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J. Numer. Anal. Approx. Theory, vol. 44 (2015) no. 1, pp. 11–24 ictp.acad.ro/jnaat

ENLARGING THE CONVERGENCE BALL OF NEWTON’S METHOD ON LIE GROUPS

IOANNIS K. ARGYROSand SA¨ID HILOUT„

Dedicated to prof. I. P˘av˘aloiu on the occasion of his 75th anniversary

Abstract. We present a local convergence analysis of Newton’s method for approximating a zero of a mapping from a Lie group into its Lie algebra. Using more precise estimates than before [55, 56] and under the same computational cost, we obtain a larger convergence ball and more precise error bounds on the distances involved. Some examples are presented to further validate the theoretical results.

MSC 2010. 65G99, 65J15, 47H17, 49M15, 90C90.

Keywords. Newton’s method, Lie group, Lie algebra, Riemannian manifold, convergence ball, Kantorovich hypothesis.

1. INTRODUCTION

In this paper, we are concerned with the problem of approximating a zero x? of C1-mappingF : G−→Q, where Gis a Lie group andQthe Lie algebra of G that is the tangent space TeGof G ate, equipped with the Lie bracket [., .] : Q×Q−→Q[7, 21, 28, 29, 53].

The study of numerical algorithms on manifolds for solving eigenvalue or optimization problems on Lie groups [1]–[12], [33]–[36], [54]–[57] is very im- portant in Computational Mathematics. Newton-type methods are the most popular iterative procedures used to solve equations, when these equations contain differentiable operators. The study about convergence matter of New- ton’s method is usually centered on two types: semilocal and local conver- gence analysis. The semilocal convergence matter is, based on the information around an initial point, to give criteria ensuring the convergence of Newton’s method; while the local one is, based on the information around a solution, to find estimates of the radii of convergence balls. There is a plethora of studies on the weakness and/or extension of the hypothesis made on the underlying operators; see for example (cf. [7, 11, 16] and references theiren). A local as

Department of Mathematical Sciences, Cameron University, Lawton, OK 73505, USA, e-mail: [email protected].

„Poitiers University, Laboratoire de Math´ematiques et Applications, Bd. Pierre et Marie Curie, T´el´eport 2, B.P. 30179, 86962 Futuroscope Chasseneuil Cedex, France, e-mail:

[email protected].

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well as a semilocal convergence of Newton-type methods has been given by several authors under various conditions [2]–[58]. Recently in [15], we pre- sented a finer semilocal convergence analysis for Newton’s method than in earlier studies [7, 32, 34, 36, 54, 55, 56].

Newton’s method with initial point x0Gwas first introduced by Owren and Welfert [43] in the form

(1.1) xn+1=xn·exp (−dFx−1n F(xn)) for each n= 0,1, . . . .

Newton’s method is undoubtedly the most popular method for generating a sequence {xn} approximating x? [7, 11, 15, 16, 32, 34, 36], [54]–[56]. In the present paper we establish a finer local convergence analysis with the advantages (Al):

(i) Larger convergence ball;

and

(ii) Tighter error bounds on the distances involved.

The necessary background on Lie groups can be found in [7, 11, 15, 16] and the references therein. The paper is organized as follows. In Section 2, we present the local convergence analysis of Newton’s method. Finally, numerical examples are given in the concluding Section 3.

2. LOCAL CONVERGENCE ANALYSIS FOR NEWTON’S METHOD

We shall study the semilocal/local convergence of Newton’s method. In the rest of the paper we assume h·,·ithe inner product and k · kon Q. As in [56]

we define a distance on Gforx, yGas follows:

m(x, y) = inf k

P

i=1

kzik: there exist k≥1 andz1,· · · , zkQ (2.1)

such that y=x·exp z1· · ·expzk

.

By convention inf ∅ = +∞. It is easy to see that m(·,·) is a distance on G and the topology induced is equivalent to the original one on G. Let wG and r >0, we denote by

U(w, r) ={y∈G : m(w, y)< r}

the open ball centered at wand of radius r. Moreover, we denote the closure ofU(w, r) byU(w, r). Let alsoL(Q) denotes the set of all linear operators on Q.

