yn ) kX = inf kxn − (u − yn )kX
| {z }
u∈Un
u∈Un
=:y∈Un
(7.1)
=
inf kxn − ykX .
y∈Un
Without loss of generality we can say
kxn k = 1
(otherwise, we could scale it). This
yields
(7.1)
sup inf ky − xk ≥ inf ky − xn k = kxn k = 1.
x∈B X
y∈Un
y∈Un
Thus taking the inmum over all such spaces
Un ,
we get
dn B X , X ≥ 1
. Together
with step 1, this yields the equality.
17
As before, we will now show some basic properties of Kolmogorov numbers, which
can be found in [Har10, Satz 3.28.] for the case of Banach spaces and which are slightly
modied here.
Theorem 20. (Properties of dn (T ))
Let X,Y and W be quasi-Banach spaces, with constants CX , CY and CW . Further let
T ∈ L (X, Y).
(Md )
(Ad )
(Pd )
(Rd )
(Nd )
(Cd )
Proof.
kT k = d1 (T ) ≥ d2 (T ) ≥ . . . ≥ 0
Let S ∈ L (X, Y), n, m ∈ N, then dm+n−1 (S + T ) ≤ CY (dm (S) + dn (T ))
%
%
%
which is equivalent to dm+n−1 (S + T ) ≤ dm (S) +dn (T ) for an equivalent
% - norm with % ∈ (0, 1].
Let S ∈ L (Y, W), n, m ∈ N. Then dm+n−1 (ST ) ≤ dm (S) dn (T ).
rank T < n =⇒ dn (T ) = 0
dim X ≥ n =⇒ dn (idX→X ) = dn (idX ) = 1
T ∈ K(X, Y) ⇐⇒ limn→∞ dn (T ) = 0
The proof of monotonicity (Md ) is similar to Remark 19 . To prove the second
fact, we take a look at an arbitrary operator
T ∈ L(X, Y).
This yields that
T BX
is
bounded, and further
d1 (T ) =
sup inf kT x − ykY = sup kT xkY = kT kY .
inf
U1 ⊂Y, dim U1 <1 ||x|| ≤1 y∈U1
X
||x||X ≤1
ε > 0, n, m ∈ N. By denition
dim Un < n, we gain the following:
To prove the additivity (Ad ), let
over all subspaces
Un ⊂ Y
with
∃Um ⊂ Y, dim Um < m
and
above, we can see
and
as inmum
kT x − vnx kY < dn (T ) + ε
Wm,n = Um + Vn ⊂ Y. We notice that dim Wm,n < n + m − 1. As
x
x
x
that for all x ∈ B X there exists wm,n = um + vn ∈ Wm,n , such that
x
k(S + T )x − wm,n
kY = kSx − uxm + T x − vnx kY
(7.2)
dn
Vn ⊂ Y, dim Vn < n ∀x ∈ B X ∃uxm ∈ Um , vnx ∈ Vn :
kSx − uxm kY < dm (S) + ε
Now we denote
of
≤ CY (kSx − uxm kY + kT x − vnx kY )
< CY (dm (S) + dn (T ) + 2ε) .
18
The inequality (7.2) is maintained if we take the supremum of all
x ∈ BX
over the
wm,n ∈ Wm,n . Because there exist such Wm,n we can also take the
inmum over all such subspaces Wm,n ⊂ Y with dim Wm,n < n + m − 1. Since ε was
arbitrary, we let ε → 0 and gain the additivity (Ad )
inmum of all such
dn+m−1 (S + T ) ≤ CY (dm (S) + dn (T )) .
As we already have established two times before, this is equivalent to
dn+m−1 (S + T )% ≤ dm (S)% + dn (T )%
and is left here unproven, since the idea is the same as in (Ae ) of Theorem 14.
We advance with the multiplicativity (Md ) and start the same way as above by
and
n, m ∈ N.
ε>0
Also with the same arguments as above, through denition of the
inmum, we have
(7.3)
∃Um ⊂ Y, dim Um < m ∀x ∈ B X ∃uxm ∈ Um : ||T x − uxm ||Y ≤ dm (T ) + ε,
(7.4) and
∃Vn ⊂ W, dim Vn < n ∀y ∈ B Y ∃vny ∈ Vn : ||Sy − vny ||W ≤ dn (S) + ε.
Now let
x ∈ BX.
With our rst premise (7.3), we gain
T x − uxm dm (T ) + ε < 1
and dene
y(x)
x
= Suxm + (dm (T ) + ε) vn
=⇒ ∃wm,n
y = y(x) :=
T x − uxm
.
dm (T ) + ε
∈ S (Um ) + Vn = Wm,n ⊂ W
x (7.4)
x
x
−
Su
w
T x − um
m,n
m
−
S
< dn (S) + ε
dm (T ) + ε
dm (T ) + ε |
|
{z
}
{z
}
y(x)
y(x)
vn
W
x ⇐⇒
ST x − wm,n W ≤ (dn (S) + ε) (dm (T ) + ε)
As above, this inequality is of course maintained, if we take the supremum over all
x ∈ BX
wm,n in this specic Wm,n . Since such Wm,n ⊂ W
< m + n − 1, we can take the inmum over them. With ε
over the inmum of all such
dim Wm,n
let ε ↓ 0 and
exist and have
arbitrary, we
the desired statement follows as
dm+n−1 (ST ) ≤ dn (S) dm (T ) .
19
The statement, which denotes rank-properties (Rd ) with respect to Kolmogorov numbers is easily obtained, since rank
get
T = dim R (T ) < n,
we simply put
Un := R (T )
dn (T ) = 0.
For proving property (Nd ), we simply use Remark 19 and the fact, that
id :
and
X −→ X ∈ L (X, X).
T :=
Hence
dk (T ) = dk T B X , X = dk B X , X = 1
for
k = 1, . . . , n.
At last we will have a look at compactness properties (Cd ). For the rst direction, let
T ∈ K (X, Y).
Then with Remark 8 we know that
we can nd a nite
ε
T BX
Un0 :=
dn0 +1 (T ) ≤ ε.
is relatively compact, hence
- net. That means
∃n0 ∈ N, x0 , . . . , xn0 ∈ X ∀x ∈ T B X :
We dene
span
{x1 , . . . , xn0 }
and since
min kx − xi kX ≤ ε.
i=1,...,n0
dim Un0 ≤ n0
we can derive that
Using the monotonicity (Md ) we get
∃n0 ∈ N ∀n > n0 : dn (T ) ≤ ε ⇐⇒ lim dn (T ) = 0.
n→∞
To prove the opposite direction, we suppose that
yields, that
T BX
is bounded in
limn→∞ dn (T ) = 0. T ∈ L (X, Y)
Y.
=⇒ d1 (T ) =
sup kxkX < ∞
x∈T (B X )
If we have a look at our premise
limn→∞ dn (T ) = 0,
we see that this means
∀ε > 0 ∃n0 ∈ N ∀n > n0 ∃Un ⊂ Y, dim Un < n ∀x ∈ T B X ∃uxn ∈ Un : kx − uxn kY < ε.
For these
uxn ∈ Un ,
we have

