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# 7 Quantitative isoperimetric inequality

As in the previous section, let (M, g,m) be a weighted Riemannian manifold with Ric 1 and m(M) = 1, fix θ (0,1) and take a Borel set A M with m(A) = θ. We employ the needle decomposition associated with f := χA −θ as in Subsection 2.3: (Q, ν), {(Xq,mq)}q∈Q, and the guiding function u with ∫

Mu dm= 0. Set Aq:=A∩Xq as in the previous section.

Put δ(A) :=P(A)− I(R,γ)(θ) and define Q :={

q ∈Qmq(Aq) =θ, P(Aq)− I(R,γ)(θ)<δ(A)}

(7.1) as a set of ‘long’ needles (recall from Lemma 2.6 that small deficit implies large diame- ter). Notice that Q is a measurable set since the function q 7→ P(Aq) is measurable by [CMM, Lemma 4.1]. We observe from Lemma 6.1 the following (similarly to the proof of Theorem 6.2).

Lemma 7.1 (Q is large) We have ν(Q)1δ(A).

For further analyzing the behavior of long needles, we define Q :={

q ∈Qmq

(Aq(−∞, rm

q(θ)])

δ(A)}

, Q+ :={

q ∈Qmq(

Aq[r+mq(θ),∞))

δ(A)}

,

(7.2) where Xq is parametrized byu and rm±q(θ)∈Xq are defined by

mq(

Xq(−∞, rmq(θ)])

=mq(

Xq[r+mq(θ),∞))

=θ

as in Proposition 4.1. The measurability of Q+ and Q can be shown as in [CMM] (see Lemma 6.1 and the paragraph following it). Then the next lemma is a consequence of Proposition4.1.

Lemma 7.2 (Q ∪Q+ is large) If δ(A) is suﬃciently small, then we have ν(

Q\(Q ∪Q+ ))

δ(A).

Proof. Recall from Theorem 2.10 that, for ν-almost every q Q, (Xq,mq) satisfies Ric1 and mq(Aq) = θ. Then we deduce from (4.1) and limδ0C5(θ, δ) = that

P(Aq)− I(Xq,mq)(θ)min {

mq(

Aq(−∞, rmq(θ)]) ,mq(

Aq[rm+q(θ),∞))}

for q∈Q provided that δ(A) is suﬃciently small. Hence

P(Aq)− I(R,γ)(θ)P(Aq)− I(Xq,mq)(θ)>δ(A) for q∈Q\(Q ∪Q+ ), and it follows from Lemma 6.1 that

δ(A)

δ(A)·ν(

Q\(Q ∪Q+)) .

2

Next we shall show that one of Q and Q+ necessarily has a small volume. This is the most technical step in this section and the structure of the proof diﬀers from that of [CMM, Proposition 6.4], due to the fact that the diameter of M is not bounded and needles can be infinitely long (cf., for example, [CMM, Proposition 5.1, Corollary 5.4]).

The following observation by virtue of (6.2) will play a crucial role. Recall that aθ R is defined by γ((−∞, aθ]) = θ.

Proposition 7.3 (u is nearly centered on most needles) Ifδ(A)is suﬃciently small, then there exists a measurable set Qc⊂Q such that ν(Qc)1−δ(A)(1−ε)/(93ε) and

max{

|aθ−rm

q(θ)|,|a1θ−rm+

q(θ)|}

≤C8(θ, ε)δ(A)(1ε)/(93ε) (7.3) for every q ∈Qc∩Q.

Proof. We setδ :=δ(A) and

a:= 2(1−ε) 3(3−ε)

for simplicity, and observe from (6.2) that the setQc⊂Q consisting of q with ( ∫

Xq

u dmq

)2

≤C7(θ, ε)δa (7.4)

satisfies ν(Qc) 1 −δ(1−ε)/(3−ε)−a. Fix a needle q Qc Q and put mq = e−ψdx, r :=rmq(θ) and r+ :=rm+q(θ) for brevity.

Since the assertion is symmetric, by reversingXqif necessary, we can assumeI(Xq,mq)(θ) = eψ(r). Then we have eψ(r) P(Aq)≤ I(R,γ)(θ) +

δ and deduce from (3.3) that ψ(x)ψg(

(x−r) +aθ)

(

ψ+ (r)−aθ)

(x−r)−ω(θ) δ

on Xq, where we recall that Xq is parametrized by u. We similarly observe from (3.4) that

ψ(x)ψg(

(x−r) +aθ)

(

ψ+(r)−aθ)

(x−r) +ω(θ)δ1/4

on [S +r−aθ, T +r−aθ]. Let us set α := aθ −r, β := ψ+(r)−aθ and observe

|β| ≤ (C2+ 1)

δ from (3.9). By (7.4) we also find that α 0 as δ 0, our goal is to make this quantitative.

