ESTADO DE ARTE

This section is motivated by linear relations of type S⁻¹T for linear operators S, T as they are used in [2]. As we have seen in Lemma 1.13, we can write every linear relation L in such a way: L = Q⁻¹_L Lop.

First, we want to characterize relations of type S⁻¹T , where S and T are operators.

Lemma 1.29 For two linear operators S : Y ⊇ dom S → Z, T : X ⊇ dom T → Z, the product L := S⁻¹T is a linear relation in X × Y , which can be characterized by

(x, y) ∈ S⁻¹T ⇐⇒ x ∈ dom T ∧ y ∈ dom S ∧ T x = Sy.

Proof: We have S⁻¹T =

(x, y) ∈ X × Y 

 ∃z ∈ Z : (x, z) ∈ T, (z, y) ∈ S⁻¹

= {(x, y) ∈ X × Y | ∃z ∈ Z : (x, z) ∈ T, (y, z) ∈ S}

= {(x, y) ∈ X × Y | x ∈ dom T, y ∈ dom S ∧ ∃z ∈ Z : T x = z, Sy = z}

= {(x, y) ∈ X × Y | x ∈ dom T ∧ y ∈ dom S ∧ T x = Sy} .  Intuitively, in a relation L = S⁻¹T with S, T as in the preceding proposition, T represents the “operator-like” behavior of L whereas S determines its multivalued part, though we have to be careful, since for L only those x ∈ X and y ∈ Y matter which are mapped into ran S ∩ ran T by T and S respectively. The first evidence for that intuition is the following lemma.

Lemma 1.30 Let S, T be linear operators as in Lemma 1.29 and let L := S⁻¹T . Then:

(i) dom L = T⁻¹[ran T ∩ ran S] ⊆ dom T , (ii) ran L = S⁻¹[ran T ∩ ran S] ⊆ dom S, (iii) ker L = ker T and

(iv) mul L = ker S.

Proof: The first two points follow immediately from Lemma 1.29. – We have ker L = {x ∈ X | (x, 0) ∈ L}

1.29= {x ∈ X | x ∈ dom T ∧ 0 ∈ ran S ∧ T x = S0 = 0}

= {x ∈ dom T | T x = 0} = ker T.

By replacing L with L⁻¹, we get mul L = ker L⁻¹= ker S, since L⁻¹= T⁻¹S.  This justifies setting

e

α(L) := inf {n ∈ N | ker L ∩ ran Lⁿ = {0}},

compare the definition of the ascent of T relative to S in [2] in section 1, page 149:

α(T : S) := inf n ∈ N

 ker T ∩ ran(S⁻¹T )ⁿ= {0}

.

Now, let L ⊆ X × Y be a linear relation. We examine to which extend L can be decomposed into a product S⁻¹T of operators S, T : X → Y .

Definition 1.8 We say a vector space Z and two linear operators T : X ⊇ dom T → Z and S : Y ⊇ dom S → Z split L with respect to the auxiliary space Z, if L = S⁻¹T . The triple (Z, S, T ) will be called split of L. In case dom T = dom L and dom S = ran L, we call (Z, S, T ) a minimal split of L with respect to Z.

Proposition 1.31 Every split (Z, S, T ) of L induces a minimal split (Zmin, Smin, Tmin) by setting Zmin:= Z, Smin:= S|ran L and Tmin:= T |dom L.

Proof: This follows immediately from Lemma 1.29. 

A trivial, yet useful observation is

Proposition 1.32 If (Z, S, T ) is a (minimal) split of L, then (Z, T, S) is a (minimal) split of L⁻¹.

Lemma 1.33 Every linear relation L is splittable. More precisely:

(_{mul L}^{ran L}, QL, Lop) is a minimal split of L, so L can be written as L = Q⁻¹_L Lop(= Q⁻¹_L QLL),

where

Lop: X ⊇ dom L → ran L mul L is the associated linear operator and

QL: Y ⊇ ran L → ran L mul L is the canonical quotient map.

Proof: This was shown in Lemma 1.13 for QL: Y → _{mul L}^Y and Lop: dom L → _{mul L}^Y , which we can obviously compress to ran L → _{mul L}^{ran L} and dom L → _{mul L}^{ran L} respectively.

This way we guarantee the minimality of the split and Lopand QL being surjective.

