I[µeq]≥lim sup
i
I∗[µni] (6.25)
≥lim inf
i I∗[µni] (6.26)
≥lim sup
N
Z
M×M
KνQ∧Ndµ⊗2 (6.27)
≥ Z
M×M
lim inf
N KνQ∧Ndµ⊗2 (6.28)
=Z
M×M
KνQdµ⊗2 (6.29)
≡I[µ] (6.30)
where 6.27 follows from 6.18-6.24. Inequality 6.28 follows from Fatou’s lemma (noting that KνQ∧N is bounded from below uniformly inN). It then follows from 6.30 and the definition of the equilibrium measure thatI[µeq]=I[µ] andµ=µeqwhich proves 6.16. Moreover, we have shown that
I[µeq]≥lim sup
i
I∗[µni]
≥lim inf
i I∗[µni]
≥I[µ]
≥I[µeq],
hence limiI∗[µni] = I[µeq]. Since{µni}i∈Nis an arbitrary subsequence of Fekete measures we have proven 6.17.
In what follows we will suppress theβ,νandQdependence in the notation forZandΠunless it is needed.
Before proceeding, we prove a technical lemma which will be used later.
Lemma 6.3.1. Let µbe a finite positive Borel measure with finite energy. Then there exists a family of absolutely continuous finite positive Borel measures{µδ}δ>0 such that the Radon-Nikodym derivative ofµδ with respect to volgis continuous,µδ(M)=µ(M),
µδ* µ. (6.31)
Moreover suppose that Qν,gis continuous in a neighborhood of the support ofµ. Then
I[µδ]→I[µ] (6.32)
asδ→0.
Proof. For notational convenience we will suppress the metric dependence and denote the geodesic ball centered atxwith radiusδsimply asB(x, δ). We will denote its volume with respect to thevolgby|B(x, δ)|.
Let
Ψδ(z)≡ Z
B(z,δ)(|B(w, δ)|)−1dµ(w).
We note thatΨis clearly continuous. We define
µδ≡Ψδvolg.
We note thatµδhas finite energy sinceµδ<Cδvolgandvolghas finite energy. Note that µδ(M)=Z
M
Ψδvolg (6.33)
=Z
M
[ Z
B(z,δ)(|B(w, δ)|)−1dµ(w)]dvolg(z) (6.34)
=Z
M
[ Z
M
1B(z,δ)(w)(|B(w, δ)|)−1dµ(w)]dvolg(z) (6.35)
=Z
M
[ Z
M
1B(w,δ)(z)(|B(w, δ)|)−1dµ(w)]dvolg(z) (6.36)
=Z
M
[ Z
M
1B(w,δ)(z)(|B(w, δ)|)−1dvolg(z)]dµ(w) (6.37)
=Z
M
dµ(w) (6.38)
=µ(M), (6.39)
where equation 6.36 used the fact the simple identity 1B(z,δ)(w)=1B(w,δ)(z), and 6.38 follows since the inner bracket is identically one. We now show
µδ* µ, (6.40)
asδ→0. Let f ∈C(M). Fix an >0, by uniform continuity, there exists aδ0>0 such that for allw,z∈M, d(w,z)< δ0 =⇒ |f(w)−f(z)|< . Forδ < δ0,
|µδ(f)−µ(f)|=| Z
M
[ Z
B(z,δ)
(|B(w, δ)|)−1dµ(w)]f(z)dvolg(z)− Z
M
f(w)dµ(w)| (6.41)
=| Z
M
[ Z
M
f(z)1B(z,δ)(w)(|B(w, δ)|)−1dvolg(z)]dµ(w)− Z
M
f(w)dµ(w)| (6.42)
=| Z
M
Z
M
(f(z)−f(w))1B(z,δ)(w)(|B(w, δ)|)−1dvolg(z)dµ(w)| (6.43)
≤ Z
M
Z
M
|f(z)−f(w)|1B(z,δ)(w)(|B(w, δ)|)−1dvolg(z)dµ(w) (6.44)
≤Z
M
[ Z
M
1B(z,δ)(w)(|B(w, δ)|)−1dvolg(z)]dµ(w) (6.45)
=µ(M), (6.46)
where to obtain 6.42, we used 1B(z,δ)(w)=1B(w,δ)(z) to bring f inside the inner integral. This completes the proof of 6.31. We are now ready to show 6.32.