kuxn kY ≤ CY (kx − uxn kY + kxkY ) ≤ CY ε +

sup kxkY  = CY (ε + d1 (T )) .
x∈T (B X )
M0 := {u ∈ Un : kukY ≤ CY (ε + d1 (T ))}, we see that M0 is bounded
x
and un ∈ M0 ⊂ Un . dim Un < n yields that M0 is relatively compact and therefore we
can nd a nite ε - net M1 := {η0 , . . . , ηm } for M0 . This specic M1 is a 2CY ε - net for
T B X , hence T B X is relatively compact and that means, that T ∈ K (X, Y).
If we dene
20
Remark
21
.
As we have already done in Remark 17 for approximation numbers, we
should point out, that we have a loss of information, when considering quasi-Banach
spaces in property (Rd ), since we have an equivalence for Banach spaces. (See [Har10,
Satz 3.28.].)
We have established various properties of Kolmogorov numbers and have already seen,
that they can be introduced in dierent ways (on arbitrary sets or operators), and we
will now show the equivalence to other denitions taken from [CS90, Sect. 2.2.] and
[Pie87, Ch. 2., 2.5.2.].
Proposition 22. Let X and Y be arbitrary quasi-Banach spaces and T ∈ L (X, Y),
n ∈ N. Then the n - th Kolmogorov number dn (T ) can be expressed as
(i)
dn (T ) = inf ε > 0 : T B X ⊂ Nε + εB Y , Nε ⊂ Y, dim Nε < n .
(ii) dn (T ) = inf kQYV T k : V ⊂ Y, dim V < n .
Proof.
In the
rst step, we show the equivalence to a denition taken by [CS90].
we will use Theorem 3 to consider
We
%
- norms instead of quasi-norms. Let
dˆn (T ) := inf ε > 0 : T B X ⊂ Nε + εB Y , Nε ⊂ Y, dim Nε < n .
now want to show that dˆn (T ) ≤ dn (T ). By denition, we can nd a
N ⊂ Y, dim N < n
Again
subspace
such that
sup inf kT x − yk ≤ dn (T ) + δ
kxk≤1 y∈N
for some
δ > 0.
That is to say, that for every element
x ∈ BX
exists a
y ∈ N,
such
that
n
o
1/%
T x ∈ y + (dn (T )% + δ % ) B Y ,
which means nothing more than
1/%
T B X ⊂ N + (dn (T )% + δ % ) B Y .
Hence be letting δ ↓ 0 we get dˆn (T ) ≤ dn (T ) . Now we
inequality. For δ > 0 we choose N ⊂ Y, dim N < n such that
T B X ⊂ N + dˆn (T ) + δ B Y ,
which means that for every
x ∈ BX
exists
21
y ∈ N,
such that
establish the opposite
n
o
ˆ
ˆ
T x ∈ y + dn (T ) + δ B Y ⇐⇒ kT x − yk ≤ dn (T ) + δ .
x ∈ B X over the inmum
subspaces and letting δ ↓ 0,
Of course this stays correct if we take the supremum over all
y ∈ N . Hence by
get dn (T ) ≤ dˆn (T ).
of all
we
In our
second step
taking the inmum over all such
we show the equivalence of (i) to (ii), where the denition in (ii)
is taken from [Pie87, Ch. 2., 2.5.2.]. So let
n - th Kolmogorov number
V ⊂ Y, V = V (ε) with dim V < n and get
denition of the
dn (T ) = ε.
According to the equivalent
established in (i), we can take a subspace
T B X ⊆ V + εB X .
By applying the quotient map of Denition 9 we get
QYV T B X ⊆ εQYV B X = εB Y/V .
kQYV T k ≤ ε and taking the inmum over all such subspaces V , and the notation
d˜n (T ) = inf kQYV T k : V ⊂ Y, dim V < n , we get d˜n (T ) ≤ dn (T ).
We now wish to show that d˜n (T ) ≥ dn (T ). We choose a δ > 0 and an arbitrary
Y
subspace V ⊂ Y with dim V < n, such that kQV T k < d˜n (T ) + δ . This inequality
Hence
implies
QYV T B X ⊆ d˜n (T ) + δ B Y/V
Lemma11
=
QYV
d˜n (T ) + δ B Y .
If we pass on to the inverse (as far as sets are concerned), we have
V + T B X ⊆ V + d˜n (T ) + δ B Y .
=⇒ T B X
Hence, with (i) we nd that
dn (T ) ≤ d˜n (T ),
˜
⊆ V + dn (T ) + δ B Y
dn (T ) ≤ d˜n (T ) + δ .
hence the equality is established.
If we let
δ ↓ 0,
we nally get
Now that we have established the three numbers with respect to operators, there
arises the question of a connection between them, on which we will have a look at in
the following part.
22
Part 3. Relationship between the Numbers
8. Relationship between Entropy and Approximation Numbers
To begin with the investigation of connections between the numbers, we shall start
with approximation- and entropy numbers.
We will compare both concerning mean
values and their limits. Since entropy numbers hold more information about the structure of the operator, one could identify them with the modulus of continuity, whereas
approximation numbers hold more information about how good an approximation can
be carried out and thus could be identied with the best approximation of continuous functions by polynomials.
Hence the idea of this part is to give inequalities of
Bernstein-Jackson type for operators and to introduce the theory of
s-scales.
Lemma 23. An operator T acting between real quasi-Banach spaces X and Y is of
rank m if and only if there exist constants C, c > 0 such that
c · 2−(n−1)/m ≤ en (T ) ≤ CkT k · 2−(n−1)/m , for n = 1, 2, 3, . . . .
If T acts between complex quasi-Banach spaces X and Y, it is of rank m if and only
if there exists a constant C, c > 0 such that
c·2−(n−1)/2m ≤ en (T ) ≤ CkT k · 2−(n−1)/2m , for n = 1, 2, 3, . . . .
Proof.
This statement is taken from [CS90, p. 21] and is slightly altered here for the
context of quasi-Banach spaces. It is used here without proof, but its validity can be
understood, by reconstructing [Har10, Satz 3.31] for general linear operators acting
between quasi-Banach spaces. (See also [Har10, Üb. III-4].)
With this lemma in mind, we can state the following proposition, which will become
useful later on. It can be found in [CS90, Sect. 3.1., Thm. 3.1.1.] for the case of Banach
spaces and is slightly modied here, to t in the context of quasi-Banach spaces.
A
similar theorem, but more general, as well as its proof can also be found in [ET96,
Subsect. 1.3.3.].
Proposition 24. Let 0 < p < ∞ and let T ∈ L (X, Y), where X and Y are arbitrary
quasi-Banach spaces with constants CX and CY . Then
sup k 1/p ek (T ) ≤ cp sup k 1/p ak (T ) for m = 1, 2, 3 . . . .
1≤k≤m
Proof.
1≤k≤m
As we have shown in Theorem 3 we can nd a proper
to the quasi-norm of
Y.
% - norm, which is equivalent
Thus it is enough to show, that the estimate stands for that case.
23
n = 2N , N ∈ N.
We proceed by looking at dyadic numbers
an (T )
of approximation numbers
rank
j
Aj < 2
According to the denition
(Denition 15), we nd operators
Aj ∈ L (X, Y)
with
, such that
kT − Aj k ≤ a2j (T ) + εj for j = 0, 1, 2, . . . , N
(8.1)
and for arbitrary
εj > 0.
Since we have no equality in Theorem 16 (Ra ), we are not
j
sure if it might happen that a2j (T ) = 0, although rank T ≥ 2 . If a2j (T ) 6= 0 we set
εj := a2j (T )
kT − Aj k ≤ 2a2j (T )
(8.2)
for
j = 0, 1, 2, . . . , N
A0 = 0. For simplicity we may and shall assume that a2j 6= 0, thus we can
set εj := a2j (T ) for all j = 0, 1, . . . , N . The argument is modied in an obvious way
otherwise. We may now take dierences Aj − Aj−1 for j = 1, . . . , N , which amounts to
j+1
rank (Aj − Aj−1 ) < 2
. With that in mind, we nd another representation of T
where
T =
N
X
(Aj − Aj−1 ) + (T − AN ) .
j=1