We will see that minimality of the split and surjectivity of the involved operators can always be achieved by modifying the operators involved marginally (in algebraic terms).

Applying the lemma to L⁻¹yields

Corollary 1.34 

dom L

ker L, Q_L−1, L⁻¹

op

 with QL⁻¹ : X ⊇ dom L → ^{dom L}_{ker L} splits L⁻¹ minimally.

Now applying Proposition 1.32 to the corollary, we get

Corollary 1.35

dom L ker L, L⁻¹

op, QL⁻¹

splits L.

But the two splits of L we constructed are actually isomorphic, in an adequate sense.

Definition 1.9 Let (Z1, S1, T1) and (Z2, S2, T2) be splits of L. A linear map J : Z1→ Z2such that

S2= JS1 and T2= JT1

will be called a morphism (Z1, S1, T1) → (Z2, S2, T2). We will also say J induces such a morphism.

Proposition 1.36 If (Z1, S1, T1) is a split of L and J : Z1 → Z2 is an injective linear map, then (Z2, S2, T1) with T2= J ◦ T1 and S2= J ◦ S1is a split of L.

Proof: We have

S₂⁻¹T2= (J ◦ S1)⁻¹(J ◦ T1) = S₁⁻¹J⁻¹JT1= S₁⁻¹T1= L,

because J⁻¹J = IZ1, see Lemma 1.9. 

Remark: A straightforward calculation verifies that Split_Lis a category given by the class Ob(Split_L) :=

(Z, S, T ) | Z F -vector space, S : Y ⊇ dom L → Z, T : X ⊇ dom T → Z linear operators such that L = S⁻¹T

of splits, together with morphisms

MorSplit_L((Z1, S1, T1), (Z2, S2, T2)) := {J : Z1→ Z2 linear map such that S2= JS1, T2 = JS2}

for two given splits (Z1, S1, T1), (Z2, S2, T2), where the composition and an identity are given by the composition of linear maps and identity on the vector space respectively. That justifies the term morphism defined above.

Proposition 1.37 The following diagram of vector spaces and linear operators com-mutes:

dom L ran Lmul L

dom Lker L ran L

Lop

L⁻¹

op

QL⁻¹ J QL

∼=

where J : _{mul L}^{ran L} → ^{dom L}_{ker L} is the canonical linear isomorphism ^(L⁻¹)_op associated with (L⁻¹)opand J⁻¹ is the canonical linear isomorphism gLopassociated with Lop.

Proof: The only thing we have to prove is J⁻¹ = gLop: let x ∈ dom L, then

Exchanging the roles of L and L⁻¹ yields the other relation gLopJ = I^{ran L}

mul L. 

Corollary 1.38 The map J from the preceding proposition induces an isomorpism

ran L

Using Lemma 1.30, we get the following import result about minimal splits (Z, S, T ) with surjective T or surjective S: they are actually all isomorphic.

Theorem 1.3 Let L ⊆ X × Y be a linear relation minimally split by (Z, S, T ). If T is surjective, then there exists an isomorphism

dom L ker L, L⁻¹

op, QL⁻¹



→ (Z, S, T ),

which is induced by the linear isomorphism associated with T : X ⊇ dom L → Z.

As consequence all splits (Z, S, T ) with surjective T are isomorphic.

The dual statement for splits (Z, S, T ) with surjective S also holds true.

Proof: Let eT : ^{dom L}_{ker T} → Z be the linear isomorphism associated with T . By Lemma 1.30, we have ker T = ker L, so actually eT : ^{dom L}_{ker L} → Z. Since

The second equality is the very definition of eT , so we only have to prove the first equality. It is more convenient to show the equivalent equation eT⁻¹S = L⁻¹

op: we have Te⁻¹S = Q_L−1T⁻¹S = Q_L−1L⁻¹= Q_L−1Q⁻¹_L−1 L⁻¹

op,

where the last equality is an immediate consequence of Corollary 1.38. Now, since QL⁻¹ is an operator, by Lemma 1.9 it holds true that QL⁻¹Q⁻¹_L−1 = I^{dom L}

ker L. But that means

Te⁻¹S = I^{dom L}

ker L L⁻¹

op= L⁻¹

op. 

