vol13_pp175-186. 203KB Jun 04 2011 12:05:53 AM

ALGEBRAICCONNECTIVITY OFTREES WITHA PENDANT

EDGE OFINFINITE WEIGHT

∗

A. BERMAN

†

AND K.-H. FÖRSTER

‡

Abstrat. Let

G

beaweighted graph. Let

v

bea vertexof

G

and let

G

v

ω

denote the graph

obtainedbyaddingavertex

u

andanedge

{

v, u

}

withweight

ω

to

G

.Thenthealgebraionnetivity

µ(G

v

ω

)

of

G

v

ω

isanondereasingfuntionof

ω

andisboundedbythealgebraionnetivity

µ(G)

of

G

.Thequestionofwhen

lim

ω

→∞

µ(G

v

ω

)

isequalto

µ(G)

isonsideredandansweredintheasethat

G

isatree.

Keywords. Weightedgraphs,Trees,Laplaianmatrix,Algebraionnetivity,Pendantedge.

AMSsubjetlassiations. 5C50,5C40,15A18.

1. Introdution. Aweightedgraphon

n

verties isanundiretedsimplegraph

G

on

n

vertiessuhthatwitheahedge

e

of

G

,thereisanassoiatedpositivenumber

ω

(

e

)

whihisalledtheweight of

e

.

TheLaplaian matrix of aweighted graph

G

on

n

verties isthe

n

×

n

matrix

L

(

G

) =

L

= (

l

ij

)

,whereforeah

i, j

= 1

, . . . , n,

l

ij

=



















−

ω

(

e

)

if

i

=

j

and

e

=

{

i, j

}

,

0

if

i

=

j

and

i

isnotadjaentto

j,

k

=

i

l

ik

if

i

=

j.

Clearly

L

is asingular

M

-matrixandpositivesemidenite, so

λ

1

(

L

) = 0

,where

forasymmetrimatrix

A

wearrangetheeigenvaluesinnondereasingorder

λ

1

(

A

)

≤

λ

2

(

A

)

≤

. . .

Fiedler [3℄ showed that

λ

2

(

L

)

is positive i

G

is onneted and alled it the

algebraionnetivity of

G

. Thealgebraionnetivityof

G

willbedenotedby

µ

(

G

)

.

Inthispaper

G

alwaysdenotesaonnetedweightedgraphwithoutloops.

Let

G

beagraphwith

n

verties. Let

v

beavertexof

G

andlet

G

v

ω

bethegraph

with

n

+ 1

vertiesobtainedbyadding to

G

avertex

u

andanedge

e

=

{

v, u

}

with

weight

ω

.

∗

Reeivedbytheeditors21July2003.Aeptedforpubliation24June2005. HandlingEditor:

StephenJ.Kirkland.

†

Department of Mathematis, Tehnion, Haifa 32000, Israel,(bermantehunix.tehnion.a.il).

Researh supportedbythe New York MetropolitanFund for Researh at the Tehnion. Mostof

theworkwasdoneduringavisittoTUBerlin;partoftheworkwasdoneinFUBerlin,Charite

-BenjaminFranklin.

‡

DepartmentofMathematis, Tehnial UniversityBerlin,Sekr. MA6-4,Strassedes17. Juni

Theorem 1.1. The algebraionnetivity

µ

(

G

v

ω

)

isa nondereasing funtion of

ω

andfor every

ω

and

n >

1

µ

(

G

v

ω

)

≤

µ

(

G

)

.

Proof. Let

L

ω

be the Laplaian matrix of

G

v

ω

and let

0

< ω

1

≤

ω

2

. Then

B

=

L

ω

2

−

L

ω

1

isasingularrankonepositivesemidenitematrix. By[7,Th. 4.3.1℄

λ

k

(

L

ω

1

)

≤

λ

k

(

L

ω

2

)

for

k

= 1

. . . n,

andfor

k

= 2

,

µ

(

G

v

ω

1

)

≤

µ

(

G

v

ω

2

)

.