On the other hand, we may successively conclude that
en1 +...+nN −(N −1) (T )
(8.3)
%
≤
N
X
enj (Aj − Aj−1 )
%
+ kT − AN k%
j=1
for
nj
natural numbers to be chosen later.
Without loss of generality, we can say, that
and because of rank
(8.4)
j+1
(Aj − Aj−1 ) < 2
j+1
C > 0.
If
T
acts between real quasi-Banach spaces
and Lemma 23 we have the estimate
enj (Aj − Aj−1 )% ≤ C · 2−(nj −1)%/2
for some
T
kAj − Aj−1 k% ,
for
j = 1, 2, . . . , N
acts between complex quasi-Banach spaces , we use the
second statement of Lemma 23.
This has no eect on the proof except that we get
another constant. We proceed for the real quasi-Banach space case, by using the triangle
inequality, and have
24
(8.2)
kAj − Aj−1 k% ≤ kAj − T k% + kT − Aj−1 k% ≤ 2%+1 (a2j−1 (T ))%
(8.5)
for
j = 1, 2, . . . , N ,
where we also used the monotonicity of the approximation num-
bers in the last estimation. If we combine (8.4) and (8.5), we can conclude that
%
j+1
enj (Aj − Aj−1 ) ≤ C2 · 2−(nj −1)%/2 (a2j−1 (T ))% ,
(8.6)
for some
C2 > 0.
j = 1, 2, . . . , N
Now we estimate (8.3) by using (8.2) and (8.6), to get
en1 +...+nN −(N −1) (T )
(8.7)
for
%
≤ C2
N
X
2−(nj −1)%/2
j+1
(a2j−1 (T ))% + (2 · a2N (T ))% .
j=1
We shall now have a look at the sum. It can easily be seen that
N
X
2−(nj
−1)%/2j+1
N
X
(a2j−1 (T ))% ≤
j=1
%
−(j−1) p%
−(nj −1) j+1
2
!
1≤j≤N
j=1
N
X
≤
%
sup 2(j−1) p (a2j−1 (T ))%
2
%
%
!
2−(nj −1) 2j+1 −(j−1) p
%
sup j p (aj (T ))% .
1≤j≤2N
j=1
%
Furthermore we can nd an upper bound for (a2N (T )) , which is given through
(a2N (T ))% ≤ 2−N %/p sup1≤j≤2N j %/p (aj (T ))% . Combining these two estimates with (8.7)
amounts to
(8.8)
%
en1 +...+nN −(N −1) (T ) ≤
C2
N
X
%
−(nj −1) j+1
−(j−1) p%
2
2
− Np%
+ 2% · 2
!
sup j %/p (aj (T ))% .
1≤j≤2N
j=1
We will now choose the still free natural numbers
nj
in a way, that the large sum can
be estimated by terms, which can easier be expressed. Therefore we choose a natural
1 + p1 ≤ K ≤ 2 + p1 and set nj = 1 + K (N − j) 2j+1 for j = 1, 2, . . . , N . We can
−(nj −1)/2j+1
see that 2
= 2−K(N −j) . Using properties of a nite geometric series, we
number
now
conclude
25
N
X
−K(N −j)%−(j−1)%/p
2
= 2
%/p −%KN
2
N
X
2(K−1/p)%j
j=1
j=1
2(K−1/p)%N − 1
2(K−1/p)% − 1
2(K−1/p)%N − 1
= 2−KN % 2K% (K−1/p)%
2
−1
−N %/p
≤ C3 2
for some C3 > 0,
= 2%/p 2−%KN 2(K−1/p)%
where the last inequality is established through
1+
1
p
≤ K ≤ 2+
1
. Thus we can
p
now estimate (8.8) through
en1 +...+nN −(N −1) (T )
%
≤ 2−N %/p · C4 sup j %/p (aj (T ))%
1≤j≤2N
C4 > 0. We now need to estimate n1 + . . . + nN − (N − 1). Since we
have chosen nj already for j = 1, . . . , N , we conclude, that n1 + . . . + nN − N = K ·
PN
PN
j+1
j+1
= 4 2N − (N + 1)
. By induction we derive that
j=1 (N − j) 2
j=1 (N − j) 2
N
is valid for N ∈ N and thus n1 + . . . + nN − (N − 1) ≤ 4 · K · 2 . Using the monotonicity
of en (T ) we obtain
for some
(e4K2N (T ))% ≤ 2−N %/p C4 · sup j %/p (aj (T ))% ,
for
N = 1, 2, . . . .
1≤j≤2N
We shall next estimate
en (T ) for an arbitrary natural number n. Let us rst consider
K given as above). We then nd a natural number N such
n ≥ 8K (with
8K2N −1 ≤ n ≤ 8K2N .
the case,
that
Hence
2−N %/p sup j %/p (aj (T ))% ≤ (8K)%/p n−%/p sup j %/p (aj (T ))% .
1≤j≤n
1≤j≤2N
If we again use the monotonicity of the entropy numbers in combination with
N
4K2
, we have
(en (T ))% ≤ (e4K2N (T ))% ≤ C4 (8K)%/p n−%/p sup j %/p (aj (T ))%
1≤j≤2N
and this brings us nally to
n%/p (en (T ))% ≤ C4 (8K)%/p sup j %/p (aj (T ))% .
1≤j≤n
We will now see, that this estimate holds in the case
26
1 ≤ n ≤ 8K ,
since
n≥
n%/p (en (T ))% ≤ (8K)%/p kT k% ≤ C4 (8K)%/p sup j %/p (aj (T ))% .
1≤j≤n
We already know by construction that
K ≤2+
1
and by taking the
p
the supremum of both sides of the inequality with respect to
n ≤ m,
%
- th root and
we nally arrive
at
sup n1/p en (T ) ≤ cp sup n1/p an (T )
1≤n≤m
for
m = 1, 2, . . . .
1≤n≤m
This inequality will be needed in the following theorem, which is again slightly modied taken from [CS90, Sect. 3.1, Thm.3.1.1.; Sect. 3.5, Prop. 3.5.3. ], to t in the
context of quasi-Banach spaces.
Theorem 25. (Relationship of an & en )
Let X and Y be quasi-Banach spaces with constants CX , CY and T ∈ L (X, Y). Furthermore let 0 < p < ∞, then we have
1/p
P
en (T ) ≤ cp n1 ni=1 (ai (T ))p
limn→∞ en (T ) ≤ limn→∞ an (T ).
(i)
(ii)
Proof.
for
n = 1, 2, . . . .
(i)
The proof of the estimate of entropy numbers by the arithmetic mean of order
p
of
the approximation numbers is an immediate consequence of Proposition 24. With the
monotonicity of approximation numbers in mind, we will rst have a look at
1/p
k 1/p ak (T ) = (kak (T )p )
≤
k
X
!1/p
ai (T )p
.
i=1
=⇒ sup k
(8.9)
1/p
n
X
ak (T ) ≤
1≤k≤n
And now for arbitrary
en (T )
≤
n∈N
ai (T )
i=1
we have
n−1/p sup k 1/p ek (T ) ≤ cp · n−1/p sup k 1/p ak (T ) .
1≤k≤n
n
(8.9)
≤
!1/p
p
cp
1X
ai (T )p
n i=1
1≤k≤n
!1/p
.
27
(ii)
limk→∞ ek (T ) ≤ an (T ) for every n ∈ N. So
for xed n ∈ N we may and shall assume that an (T ) 6= 0, since an (T ) = 0 would
mean, that limn→∞ an (T ) = 0 and therefore T ∈ K (X, Y) . This however would result
in limn→∞ en (T ) = 0, which is the desired estimate. So let us choose an arbitrary
δ > an (T ). By denition of the approximation numbers, we can nd an operator
A ∈ L (X, Y) with rank A < n such that
Our rst step will be, to show that
kT − Ak < δ.
We nd that the following inclusion is valid:
T B X ⊆ (kT − Ak% + kAk% )1/% B Y .
Hence if we nd a covering of the set A B X , we have also found one for T B X . A
is a nite rank operator, because rank A < n, hence it is compact. Therefore for any
given ε > 0 we can nd nitely many elements yj ∈ Y, for j = 1, . . . , m, such that
(8.10)
A BX ⊆
m
[
yj + εB Y
j=1
indeed is a covering. With (8.10) we can now conclude that
T BX
m n
o
[
⊆
yj + (δ % + ε% )1/% B Y .
j=1
By denition and monotonicity of the entropy numbers we get
lim ek (T ) ≤ en (T ) ≤ (δ % + ε% )1/% .
k→∞
Since
ε > 0
and hence by taking the inmum over all
n = 1, 2, . . ..
ε ↓ 0
such δ , we
was arbitrarily chosen, we let
limk→∞ ek (T ) ≤ δ
limk→∞ ek (T ) ≤ an (T ) for
and obtain
get
Therefore we get
lim ek (T ) ≤ lim ak (T ) .
k→∞
k→∞
28
Remark
26
.
If
H1
and
sup 2−n/k
1≤k<∞
H2
k
Y
are Hilbert spaces and
T ∈ L (H1 , H2 ),
!1/k
≤ en (T ) ≤ 14 · sup 2−n/k
ai (T )
1≤k<∞
i=1
then
k
Y
!1/k
ai (T )
.
i=1
The proof can be found in [CS90, Sect. 3.4.; Thm. 3.4.2.].
9. Relationship between Approximation and Kolmogorov Numbers
Theorem 27. (Relationship of an & dn )
Let X and Y be quasi-Banach spaces and T ∈ L (X, Y), n ∈ N. Then
dn (T ) ≤ an (T ) .
Proof.
The proof is taken from [CS90, Rem. p.50] and [Har10, Lem. 3.34] for the case
of Banach spaces and is slightly altered here again. It starts with an arbitrary
and
n ∈ N.
We know the following