I[µδ]−I[µ]=Z
M×M
KνQdµ⊗2δ − Z
M×M
KνQdµ⊗2 (6.47) Z
M×M
[Gg(z,w)+Qν,g(z)+Qν,g(w)]dµδ(z)dµδ(w)− Z
M×M
[Gg(z,w)+Qnu,g(z)+Qν,g(w)]dµ(z)dµ(w) (6.48)
=[ Z
M×M
Gg(z,w)dµδ(z)dµδ(w)− Z
M×M
Gg(z,w)dµ(z)dµ(w)]+2µδ(M) Z
M
Qν,gdµδ−2µ(M)Z
M
Qν,gdµ (6.49)
=[ Z
M×M
Ggdµ⊗2δ − Z
M×M
Ggdµ⊗2]+2µ(M)[Z
M
Qν,gdµδ− Z
M
Qν,gdµ], (6.50) where 6.50 uses the fact thatµδ(M) = µ(M). Note that sinceµandµδ have finite energy we can split the integrand in 6.49. As we have shown thatµδ* µasδ→0+, we have
δ→0lim+2µ(M)[Z
M
Qν,gdµδ− Z
M
Qν,gdµ]=0, thus it suffices to prove that
δ→0lim+ Z
M×M
Ggdµ⊗2δ − Z
M×M
Ggdµ⊗2=0. (6.51)
By an argument similar to that used in equations 6.41-6.46 we have Z
M×M
Ggdµ⊗2δ − Z
M×M
Ggdµ⊗2 (6.52)
=Z
M×M
[ Z
M×M
(Gg(z,w)−Gg(x,y))1B(z,δ)(x)
|B(x, δ)|
1B(w,δ)(y)
|B(y, δ)| dvolg(z)dvolg(w)]dµ(x)dµ(y). (6.53) Let
Φδ(x,y)=Z
M×M
(Gg(z,w)−Gg(x,y))1B(z,δ)(x)
|B(x, δ)|
1B(w,δ)(y)
|B(y, δ)| dvolg(z)dvolg(w) (6.54) denote the integrand in 6.54. Notice that
Φδ
|Gg|+1
is uniformly bounded. Moreover, limδΦδ =0. Sinceµhas finite energy,|G|g+1 is integrable, and thus by the dominated convergence theorem
δ→0lim+ Z
M×M
Ggdµ⊗2δ − Z
M×M
Ggdµ⊗2
= lim
δ→0+
Z
M×M
Φδdµ⊗2=0.
The proof is complete.
Proposition 6.3.2. Existence of free energy on Riemann surfaces
Let Q be an admissible potential and let Qν,gbe continuous in a neighborhood of S . Then
n→∞lim
−1
n2logZnβ= β 2V.
Proof. SinceQis admissible, the equilibrium measureµeqexists. Since by hypothesisQν,g is continuous in
a neighborhood ofS, we can apply Lemma 6.3.1 to obtain a family of measures{µδ}δ>0satisfying 6.31 and 6.32 whereµδ= Ψδvolg. LetEδ≡ {Ψδ>0}.
Zβn≡ Z
Mn
e−βHndvol⊗ng (6.55)
≥ Z
Enδ
e−βHndvol⊗ng (6.56)
=Z
Enδ
e−βHn(z)−Pni=1log(Ψδ(zi))
n
Y
i=1
Ψδ(zi)dvolg(z1)...dvolg(zn). (6.57)
By Jenson’s inequality we have
logZnβ≥ Z
Enδ
[−βHn(z)−
n
X
i=1
log(Ψδ(zi))]
n
Y
i=1
Ψδ(zi)dvolg(z1)...dvolg(zn)
=−n(n−1)
2 βZ
E2δ
KνQ(z,w)Ψδ(z)dvolg(z)Ψδ(w)dvolg(w)−n Z
Eδ
log(Ψδ(z))Ψδ(z)dvolg(z).
We then have
lim sup
n
−1
n2 logZnβ≤β
2I[Ψδvolg]. (6.58)
Since by hypothesis
δ→0lim+I[Ψδvolg]=I[µeq] we obtain:
lim sup
n
−1
n2 logZnβ≤ lim
δ→0+
β
2I[Ψδvolg] (6.59)
= β
2I[µeq] (6.60)
≡ β
2V. (6.61)
On the other hand,
Znβ≤ |M|ngsup
z∈Mn
e−βHn(z)=|M|ngexp(−βn2 2 I∗[µn]), We then have
lim inf
n
−1
n2 logZnβ≥β 2 lim
n→∞I∗[µn]=β
2V, (6.62)
where the equality above is due to Proposition 6.2.3. By 6.61 and 6.62,
n→∞lim
−1
n2 logZnβ= β
2V. (6.63)
The proof is complete.
Recall that thek-th marginal measure ofΠnis defined by
Πn,k(A)= Πn(A×Mn−k).