Toshowthat

µ

(

G

v

ω

)

isbounded, write

L

ω

asthesumoftwoblokmatries

L

ω

=







L

(

G

) 0

0

0







+







0

0

0

ω

−

ω

−

ω

ω







,

where

L

(

G

)

is

n

×

n

andtheleftupperzeroblokintheseondmatrixis

(

n

−

1)

×

(

n

−

1)

.

By[7,Th. 4.3.4(a),thease

k

= 2

℄,

µ

(

G

v

ω

) =

λ

2

(

L

ω

)

≤

λ

3

(

L

(

G

)

⊕

(0))

=

λ

2

(

L

(

G

)) =

µ

(

G

)

.

Remark 1.2. Thetheorem is essentiallyaonsequeneof Cor. 4.2 of[6℄. It is

provedfortreesin [8℄.

Example 1.3. For theompletegraphs

K

n

, n >

1

,

withallweightsequalto 1

lim

ω→∞

µ

((

K

n

)

v

ω

) =

n

+ 1

2

< n

=

µ

(

K

n

)

.

Example 1.4. Fortheyles

C

n

, n >

2

,

withweightsequalto1

lim

ω→∞

µ

((

C

n

)

v

ω

) =

µ

(

C

n

+1

)

< µ

(

C

n

)

.

Example 1.5. Let

G

be thegraphobtainedfrom

K

4

bydeleting an edge and

letalltheweightsof

G

beequalto1. Ifthedegreeof

v

is3,

lim

ω→∞

µ

(

G

v

ω

) = 2 =

µ

(

G

)

.

Ifthedegreeof

v

is2

lim

ω→∞

µ

(

G

v

ω

)

< µ

(

G

)

.

Sine

µ

(

G

v

ω

)

isboundedby

µ

(

G

)

,itisnaturaltoaskwhendoes

lim

ω→∞

µ

(

G

v

ω

) =

µ

(

G

)

.

2. Results on trees. Our paper relies heavily on the work of [12℄ so in this

setionwedesribetheirmainresultsandbasibakgroundontreesneededforthese

resultsandforthenextsetion. Insomeaseswehangethenotationof[12℄.

Theorem 2.1. [4,Th. 3.11℄Let

T

be a weighted tree with Laplaian matrix

L

and algebrai onnetivity

µ

. Let

y

be an eigenvetor of

L

assoiated with

µ

. Then

exatlyone ofthe followingtwoasesour:

(a) Someentryof

y

is

0

.

(b) Allentries of

y

arenonzero.

In the rst asethere exists aunique vertex

c

suhthat

y

c

= 0

and

c

isadjaent to

avertex

d

with

y

d

= 0

. In the seondasethereis aunique pair of verties

i

and

j

adjaent in

T

suhthat

y

i

y

j

<

0

.

Definition 2.2. Aweightedtree

T

is saidtobeoftypeI withaharateristi

vertex

c

ifase

(

a

)

ofTheorem2.1holds,andoftypeII withharateristiverties

i

and

j

inase

(

b

)

. Weusealsothenotation

I

c

in therstaseand

II

i,j

intheseond

ase.

Thenameharateristivertieswasoinedin[11℄byR.Merriswhoshowedthat

if

µ

isnotasimpleeigenvalue,thenalltheorrespondingeigenvetorsyieldthesame

typeoftreeandthesameharateristiverties.

Definition 2.3. Let

v

beavertex ofatree

T

. Let

L

v

bethematrixobtained

bydeleting therowandolumn of theLaplaianmatrixof

T

that orrespond to

v

.

Thematrix

M

v,T

:=

L

−

1

v

isalled thebottlenekmatrixof

T

at

v

.