We need the definition of Lipschitz continuity for a mapping.

Definition 2.1. Let M : G−→L(Q), x?Gand r >0. We say that M satisfies the L-Lipschitz condition on U(x?, r) if

(2.2) kM(x·exp u)M(x)k≤L kuk

holds for any uQ andxU(x?, r) such thatkuk+m(x, x?)< r .

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It follows from (2.2) that there exists L? such that

(2.3) kM(x·exp u)M(x?)k≤L?(kuk+m(x, x?)) holds for any uQ andxU(x?, r) such thatkuk+m(x, x?)< r.

We then say M satisfies the L?-center Lipschitz condition at x?G on U(x?, r). Note that ifM satisfies theL-Lipschitz condition onU(x?, r), then it also satisfies theL?-center Lipschitz condition at x?GonU(x?, r). Clearly,

(2.4) L?L

holds in general andL/L? can be arbitrarily large [4]–[16].

Let us show that (2.2) indeed implies (2.3). If (2.2) holds, then for x=x? there existsK ∈(0, L] such that

kM(x?·expu)−M(x?)k≤K kuk for any uQ such that kuk< r.

There exist k ≥1 andz0, z1,· · · , zkQ such that x =x?·expz0· · ·expzk. Let y0 =x?,yi+1 =yi·exp ui,i= 0,1,· · ·, k. Then, we have that x =yk+1. We can write the identity

M(x·exp u)M(x?) =

= (M(x·expu)M(x)) + (M(x)−M(x?))

= (M(x·expu)M(x)) + (M(yk·expuk)−M(yk)) + (M(yk)−M(x?))

= (M(x·expu)M(x)) + (M(yk·expuk)−M(yk)) +(M(yk−1·exp uk−1)−M(x?))

= (M(x·expu)M(x)) + (M(yk·expuk)−M(yk)) +· · ·+ +(M(y1·exp u1)−M(y1)) + (M(x?·exp u0)−M(x?)).

Using (2.2) and the triangle inequality, we obtain in turn that kM(x·exp u)M(x?)k≤

≤kM(x·expu)M(x)k+kM(yk·expuk)−M(yk)k+· · · +kM(y1·expu1)−M(y1)k+kM(x?·exp u0)−M(x?)k

L(kuk+kuk k+· · ·+ku1 k) +K ku0k

=L(kuk+kuk k+· · ·+ku0 k) + (K−L) ku0 k

L(kuk+kuk k+· · ·+ku0 k), which implies (2.3).

We need a Banach type lemma on invertible mappings.

Lemma 2.2. Suppose that dFx−1? exists and let 0 < rL1

?. Suppose dFx−1? dF satisfies the L?-center Lipschitz condition at x?G on U(x?, r);

for xU(x?, r), there exist k ≥ 1 and z0, z1,· · · , zkQ, such that x = x?·expz0· · ·expzk and

k

P

i=0

kzi k< r. Then, linear mapping dFx−1 exists and

(2.5) kdFx−1dFx? k≤

1−L?

k

P

i=0

kzi k −1

.

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Proof. Let yk = x? ·expz0 ·expz1 · · ·exp zk−1. Then, we deduce ykU(x?, r), since

k−1

P

i=0

k zi k< r. Note that x = yk·expzk. Using (2.3) for M =dFx−1? dF, we get in turn

(2.6)

kdFx−1? (dFxdFx?)k≤L?(kzkk+m(yk, x?))≤L? Pk

i=0

kzik< L?r≤1.

It follows from (2.6) and the Banach lemma on invertible operators [7], [32]

thatdFx−1exists and (2.5) holds. That completes the proof of Lemma 2.2.

Next, we study the convergence domain of Newton’s method around a zero x? of mapping F. First from Lemma 2.3 until Corollary 2.9, we present the local result for Newton’s method when G is an Abelian group. Then, the corresponding local results follow whenGis not necessarily an Abelian group.