∃L ∈ L (X, Y) ,
since
R (L)
ε>0
an (T )
rank
L < n : kT − Lk ≤ an (T ) + ε,
is given through the best approximating operator. Now we dene
Un :=
and see through denition of the norm of an operator
∃Un , dim Un < n ∀x ∈ B X : kT x − LxkY ≤ an (T ) + ε.
Furthermore we get
Lx =: y ∈ R (L) = Un ,
and therefore
∃Un , dim Un < n ∀x ∈ B X ∃y ∈ Un : kT x − ykY ≤ an (T ) + ε.
We have seen, that there exist such
x ∈ BX,
y
and since the inequality is valid for arbitrary
we can take the supremum of all
x ∈ BX
over the inmum of these
y,
which
results to
∃Un , dim Un < n : sup inf kT x − ykY ≤ an (T ) + ε.
kxkX ≤1 y∈Un
Now taking the inmum over all such subspaces
Un
with
dim Un < n,
yields
dn (T ) < an (T ) + ε.
And since
ε>0
was arbitrary, we let
ε ↓ 0,
which is our wanted result.
29
Remark
28
.
Under the same conditions as in Theorem 27, we nd that
an (T ) ≤ (2n)1/2 dn (T )
is also valid. This inequality is taken from [CS90, Sect. 2.4. Prop. 2.4.6.] in the case
of Banach spaces. It is proved there with the help of the lifting of an operator and their
corresponding lifting constants, which are, in case of Banach spaces, bounded. Since
the proof of the quasi-Banach space case would be completely analogue, we only refer
to the proposition of the above authors.
Remark
29
.
case, where
For the matter of completeness, we will add without proof, that in the
X
is a Banach space,
H
a Hilbert space and
T ∈ L (X, H),
we get even an
equation
dn (T ) = an (T ) .
A proof can be found in [Pie78, Sect. 11., Prop. 11.6.2.].
10. Relationship between Entropy and Kolmogorov Numbers
At last we examine entropy numbers and Kolmogorov numbers. Equivalent to Proposition 24 we can make the following statement, taken from [Vyb08, Lem 4.4.], which is
altered here only in the exponent.
Proposition 30. Let α > 0, 0 < p < ∞ and X and Y be two quasi Banach spaces with
constants CX and CY . Further let T ∈ L (X, Y). Then for n ∈ N there exists a constant
c > 0 such that
sup k α ek (T ) ≤ cp,α · sup k α dk (T ) .
1≤k≤n
Proof.
1≤k≤n
According to [Car81, Thm. 1.] it is enough to show that
sup k α ek (T ) ≤ cp,α · sup k α sk (T )
1≤k≤n
in the case of Banach spaces, where
1≤k≤n
sk (T ) either
denotes Kolmogorov or approxima-
tion numbers, since this statement would be valid for the corresponding other number
as well. We have already shown this relation for approximation numbers in Proposition
24, thus the proof is done for Banach spaces.
To extend this result for the desired
quasi-Banach spaces, we refer to [BBP95, Lem. 1.].
30
Corollary 31. (Relationship of en & dn )
Let X and Y be quasi-Banach spaces with constants CX , CY and T ∈ L (X, Y). Furthermore let 0 < p < ∞, then we have
n
en (T ) ≤ cp
Proof.
1X
(di (T ))p
n i=1
Based on Proposition 30 with
α=
!1/p
for n = 1, 2, . . . .
1
, the proof follows Theorem 25 (i) analop
gously.
11. Axiomatic Theory of
s-Numbers
If we compare Theorems 14, 16 and 20 we nd many similarities, like the monotonicity
or additivity of the corresponding numbers. There is however another way to introduce
numbers like approximation- and Kolmogorov numbers. This axiomatic way and goes
back to Albrecht Pietsch. For further detail one may have a look at [Pie87, Sect. 2.2.].
To t in the context of quasi-Banach spaces we will follow the notation of [Vyb08, Sect.
2.4.].
Denition 32. (s - numbers )
W,X,Y and Z be quasi-Banach spaces and T ∈ L (X, Y). A rule s : T →
(sn (T ))n∈N , which assigns to every operator a scalar sequence is called an s - scale if it
Let
satises the following conditions
kT k = s1 (T ) ≥ s2 (T ) ≥ . . . ≥ 0
sm+n−1 (T ) ≤ CY (sm (T ) + sn (T )) where CY is the constant of the quasi%
%
%
Banach space Y or likewise sm+n−1 (T ) ≤ sm (T ) + sn (T ) , for an equivalent % - norm with % ∈ (0, 1]. S ∈ L (X, Y) and n, m ∈ N.
sn (RT U ) ≤ kRksn (T ) kU k for all U ∈ L (W, X) , T ∈ L (X, Y), R ∈
L (Y, Z) and n ∈ N.
rank T < n =⇒ sn (T ) = 0
sn (id : `n2 → `n2 ) = 1
(Ms )
(As )
(Ss )
(Rs )
(Is )
Furthermore an
- scale is said to be multiplicative if it also satises
sm+n−1 (RT ) ≤ sm (R) sn (T )
(Ps )
Remark
s
33
.
for every
R ∈ L (Y, Z)
and
m, n ∈ N.
At rst we see, that entropy numbers do not t in this construct, since
they do not satisfy (Rs ) as we can see in Lemma 23.
As stated above, this is another approach to the theory of approximation- and Kolmogorov numbers. We can easily make sure, that these numbers are indeed
31
s
- scales,
since we have already shown all of the necessary properties. (Theorem 16 and Theorem
20)
Although they will not be a part of this bachelor's thesis we denote that there are
many other concepts of
s
- scales, as for example
cn (T ) := inf kT SVX k : codim V < n , where V
subspace of the Banach space X.
Weyl numbers : xn (T ) := sup {an (T S) : S ∈ L (`2 , X) , kSk ≤ 1}
(i)
Gelfand numbers :
(ii)
The proof that these numbers are
s
is a
- scales indeed, can be found in [Pie87, Sect. 2.4.
Thm. 2.4.3.*, Thm. 2.4.14.*] and is left out here, since it would go beyond the intended
scope of this bachelor's thesis.
Part 4. Compact Embeddings
12. id :
`np → `nq
and entropy numbers
T := id `np → `nq
denote ek := ek (T ),
In this part we will only deal with one specic operator, which is
0 < p, q ≤ ∞ and n ∈ N. For reasons of abbreviation we
ak := ak (T ) and dk = dk (T ) for the whole following part. But before
for given
we investigate
the embeddings, we need to have a look at the following proposition, taken from [ET96,
Subsect. 3.2.1., Prop.]. Since we introduced the sequence spaces
not clear whether we deal with real or complex elements, but
2n
R
n
C
`p
over
R
or
C,
it is
may be identied with
. Using this interpretation we make the convention, that with the volume volB `n
,
p
n
o
p
Pn
2
2 2
x ∈ R2n :
x
+
x
≤
1
to cover
2j−1
2j
j=1
identify x2j−1 and x2j with real and imaginary part of
we mean the Lebesgue-2n-measure of
both cases. Our aim here, is to
a complex number, but we reduce it to two real values.
Proposition 34. Let n ∈ N, then
n
0<p≤∞
(i)
If
, then the volume of the unit ball in
(ii)
There exists a function
such that for all
volB `n
p
Proof.
1
θ : (0, ∞) → R,
p ∈ (0, ∞)
= 2n−1 π 2 (3n−1) p
−(n−1)
2
2n
with
`np
is volB `n
p
0 < θ (x) <
1
12
Γ(1+ 2 )
= π n Γ 1+ p2n
( p)
for all x > 0,
1
n− p − 2 exp (nθ (2/p) p/2 − θ (2n/p) p/2n) .
Since this statement is not crucial for the topic of this bachelor's thesis and
will be needed later on only for a matter of constants, it will be used without proof.
Nevertheless the proof can be found in [ET96, Subsect. 3.2.1., Prop.].
32
The following proposition is taken from [ET96, Subsect. 3.2.2., Prop.].
Proposition 35. (ek of id : `np −→ `nq , upper estimate)
Let 0 < p ≤ q ≤ ∞ then
ek ≤ cp,q ·
(12.1)



