In this section, we give the proofs of the lemmas from Section 5.2. We prove Lemma 5.3 (and later Lemma 5.5) with Janson’s inequality, using similar calculations to Ferber, Luh, and Nguyen [54].
Proof of Lemma 5.3
Fixingh, note that there are certainly at most2n·2nchoices forSandU. Therefore, it is sufficient to prove, for fixedS ⊆ ShandU ⊆V(Gα)\fh−1(V(Fh−10 ))with|S| ≥ sεn2
hk and|U| ≥ sεn
hk, the property in the lemma holds with probability1−e−ω(n).
Let s = sh. Pick someS0 ∈ S, so that, by P3, each graph in S is isomorphic toS0, and label V(S0) ={v1, . . . , vs}. LetHbe the set of copies ofS0in the complete graph with vertex set inU. Note that|U|= Ω(n)and|H|= Ω(ns).
For eachS ∈ S and H ∈ H, label V(S) = {zS,1, . . . , zS,s} and V(H) = {vH,1, . . . , vH,s} so that vi 7→zS,iandvi 7→vH,iare embeddings ofS0. We now distinguish two cases: Case I where there is some edge betweenS0andFh−10 inFh0 and Case II where there is no such edge.
Let us assume first that we are in Case I. For eachS ∈ S, letWS =fh−1(S
v∈V(S)NF0
h−1(v))be the
5.3 Proofs of the lemmas
images of the already embedded neighbours of vertices inS. Note that these setsWS are non-empty by the definition of Case I and byP5disjoint. For eachH ∈ HandS ∈ S, letH⊕WS be the graph with vertex setV(H)∪WScontaining exactly those edges that we would need in order to extend the partial embedding we have so far toSembedded intoH. That is,H⊕WShas edge set
E(H)∪ {vH,iv: 1≤i≤s, v∈fh−1(NF0
h−1(zS,i))}.
For eachS ∈ S,H ∈ HandJ ⊆H, letJ ⊕WS = (H⊕WS)[V(J)∪WS]. LetH+ ={H⊕WS :H ∈ H, S∈ S}, and note that if any graph fromH+appears inGhthen we can indeed extend our current embedding to one more dense spot inS, and hence are done.
LetJ ={H∩H0 :H, H0∈ H, e(H∩H0)>0}andJ0={H∩H0:H, H0∈ H, H6=H0} \ ∅. We will
5. Randomly perturbed graphs
=o X
J∈J
n−2|J|n|J|−1+ X
J∈J0
n−2|J|−1n|J|
=o(n−1).
Therefore, asµ=ω(n)and µδ2 =o(n−1), by Theorem 2.18, the probability that there is no graph in H+inGhis at mostexp(−4(µ+δ)µ2 ) = exp(−ω(n)), as required. For Case I, it is left then only to prove Claim 5.6.
Proof of Claim 5.6. For (i) letH, H0 ∈ Hbe such thatH ∩H0 =J. IfJ 6=H, and|J| ≥3, then, byP2, we have2e(J)≤(∆ + 1)(|J| −2)<(∆ + 1)(|J| −1), as required. If|J|= 2, then(∆ + 1)(|J| −1) =
∆ + 1>2≥e(J).
Suppose then thatJ =H, so|J|=s. Ifs≤∆, then2e(J)≤s(s−1)<(s+1)(s−1)≤(∆+ 1)(s−1), and ifs >∆ + 1, then2e(J)≤s∆< s∆ +s−∆ + 1 = (∆ + 1)(s−1), as required. Ifs= ∆ + 1, note that, as there is some edge betweenS0andFh−1inFh, we have thatS0, and henceJ =H, is not a clique with∆ + 1vertices. Thus,2e(J)< s(s−1) = (∆ + 1)(s−1).