Lemma 2.3. Let G be an Abelian group and 0< rL1

?. Let F : G−→Q be a C1-mapping. Let x?G be a zero of mapping F. Suppose dFx−1? exists and let L? >0; there exist j≥1 and z1, z2,· · · , zjQ such that

(2.7) x0=x?·expz1· · ·expzj f or

j

P

i=1

kzi k< r;

dFx−1? dF satisfies the L?-center Lipschitz condition at x? on U(x?, r). Then, linear mapping dFx−10 exists and

(2.8) kdFx−10 F(x0)k≤

2+L?

j

P

i=1

kzik j

P

i=1

kzik

2

1−L?

j

P

i=1

kzi k

.

Proof. Using hypothesisdFx−1? dF satisfies theL?-center Lipschitz condition atx? on U(x?, 1

L?

), we have that

(2.9) kdFx−1? (dFx·expudFx?)k≤L?(kuk+m(x, x?))

for all uQ, xU(x?, r) such that k u k +m(x, x?) < r. Let z = z1 + z2+· · ·+zj. Then, since G is an Abelian group, expz1·expz2· · ·expzj = exp (z1+z2+· · ·+zj) = expz, so, we can writex0 =x?·expz. It then follows from Lemma 2.2 that dFx−10 exists and

(2.10) kdFx−10 dFx?)k≤

1−L?

j

P

i=1

kzi k −1

. We also have the following identity

(2.11)

F(x0) = F(x?·expz)F(x?)

= Z 1

0

dFx?·exp (tz)z dt= Z 1

0

(dFx?·exp (tz)dFx?)z dt+dFx?z.

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In view of (2.9) and (2.11), we get that (2.12)

kdFx−1? F(x0)k ≤ Z 1

0

kdFx−1? (dFx?·exp (t z)dFx?)k kzk dt+kzk

Z 1

0

kL?(t kzk) kzk dt+kzk

L?

2 j

P

i=1

kzi k+1 j

P

i=1

kzi k. Moreover, by (2.10) and (2.12), we obtain that

kdFx−10 F(x0)k≤kdFx−10 dFx? k kdFx−1? F(x0)k≤

2+L?

j

P

i=1

kzi k j

P

i=1

kzi k

2

1−L?

j

P

i=1

kzik

.

That completes the proof of Lemma 2.3.

Remark2.4. The proof of Lemma 2.3 reduces to [56, Lemma 3.2] ifL? =L.

Otherwise (i.e., if L? < L) it constitutes an improvement. We have also included the proof although similar to the corresponding one in [56] because it is not straightforward to see thatL? can replace L in the derivation of the crucial estimate (2.8). Note also that (2.9) can holds for some L1? ∈ (0, L?].

If L1? < L?, then according to the proof of Lemma 2.3, L1? can replace L? in (2.8).

Let us define parameter α forβ = LL? by

(2.13) α=

4

1+3β−(1+β)

β(1−β) , if L? 6=L

1, if L? =L.

Then, it is can easily be seen that

(2.14) α≥1.

We have the local convergence result for Newton’s method.

Theorem2.5. LetGbe an Abelian group. Let0< r4αL, whereαin given in (2.13). LetF : G−→Qbe aC1-mapping. Suppose there existsx?Gsuch that F(x?) = 0, dFx−1? exists and dFx−1? dF satisfies the L-Lipschitz condition onU(x?,1−L3r

?r). Then, sequence {xn}(n≥0) generated by Newton’s method starting at x0U(x?, r)is well defined, remains inU(x?,1−L3r

?r) for alln≥0 and converges to a zero y? of mapping F such that

m(x?, y?)< 1−L3r

?r.

Proof. It follows from hypothesis x0U(x?, r) that there exist j ≥ 1 and z1, z2,· · ·, zjQ such that (2.7) holds. If dFx−1? dF is L-Lipschitz on U(x?,1−L3r

?r), then it is L?-center Lipschitz atx? on U(x?,1−L3r

?r). Note also

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that 4αLL1

? by the choice of α. It follows from Lemma 2.3 that linear mappingdFx−10 exists and

(2.15) η=kdFx−10 F(x0)k≤

2+L?

j

P

i=1

kzi k j

P

i=1

kzik

2

1−L?

j

P

i=1

kzi k

.

Set

L= L

1−L? j

P

i=1

kzik

and r = (2+L1−L?r)r

?r .