We have

Xq

u dmq =

Xq

xmq(dx)

0

xexp

(ψg(x+α)−β(x−r) +ω√ δ

) dx

+

0

Sα

xexp

(ψg(x+α)−β(x−r)−ωδ1/4 )

dx

= 1

2π

0

xexp (

(x+α+β)2

2 +αβ+ β2

2 +βr+ω√ δ

) dx

+ 1

2π

0 Sα

xexp (

(x+α+β)2

2 +αβ +β2

2 +βr−ωδ1/4 )

dx

= exp (

αβ+β2

2 +βr+ω√ δ

) ∫

α+β

(x−α−β)γ(dx) + exp

(

αβ+ β2

2 +βr−ωδ1/4

) ∫ α+β S+β

(x−α−β)γ(dx).

Since |β| ≤(C2+ 1)

δ and α→0 as δ 0, we find exp

(

αβ+ β2

2 +βr+ω√ δ

)

1 +C(θ) δ,

exp (

αβ+β2

2 +βr−ωδ1/4 )

1−C(θ)δ1/4. Moreover, we observe

α+β

S+β

(x−α−β)γ(dx) =

α+β

−∞

(x−α−β)γ(dx)

S+β

−∞

(x−α−β)γ(dx)

=

α+β

−∞

(x−α−β)γ(dx) + 1

2π

[ex2/2]S+β

−∞ + (α+β)γ(

(−∞, S+β]) and, assuming that δ is suﬃciently small,

e(S+β)2/2 e(1ε)S2/2 = (1−S)(1ε)2eε(1ε)S2/2

(eS2/2 1−S

)(1−ε)2

≤C(θ, ε)δ(1ε)2/4 and

γ(

(−∞, S+β])

γ(

(−∞, S])

+√|β|

2π ≤C(θ)δ1/4 by (3.19) and (7.1). Therefore we obtain

Xq

u dmq

−∞

(x−α−β)γ(dx) +C(θ, ε)δ(1ε)2/4

=−α−β+C(θ, ε)δ(1ε)2/4

≤ −α+C(θ, ε)δ(1−ε)2/4.

A similar calculation shows

Xq

u dmq ≥ −α−C(θ, ε)δ(1ε)2/4

as well. Combining these with (7.4) yields (provided that a/2(1−ε)2/4)

|α| ≤ α+

Xq

u dmq +

Xq

u dmq

≤C(θ, ε)δa/2. (7.5)

In order to bound |a1θ−r+|, let us recall eψ(x) 1

2πexp (

αβ +β2

2 +βr+ω√ δ

) exp

(

(x+α+β)2 2

)

(

1 +C(θ) δ)

eψg(x+α+β)

onXq. Therefore, on one hand, for Θ>−(α+β) with eψg(a1−θ+Θ+α+β) eψg(a1−θ)/2, mq(

[a1θ+ Θ,∞))

(

1 +C(θ) δ)

γ(

[a1θ+ Θ +α+β,∞))

(

1 +C(θ) δ)(

θ− eψg(a1−θ)

2 (Θ +α+β) )

.

Then choosing

Θ = 2eψg(a1−θ)θC(θ)

δ−α−β implies mq([a1θ+ Θ,∞))< θ and hence

r+ < a1θ+ Θ≤a1θ+C(θ, ε)δa/2,

where we used (7.5). On the other hand, for Ξ > α + β with eψg(a1θΞ+α+β) eψg(a1−θ)/2, we observe

mq

([a1θΞ,∞))

1(

1 +C(θ) δ)

γ(

(−∞, a1θΞ +α+β])

1(

1 +C(θ) δ)(

(1−θ) eψg(a1θ)

2 (Ξ−α−β) )

.

This yieldsmq([a1θΞ,∞))> θ with Ξ = 2eψg(a1θ)(1−θ)C(θ)

δ+α+β, and hence r+ > a1θΞ≥a1θ−C(θ, ε)δa/2.