1
if 1 ≤ k ≤ log2 (2n)
k −1 log2 1 +



k
− 2n
2
(2n)
p1 − 1q
2n
if log2 (2n) ≤ k ≤ 2n
k
1
− p1
q
if k ≥ 2n
where cp,q > 0 is a constant independent of n and k .
Proof.
First of all, for a matter of abbreviation we introduce the notation
n
B p := B `np .
rst
To proof the whole estimate, we are going to need four steps. We begin with the
1
−1
− k
, which deals with large k ≥ 2n and 0 < p ≤ q ≤ 1. We set r = 2 2n (2n) q p and
n
j
j
m
furthermore K = K(r) be the maximal number of points y ∈ B p with ky − y kq > r
n
if j 6= m. Now for given z ∈ B q and by choice of r , we can show that for p ≤ 1
step
ky j + rzkp`np ≤ 1 + rp kzkp`np
1
1
≤ 1 + rp kzkp`nq np( p − q ) ≤ 2,
(12.2)
where the second estimate is obtained through Hölder's inequality.
Now let
{y j : j = 1, . . . , K} be such a set, which is maximal in the above sense.
Then
clearly we get
n
Bp
(12.3)
⊂
K
[
n (12.2)
y j + rB q
1
n
⊂ 2 p Bp .
j=1
1
exists an element
(12.4)
z
n
y j + 2− q rB q
If we have a look at the balls
for
j = 1, . . . , K
and assume, that there
which is in two of these balls, then
ky j − y m kq`nq ≤ ky j − zkq`nq + ky m − zkq`nq ≤ rq ,
| {z } | {z }
≤2−1 rq
which is by choice of the
yj
m = j,
Together with (12.3) this yields
−2n
q
q ≤ 1,
≤2−1 rq
only possible, if
K2
for
n
2n
r2n volB q ≤ 2 p
33
n
volB p ,
hence the balls are disjoint.
Cn with R2n . With
c = c(p, q) > 0, independent
where the constants arise, because of our convention of identifying
Proposition 34 (ii) we see that for some positive constant
of
k, n
1
1
≤ c2n (2n)−2n( p − q ) volB q
n
volB p
n
where we again used Hölder's inequality. Combining this estimate with the preceding
r
one and our choice of
yields
K ≤ 2k+cn
and hence
1
−k
1
ek+cn ≤ r = 2 2n (2n) q − p
(12.5)
by denition of
ek .
Since
k + cn
will from now on use the notation
if
k ≥ 2n,
does not necessarily has to be a natural number, we
eλ = ebλc+1
if
λ ≥ 1,
where
bλc
denotes the smallest
λ.
integer bigger than
ek + cn = ek̃ ≤ 2
| {z }
−(k̃−cn)
2n
1
1
c
1
−k̃
1
(2n) q − p = 2 2 · 2 2n (2n) q − p
k̃
k ≥ c1 n
0 < p ≤ q ≤ 1.
Thus if
Our
k ≥ 2n
second step
for
c1 > 1
independent of
n
and
k,
we have proved (12.1) for
will be only a modication of the rst one.
is large and notice, that the argument above holds for all
we only used properties of the
p
- and accordingly the
q -norms,
We still assume that
0 < p ≤ q ≤ ∞,
since
thus we only need to
alter these points by using the triangle inequality in (12.2) and (12.4). The rest of the
proof proceeds analogously. In particular, we consider the case
know by Theorem 14, that
kid
:
ek (T ) ≤ kT k
small
ek ≤ 1
. We
and since
p( p1 − 1q )
`np → `nq k = sup kxk`nq ≤ sup n
| {z }kxk`np ≤ 1,
kxk n ≤1
kxk n ≤1
`p
hence
0<p=q≤∞
for all
k ∈ N.
`p
=1
This proves the statement for
p=q
for mid-ranged and
k.
We proceed with the
third step,
which covers the case
k ≤ c1 n. Here c1 has the same meaning
−1/p
c2 > c11 log2 1 + c11
and set
34
as before.
0 < p < q = ∞
and
1 ≤
We choose a second constant
1/p
n
1/p
n
1
−1
−1/p n
+1
log2
+1
(12.6)
σ := c2 k log2
= c2 n
> n− p ,
k
k
k
where the last estimation occurs by choice of c2 . We dene nσ as the maximal number
n
of components yn , which a point y = (y1 , y2 , . . . , yn ) ∈ B p may have, for which |yn | > σ .
n
By the preceding estimate (12.6), we have nσ < n, otherwise y ∈
/ B p . Furthermore we
p
−p
−p
get nσ σ ≤ 1, hence nσ ≤ σ
. Let us now assume that σ
∈ N and nσ σ p = 1.
This is possible, because we can always nd such a number σ which satises the above
conditions. We set
(σ)
ek := ek
id
: `np σ → `n∞σ
and with (12.5) from the rst step, where we set
k = c1 σ −p ,
we know that
(σ)
ec1 σ−p ≤ c3 n−1/p
= c3 σ,
σ
(12.7)
−p
c1 ≥ 1 and c3 ≥ 1. This estimate means, that we need 2c1 σ balls in `n∞ with
nσ
n
radius c3 σ to cover B p . But since we want to cover the whole B p , we need to know
in how many ways we can select nσ coordinates out of n. The number of possibilities
n
c σ −p n
n
is given through
and therefore we know, that we need 2 1
balls in `∞ with
nσ
nσ
n
radius c3 σ to cover B p . To estimate further, we need the fact, that for given natural
numbers N, K ∈ N0 , with N ≥ K we have an upper bound for the binomial coecient
NK
N
through
≤ K! . By using this and properties of the logarithm, we get
K
for
log2
n
nσ
n
≤ log2
σ
X
nnσ
= nσ log2 n −
log2 j
nσ !
j=1
≤ nσ log2 n − nσ log2 nσ + cnσ
n
0
≤ c nσ log2
+1
nσ
where
c
and
c0
denote some positive constants. Combining this estimate, with the
above construction, we know, that we need
2c4 σ
(12.8)
balls in
`n∞
with radius
c3 σ
−p
log2 ( nn +1)
σ
to cover
= 2c4 σ
n
Bp ,
−p
where
log2 (nσ p +1)
c4 > 0
is independent of
Furthermore with (12.6) and the logarithm properties, we get
35
k
and
n.
n
n
log2 cp2 log2
+1 +1
k
k"
n
log cp + log n + 1 + log log
2 k
2 2
2 2
≤
c00 · log2
+1
k
log2 nk + 1



p
n

n
log
c
+
log
log
+
1


2
2
2
2
k
+ 1 1 +
= c00 · log2

n
k


log2 k + 1
{z
}
|
log2 (nσ p + 1)
=
n
k
+1
#
≤c000
≤
c̃ · log2
n
k
+1
c00 , c000 and c̃. Hence we can estimate (12.8) from above with 2c5 k
constant c5 ≥ 1 independent of n and k by using (12.6) . Hence with
for some constants
for a positive
(12.7) we nally get
ec5 k
(12.9)
σ
≤ c6 = c6
c2
n
1
log2
+1
k
k
p1
if
1 ≤ k ≤ c1 n,
c6 > 0 is also independent of n and k . (12.1) follows, if we have in mind that
ek ≤ 1 for all k ∈ N always assuming that 0 < p < ∞ and q = ∞.
We now advance to the fourth step, which deals with 0 < p < q < ∞ and 1 ≤ k ≤ c1 n,
where c1 has still the same meaning as in the rst step. We show (12.1) by using [ET96,
Subsect. 1.3.2., Thm. 1.], with the following cast and θ ∈ (0, 1)
where
A = B0 = `np , B1 = `n∞ , Bθ = `nq ,
where
(1 − θ)
1
=
.
q
p
`np ∩ `n∞ ⊂
`np ∩ `n∞ = `np and
Now if we consider the premises of the theorem, we have to check that
`nq ⊂ `np + `n∞ , which can be rewritten as `np ⊂ `nq ⊂ `n∞ since we have
`np + `n∞ = `n∞ . The statement follows from the monotone alignment
of the sequence
spaces.
We may now conclude, that
ek+1−1
id :
`np → `nq
1
≤ 2 p e1−θ
|1
1
p
≤ 2 eθk
id :
36
`np → `np eθk
{z
}
id :
≤1
`np → `n∞ .
id :
`np → `n∞
ek
This yields
n
id : `p
→ `nq ≤ ceθk
: `np → `n∞
id
θ
p
and we use (12.9) with
=
1
p
−
1
q
.
This, and the fact that
ek ≤ 1 ∀k ∈ N
We have established an upper estimate for the
operator between the two spaces
`np
proves the theorem.
and
`nq .
k
- th entropy number of the identity
As we already know, this operator is
compact, since it maps between two nite dimensional quasi-Banach spaces. This can
also be seen if
k → ∞,
as we have shown in the properties of entropy numbers. We
will now go on by giving a lower estimate for both large and small
theorem, taken from [Tri97, Sect.
ranged
k
7, Prop.
7.2, Thm.
7.3].
k
by the following
We will deal with mid-
later on.
Proposition 36. (ek of id : `np −→ `nq , lower estimate I)
Let 0 < p ≤ q ≤ ∞ and k ∈ N then
ek ≥ c ·