In[9℄and[10℄,itisshownthattheentryof

M

v,T

thatorrespondstotheverties

k

and

l

is

m

kl

=

1

ω

(

g

)

wherethesummationisonalledges

g

thatlieontheintersetionofthepathbetween

k

and

v

and thepathbetween

l

and

v

. Thematrix

M

v,T

is permutationally similar

to ablok diagonalmatrix, wherethe numberofbloksis thedegreeof

v

and eah

blokisapositivematrixwhihorrespondsto auniquebranhat

v

.

Forverties

u, v

ofatree

T

let

v

→

u

denotethebranhof

T

at

v

,thatontains

u

.

Wedenoteby

M

v→u,T

theblokof

M

v,T

thatorrespondsto

v

→

u

,andby

M

v→u,T

thematrixobtainedfrom

M

v,T

bydeleting therowsandtheolumns orresponding

to

M

v→u,T

.

Definition 2.4. A diagonal blok of

M

v,T

whose spetral radius is equal to

ρ

(

M

v,T

)

, where

ρ

(

A

)

denotes thespetralradiusof thematrix

A

, isalled aPerron

blokandtheorrespondingbranhof

T

at

v

isalled aPerronbranh.

Theorem2.5. [9,Cor. 2.1℄Let

T

beaweightedtree. Then

T

isoftypeIwitha

harateristi vertex

c

,if andonlyif at

c

,

T

has morethanone Perron branh.

Inthis ase,

µ

(

T

)

,the algebrai onnetivity of

T

isequal to

1

ρ

(

M

c,T

)

.

Let

e

be an edge of a graph

G

. Replae the weightat

e

by

ω

and denote the

resultinggraphby

G

e

ω

. Observethat sine

e

=

{

v, u

}

isapendantedgeof

G

v

ω

,then

(

G

v

ω

)

e

ω

=

G

v

ω

. Let

G

e

∞

denotethefamilyofweightedgraphs

{

G

e

ω

, ω >

0

}

,andlet

G

v

∞

denotethefamilyofweightedgraphs

{

G

v

Theorem 2.6. [12,Corollary1.1℄Let

T

bea weighted tree andlet

e

be anedge

of

T

. Then thereexistsapositive number

ω

0

suhthatallthe trees

T

e

ω

, ω

0

< ω <

∞

,

areof the sametypeandhave the sameharateristiverties.

Thefollowingdenitionsareusedin [12℄.

Definition 2.7. Thefamilyoftrees

T

e

∞

isatypeItreeatinnitywith

hara-teristivertex

c

if thereexistsan

ω

0

>

0

suhthat forall

ω

∈

[

ω

0

,

∞

)

,

T

e

ω

isoftype

I

c

. Similarly,

T

e

∞

is a typeII tree at innity with harateristi verties

i

and

j

if

thereexists an

ω

0

>

0

suhthat forall

ω

∈

[

ω

0

,

∞

)

,

T

e

ω

is oftype

II

i,j

.

Wenowanstatethemainresultof[12℄.

Theorem 2.8. [12,Th.1.8℄Let

e

=

{

v, u

}

be anedge thatis notapendantedge

of a tree

T

. Let

T

1

and

T

2

be the resulting omponents arising from the deletion of

e

. Suppose

v

∈

T

1

,

u

∈

T

2

and

µ

(

T

1

)

≤

µ

(

T

2

)

. Then

lim

ω→∞

µ

(

T

e

ω

) =

µ

(

T

1

)

i

T

1

isa

treeoftypeIwith aharateristivertex, say,

c

,and oneof thefollowing onditions

holds:

(a)

T

e

∞

isoftype Iwithaharateristivertex

c

.

(b)

c

isinident to

e

and

ρ

(

M

u,T

2

)

≤

1

µ

(

T

1

)

.

Weonludethe bakgroundresultswiththe analogueofTheorem 2.5 fortype

IItreesandtwopropositionsthatwillbeusedinprovingthemainresult.