This completes the proof. 2

Let us explain the geometric intuition of the proof of the next proposition. If both ν(Q) andν(Q+ ) have a certain volume, then the strict concavity ofI(R,γ)implies that the sum of the perimeters of regions A and A+ corresponding to Q and Q+ , respectively, is larger thanI(R,γ)(θ). This contradicts the assumed small deficit when the gap between P(A) and P(A) +P(A+) is suﬃciently small. In order to construct such a decomposition

̸

Proposition 7.4 (One of Q and Q+ is small) Assume θ ̸= 1/2. Then we have min(Q), ν(Q+)} ≤C9(θ)δ(A)(1ε)/(93ε),

provided that δ(A) is suﬃciently small.

Proof. Put δ = δ(A) again in this proof. Let us first assume θ (0,1/2) and consider the decomposition of A,

Ar :=A∩ {u≤r}, A+r :=A∩ {u≥r}, for r∈(r1, r2) with

r1 := 2 3aθ+1

3a1θ, r2 := 1 3aθ+ 2

3a1θ.

Note that aθ < 0 < a1θ = −aθ since θ < 1/2. Moreover, letting δ smaller if necessary, we find from (7.3) thatr1 ≥rmq(θ) holds forq ∈Qc∩Q.

Since|∇u|= 1 almost everywhere, we obtain from the coarea formula (see, e.g., [Cha]) that

m(

A∩ {r1 < u < r2})

=

A∩ {r1<u<r2}|∇u|dm=

r2

r1

|A∩u1(r)|dr,

where | · | denotes the (n−1)-dimensional measure induced from m (precisely, eΨHn1 where Hn1 is the (n−1)-dimensional Hausdorﬀ measure). For q∈Qc∩Q , we deduce fromr1 ≥rm

q(θ) and (7.2) that mq(

Aq(−∞, r1])

=mq(Aq)mq(

Aq\(−∞, r1])

≥θ−√

δ. (7.6)

Similarly mq(Aq [r2,∞)) θ

δ holds for q Qc Q+. Then it follows from Theorem 2.10(i), Lemmas 7.1,7.2 and Proposition7.3 that

m(Ar1 ∪A+r2)

QcQ

mq(

Aq(−∞, r1])

ν(dq) +

QcQ+

mq(

Aq[r2,∞)) ν(dq)

(θ−√ δ)ν(

Qc(Q ∪Q+ ))

(θ−√

δ)(12

δ−δ(1ε)/(93ε))

≥θ−(1 + 2θ)

δ−θδ(1ε)/(93ε)

≥θ−δ(1ε)/(93ε). Therefore we obtain

r2

r1

|A∩u1(r)|dr=θ−m(Ar1 ∪A+r2)≤δ(1ε)/(93ε), and we can choose some ˆr∈(r1, r2) satisfying

|A∩u1r)| ≤ δ(1ε)/(93ε)

r2−r1 = 3δ(1ε)/(93ε)

a1θ−aθ = 3δ(1ε)/(93ε) 2|aθ| .

This yields that

P(Arˆ) +P(A+rˆ)P(A)2|A∩u1r)| ≤ 3δ(1ε)/(93ε)

|aθ| . (7.7)

In the first inequality, take a sequence i}iN of Lipschitz functions such that 0 ≤ϕi χA, ϕi χA in L1(m) and limi→∞

M|∇ϕi|dm= P(A) (recall (2.2) for the definition of P(A)), and put

ρ+i (x) := min{

max{u(x)−r,ˆ 0},1}

, ρi (x) := 1−ρ+i (x).

Then ρ±i ϕi →χA± ˆ

r in L1(m) and P(Aˆr) +P(A+ˆr)lim inf

i→∞

M

(|∇(ρi ϕi)|+|∇(ρ+i ϕi)|) dm

lim

i→∞

M

(ρi +ρ+i )|∇ϕi|dm+ lim inf

i→∞

M

(|∇ρi |+|∇ρ+i |) ϕidm

P(A) + lim

i→∞

A∩{ˆr<u<ˆr+i1}

2i dm

=P(A) + 2|A∩u1r)|.

Now, it follows from Lemma 2.7 that I(′′R,γ) ≤ −I(R,γ)(θ)1 on (0, θ] (since θ < 1/2), which implies

P(Aˆr)≥ I(R,γ)

(m(Aˆr))

m(Arˆ)

θ I(R,γ)(θ) + 1 2I(R,γ)(θ)

(

1 m(Arˆ) θ

)m(Arˆ) θ θ2. Concerning the second term in the RHS, on one hand, we observe from (7.6) that

m(Aˆr)(θ−√

δ)ν(Qc∩Q) θ

2ν(Qc∩Q).

On the other hand, we similarly find

m(Aˆr) = θ−m(A+rˆ)≤θ− θ

2ν(Qc∩Q+ ).