We shall show mappingdFx−10 dF satisfies theL-Lipschitz condition onU(x0, r).

Indeed, let xU(x0, r) and zQ be such that k zk +m(x0, x) < r. Then, we get that

kzk+m(x, x?)<kzk+m(x, x0) +m(x0, x?)< r+r1−L3r

?r. Using Lemma 2.2, we have that

kdFx−10 (dFx·expzdFx)k ≤ kdFx−10 dFx? k kdFx−1? (dFx·expzdFx)k

Lkzk

1−L?

j

P

i=1

kzik

=L kzk.

Set

(2.16) h=L η and r1 = 2η

1+

1−2h

≤2η.

Then, by (2.15), we obtain that

(2.17) r1

2+L?

j

P

i=1

kzik j

P

i=1

kzik

1−L?

j

P

i=1

kzi k

(2+L1−L?r)r

?r =r and

(2.18) h(2+L2 (1−L?r)L r

?r)212,

by the choice of r and α. By standard majorization techniques [7], {xk} converges to some zeroy? of mappingF andm(x0, y?)≤r1. Furthermore, we obtain that

m(x?, y?)≤m(x?, x0) +m(x0, y?)≤r+r1r+r1−L3r

?r.

That completes the proof of Theorem 2.5.

The proofs of remaining results in this Section are omitted, since they can be obtained from the corresponding ones in [56].

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Corollary 2.6. Let G be an Abelian group. Let F : G −→ Q be a C1-mapping. Let x?G be a zero of mapping F. Suppose dFx−1? exists and dFx−1? dF satisfies the L-Lipschitz condition on U(x?,L1

?). Let also x0U(x?,4αL). Then, sequence{xn} (n≥0) generated by Newton’s method start- ing at x0 is well defined, remains in U(x?,4αL) for all n≥0 and converges to a zero y? of mapping F such that m(x?, y?)< L1

?. We denote in the following Corollary

B(0, r) ={z∈Q :kzk< r}.

Corollary 2.7. Let G be an Abelian group. Let F : G −→ Q be a C1-mapping. Let x?G be a zero of mapping F. Suppose dFx−1? exists and dFx−1? dF satisfies the L-Lipschitz condition on U(x?,L1

?). Let also x0U(x?,L1

?). Let σ be the maximal number such that U(e, σ) ⊆exp (B(0,L1

?)).

Set r = minn3+Lσ

?σ, 4αLo and N(x?, r) = x?exp(B(0, r)). Then, sequence {xn} (n ≥ 0) generated by Newton’s method starting at x0N(x?, r) con- verges tox?.

Corollary 2.8. Let G be an Abelian group. Let F : G −→ Q be a C1-mapping, where G is a compact connected Lie group equipped with a bi- invariant Riemannian metric. Letx?Gbe a zero of mappingF and0< r <

α

4L. Suppose dFx−1? exists and dFx−1? dF satisfies the L-Lipschitz condition on U(x?,1−L3r

?r). Let also x0U(x?, r). Then, sequence {xn} (n≥0) generated by Newton’s method starting at x0, remains inU(x?,1−L3r

?r) for all n≥0 and converges to x?.

Corollary 2.9. Let G be an Abelian group. Let F : G −→ Q be a C1-mapping, where G is a compact connected Lie group equipped with a bi- invariant Riemannian metric. Let x?G be a zero of mapping F. Suppose dFx−1? exists and dFx−1? dF satisfies the L-Lipschitz condition on U(x?,L1

?).

Let also x0U(x?,4αL). Then, sequence {xn} (n≥0) generated by Newton’s method starting at x0, remains in U(x?,4αL) for all n ≥ 0 and converges to x?.

Lemma 2.10. Let 0 < rL1. Let F : G −→ Q be a C1-mapping. Let x?G be a zero of mapping F. Suppose dFx−1? exists; there exist j ≥ 1 and z1, z2,· · · , zjQ such that x0 = x? ·expz1· · ·expzj for

j

P

i=1

k zi k< r and dFx−1? dF satisfies the L-Lipschitz condition atx? onU(x?, r). Then, the following assertions hold

(i)

kdFx−1? (dFx?·expzdFx?)k≤K? kzk for each zQ withkzk< r and some K? ∈(0, L].