In Proposition 6.2.3 we showed that the Fekete measures converge weakly to the equilibrium measure. We now prove that thek-th marginal measures converge weakly to the theµ⊗keq. This is an analogue of the well- known theorem by Johansson inR([18]).
Theorem 6.3.3. Johansson’s marginal measure theorem on Riemann surfaces Let Q be admissible and let Qν,gbe continuous in a neighborhood of S . Then
Πn,k* µ⊗keq,
as n→ ∞.
Proof. LetAn,η ≡ {z∈ Mn| 2
n2Hn(z)≤V+ηn}forη >0. AsHnis lower semi-continuousAn,ηis closed, and thus compact.
By Proposition 6.3.2 , for a fixedη >0, there existsNsuch thatn>Nimplies
Znβ≤e−n2β2(V−ηn). (6.64)
Then using 6.64 and the definition ofAn,ηwe have:
Πn[Acn,η]≡ 1 Zβn
Z
Acn,η
e−βHndvol⊗ng ≤en2β2(V−ηn)e−n2β2(V+ηn) Z
Rn
dvol⊗ng =e−nβη|M|ng. (6.65)
Let f ∈C(Mk). We define then-symmetrization of f,S ymn,k[f] :Mn→R, by
S ymn,k[f](z1, ...,zn)= 1 nk
X
1≤i1,..,ik≤n
f(zi1, ...,zik).
By the symmetry of the measureΠn, we have
Πn,k[f]= Πn[S ymn,k(f)]. (6.66)
Chooseη≥ 1β(log|M|g+1), then from 6.65:
Πn[Acn,η]≤e−n. (6.67)
It is easy to verify that
||S ymn,k[f]||∞=||f||∞. (6.68) It follows from 6.67 and 6.68, that
n→∞limΠn[S ymn,k(f)1Acn,η]=0, (6.69) so from 6.66 and 6.69, the proof is complete once we show:
n→∞limΠn[S ymn,k(f)1An,η]=Z
Mk
f dµ⊗keq. (6.70)
The remaining portion of the proof is dedicated to proving 6.70. We argued above thatAn,ηis compact, and thus so isAnn,η. SinceS ymn,k(f) is continuous, it attains its maximum and minimum onAnn,η. Letan,ηandbn,η
denote such points. Recall for az∈Mn, we definedµzto be:
µz≡1 n
n
X
i=1
δzi. (6.71)
Let
µ(n,η)max ≡µan,η, (6.72)
µ(n,η)min ≡µbn,η. (6.73)
From the definitions ofµ(n,η)max andµ(n,η)min and the symmetry ofS ymn,k(f) we have:
Z
Mk
f d(µn,ηmin)⊗k=Z
Mn
S ymn,k(f)d(µn,ηmin)⊗n≤Πn[S ymn,k(f)1An,η]≤ Z
Mn
S ymn,k(f)d(µn,ηmax)⊗n=Z
Mn
f d(µn,ηmax)⊗k. (6.74)
It thus suffices to show
(µ(n,η)max)⊗k* µ⊗keq (6.75)
asn→ ∞and similarly for (µ(n,η)min)⊗k. The proof forµminis identical, so we proceed to show 6.75. To show 6.75, it suffices to simply show that µ(n,η)max * µeq as n → ∞. It then clearly suffices to show that for any sequence of measures{µzn}n∈Nwherezn∈An,ηwe haveµzn* µeqasn→ ∞(sinceµn,ηmax∈Pn(An,η)).
Let{µn}n∈Nbe an arbitrary sequence of measures withµn ∈ P(An,η). It suffices to show, that for any sub- sequence{µni}i∈N, there exists a further subsequence which converges weakly toµeq. AsMis compact, by Prokhorov’s theorem there exists a convergent subsequence of{µni}i∈Nconverging to a probability measureµ.
By relabeling if necessary, we denote this further subsequence simply by{µni}i∈N. We will show thatµ=µeq. LetN∈Z+. Now asµni * µasi→ ∞and
Z
M×M
KνQ∧Ndµ⊗2=lim
i→∞
Z
M×M
KνQ∧Ndµ⊗2ni (6.76)
=lim
i→∞
Z
M×M\∆
KνQ∧Ndµ⊗2n
i +N
ni
(6.77)
=lim
i→∞
1
n2i Hni(zni)+N ni
(6.78)
≤lim
i→∞V+ 1 ni +N
ni =V, (6.79)
by the monotone convergence theorem Z
M×M
KνQdµ⊗2= lim
N→∞
Z
M×M
KνQ∧Ndµ⊗2 ≤V (6.80)
which shows thatI[µ]≤Vand thusµ=µeq, completing the proof.