Therefore, setting V := min(Qc∩Q), ν(Qc∩Q+ )} ≤1/2, we obtain P(Aˆr) m(Aˆr)

θ I(R,γ)(θ) + 1 2I(R,γ)(θ)

( 1 V

2 )V

2θ2.

We have a similar inequality for A+rˆ in the same way. Summing up, we obtain P(Aˆr) +P(A+ˆr)≥ I(R,γ)(θ) + 1

I(R,γ)(θ) (

1 V 2

)V

2θ2 ≥ I(R,γ)(θ) +c(θ)V.

Combining this with (7.7) and I(R,γ)(θ) =P(A)−δ yields 3δ(1ε)/(93ε)

P(A) +P(A+)P(A)≥c(θ)V −δ

and hence, by Proposition 7.3,

min(Q), ν(Q+ )} ≤V +δ(1−ε)/(93ε)

1 c(θ)

(3δ(1ε)/(93ε)

|aθ| +δ )

+δ(1ε)/(93ε)

≤C(θ)δ(1ε)/(93ε). This completes the proof for θ <1/2.

When θ >1/2, the complement Ac of A satisfies P(Ac) =P(A) and m(Ac) = 1−θ <

1/2. Note also thatI(R,γ)(θ) = I(R,γ)(1−θ) andrmq(θ) =r+mq(1−θ),rm+q(θ) =rmq(1−θ).

Hence we have, sinceE\F =E∩Fc=Fc\Ec,

Aq(−∞, rmq(θ)] =Acq(rmq(θ),∞) =Acq(r+mq(1−θ),∞)

and similarly Aq[rm+q(θ),∞) = Acq(−∞, rmq(1−θ)). Therefore we can obtain the

claim for A by applying the above argument to Ac. 2

From the proof of Proposition7.4, we find thatC9(1−θ) =C9(θ) and limθ1/2C9(θ) =

(since a1/2 = 0). Hence the case ofθ = 1/2 is not covered.

We finally prove our main theorem. We employ the sub-level and super-level sets of the guiding functionu instead of balls in [CMM].

Theorem 7.5 (Quantitative isoperimetry) Let(M, g,m)be a complete weighted Rie- mannian manifold such thatRic1andm(M) = 1. Fixθ (0,1)\{1/2}andε∈(0,1), take a Borel set A M with m(A) = θ, and assume that P(A)≤ I(R,γ)(θ) +δ holds for suﬃciently small δ >0 (relative to θ and ε). Then, for the guiding function u associated with A such that

Mu dm= 0, we have min

{ m(

A△ {u≤aθ}) ,m(

A△ {u≥a1θ})}

≤C(θ, ε)δ(1ε)/(93ε). (7.8) Proof. We set again δ = δ(A). Thanks to Proposition 7.4, we first assume ν(Q+ ) C9(θ)δ(1ε)/(93ε). Then we deduce from Lemmas 7.1 and 7.2 that

ν(Q\Q) = ν(Q\Q) +ν(

Q\(Q ∪Q+ ))

+ν(Q+ )2

δ+C9(θ)δ(1ε)/(93ε). Therefore we obtain

m(

A△ {u≤aθ})

Q

mq

(Aq(−∞, aθ])

ν(dq) +ν(Q\Q )

Q

mq(

Aq(−∞, rmq(θ)])

ν(dq) +

Q

mq(

(−∞, aθ](−∞, rmq(θ)]) ν(dq)

+ν(Q\Q )

Q

mq(

(−∞, aθ](−∞, rmq(θ)])

ν(dq) + 3

δ+C9(θ)δ(1ε)/(93ε). (7.9)

In order to estimate the first term, we recall from Proposition 7.3 that |aθ−rmq(θ)| ≤ C8(θ, ε)δ(1−ε)/(93ε) for q∈Qc∩Q. This implies

mq

((−∞, aθ](−∞, rm

q(θ)])

=mq

((min{aθ, rm

q(θ)},max{aθ, rm

q(θ)}])

≤C(θ, ε)δ(1ε)/(93ε) for q∈Qc∩Q. Substituting this into (7.9), we obtain

m(

A△ {u≤aθ})

≤C(θ, ε)δ(1ε)/(93ε)+ν(Q \Qc) + 3

δ+C9(θ)δ(1ε)/(93ε)

≤C(θ, ε)δ(1ε)/(93ε).