For (ii) supposes≥∆ + 1. AsH is dense we have2e(H)>(∆ + 1)(s−2), and thus
2e(H⊕S)≤2∆s−2e(H)<2∆s−(∆ + 1)(s−2) = (∆ + 1)s+ 2(∆ + 1−s)≤(∆ + 1)s, as required.
So suppose thats≤∆. If4≤s≤∆−1, then, as2e(H)>(∆ + 1)(s−2), we must have s(s−1)>(∆ + 1)(s−2)≥(s+ 2)(s−2) =s(s−1) +s−4≥s(s−1), a contradiction. Ifs= 3, then2e(H)>∆ + 1contradicts∆≥5.
Finally, ifs = ∆, thenH must be the clique on ∆ vertices because2e(H) > (∆ + 1)(∆−2) =
∆(∆−1)−2. Therefore,
2e(H⊕S)≤2∆2−2e(H) = ∆(∆ + 1) = (∆ + 1)s.
For (iii) letH, H0 ∈ Hbe such thatH ∩H0 =J andH 6=H0, which exist by the definition ofJ0. Observe that|J|< s. LetI=H−V(J), and lete(I, J)be the number of edges betweenIandJinH. Then,
2e(J⊕WS)≤2(∆|J| −e(J)−e(J, I)) = 2(∆|J| −e(H) +e(I))
= (∆ + 1)|J|+ (∆−1)|J| −2e(H) + 2e(I).
Thus, to prove the claim it is sufficient to show that(∆−1)|J|<2e(H)−2e(I).
AsH is dense, we have2e(H)>(∆ + 1)(|J|+|I| −2). If|I| ≥3, then, fromP2, we have2e(H)>
(∆ + 1)|J|+ 2e(I). If|I|= 2, then2e(H)>(∆ + 1)|J| ≥(∆−1)|J|+ 2e(I). Finally, if|I|= 1then e(I) = 0and, since by (ii),|J|=s−1≥∆−1and∆≥3, we have2e(H)>(∆−1)|J|+ 2e(I).
So in each case,2e(H)−2e(I)>(∆−1)|J|as required.
It remains to consider Case II. In this case the graphs inShhave no edges toFh−1. Therefore, it is sufficient for some graph inHto exist. Letmbe the size of each (isomorphic) graph inH, and note
5.3 Proofs of the lemmas
The proof of this lemma is based on the fact that the setsB(v)arerandomsets. The reason for this is thatFˆis a random copy ofF∗inGand thatGis itself a random graph.
Recall that we haveF∗ ⊂F, and a copyFˆ ofF∗ chosen randomly inG. SinceGis the union of random graphs it follows thatV( ˆF)⊆ V(Gα)is chosen uniformly at random inV(Gα), which will be crucial in the following. Recall also that we call the resulting embeddingg0 :F∗ → Fˆ. Further, recall that we have an independent setV0 ⊆V(F∗)with|V0| ≥ 2∆n , andW0 =g0(V0). Again,W0is a
5. Randomly perturbed graphs
ifm=|Zwi|andFiis a random variable recording the location of the vertices in{wj}∪Zwjwithj < i, then
E[Iwi|Fi]≥ αn/2· αn/2−1m
(m+ 1)n mn ≥ (α4)m+1
(m+ 1)2 ≥ α∆+1
4∆+1(∆ + 1)2 =:δ. (5.7) Therefore, by Lemma 2.17 and (5.6), with probability1−exp(−Ω(δr)) = 1−o(n−2)we have
|NGα(u)∩B(v)| ≥δr/2≥ α∆+2n
4∆+2(∆ + 1)5 ≥4 α 4∆
2∆
n= 4εn.
Therefore, with probability1−o(1),|NGα(u)∩B(v)| ≥4εnfor eachu, v ∈V(Gα).
Proof of Lemma 5.5
Again, we use Janson’s inequality and similar calculations to Ferber, Luh and Nguyen [54].