Corollary 1.39 For every minimal split (Z, S, T ) of L we have: T is surjective ⇐⇒

S is surjective.

Proof: Consider “⇒”: This is obvious for the split (_{mul L}^{ran L}, QL, Lop), and for every other minimal split (Z, S, T ) of L we have an isomorphism J : _{mul L}^{ran L} → Z such that S = JQL, which immediately implies the surjectivity of S. – The other implication follows from

duality. 

Definition 1.10 A minimal split (Z, S, T ) with surjective S or T will itself hence-forth be called surjective.

For algebraic considerations, it suffices to consider surjective minimal splits, since every split (Z, S, T ) can be made minimal by restricting the domains of S and T , and it can be made surjective by replacing the auxiliary space Z by the range of S, which coincides with the range of T .

Lemma 1.40 Let (Z, S, T ) be a minimal split of L, then ran S = ran T =: Z1. The maps S1: Y ⊇ ran L → Z2, T1: X ⊇ dom L → Z2split L in Z2, and the embedding Z2֒→ Z1 is a morphism (Z2, S1, T1) → (Z, S, T ).

Proof: Let z ∈ ran T , then there exists an x ∈ dom L with T x = z and (x, y) ∈ L for some y ∈ ran L. But since L = S⁻¹T , the latter means Sy = T x = z, so z ∈ ran S.

The converse inclusion follows from using what we just showed on the split (Z, T, S)

of L⁻¹. The rest of the lemma is trivial now. 

Remark: At this point it becomes clear that all results of [2] using only numbers and spaces concerning S⁻¹T with operators S, T can be applied to arbitrary relations.

Suppose we have two linear relations L, K ⊆ X × Y with L ⊆ K, and let K be split by (Z, S, T ). One could surmise that now (Z1, S1, T1) splits L, where Z1 := Z, S1:= S|ran L and T1 := T |dom L. But, unfortunately, this is not true in general. For instance, Lemma 1.30 tells us that

ker L = ker T1= ker T ∩ dom L = ker K ∩ dom L

and analogously mul L = mul K ∩ ran L are necessary conditions, though in general, we only have “⊆” between these spaces.

Example 1.4 Let X = Y = R², and consider K := span {(e1, 0), (0, e1), (0, e2)} and L := span {(e1, e2)}. Then L ⊆ K, ker L = {0}, but

ker K ∩ dom L = span {e1} ∩ span {e1} = span {e1} ) {0} .

It can be immediately verified that K = S⁻¹T , where S : R² → R², S = 0 and T : span {e1} → R², T = 0. But L is already an operator, so (R², IR², L) splits L.

Obviously, it is impossible to construct a split isomorphic to that formed by S and T . But if we have ker L = ker K ∩ dom L or mul L = mul K ∩ ran L, then we have both equalities, and (Z1, S1, T1), as defined above, splits L.

Lemma 1.41 Let L, K ⊆ X × Y be linear relations with L ⊆ K, and let K be split by (Z, S, T ). If additionally

ker L = ker K ∩ dom L (1.3)

or

mul L = mul K ∩ ran L, (1.4)

then both equalities hold and (Z1, S1, T1) splits L, where Z1:= Z, S1:= S|ran L and T1:= T |dom L).

Proof: We have

dom T |dom L= dom T ∩ dom L = dom L,

since dom T ⊇ dom K ⊇ dom L, and dom S|ran L= ran L analogously. Furthermore, (x, y) ∈ L =⇒ x ∈ dom L ∧ y ∈ ran L ∧ (x, y) ∈ K = S⁻¹T

⇐⇒ x ∈ dom L ∧ y ∈ ran L ∧ T x = Sy

⇐⇒ x ∈ dom T |dom L ∧ y ∈ dom S|ran L ∧ T |dom L(x) = S|ran L(y)

⇐⇒ (x, y) ∈ S₁⁻¹T1. This shows L ⊆ S₁⁻¹T1.

Now, suppose (1.3) holds true. We will first show that equality (1.4) also holds. Let y ∈ mul K ∩ ran L, so (0, y) ∈ K and there is some x ∈ dom L such that (x, y) ∈ L.

Then (x, 0) ∈ K, so x ∈ ker K ∩ dom L = ker L, which means (x, 0) ∈ L. Together with (x, y) ∈ L this implies (0, y) ∈ L, so y ∈ mul L. The converse inclusion is trivial.