Theorem 2.9. [9,Th.1℄Aweightedtree

T

isof typeIIiateveryvertex

T

has

auniquePerronbranh. Ifthe harateristiverties,

i

and

j

,of

T

arejoinedby an

edge ofweight

θ

,thenthereexistsanumber

0

< γ <

1

,suhthat

ρ

(

M

i→j,T

−

γ

θ

J

) =

ρ

(

M

j→i,T

−

1

−

γ

θ

J

)

, and

µ

(

T

) =

1

ρ

(

M

i→j,T

−

γ

θ

J

)

=

1

ρ

(

M

j→i,T

−

1

−γ

θ

J

)

,

whereJdenotes an allonesmatrix.

Proposition2.10. [8,Cor. 1.1℄Theharateristivertiesof

T

v

ω

lieonthe path

between theharateristivertiesof Tand

u

.

Proposition 2.11. [12, Claim 3.2℄ Let Tbea tree. Let

{

i

k

, j

k}

beedges in T

withweights

α

k

,for

k

= 1

,

2

,suhthatthepathfrom

i

1

to

j

2

ontains

j

1

and

i

2

,and let

0

< γ

1

, γ

2

<

1

. Then

ρ

(

M

j

1

→i

1

,T

−

γ

1

α

1

J

)

< ρ

(

M

j

1

→i

1

,T

)

< ρ

(

M

j

2

→i

2

,T

−

γ

2

α

2

J

)

.

3. Assigninganarbitrarilylargeweighttoapendantedgeofatree. In

thissetionweonsidertheasewhere

T

isatreeand

u

isapendantvertexof

T

∪

e

where

e

=

{

v, u

}

and

v

∈

T

.

In some sense the question in this ase may be onsidered asa speial ase of

thedisussion in [12℄. Todo this,

{

u

}

isto beonsidered asa "treewith algebrai

Ourdisussionisbasedontheanalysisofthelimitsofthebottlenekmatriesof

T

v

ω

when

ω

inreasesto

∞

;namely

M

v,T

v

ω

=

M

v,T

⊕

1

ω

,

M

u,T

v

ω

= (

M

v,T

⊕

(0)) +

1

ω

J

andif

s

=

v

isavertexof

T

,

M

s,T

v

ω

=

M

s,T

M

(

v

)

M

(

v

)

t

m

vv

+

ω

1

,

where

M

(

v

)

istheolumnof

M

s,T

orrespondingto

v

and

m

vv

isthediagonalentry

of

M

s,T

, orresponding to

v

. In partiular, for all the branhes of

T

at

s

that do

not ontain

v

, the diagonal bloks of

M

s,T

v

ω

and of

M

s,T

are the same. Denoting

lim

ω→∞

M

s,T

v

ω

by

M

s,T

v

∞

weseethatfor

s /

∈

e

M

s,T

v

∞

=





M

s,T

M

(

v

)

M

(

v

)

t

m

vv





(3.1)

and

M

v,T

v

∞

=

M

u,T

v

∞

=

M

v,T

⊕

(0)

.

(3.2)

Thereadershouldnotbeonfusedbythefatthat

T

v

∞

denotesafamilyoftrees

while

M

s,T

v

∞

denotesasinglematrix(upto permutationsimilarity).

Example3.1.

◦

1

/

2

a

@

@

@

@

@

@

@

◦

c

1

◦

v

ω

◦

u

◦

1

/

2

b

~

~

~

~

~

~

~

M

v,T

v

∞

=

M

u,T

v

∞

=









3 1

1 0

1 3

1 0

1 1

1 0

0 0

0 0









,

M

c,T

v

∞

=









2 0

0 0

0 2

0 0

0 0

1 1

0 0

1 1



M

a,T

v

∞

=

M

b,T

v

∞

=









4

2 2

2

2

2 2

2

2

2 3

3

2

2 3

3









.

Remark 3.2. Thematries

M

s,T

v

∞

are ofourse singular, but theydo ontain

informationon

lim

ω→∞

µ

(

T

v

ω

)

.