Recall thatWh−1 ={g0(v) :gh−1(v) =g0(v), v∈V0}is the set of vertices ofFˆ inW0that have not been switched. Let1 ≤ h ≤k,1 ≤ r ≤ |Sh0| ≤ εn/s2hk,S ⊆ Sh0 andU ⊆ Wh−1 with|S| = rand
|U| ≤s2hr. Note that, as|U| ≤s2hr≤εn, we have|U∪(W0\Wh−1)| ≤2εn. Therefore, by the property from Lemma 5.4, for eachu, v∈V(Gα), we have
|NGα(u, B(v))∩(Wh−1\U)| ≥2εn, (5.8) and, in particular,|B(v)∩(Wh−1\U)| ≥2εn.
Lets = sh. PickS0 ∈ S, so that, byP3, each graph inS is isomorphic toS0. As in the proof of Lemma 5.3, we will consider two cases: Case I where there is some edge betweenS0andFh−1inFh, and Case II where there is no such edge.
Suppose first we are in Case I. LabelV(S0) ={v1, . . . , vs}so thatv1has a neighbour inFh−1. Recall that forS ∈ S we labelledV(S) = {zS,1, . . . , zS,sh}. Without loss of generality, we can assume for eachS ∈ Sthatvi 7→zS,iis an isomorphism fromS0intoS. LetHbe the set of copies ofS0in the complete graph with vertex setWh−1\U. For eachH ∈ H, labelV(H) = {vH,1, . . . , vH,s}so that vi7→vH,iis an isomorphism ofS0toH.
For eachS ∈ S, pick the imagewS of an already embedded neighbour of the vertexzS,1 corre-sponding tov1, that is, pickwS ∈gh−1(NFh(zS,1)). For eachS ∈ S, let
HS ={H ∈ H:vH,1∈NGα(wS)andvH,i∈B(vS,i)for each1≤i≤h}.
These are precisely those copies ofS0where the vertex previously embedded tovH,ican be replaced byvS,ifor1 ≤ i ≤ hand the edge fromvH,1towS is already present inGα. For each S ∈ S, note that, from (5.8), we have|HS|= Ω(ns). For eachS ∈ S, letWS =gh−1(S
v∈V(S)NFh(v))be the set of images of already embedded neighbours of vertices inS.
For eachH ∈ HS andS∈ S, letH⊕WSbe the graph with vertex setV(H)∪WS and edge set E(H)∪({vH,iv: 1≤i≤s, v∈gh−1(NFh(wS,i))} \ {vH,1wS}).
These are exactly the edges we need in order to extend our embedding ofFh−10 to containSembedded
5.3 Proofs of the lemmas
intoH. For eachS∈ S,H ∈ HandJ ⊆H, letJ⊕WS = (H⊕WS)[V(J)∪WS]. LetH+={H⊕WS : S∈ S, H∈ HS}, and note that if any graph fromH+appears inG0hthen we are done.
LetJ ={H∩H0 :H, H0 ∈ H, e(H∩H0)>0}andJ0 ={H∩H0 :H, H0 ∈ H, H 6=H0} \ ∅. Note that (i) and (iii) of Claim 5.6 hold here as well. For eachH ∈ Hand S ∈ S,E(H ⊕WS)does not includevH,1wS, and, therefore, in place of (ii), the following holds.
(ii’) For eachS∈ SandH ∈ HS,2e(H⊕WS)≤(∆ + 1)s−2.
Note that, byP3, each graph inH+has the same number of edges,msay. Note that, as the property we are looking for is monotone, we may assume thatq−1/2=ω(lnn). Using (ii’) from above, let
µ:=X
Definingδas follows, and using similar deductions to those used to reach (5.5), we have δ:= X so, with very similar calculations to Case II in the proof of Lemma 5.3, we have that the probability that there exists no graph fromH0inG0his at mostexp(−ω(n))≤exp(−ω(rln(n/r)), as required.
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5. Randomly perturbed graphs