1
if 1 ≤ k ≤ log2 (2n)
k
− 2n
2
(2n)
1
− p1
q
if k ∈ N
for some positive constant c, which is independent of k and n but may depend on p
and q .
Proof.
In the
rst step
of this proof, we will show the rst inequality. Let
all components are zero except for one, which is either
there exist
2n
such elements in
us now assume that
y
1
and
y
2
`np ,
1
or
−1.
y ∈ `np
where
Then we know, that
which also happen to belong to
B `np
and
are two such elements belonging to the same
B `nq . Let
ε- ball in
`nq . That means
y 1 ∈ x + εB `nq
and
y 2 ∈ x + εB `nq
for some x
∈ `nq .
0 < q < 1 and 1 ≤ q ≤ ∞, let q = min {1, q}.
c > 0, which is independent of n and q , and satises
Since we want to cover the cases where
Next we consider a constant
c ≤ ky 1 − y 2 kq`nq ≤ ky 1 − xkq`nq + kx − y 2 kq`nq ≤ 2εq .
The wanted estimate follows by denition of the entropy numbers (since we unify all
ε- balls containing 2 elements and take the inmum over all such ε) the preceding
k−1
estimate and the fact that k ≤ log2 2n implies 2
< 2n.
these
We proceed with our
second step, which deals with the second inequality.
k−1
n
balls in `q with radius
ε > 0 such, that B `np is covered by 2
n
2n
that C is identied with R , this yields
for appropriate
37
ε
ε.
We choose
With the convention
≤ 2k−1 ε2n volB `nq
volB `n
p
≤ 2k e2n
.
k volB `n
q
(12.10)
Now according to [Tri97, Sect.
constants
c1 , c2
7, 7.1] for
0 < p ≤ ∞
there exist two positive
such that
1
c1 n− p ≤
volB `n
p
2n1
1
≤ c2 n − p .
If we combine this with (12.10), we obtain the desired inequality.
Since there is only one estimate missing for the case of mid-ranged
k,
we will now
have a look at this case. Therefore we follow the results of a paper by [Küh01], which
deals exactly with this missing case.
The results can almost directly be transferred,
`p
except for the fact, that the sequence spaces
are complex in this bachelor's thesis.
Lemma 37. (ek of id : `np −→ `nq , lower estimate II)
Let 0 < p ≤ q ≤ ∞. Then
p1 − 1q
2n
ek ≥ c · k log2 1 +
k
for some positive constant c which is independent of n and k but may depend on p
and q .
Proof.
−1
`np and `nq as real valued and begin
n ≥ 4 and 1 ≤ m ≤ n4 and dene the set
In this proof, we will rst consider the spaces
with two arbitrary integers
n, m ∈ N
with
(
S :=
x = (xj )nj=1 ∈ {−1, 0, 1}n :
n
X
)
|xj | = 2m .
j=1
It is plain to see, that
#S =
n
2m
2m
·2
, since we need exactly
element, which are not zero. There are
we can only choose between
Furthermore we notice, that
the Hamming distance on
S,
n
2m
2m components of every
ways to choose from them and because
−1 and 1, there are 22m ways
(2m)−1/p S is contained in the
to design such an element.
unit sphere of
which is
h (x, y) := # {j ∈ {1, . . . , n} : xj 6= yj } .
We observe, that for xed
x∈S
we get an upper bound for
38
`np .
Let
h
be
n
# {y ∈ S : h (x, y) ≤ m} ≤
· 3m ,
m
since we can obtain every element y ∈ S with h (x, y) ≤ m as follows: If we choose
an arbitrary set J ⊂ {1, . . . , n} with #J = m, then we set yj = xj for j ∈
/ J and choose
yj ∈ {−1, 0, 1} arbitrarily for j ∈ J . We proceed by dening an arbitrary subset A ⊂ S
n
n
with a cardinality not exceeding a :=
/ m . Hence
2m
# {y ∈ S : ∃x ∈ A with
n
h (x, y) ≤ m} ≤ #A ·
· 3m
m
n
≤
· 3m < #S.
2m
y ∈ S with
h (x, y) > m for all x ∈ A. Therefore we may inductively construct a subset à ⊆ S with
#Ã > a and h (x, y) > m for x, y ∈ Ã, x 6= y . For such x, y we conclude kx−ykq > m1/q
−1/p
. Now we see that (2m)
à ⊂ B `np . As we have already established, this set has a
cardinality larger than a. Furthermore we see, that the elements of this set have a
−1/p
distance of kx − ykq > (2m)
· m1/q =: ε. If we now set k := log2 a and use the
notation of entropy numbers eλ with λ ≥ 1 of the rst step of Proposition 35 we see
Through this estimate, it has been shown, that we can nd an element
where
c1 > 0
is independent of
n
2m
n
m
a=
By our choice of
n
and
Therefore we can estimate
(12.12)
1
1
ε
= c1 m q − p ,
2
n. Now we have
ek ≥
(12.11)
c2 m log2
n
m
k
or
a closer look at
a,
which is
m
Y n − 2m + j
m! (n − m)!
=
=
.
(2m)! (n − 2m)! j=1 m + j
decreases
m we notice that f (x) = n−2m+x
m+x
m
m
≤ a ≤ n−2m
and get
a through n−m
2m
m
≤ m log2
n−m
2m
≤ k ≤ m log2
n − 2m
m
for
≤ m log2
x > 0.
n
m
c2 > 0, which is independent of n and m. Furthermore we see that the
n
n
function g (x) = x · log2
is strictly increasing on 1;
and maps this interval on
x
4
n
log2 n; 2 . Since this function is strictly increasing and continuous, its inverse exists
y
on the latter interval and we see that x ≤
, by
log2 ( n
y)
for some
39
n
n
1
= ·
x
y
x log2 nx
n
n
n
= log2
− log2 log2
=⇒ log2
y
x
x
log n
y
2 x
= x·
≥ x.
=⇒
log2 log2 ( n
n
x)
log2 ny
log2 x 1 − log n
2( x )
n
c n
. We now consider log2 n ≤ k ≤ 2 and
The last inequality occurs because x ∈ 1;
4
2
n
k
,
because
set x = m as well as y = k and get m ≤ 2 ·
≥
2
and
therefore
k
log2 ( n
+1)
k
n
n
2 · log2 k ≥ log2 k + 1 .
y = x · log2
n
⇐⇒
Furthermore we conclude with (12.11) and (12.12)
1/p−1/q

 log 1 + n n
+1 
log2 m
k
 2
k
·

·
≥ c·

n
n
k

m log2 m + 1 log2 k + 1 
|
{z
} |
{z
}
ek
≥c3
≥ c0
for
c0 > 0,
log2 1 +
k
≥c4
!1/p−1/q
n
k
log2 n ≤ k ≤
for
which is independent of
n
and
k.
c2 n
2
The lower estimate
c4
arises from the
following
log2
log2
n
m
n
k
+1
log n
log2 n
log2 n
≥
≥ c̃ ·
2
= c̃ ·
≥ c4 .
log2 n − log2 log2 n
+1
log2 logn n + 1
log2 logn n
c n
The case 2
2
2
≤k≤n
2
follows from the monotonicity of entropy numbers and a lower
estimate, which we get from (12.11) through
1
1
en ≥ c00 n q − p ,
for a certain
c00 > 0
(also independent of
ek ≥ c000 ·
n
and
log2 1 +
k
is valid.
40
n
k
k ),
since for these
!1/p−1/q
Of course this was only the proof for real-valued sequence spaces
analogously, if we set
n = 2l
`np
but the proof is
and consider complex-valued sequence spaces.
Now we estimated every case and nish this part by summarizing the results in the
following theorem.
Theorem 38. (Behavior of ek id : `np → `nq )
Let 0 < p ≤ q ≤ ∞. Then
ek id :
`np
→
`nq
∼



1
k −1 log2 1 +



Proof.
if 1 ≤ k ≤ log2 (2n)
k
− 2n
2
(2n)
1
− p1
q
p1 − 1q
2n
k
if log2 (2n) ≤ k ≤ 2n
if k ≥ 2n.
The proof rests on Propositions 35, 36 and on Lemma 37.
13. id :
`np → `nq
and approximation numbers
Next we will give upper and lower estimates of the numbers
ak
`np → `nq
id :
. Here,
we will mostly follow [ET96, Subsect. 3.2.3.] and as these authors have done, we will denote real valued sequence spaces by
`n,R
p
and correspondingly
aRk :=
id :
→ `n,R
`n,R
q
p
the approximation numbers of the identity operator, acting between real valued sequence spaces. We will rst mention an estimate for the latter ones and proceed by
giving a relation between approximation numbers of the identity operator acting between real and complex valued sequence spaces later on.
Theorem 39. Let k ≤ n. We dene

n
o p11 − 11q

1
1

−

2−q
min 1, n q k 2
if 2 ≤ p < q ≤ ∞



q
o
o
n
n

1
1

1
− p1
q
1 − nk
if 1 ≤ p < 2 ≤ q ≤ ∞
, min 1, n q k − 2
Φ (n, k, p, q) = max n

1−1


 1 1 q
p1 1q 


−
−p
k

q

,
max n
1− n p 2
if 1 ≤ p < q ≤ 2




and
Ψ (n, k, p, q) =

Φ (n, k, p, q)
if 1 ≤ p < q < p0
Φ (n, k, q 0 , p0 ) if max {p, p0 } < q ≤ ∞
where for given 1 ≤ p, q ≤ ∞, p0 and q 0 are given through
respectively.
41
1
p
+
1
p0
= 1 and
1
q
+
1
q0
=1
(i) If we assume that 1 ≤ p < q ≤ ∞ and (p, q) 6= (1, ∞), Then
aRk ∼ Ψ (n, k, p, q) ,
where the constants of equivalence only depend on p and q .
(ii) If 1 ≤ p ≤ q ≤ 2 or 2 ≤ p ≤ q ≤ ∞, then
s
k
R
ak ≥
1−
.
n
Proof.
A reference for the proof is given in [ET96, Subsect. 3.2.3, Thm. 1].
Since these were only the approximation numbers for real valued sequence spaces, we
want to give a relationship to the complex ones, as it is done in [ET96, Subsect. 3.2.3.,
Prop.]
Lemma 40. Let k ∈ N, k ≤ n and suppose that p, q ∈ [1, ∞]. Then
aR2k−1 ≤ ak ≤ 2aR2k
(where αkR = 0 if k > n).
Proof.
Again we shall use this statement without proof in this bachelor's thesis and
refer to the above mentioned proposition.
For a matter of completeness we shall add the following two lemmata.
They are
directly taken from [Vyb08, Lem 3.3., Lem 3.4.].
Lemma 41. If 1 ≤ k ≤ n < ∞ and 0 < q ≤ p ≤ ∞, then
ak = (n − k)1/q−1/p .
Proof.
According to the author, this statement is a generalization of the proof for
q≤p≤∞
1≤
from [Pie78, Subsect. 11.11.5., Lem.]. It is also left without proof in this
bachelor's thesis.
Lemma 42. Let 0 < p ≤ 1.
(i) Let 0 < λ < 1. Then there exists a number cλ > 0 such that for all k, n ∈ N with
nλ < k ≤ n, we have
cλ
ak id : `np → `n∞ ≤ √ .
k
(ii) There is a number c > 0 such that for n ≥ 1
42
Proof.
Again we only refer to [Vyb08, Lem. 3.4.] for the proof.
Corollary 43. Let n ∈ N, 1 ≤ p, q ≤ ∞ and k ≤ n4 , then