As in theaseofnonsingularbottlenek matrieswe allthe diagonalbloks of

M

s,T

v

∞

whosespetralradiusismaximal,Perronbloks. Theorrespondingbranhes

of

T

v

ω

donotdependon

ω

;theywillbealledthePerronbranhesofthefamily

T

v

∞

.

Lemma 3.3. If

M

s,T

v

∞

hasmorethanonePerronblok, then

lim

ω→∞

µ

(

T

v

ω

) =

1

ρ

(

M

s,T

v

∞

)

.

Proof. Considertheprinipal submatrix

L

s,T

v

ω

obtainedfromtheLaplaian

ma-trixof

T

v

ω

bydeleting therowandolumn orrespondingto

s

. Then

M

s,T

v

∞

= lim

ω→∞

(

L

s,T

v

ω

)

−

1

.

By[7,Th. 4.3.15℄for

r

=

n

−

1

, k

= 1

and

k

= 2

,

λ

1

(

L

s,T

v

ω

)

≤

µ

(

T

v

ω

)

≤

λ

2

(

L

s,T

v

ω

)

,

so

lim

ω→∞

λ

1

(

L

s,T

v

ω

)

≤

ω→∞

lim

µ

(

T

v

ω

)

≤

ω→∞

lim

λ

2

(

L

s,T

v

ω

)

,

Sine

M

s,T

v

∞

hasatleasttwoPerronbloksweobtain

lim

ω→∞

λ

2

(

L

s,T

v

ω

) = lim

ω→∞

λ

1

(

L

s,T

ω

v

) =

ρ

(

M

s,T

∞

v

)

.

Remark 3.4. Ifthere existsan

ω

0

suh thattwoofthePerronbloksof

M

s,T

v

∞

arePerronbloksof

M

s,T

v

ω

for

ω

≥

ω

0

then

λ

2

(

L

s,T

v

ω

) =

λ

1

(

L

s,T

ω

v

)

for

ω

≥

ω

0

.

Lemma 3.5. Let

s

be a vertex of

T

. Suppose

M

s,T

v

∞

has at least two Perron

bloks andlet

t

beanother vertexof

T

. Then

ρ

(

M

t,T

v

Proof. Byassumption, thefamily

T

v

∞

has atleast twoPerron branhesat

s

, so

oneofthem,say

s

→

x

,doesnotontain

t

. Let

t

→

s

bethebranhat

t

thatontains

s

. Thenitontainsthebranh

s

→

x

,andweobtain

ρ

(

M

t,T

v

∞

)

≥

ρ

(

M

t→s,T

v

∞

)

> ρ

(

M

s→x,T

v

∞

) =

ρ

(

M

s,T

v

∞

)

,

wherethestritinequalityfollowsfrom [1,Cor. 2.1.5℄andthefat that

M

s→x,T

v

∞

is

asubmatrixof

M

t→s,T

v

∞

,whihispositive.

Corollary3.6. Thereisatmost onevertex, say

c

,suhthat

M

c,T

v

∞

has more

thanonePerronblok.

Definition 3.7. Intheasethatthere isavertex

c

suhthat

M

c,T

v

∞

hasmore

thanonePerronblok,wewillsaythatthefamilyoftrees

T

v

∞

isatrii(treeininnity)

oftypeI. Ifnosuh

c

existswesaythat

T

v

∞

isatriioftypeII.

Remark 3.8.

(a)If thetrees

T

v

ω

are oftypeI

c

for allsuientlylarge

ω

, then thefamily

T

v

∞

isatriioftypeI

c

(andalsoatypeI treeat innitywithharateristivertex

c

). In

otherwords,if

T

v

∞

isatriioftypeII,thenforall

ω

largeenough,

T

v

ω

aretreesoftype

II.