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(ii)

kdFx−1? (F0(x0)−F0(x?))k≤L? j

P

i=1

kzi k f or some L?∈(0, L].

(iii) Linear mappingdFx−10 exists and

kdFx−10 F(x0)k≤

2+L j

P

i=1

kzi k j

P

i=1

kzi k

2

1−L?

j

P

i=1

kzi k .

Proof. (i) This assertion follows from the hypothesis thatdFx−1? dF sat- isfies theL-Lipschitz condition atx? on U(x?, r).

(ii) Let w0 =x?, wi+1 =wi·expzi+1, i= 1,2,· · ·, j−1. Then, we have wj = wj−1·expzj = x0. Using the L-Lipschitz condition, we get in turn that

kdFx−1? (dFwjdFx?)k

≤kdFx−1? (dFwj−1·expzjdFwj−1)k+· · ·+kdFx−1? (dFw1·expz2dFw1)k+ kdFx−1? (dFx?·expz1dFx?)k

L(kzj k+· · ·+kz2k) +K? kz1 k

=L(kzj k+· · ·+kz1 k) + (K?L) kz1 k≤L(kzj k+· · ·+kz1 k) which shows (ii).

(iii) We have by (ii) that

kdFx−1? (dFx0dFx?)k≤L?(kzj k+· · ·+kz1 k)≤L r <1.

Hence,dFx−10 exists and (2.19) kdFx−10 dFx? k≤

1−L?

j

P

i=1

kzi k −1

. We have that

dFx−1? (F(wi)−F(wi−1)) =dFx−1?

Z 1 0

dFwi−1·exp (t zi)zidt

= Z 1

0

dFx−1? (dFwi−1·exp (t zi)dFx?)zidt+zi. Hence, we get that

kdFx−1? (dFwk−1·exp (zk)dFwk−1)k≤L kzkk for each k= 1,2,· · ·, j.

Therefore, we obtain that kdFx−1? (dFwi−1·exp (t zi)dFx?)k ≤

i−1P

k=1

kdFx−1? (dFwk−1·exp(zk)dFwk−1)k+kdFx−1? (dFwi−1·exp(twi)dFwi−1)k

L

i−1

P

k=1

kzkk+Ltkzik.

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We have thatF(x0) =

j

P

i=1

(F(wi)−F(wi−1)). That is, we can get

(2.20)

kdFx−1? F(x0)k

j

P

i=1

Z 1

0

kdFx−1? (dFwi−1·exp (t zi)dFx?)k kzi k dt+kzik

j

P

i=1

Z 1

0

L i−1

P

k=1

kzkk+t kzik

kzik dt+kzik

L 2

j

P

i=1

kzi k+1 j

P

i=1

kzi k.

The result now follows from (2.19), (2.20) and

kdFx−10 F(x0)k≤kdFx−10 dFx?k kdFx−1? F(x0)k≤

2+L j

P

i=1

kzi k j

P

i=1

kzi k

2

1−L? j

P

i=1

kzik .

The proof of Lemma 2.10 is complete.

Let us define parameter δ by

(2.21) δ=

( 4

2 (1+β)−(1+β)

1−β2 , if L? 6=L

1, if L? =L.

We have that

δ ≥1.

Theorem 2.11. Let 0 < r4δL, where δ in given in (2.21). Let F : G−→Q be a C1-mapping. Suppose there exists x?G such that F(x?) = 0, dFx−1? exists and dFx−1? dF satisfies the L-Lipschitz condition on U(x?, R =

3+(L−L?)r

1−L?r r). Then, sequence {xn} (n ≥ 0) generated by Newton’s method starting at x0U(x?, r) is well defined, remains in U(x?, R) for all n ≥ 0 and converges to a zero y? of mapping F such thatm(x?, y?)< R.

Proof. The proof is similar to the proof of Theorem 2.5. Note thatr in the proof of Theorem 2.5 is replaced by

r = (2+L r)1−L r

?r

and η satisfies the new condition

η

2+L j

P

i=1

kzik j

P

i=1

kzik

2

1−L? j

P

i=1

kzi k .