1




min 1, n 1q k − 12
ak ∼
1
1

p0 k − 2

min
1,
n




1
if 1 ≤ p < q ≤ 2
if 1 ≤ p < 2 ≤ q < p0
if 1 ≤ p ≤ 2 ≤ p0 ≤ q ≤ ∞ and (p, q) 6= (1, ∞)
if 2 ≤ p ≤ q ≤ ∞
where p0 is dened through the equation
Proof.
1
p
+
1
p0
= 1.
The proof rests upon Theorem 39 and Lemma 40.
We proceed by extending these results for cases, where
p, q ∈ (0, 1)
in the following
theorem. (See [ET96, Subsect. 3.2.3, Thm. 2].)
Theorem 44. Let n ∈ N, then
(i) If 0 < p ≤ q ≤ 2 and k ≤ n4 , then ak ∼ 1.
1
1
1
(ii) If 0 < q ≤ p ≤ ∞ and n = 2k , then ak ≥ 2− q n q− p .
1
n
− 12
0
q
.
(iii) If 0 < p < 2 < q < p and k ≤ 4 ,then ak ∼ min 1, n k
Proof.
In the
rst step, we prove (iii).
th component
δjr
for
j, r = 1, . . . , n
βk := ak
Let
id :
`n1 → `nq
and
xr ∈ `n1
with
j
-
(which denotes the Kronecker delta). It is plain to
see, that
(
B `n1 =
x=
n
X
λr xr :
r=1
So let
T :
`n1
→
n
X
r=1
=⇒
βk ≤
)
|λr | = 1 .
r=1
`nq be linear and with rank
x − Tx =
n
X
T < k,
then we get for
λr (xr − T xr ) =
| {z }
=:wr
n
X
x ∈ B `n1
λr wr .
r=1
sup kx − T xkq ≤ sup kwr kq
r=1,...,n
x∈B `n
1
sup k
=
id :
r=1,...,n
(13.1)
≤
k
id :
`np → `nq − T xr kq
`np → `nq − T k.
Thus by taking the inmum over all legitimate operators
we arrive at
43
T
and with Corollary 43
1
1
min 1, n q k − 2 ∼ βk ≤ ak .
(13.2)
On the other hand, we obtain the opposite inequality by the fact that in this case
p < 1
and the monotonicity of the
`np
spaces.
In particular we use
`np ,→ `n1 .
This
completes step one.
We proceed with the
second step,
in which we prove (i).
At rst, we assume that
0 < p < 1 < q ≤ 2 from which we get (13.2) with 1 on the left-hand side with Corollary
n
43. Furthermore we get ak ≤ βk , since p < 1 and the monotonicity of the `p spaces (As
n
n
above `p ,→ `1 ). Hence ak ∼ 1. We now investigate the remaining case 0 < p ≤ q ≤ 1
n
n
and set βk = ak id : `q → `q . As we have shown earlier in Theorem 16 (Na ), we know
that βk = 1 for those k admissible here. The analogue to (13.2) is
1 ≤ k
n
X
λr wr kqq
≤
sup
r=1,...,n
≤
|λr |q kwr kqq
r=1
r=1
≤
n
X
kwr kqq
sup k
id :
kxkp ≤1
= sup kxr − T xr kqq
r=1,...,n
`np → `nq − T k.
T , we get ak id : `np → `nq ≥
1. We conclude the second step by using the monotonicity of the `np spaces again, which
nally yields ak ∼ 1.
Hence, by taking the inmum over all legitimate operators
The third step will be, to prove the last remaining part of this theorem, which is (ii).
T : `np → `nq be represented as an n x n matrix with rank T < k = n2 . We
n
know that dim ker T > k =
and use V.D. Milman's lemma (see [Pie87, Sect. 2.9., Lem.
2
2.9.6.]) . From there, it follows that there exists an element x = (x1 , . . . , xn ) ∈ ker T
with |xj | ≤ 1 for all j = 1, . . . , n and, (for example) |x1 | = . . . = |xn/2 | = 1. For this x
Therefore let
we know
kxkq ≥ (n/2)1/q
(13.3)
and
kxkp ≤ n1/p .
Therefore we conclude
kxkq = k (I − T ) xkq ≤ k (I − T ) kkxkp .
44
Here
I
stands for
same with
I.
id :
`np → `nq
, but since we identied
By taking the inmum over all such operators
T
T
as a matrix, we do the
and the above estimate,
we nally get
ak
kxkq (13.3) − 1 1 − 1
`np → `nq ≥
= 2 q nq p ,
kxkp
id :
which concludes the third step, as well as the theorem.
As far as approximation numbers of the identity operator between nite sequence
spaces are concerned, there is one last corollary, that we will add in this bachelor's
thesis, which is a conclusion of Corollary 43. It is taken from [Cae98, Cor. 2.2.].
Corollary 45. Let 0 < p ≤ 2 ≤ q < ∞ (or 1 < p ≤ 2 < q = ∞ ). Then
(i) there exists c > 0 such that for all k, n ∈ N
1
0
ak ≤ cn1/ min{p ,q} k − 2 .
(ii) there exists c > 0 such that for all k, n ∈ N with k ≤ 14 n2/ min{p ,q}
0
ak ≥ c.
Proof.
To prove (i) we will rst have a look at
16 we then have
ak = 0
k > n.
As we have shown in Theorem
and therefore the required estimate is valid. We now consider
the remaining case, which is
k ≤ n.
Our aim is to use Corollary 43 and therefore we
have a look at the composition
J

We dene
(ξ1 , . . . , ξn )
id
J (ξ) = ξ1 , . . . , ξn , 0, . . . , 0 
| {z }
for
ξ ∈ `4n
q .
ak
id :
P
4n
n
`np −→ `4n
p −→ `q −→ `q .

for
ξ = (ξ1 , . . . , ξn ) ∈ `np
as well as
P (ξ) =
3n times
This yields
`np → `nq
= ak+1−1 P ·
id :
≤ a1 (P ) · ak+1−1
≤ kP kkJkak
id :
4n
·J
`4n
→
`
q
p
4n
4n
id : `p → `q
·J
4n
`4n
,
p → `q
where we used the properties (Ma ) and (Pa ) of Theorem 16. Obviously
P
and
J
are
bounded operators and we are now in position to use Corollary 43. This amounts to
45
ak
id :
1
0
`np → `nq ≤ c · n1/ min{p ,q} k − 2 ,
0<p≤2≤q<∞
0
(or completely analogue 1 < p ≤ q = ∞). Therefore we have min {p , q} ≥ 2, since the
1 2/ min{p0 ,q}
0
0
conjugate index p is set p = ∞ if 0 < p ≤ 1. However k ≤ n
, amounts to
4
k ≤ n4 , which brings us again in position to use Corollary 43. Furthermore we get
which is (i). The next step will be proving (ii). Our premise is still
n
1/ min{p0 ,q} − 21
≥n
k
which nally yields
ak ≥ c
1/ min{p0 ,q}
for some
·
c>0
1 2/ min{p0 ,q}
n
2
− 12
independent of
=
k
√
and
2 > 1,
n.
14. id :
`np → `nq
and Kolmogorov numbers
The last remaining numbers in this bachelor's thesis, concerning id :
Kolmogorov numbers.
`np → `nq
are
With our investigation we will mostly follow the notations of
[Vyb08, Sect. 4.] and start with a lemma taken from there. ([Vyb08, Lem. 4.2.])
Lemma 46. Let 1 ≤ k ≤ n < ∞ and 1 ≤ p, q ≤ ∞. We dene