(b)Supposethetrees

T

v

ω

areoftypeII

p,q

forallsuientlylarge

ω

,thenbythe

representationof

L

ω

intheproofofTheorem1.1 andbyTheorem2.1,

{

p, q

}

annot

bethependantedge

{

v, u

}

.

() It is possible that

T

v

ω

are of type II for all suiently large

ω

(so

T

v

∞

is a

typeII treeat innity) but

T

v

ω

is atriiof typeI;see Lemma3.10and Subase 4of

Example3.13in thefollowingdisussion.

Remark3.9. TheproofofLemma3.5showsthatif

T

isatreeoftypeIwitha

harateristivertex

c

, thenforanyothervertex

s

of

T

ρ

(

M

s,T

)

> ρ

(

M

c,T

)

.

(ThishasalreadybeenestablishedinProposition2of[9℄.)

ConsiderTheorem2.9 where

T

v

ω

isoftypeII

i,j

andtheweightoftheedge

{

i, j

}

is

θ

. Thenforevery

ω

(suientlylarge)there existanumber

γ

ω

,between0and1,

suhthat

µ

(

T

ω

v

) =

1

ρ

(

M

i→j,T

v

ω

−

γ

ω

θ

J

)

=

1

ρ

(

M

j→i,T

v

ω

−

1

−γ

ω

θ

J

)

.

Whathappenstothethenumber

γ

ω

when

ω

goesto

∞

? Welaimthat

lim

ω→∞

γ

ω

exists. Indeed,oneofthebranhesorrespondingto

M

i→j,T

v

ω

and

M

j→i,T

v

ω

doesnot

ontain

u

. Suppose it is the seond, so

M

j→i,T

v

ω

=

M

j→i,T

. The numbers

µ

(

T

v

ω

)

inreasetoalimit,seeTheorem1.1,sothenumbers

ρ

(

M

j→i,T

v

ω

−

1

−γ

ω

θ

J

)

dereaseto

alimit,whihmeansthatthenumbers1-

γ

ω

inreasetoalimit. Thislimitisatmost

1sine

0

< γ

ω

<

1

.

Lemma 3.10. If the trees

T

v

ω

are of type II, with harateristi verties

i, j

for

ω

0

< ω <

∞

,andif

γ

= lim

ω→∞

γ

ω

= 0

,where

ρ

(

M

i→j,T

v

ω

−

γ

ω

θ

J

) =

ρ

(

M

j→i,T

ω

v

−

1

−γ

ω

and

θ

isthe weightof theedge

{

i, j

}

,then

T

v

∞

isatrii oftypeI

i

. Similarly,if

γ

= 1

then

T

v

∞

isatrii oftype I

j

.

Proof.

M

j→i,T

v

∞

= (

M

i

j,T

v

∞

⊕

(0)) +

1

θ

J

so if

γ

= 0

, then

ρ

(

M

i→j,T

v

∞

) =

ρ

(

M

j→i,T

v

∞

−

1

θ

J

) =

ρ

(

M

i

j,T

∞

v

)

so

T

e

∞

isatriioftypeI

i

.

Corollary 3.11. If the trees

T

v

ω

are of type II and if

T

v

∞

is atrii of type II, then

0

< γ

= lim

ω→∞

γ

ω

<

1

.

Remark 3.12. Thetree

T

anbeatree oftypeI with aharateristivertex,

say

c

,or atreeoftypeII.Intherstasethereare3possibilities:

1

T

v

∞

is atriioftypeI

c

,

2

T

v

∞

is atriioftypeI

s

, where

s

=

c

,

3

T

v

∞

is atriioftypeII.

Intheseond asethere aretwopossibilities:

4

T

v

∞

is atriioftypeI

s

forsome

s

,

5

T

v

∞

is atriioftypeII.

Thefollowingexampledemonstratesthatallvesubasesarepossible.

Example 3.13.

