(1)



Lemma4[14]LetGbe a graph withnvertices andmedges,anddvdenote the degree of a vertexvinG,then

ii) the degree inL(G) of an edgeuvofGisd(u)+d(v)-2.
2 Bounds for GA1 index of line graph
Theorem1LetGbe a simple connected graph of ordernwithmedges, maximum vertex degreeΔand minimum vertex degreeδ,δ≥2, then
GA1(L(G))≥GA1(G),
(2)
with equality holding if and only ifG≅Cn.
ProofDefine the index
For 2≤d(u1),d(v),d(u2)≤Δ, we have
(3)
with equality holding if and only ifd(u2)=d(v)=Δ=2.Thus
(4)
Forv∈V(G), we have
(5)
Thus
(6)
Now suppose that equality holds in (2),then all inequalities in the above argument must be equalities, from equality (5) and (6), we have forv∈V(G),d(v)=2, that isG≅Cn.
Conversely, we can see easily that the equality holds in (2) forG≅Cn.
Theorem2LetGbe a simple connected graph of ordernwithmedges,ppendent edges, maximum vertex degreeΔ, minimum vertex degreeδ1of non-pendent edge, (d1,d2,…,dm) be a degree sequence ofL(G). Then
(7)
with equality holding if and only ifGis isomorphic to a regular graph orG∈Γ.
ProofDefine the index
For brevity, write
Forδ1≤d(u1),d(v),d(u2), we have
This implies
and finally we get
(8)
Now since
Note that |E′-P|=e(L(G))-p, using (1), we get
(9)
BecauseP={(u1,v,u2)|u1v,u2v∈E(G),d(u1)=1,d(v)=2}, then |P|=p, we have
(10)
Thus
(11)

(12)
with equality holding if and only ifΔ=δ1(for allv∈V″, thend(v)=(2m-3p)/(n-2p)).
Forδ1≤d(u2)≤Δ, we have
(13)
From (10)-(13), we get (7).
Suppose that equality holds in (7), then all inequalities in the above argument must be equalities. Now we consider two case:
i) ifp=0, from equality in (8), we getΔ=δ1. ThenGis isomorphic to a regular graph.
ii) ifp>0, from equality in (10) and (13), we getd(u1)=1,d(v)=2,d(u2)=Δ=δ1, thusG∈Γ.
Conversely, one can easily see that the equality holds in (7) for regular graph orG∈Γ.
Theorem3LetGbe a simple connected graph of ordernwithmedges,ppendant edge. Then
(14)
with equality holding in (14) if and only ifGis isomorphic to the complete graphK3orG≅P3orG≅P4.
ProofFor each pendant edge (u1,v,u2)∈E′,we have eitherd(u1)=1,d(v)=2 ord(u2)=1,d(v)=2.Thus
and finally we get
(15)
with equality holds in (15) if and only ifd(u1)=1,d(v)=2,d(u2)=n-2.
For each non-pendant edge (u1,v,u2)∈E′, we have 2≤d(u1),d(v),d(u2)≤n-1. Thus
then
(16)
with equality holds in (16) if and only ifd(u1)=2,d(v)=2,d(u2)=n-1.Thus
Suppose that equality holds in (14), then all inequalities in the above argument must be equalities. Now we consider two case
i) ifp=0, from equality (16), we getd(u1)=2,d(v)=2,d(u2)=n-1. Then we have a common neighborv. ThusGis isomorphic to the complete graphK3.
ii) ifp>0, from equality (16), we getd(u1)=1,d(v)=2,d(u2)=n-2. Firstly, assume thate(L(G)) =p, then all the edges are pendant edges inL(G), and henceGis isomorphic toP3orP4. Next, assume thate(L(G))>p. In this case, the maximum degree vertex, sayu2,by equality in (16) has degreen-1. There exists at least one non-pendant edge inG, and hence two vertices, sayviandvj, are adjacent to vertexu2. By equality in (16), for the non-pendant edgevivj∈E(G), eitherdi=n-1 ordj=n-1. Thus do not have any pendant vertex inGasdi=n-1, that is a contradiction.
Conversely, we can easily see that the equality holds in (14) for the complete graphK3orG≅P3orG≅P4.
Theorem4LetGbe a simple connected graph withnvertices,medges,ppendant edges,maximum vertex degreeΔand minimum vertex degreeδ1of non-pendant edge,(d1,d2,…,dm) be a degree sequence ofL(G).Then
(17)
with equality holding in (17) if and only ifGis isomorphic to a regular graph orG∈Γ.
ProofFor each pendant edge (u1,v,u2)∈E′,we have eitherd(u1)=1,d(v)=2 ord(u2)=1,d(v)=2.By Lemma 1,we have
(18)
with equality holds in (18) if and only ifd(u1)=1,d(v)=2,d(u2)=δ1.
Sincee(L(G))-pis the number of non-pendant edges inL(G). By Cauchy-Schwarz Inequality, we get
(19)
Note that |V″|=n-2p, by lemma 4, we have
(20)
with equality holds in (20) if and only if for ∀v∈V″,d(v)=Δ.
From (18),(19) and (20), we get (17).
Suppose that equality holds in (17), then all inequalities in the above argument must be equalities. Now we consider two case:
i) ifp=0, from equality (19),(20), we getd(u1)=d(v)=d(u2)=Δ=δ1.That is,Gis isomorphic to a regular graph.
ii) ifp>0, from equality (18)-(20), we getd(u1)=1,d(v)=2,d(u2)=Δ=δ1. That is,G∈Γ.
Conversely, one can easily see that the equality holds in (17) for a regular graph orG∈Γ.
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