– If conversely (1.4) holds true, we have

ker L⁻¹= ker K⁻¹∩ dom L⁻¹,

which by what we just showed implies mul L⁻¹= mul K⁻¹∩ ran L⁻¹, or equivalently ker L = ker K ∩ dom L.

Again, assume (1.3), in order to show S₁⁻¹T1 ⊆ L. Using the equivalences we have proved in the beginning, we examine (x, y) ∈ (dom L) × (ran L) with (x, y) ∈ K. We must have some x^′ ∈ dom L such that (x^′, y) ∈ L. But then (x − x^′, 0) ∈ K, which means x − x^′ ∈ ker K ∩ dom L = ker L. So actually (x − x^′, 0) ∈ L, which implies

(x, y) = (x − x^′, 0) + (x^′, y) ∈ L.  In [2], the authors also make use of T S⁻¹for operators S, T . In the following para-graphs, we want to concern ourselves with this “commutation” of the split components.

This is, of course, only possible for the case X = Y , so in the following we will always consider a linear relation L ⊆ X × X.

Definition 1.11 Let (Y, S, T ) be a split of L. The relation

Lc:= T S⁻¹= {(y, y^′) ∈ Y × Y | ∃ x ∈ dom S ∩ dom T : Sx = y ∧ T x = y^′}

= {(Sx, T x) | x ∈ dom S ∩ dom T }

is called L commutated with respect to (Y, S, T ).

In case of a surjective minimal split (Y, S, T ), Lc is unique up to an isomorphism of splits, so we just call it “L commutated”.

Right from the definition, we see that Lcdoes only depend on the behavior of S and T on dom S ∩ dom T .

Proposition 1.42 Let S, T be linear operators in X to Y . Then T S⁻¹= T |dom S∩dom T(S|dom S∩dom T)⁻¹.

Again, it is easy to see, but useful, that commutating a relation commutes with inverting.

Proposition 1.43 We always have L⁻¹_c = (L⁻¹)c.

Proof: L⁻¹_c = (T S⁻¹)⁻¹= ST⁻¹= (L⁻¹)c, since L⁻¹ is split by (Z, T, S).  The usual linear subspaces ker, ran, dom associated with the powers of S⁻¹T and T S⁻¹ are the basis of all considerations in [2], cf. section 1, page 149 ibid. We will partly recover their notions in the more general context of linear relations here.

Following [2], later we will only consider relations L and splits (Z, S, T ) of those with dom T ⊆ dom S. Dual to Lemma 1.30 we have

Lemma 1.44 Let L be splitted by (Y, S, T ), then we have

(i) dom Lc = S[dom T ], and dom Lc = ran S if and only if dom S ⊆ dom T + ker S, (ii) ran Lc= T [dom S], and ran Lc= ran T if and only if dom T ⊆ dom S + ker T , (iii) ker Lc= S[ker L] = S[ker T ] and

(iv) mul Lc= T [mul L] = T [ker T ].

Proof: We have

dom Lc= {y ∈ Y | ∃ z ∈ Y : (y, z) ∈ Lc}

= {y ∈ Y | ∃ z ∈ Y ∃ x ∈ dom S ∩ dom T : Sx = y ∧ T x = z}

= {y ∈ Y | ∃ x ∈ dom S ∩ dom T : Sx = y} = S[dom T ].

Passing to the inverse, we get the following by what we have just shown:

ran Lc= dom L⁻¹_c = dom(L⁻¹)c= T [dom S](⊆ ran T ).

If dom T ⊆ dom S + ker T , then

ran T = T [dom T ] ⊆ T [dom S + ker T ] = T [dom S] = ran Lc. If on the other hand ran T ⊆ ran Lc, then

dom T = T⁻¹[ran T ] ⊆ T⁻¹[ran Lc] = T⁻¹[T [dom S]].

So, let x ∈ dom T , then T x = T v for some v ∈ dom S, which menas x − v ∈ ker T and thus

x = v + (x − v) ∈ dom S + ker T.

Again, exchanging the roles of S and T via inverting Lc yields dom Lc= ran S ⇐⇒ dom S ⊆ dom T + ker S.