Subase1

◦

x

@

@

@

@

@

@

@

◦

c

1

◦

v

ω

◦

u

◦

x

~

~

~

~

~

~

~

0

< x

≤

1

2

.

Here

M

c,T

v

ω

=









1

/x

0

0

0

0

1

/x

0

0

0

0

1

1

0

0

1 1 +

ω

1









,

sofor

x <

1

2

, T

v

∞

isatreeoftypeI withaharateristivertex

c

andfor

x

=

1

2

,

itis onlyatriioftype

I

c

.

Anotherexampleiswhen

c

=

v

◦

x

@

@

@

@

@

@

@

◦

c

ω

◦

u

◦

x

~

~

~

~

~

~

~

Subase2

◦

x

@

@

@

@

@

@

@

◦

c

10

◦

s

1

◦

v

ω

◦

u

◦

x

~

~

~

~

~

~

~

where

ρ









1

/x

+ 0

.

1

0

.

1

0

.

1

0

.

1

1

/x

+ 0

.

1

0

.

1

0

.

1

0

.

1

0

.

1









= 2

.

Subase3

◦

x

@

@

@

@

@

@

@

◦

c

1

◦

v

ω

◦

u

◦

x

~

~

~

~

~

~

~

1

2

< x

≤

1

,

or

◦

1

◦

c

1

◦

v

ω

◦

.

Subase4

◦

1

◦

x

◦

s

1

◦

v

ω

◦

, ρ

1 + 1

/x

1

/x

1

/x

1

/x

= 2

.

Subase5

Herewesuggest3examples:

◦

x

@

@

@

@

@

@

@

◦

1

◦

ω

◦

◦

x

~

~

~

~

~

~

~

x >

1

,

◦

x

◦

1

◦

ω

◦

◦

1

@

@

@

@

@

@

@

◦

v

ω

◦

u

◦

x

~

~

~

~

~

~

~

x

= 1

.

Wearenowreadytostateandprovethemainresult.

Theorem 3.14. Let

T

beatree. Then

lim

ω→∞

µ

(

T

v

ω

) =

µ

(

T

)

(3.3)

if andonlyif

(a)

T

isatreeof typeIwith harateristi vertexsay

c

,

and

(b)

ρ

(

M

c→u,T

v

∞

)

≤

ρ

(

M

c,T

) =

1

µ

(

T

)

.

Proof. We provethetheorem byonsidering the vesubases of Remark 3.12,

andshowingthat(3), (a)and(b)holdinSubase1andonlyinthisase,i.e. ifand

onlyif

T

and

T

v

∞

areoftypeI

c

forsomevertex

c

.

Subase 1: Obviously (a) holds. From (1) and (2) follows that

ρ

(

M

c,T

v

ω

)

≥

ρ

(

M

c,T

)

. Butif

T

and

T

v

ω

areoftype

I

c

,thenequalityholds. Thus

ρ

(

M

c→u,T

v

∞

)

≤

ρ

(

M

c,T

)

,

proving(b),and

µ

(

T

) =

1

ρ

(

M

c,T

)

=

1

ρ

(

M

c,T

v

ω

)

= lim

ω→∞

µ

(

T

v

ω

)

,

proving(3). ThisompletestheproofinSubase1.

If

T

is atreeof type

I

c

and(b) holds, then itfollowseasily that

T

v

ω

is atrii of

typeI

c

. Therefore(b) doesnot hold in Subases 2and 3,while (a) obviouslydoes

nothold in Subases4 and5. Nowwewill provethat (3) doesnothold in thelast

foursubases.

Subase2: Wehaveto showthat(3)doesnothold. Indeed

µ

(

T

) =

1

ρ

(

M

c,T

)

>

1

ρ

(

M

s,T

)

=

1

ρ

(

M

s,T

v

ω

)

,byLemma3.5

= lim

ω→∞

µ

(

T

v

ω

)

, byLemma 3.3.