The proof of Theorem 2.11 is complete.

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Corollary 2.12. Let F : G−→Qbe aC1-mapping. Letx?Gbe a zero of mapping F. Suppose dFx−1? exists and dFx−1? dF satisfies the L-Lipschitz condition on U(x?,L1

?). Let also x0U(x?,4δL). Then, the conclusions of Corollary 2.6 hold.

Corollary 2.13. Let F : G−→Qbe aC1-mapping. Letx?Gbe a zero of mapping F. Suppose dFx−1? exists and dFx−1? dF satisfies the L-Lipschitz condition on U(x?,L1

?). Let also x0U(x?,L1

?). Let σ and N(x?, r) as in Corollary 2.7. Set r= minnr0, r1, 4δLo, where

r0=

−(L?+3L)+

(L?+3L)2+4L(L−L?)

2L(L−L?) , if L? 6=L

1

4L, if L? =L

and

r1=

−(3+σ L?)+

(3+σ L?)2+4σ(L−L?)

2 (L−L?) , if L?6=L

σ

3+σ L, if L?=L.

Then, sequence {xn} (n≥0) generated by Newton’s method starting at x0N(x?, r) converges to x?.

Corollary 2.14. Let F : G −→ Q be a C1-mapping, where G is a com- pact connected Lie group equipped with a bi-invariant Riemannian metric. Let x?G be a zero of mapping F and 0 < r < 4δL. Suppose dFx−1? exists and dFx−1? dF satisfies theL-Lipschitz condition on U(x?, R), whereR is defined in Theorem2.11. Let alsox0U(x?, r). Then, sequence{xn} (n≥0) generated by Newton’s method starting at x0, remains in U(x?, R) for all n ≥ 0 and converges to x?.

Corollary 2.15. Let F : G −→ Q be a C1-mapping, where G is a com- pact connected Lie group equipped with a bi-invariant Riemannian metric. Let x?G be a zero of mapping F. Suppose dFx−1? exists and dFx−1? dF satisfies the L-Lipschitz condition on U(x?,L1

?). Let also x0U(x?,4δL). Then, se- quence {xn} (n≥0) generated by Newton’s method starting at x0, remains in U(x?,4δL) for all n≥0 and converges tox?.

Remark 2.16. (a) The local results reduce to the corresponding ones in [56] if L =L?. Otherwise (i.e., if L? < L), they constitute an improvement under the same computational cost with advantages as already stated in the Introduction of this study. Note also thatα >1,δ >1 ifL? < Landα−→ ∞, δ −→ ∞if LL

? −→ ∞.

(b) The local results if G is an Abelian group are weaker and tighter than the ones when G is not necessarily an Abelian group. We have for example thatr < r,δ < α, 1−L3r

?r < Rand the upper bound onη is smaller if L? < L.

(c) It is obvious that finer results can be immediately obtained if similar conditions as the semilocal case (see [15, Section 1, L? instead L0]) are used

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instead of Kantorovich condition for L? < L. However, we decided to leave this part of analysis to the motivated reader. We refer the reader to [14] for such results involving nonlinear equations in a Banach space setting. We also refer the reader to [7, 11, 15, 16] for examples.

3. NUMERICAL EXAMPLES

In this Section we present two numerical examples in the more general setting of a nonlinear equation on a Hilbert space where L?< L.

Example 3.1. Let X =Y =R, x? = 0. Define F by F(x) = −d2sin 1 + d1x +d2 sin ed3x, where d1, d2, d3 are given real numbers. We have that F(x?) = 0. Moreover, ifd3 is sufficiently large andd2 sufficiently small,L/L?

can be arbitrarily large.

Example 3.2. Let X = Y = R3, D = U(0,1) and x? = (0,0,0). Define functionF on D forw= (x, y, z) by

(3.1) F(w) = (ex−1,e−12 y2+y, z).

Then, the Fr´echet derivative ofF is given by F0(w) =

ex 0 0

0 (e−1)y+ 1 0

0 0 1

Notice that we have F(x?) = 0, F0(x?) = F0(x?)−1 = diag{1,1,1} and L? = e−1< L= e.

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