Concerning ker Lc:

ker Lc = {y ∈ Y | ∃ x ∈ dom S ∩ dom T : Sx = y, T x = 0 = S0}

= S[{x ∈ dom S ∩ dom T | (x, 0) ∈ L}] = S[ker L] = S[ker T ].

Also:

mul Lc= ker(Lc)⁻¹= ker L⁻¹_c = T [ker L⁻¹] = T [mul L] = T [ker S].  This justifies setting

eδ(Lc) := inf {n ∈ N | ran Lc+ ker Lⁿ_c = Y },

compare the definition of the descent of T relative to S in [2], section 1, page 149:

δ(T : S) := inf n ∈ N

 ran T + ker(T S⁻¹)ⁿ = Y .

Remark: Without the additional assumption dom T ⊆ dom S, we would have to consider the relation (T S⁻¹)|ran T, if we wanted to reconstruct exactly the same spaces R_m^′ from [2].

Indeed, we have

R₁^′ = T X = ran T, R_m^′ = ran((T S⁻¹)|ran T)^m−1 ∀ m ≥ 2.

But we see, that the case m = 1 seems to be an exception, and the relation which we have to consider for m ≥ 2 is rather cumbersome. A closer examination can improve the situation:

It is easy to see that ((T S⁻¹)|ran T)^m= ((T S⁻¹)^m)|ran T for all m ≥ 1. Concerning the first flaw, we observe

ran((T S⁻¹)⁰)|ran T = ran Iran T = ran T.

So we have

R_m^′ = ran((T S⁻¹)^m)|ran T ∀ m ≥ 1,

which is compact enough, but we would no longer deal with ranges of powers of a relation.

On the other hand, note that dom T ⊆ dom S implies

ker L^p+ ran L ⊆ dom L + ran L ⊆ dom T + dom S ⊆ dom S.

So unless dom S 6= X, eδ(L) < ∞ is impossible.

The “pointwise characterization” of Lc is by far not as handy as it was in the case of S⁻¹T . But powers of Lc are closely related to those of L.

Proposition 1.45 Let n ∈ N, n ≥ 1, then

Lⁿ= S⁻¹Lⁿ⁻¹_c T and Lⁿ_c = T Lⁿ⁻¹S⁻¹, or in terms of pairs: (x1, xn+1) ∈ Lⁿ if and only if

∃ (y1, yn) ∈ Lⁿ⁻¹_c : T x1= y1 ∧ yn= Sxn+1, so if and only if (T x1, Sxn+1) ∈ Lⁿ⁻¹_c .

Respectively: (y1, yn+1) ∈ Lⁿ_c if and only if

∃ x1∈ dom S, xn∈ dom T : (x1, xn) ∈ Lⁿ⁻¹ ∧ y1= Sx1 ∧ T xn= yn+1.

Proof: Follows easily by induction. 

Corollary 1.46 Let n ∈ N, n ≥ 1, then (i) ker Lⁿ= T⁻¹[ker Lⁿ⁻¹_c ],

(ii) ker Lⁿ_c = S[ker Lⁿ].

If additionally dom T ⊆ dom S, we have for n ≥ 1:

(iii) ran Lⁿ= S⁻¹[ran Lⁿ_c] and (iv) ran Lⁿ_c = T [ran Lⁿ⁻¹].

Proof: For n ≥ 1 argue using the preceding proposition:

x ∈ ker Lⁿ ⇐⇒ (x, 0) ∈ Lⁿ

⇐⇒ (T x, S0) = (T x, 0) ∈ Lⁿ⁻¹_c

⇐⇒ x ∈ T⁻¹[ker Lⁿ⁻¹_c ];

also:

x ∈ S[ker Lⁿ]

⇐⇒ ∃ z ∈ dom S ∩ ker Lⁿ : Sz = x

⇐⇒ ∃ z ∈ dom S ∩ dom T : (z, 0) ∈ Lⁿ ∧ Sz = x

⇐⇒ ∃ z ∈ dom S ∩ dom T ∃ (y1, yn) ∈ Lⁿ⁻¹_c : T z = y1, yn= S0 = 0, Sz = x

⇐⇒ ∃ z ∈ dom S ∩ dom T ∃ (y1, 0) ∈ Lⁿ⁻¹_c : T z = y1, Sz = x

⇐⇒ ∃ y1∈ Y : (x, y1) ∈ Lc, (y1, 0) ∈ Lⁿ⁻¹_c

⇐⇒ (x, 0) ∈ Lⁿ_c

⇐⇒ x ∈ ker Lⁿ_c. Concerning (iii):