Subase3: ByRemark 3.8(a)thetrees

T

v

ω

areforsuientlylarge

ω

oftype

II

,

3.8(b)itdoesnotdependon

ω

. Withoutlossofgenerality,

p

liesonthepathbetween

q

and

c

.

ByProposition2.10theverties

p

and

q

lie onthepathbetween

c

and

u

. Let

i

beaneighborof

c

suhthat

c, p

and

q

lieonthepathbetween

i

and

u

and

c

→

i

isa

Perronbranhof

T

. Thenweobtain

lim

ω→∞

µ

(

T

v

ω

) = lim

ω→∞

1

ρ

(

M

q→p,T

v

ω

−

γ

ω

θ

J

)

,byTheorem2.9,

= lim

ω→∞

1

ρ

(

M

q→p,T

−

γ

θ

ω

J

)

,sine

q

→

p

isin

T

,

=

1

ρ

(

M

q→p,T

−

γ

θ

J

)

,where

0

< γ <

1

,byCorollary3.11,

<

1

ρ

(

M

c→i,T

)

,byProposition2.11,

=

1

ρ

(

M

c,T

)

, sine

c

→

i

isaPerronbranhof

T,

=

µ

(

T

)

,sine

T

isoftype

I

c

,

so(3)doesnothold.

Subase4: Here againwehaveto showthat (3)doesnothold. Suppose

T

is of

type

II

ij

, where

j

lies on thepathfrom

i

to

u

. Let

θ

and

γ

beasin Theorem 2.9.

Sine

T

v

ω

is atriioftype

I

s

, thebyProposition 2.10,

s

lies onthepathfrom

i

to

u

. Therefore

lim

ω→∞

µ

(

T

v

ω

) = lim

ω→∞

1

ρ

(

M

s→i,T

v

ω

)

= lim

ω→∞

1

ρ

(

M

s→i,T

)

≤

1

ρ

(

M

j→i,T

)

<

1

ρ

(

M

j→i,T

−

γ

θ

J

)

=

µ

(

T

)

.

Subase5: HereTisof,say,typeII

ij

andfor

ω

largeenough,

T

v

ω

areof,say,type

II

pq

, whereby Proposition 2.10,we maytake, withoutlossofgenerality,

p

and

q

to

liebetween

i

and

q

. Let

θ

and

γ

beas in Theorem2.9 forthe edge

{

i, j

}

in

T

and

let

θ

ˆ

and

γ

ω

betheorrespondingpairfortheedge

{

p, q

}

in

T

v

ω

. Observethat

θ

ˆ

does

notdependon

ω

byRemark3.8(b). Let

γ

ˆ

= lim

ω→∞

γ

ω

. ByCorollary3.13wehave

0

< γ <

1

. Now

lim

ω→∞

µ

(

T

v

ω

) = lim

ω→∞

1

ρ

(

M

q→p,T

v

ω

−

γ

ω

ˆ

θ

J

)

= lim

ω→∞

1

ρ

(

M

q→p,T

−

γ

θ

ˆ

ω

J

)

=

1

ρ

(

M

q→p,T

−

γ

ˆ

θ

ˆ

J

)

<

1

ρ

(

M

j→i,T

−

γ

θ

J

)

=

µ

(

T

)

,

wheretheinequalityfollowsfromProposition2.11.

Aknowledgments. We are indebted to Mr. Felix Goldberg [5℄ for suggesting

Example1.5 whih showsthat

G

doesnothaveto beatree for

lim

ω→∞

µ

(

G

v

tohold. Thequestionofwhen does

lim

ω→∞

µ

(

G

v

ω

) =

µ

(

G

)

, when

G

isageneralgraph,

seems to be muh more diult than the one in the ase that

G

is a tree. We

are grateful to thereferee for his orherimportant remarks and forsuggestingthat

Propositions 1.3and1.4, as wellasLemma 2.2of[2℄, maybeusefulin dealingwith

thegeneralase.

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