In the case of ν(Q ) C9(θ)δ(1ε)/(93ε), we similarly have m(A△ {u a1θ}) C(θ, ε)δ(1ε)/(93ε). This completes the proof. 2

We conclude with several remarks and open problems related to Theorem 7.5.

Remark 7.6 (a) If we assert only the existence of ‘some’ 1-Lipschitz function u enjoy- ing (7.8), then one can merely take u(x) := d(A, x) +aθ. Therefore the novelty of Theorem 7.5 lies in the construction of u as the guiding function of the needle de- composition. By construction the guiding function u seems closely related to the Busemann function. When there is a straight line η : R −→ M (meaning that d(η(s), η(t)) =|s−t|for alls, t∈R), the associatedBusemann function b:M −→R is defined by

b(x) := lim

t→∞

{t−d(

x, η(t))}

.

By constructionbis 1-Lipschitz and sometimes regarded as ‘a distance function from infinity’. In Cheeger–Gromoll-type splitting theorems (under RicN 0, see also (b) below), we show that b is totally geodesic and M is split into R ×Σ, where {t} × Σ = b1(t) and ηx(t) := (t, x) is a straight line for every x Σ. This is a similar phenomenon to the rigidity of the Bakry–Ledoux isoperimetric inequality (under Ric K > 0) in Theorem 2.8, where the guiding function plays a similar role to the Busemann function (see [Ma2] for details). Going back to our quantitative investigation, the guiding function u shares several properties with the Busemann function: u is 1-Lipschitz, most needles are long in both directions (limδ0S =−∞

and limδ0T =in Proposition3.2), and the direction of most needles are the same (Proposition 7.4). When, for instance, some needle is a straight line, one may relate the associated Busemann function with the guiding function and obtain (7.8) in terms of that Busemann function. In this direction, moreover, one could expect an ‘almost splitting theorem’ as metric measure spaces, namely (M, g,m) is close to the product space (R,| · |,γ)×Y in some sense (even when there is no infinite needle). This is an interesting and challenging problem, let us recall that Gromov’s precompactness theorem ([Gr, §5.A]) does not apply under Ric ≥K >0.

(b) In comparison with the Cheeger–Gromoll-type splitting theorem under Ric 0 in

assumed in Theorem 7.5. In the splitting theorem we claim that the space splits oﬀ the real line endowed with the Lebesgue measure, and hence an upper bound of Ψ is necessary to rule out Gaussian spaces (and hyperbolic spaces with very convex weight functions). Compare this with the rigidity results under Ric K > 0 in Theorems 2.4,2.8.

(c) Since the needle decomposition is available also for Finsler manifolds by [CM, Oh3], one can prove the analogue of Theorem 7.5 for reversible Finsler manifolds verbatim.

In the non-reversible case, however, the needle decomposition does not provide the sharp isoperimetric inequality and it is unclear if one can generalize Theorem7.5. See [Oh3] for more details on the non-reversible situation, and [Oh4] for a derivation of the sharp Bakry–Ledoux isoperimetric inequality for non-reversible Finsler manifolds.

(d) In Theorem 7.5 we restrict ourselves to weighted Riemannian manifolds since the needle decomposition is not yet known for metric measure spaces satisfying CD(1,∞) or RCD(1,∞). We refer to [AM] for the Bakry–Ledoux isoperimetric inequality on RCD(1,∞)-spaces.

(e) There are two open problems related to Theorem 7.5. The first one is the case of θ = 1/2. The condition θ ̸= 1/2 was used only in Proposition 7.4, where we showed that one of Q and Q+ has a small volume. If this step is established in some other way, then all the other steps of the proof work and we can obtain Theorem 7.5 for θ = 1/2.

(f) Another open problem is the optimal order of δ in (7.8). Our estimate δ(1ε)/(93ε) seems not optimal at all and, compared with the case of Gaussian spaces (recall (1.1)), the optimal order is likely

δ. We remark that the optimal order is not known also for CD(N 1, N)-spaces studied in [CMM] (N (1,∞)), where they obtained δN/(N2+2N1) depending on N (recall (1.2)).

(g) Inspired by [DF, CF], we expect that the push-forward measure um is close to γ in the Wasserstein distance W1 or W2 over R. We may make use of the Talagrand inequality W2(um,γ)2 2 Entγ(um) (recall Subsection 6.3).

Acknowledgements. We thank Fabio Cavalletti and Max Fathi for discussions during the workshop “Geometry and Probability” in Osaka (2019). SO was supported in part by JSPS Grant-in-Aid for Scientific Research (KAKENHI) 19H01786.