1
1

(n − k + 1) q − p
if 1 ≤ q ≤ p ≤ ∞



1−1


n
o p1 1q

1
1


2−q
q k− 2
min
1;
n
if 2 ≤ p < q ≤ ∞




1−1
p q 
Φ (n, k, p, q) :=
 1 1 q
1 1

−p
k p−2

q
n
max
if 1 ≤ p < q ≤ 2
;
1
−


n






n
n
o q
o


max n 1q − p1 , min 1, n 1q k − 12 · 1 − k
if 1 ≤ p < 2 < q ≤ ∞.
n
Then dk id : `np → `nq ∼ Φ (n, k, p, q) if q < ∞, where the constants of equivalence
are independent of k and n but may depend on p and q .
Furthermore there exist cp , Cp > 0 such that
cp Φ (n, k, p, ∞) ≤ dk id :
`np
→
`n∞
for 1 ≤ p ≤ ∞.
Proof.
≤ Cp Φ (n, k, p, ∞) log
en 3/2
k
As it is done in [Vyb08, Lem. 4.2.], we refer to [Glu84, Thm. 1].
46
1 ≤ p, q ≤ ∞.
The above Lemma only contained cases, in which
some estimates which apply to quasi-Banach spaces.
We shall now add
Again, the following Lemma is
taken from [Vyb08, Lem. 4.3.].
Lemma 47. If 0 < q ≤ p ≤ ∞, then there exists a constant c > 0 such that
1
1
2n
ddcne+1 id : `2n
& n q − p , for k ∈ N
p → `q
where dcne denotes the upper integer part of cn.
Proof.
First, we consider the case in which
q ≥ 1.
Then we have a special case of [Pie78,
Subsect 11.11.4., Lem. 1.] , which states
dn
id :
1
1
m
`m
& (m − n + 1) q − p
p → `q
Since this argument does not stand for
q < 1
for
1 ≤ n ≤ m.
we recall two facts.
The rst is
Proposition 36, where we have shown, that
ek
for some constant
c1 > 0
1
1
k
2n
`2n
≥ c1 · 2− 4n (4n) p − q
p → `q
id :
depending only of
q
and
p.
id :
2n
`2n
.
p → `q
The second fact is Proposition
30 with which we derive that
1
1
c2 · nα n q − p ≤ sup k α dk
1≤k≤n
That means, that for every
n∈N
there exists
kn ≤ n
1
1
c2 · nα n q − p ≤ knα dkn
Furthermore there exists a constant
id :
c ∈ (0, 1],
such that
2n
`2n
.
p → `q
such that for all
n ∈ N n ≥ k ≥ cn.
Combining this conclusion with the preceding estimate nally amounts to
1
1
c2 · n q − p ≤ d[cn]+1
id :
2n
.
`2n
p → `q
47
Part 5. Relationship to Spectral Theory
15. Preliminary Considerations
This last part will give a connection between entropy numbers as well as approximation numbers and eigenvalues of compact operators of innitely dimensional Hilbert
spaces by the inequalities of Carl and Weyl.
But when it comes to the relationship
between the here considered quantities, we cannot avoid certain denitions.
Denition 48.
Let
X
be an arbitrary, complex quasi-Banach space and
T ∈ L (X).
We then call
% (T ) = λ ∈ C : ∃ (T − λidX )−1 ∈ L (X) the resolvent set of T .
(ii) σ (T ) = C\% (T ) the spectrum of T .
(iii) λ an eigenvalue of T , if there exists an element x ∈ X, x 6= 0 with T x = λx.
p
n
(iv) r(T ) = limn→∞
kT n k the spectral radius of T .
(i)
A result, following from this denition, is the next proposition.
To maintain the
intended scope of this bachelor's thesis we shall only give the statement without proving
it. A proof however can be found in [ET96, Sect. 1.2., Thm.].
Proposition 49. Let X be a complex innite-dimensional quasi-Banach space and T ∈
K (X). Then the spectrum σ (T ) consists only of {0} and at most countably innite
number of eigenvalues of nite algebraic multiplicity, which accumulate only at 0. That
is
σ (T ) = {0} ∪ (λk )k∈N ⊂ C : λk 6= 0, λk eigenvalue of T, dim N (T − λk id) < ∞ .
Proof.
Without proof. (See reference above.)
Because of this proposition, we can construct a sequence out of all non-zero eigenvalues
(λk )k∈N ,
(15.1)
such that
|λ1 (T )| ≥ |λ2 (T )| ≥ . . . ≥ 0,
where we repeated and ordered the
λk
according to their algebraic multiplicity. If
T
m (< ∞) distinct eigenvalues and M is the sum of their algebraic multiplicities,
then we simply put λn (T ) = 0 for every n > m. From now on, we will refer to this
ordered sequence as the eigenvalue sequence of T .
has only
48
16. The Inequality of Carl
The perhaps most useful connection to the here concerned quantities is the following
inequality, which was rst stated in [CT80].
Later on it was extended to t in the
context of quasi-Banach spaces by [ET96].
Theorem 50. (Inequality of Carl)
Let X be an arbitrary complex quasi-Banach space and T ∈ K (X) with its eigenvalue
sequence λ1 (T ) , λ2 (T ) , . . . , λn (T ) , . . .. Then
! k1
k
Y
≤ inf 2 2k en (T ) for k ∈ N.
n
|λm (T ) |
n∈N
m=1
Proof.
As we have mentioned above, the proof can be found in [ET96, Subsect. 1.3.4.,
Thm.].
From this useful theorem, we can draw an immediate conclusion, which is the following corollary.
Corollary 51. Let T be as above. For all k ∈ N we have
|λk (T )| ≤
Proof.
k ∈ N.
Let
√
2ek (T )
With (15.1) we get
k· k1
|λk (T ) | = |λk (T ) |
=
k
Y
!1/k
≤
|λk (T ) |
52
.
!1/k
|λm (T ) |
.
m=1
i=1
The estimate follows immediately, if we set
Remark
k
Y
k=n
in Theorem 50.
The above inequality was rst discovered by [Car81, Thm. 4] in the context
of Banach spaces. Using this result it can be shown that
lim (ek (T n ))1/n = r(T )
n→∞
for
k ∈ N.
(See [ET96, Subsect. 1.3.4., Rem. 1].)
Let us now examine the relationship between eigenvalues of an operator
where
X
T ∈ K (X),
denotes an arbitrary complex Banach space and the approximation numbers.
It has been shown by [Kön86, Prop. 2.d.6., p. 134] that for
|λn (T ) | = lim a1/k
Tk ,
n
k→∞
49
n∈N
as well as for
p ∈ (0, ∞),
N
X
that for some constant
!1/p
|λk (T ) |p
Kp
N
X
≤ Kp
k=1
!1/p
p
[ak (T )]
.
k=1
17. Hilbert Space Setting and the Inequality of Weyl
At last, let us study the Hilbert space setting. It is only natural that better results
arise in this case.
Since we are now dealing with Hilbert spaces, we can talk about
inner products and therefore about adjoint operators. (For further information about
the adjoint operator, see [Tri92, Subsect. 2.2.3.]).
Denition 53.
operator of
T
Let
with
H
T∗
T ∈ L (H).
∗
if T = T .
be a Hilbert space and
and call
T
self-adjoint
We will denote the adjoint
Proposition 54. Let H be a Hilbert space and T ∈ K (H)be self-adjoint. For given
n ∈ N, we have
|λn (T ) | = an (T ) ,
where λn (T ) is the n - th eigenvalue of the eigenvalue-sequence of T .
Proof.
For the proof, see [CS90, Sect. 4.4., Prop. 4.4.1.].
Remark
say, that
55
.
T
If we additionally demand that
hT x, xi ≥ 0
for all
x ∈ H)
is a non-negative operator (that is to
we even get the equality
λn (T ) = an (T ).
(See
[ET96, Subsect. 1.3.4., Rem. 1.])
The perhaps most popular inequality concerning a relationship between approximation numbers and eigenvalues of an operator is the inequality of Weyl.
Theorem 56. (Inequality of Weyl)
Let H be a Hilbert space and T ∈ K (H) with its eigenvalue sequence (λk (T ))k∈N .
Then for given n ∈ N we have
n
Y
|λi (T ) | ≤
i=1
n
Y
ai (T )
i=1
and even equality if dim H = n < ∞. Furthermore, for given p ∈ (0, ∞), we get
n
X
|λi (T ) |p ≤
i=1
Proof.
n
X
[ai (T )]p .
i=1
For the proof see [CS90, Sect. 4.4., Prop. 4.4.2.].
50
References
[BBP95]
[Car81]
[CS90]
[CT80]
[Cae98]
[DL93]
[ET96]
[Glu84]
[Har10]
[Har11]
[Kön86]
[Küh01]
[Pie78]
[Pie87]
[Pis89]
[Tri92]
[Tri97]
[Vyb08]
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51
Declaration of Authorship
I declare that this document has been composed by myself, and describes my own
work, unless otherwise acknowledged in the text.
Furthermore I accept, that this bachelor's thesis is free to use in the university archive.
___________________________________________
(Place, Date, Signature)