x ∈ ran Lⁿ ⇐⇒ ∃ x1∈ dom T : (x1, x) ∈ Lⁿ

⇐⇒ ∃ x1∈ dom T : (T x1, Sx) ∈ Lⁿ⁻¹c ,

where the last step is possible, because x ∈ ran Lⁿ⊆ ran L ⊆ dom S. Since we assumed dom T ⊆ dom S, we can also apply S to x1, so that the last line is equivalent to

∃ x1∈ dom T ∩ dom S : (Sx1, T x1) ∈ Lc, (T x1, Sx) ∈ Lⁿ⁻¹_c

⇐⇒ ∃ x1∈ dom S ∩ dom T : (Sx1, Sx) ∈ Lⁿ_c

⇐⇒ Sx ∈ ran Lⁿ_c ⇐⇒ x ∈ S⁻¹[ran Lⁿ_c].

And for (iv):

y ∈ ran Lⁿ_c

⇐⇒ ∃ z ∈ Y : (z, y) ∈ Lⁿ_c

⇐⇒ ∃ z ∈ Y, x1∈ dom S, xn∈ dom T : (x1, xn) ∈ Lⁿ⁻¹, z = Sx1, T xn = y

⇐⇒ ∃ x1∈ dom S, xn ∈ dom T : (x1, xn) ∈ Lⁿ⁻¹, T xn= y.

(x1, xn) ∈ Lⁿ⁻¹ already implies x1 ∈ dom T , and on the other hand, because of the additional condition dom T ⊆ dom S, the last line in the chain of equivalences is in fact equivalent to

∃ x1∈ dom T, xn ∈ dom T : (x1, xn) ∈ Lⁿ⁻¹, T xn= y

⇐⇒ y ∈ T [dom T ∩ ran Lⁿ⁻¹] = T [ran Lⁿ⁻¹]. 

Remark: Note that (iv) is in general not true for n = 0. We only have ran L⁰= X ⊇ dom S = S⁻¹[Y ] = S⁻¹[ran L⁰c].

The following lemma condenses Lemma 1.3 and Lemma 1.4 from [2]. We will need it later on to prove a pertubation result.

Lemma 1.47 Let λ 6= 0 and m ≥ 1. Then

(i) ker L^m_c = (λS − T )T⁻¹[ker L^m−1_c ] = (λS − T )[ker L^m], (ii) ker L^m= T⁻¹(λS − T )[ker L^m−1],

(iii) ran L^m∩ dom T = (λS − T )⁻¹T [ran L^m−1] = (λS − T )⁻¹[ran L^m_c ] and (iv) ran L^m_c = T (λS − T )⁻¹[ran L^m−1_c ].

Proof:

(i) Using Corollary 1.46 as well as Lemmas 1.6, 1.9 and 1.5, we get (λS − T )[ker L^m] ^1.46= (λS − T )T⁻¹[ker L^m−1_c ]

1.6 + 1.9

⊆ (λST⁻¹− Iran T)[ker L^m−1_c ]

1.5⊆ L⁻¹_c [ker L^m−1_c ] + ker L^m−1_c

1.5= ker L^m_c + ker L^m−1_c ^1.5= ker L^m_c .

For the converse inclusion, suppose y ∈ ker L^mc = S[ker L^m], so there exists some x0∈ ker L^msuch that Sx0= y. x0∈ ker L^m means we find an L-chain

(x0, x1, . . . , xn−1, 0).

In particular: xn−1∈ ker L = ker T . Set the converse inclusion by induction. For m = 1 let

x ∈ (λS − T )⁻¹T [ran L^m−1] = (λS − T )⁻¹T [Y ]

where the last step follows from Lemma 1.44. For all other natural numbers, we use (iii):

ran L^m+1_c = T [ran L^m∩ dom T ]

= T [(λS − T )⁻¹[ran L^m_c ]]

= T (λS − T )⁻¹[ran L^m_c ] ∀ m ≥ 1. 

In document UNIVERSIDAD NACIONAL DE SAN ANTONIO ABAD DEL CUSCO (página 41-55)