Poincar inequalities that fail to constitute an open-ended - - PowerPoint PPT Presentation

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Poincaré inequalities that fail

to constitute an open-ended condition Lukáš Malý

Workshop on Geometric Measure Theory

July 14, 2017

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Poincaré inequalities

Setting Let (X, d, µ) be a complete metric space endowed with a doubling measure. Definition (Heinonen–Koskela, 1996) A Borel function g ∶ X → [0, ∞] is an upper gradient of u ∶ X → R if ∣u(γ(0)) − u(γ(lγ))∣ ≤ ∫γ g ds for every rectifiable curve γ ∶ [0, lγ] → X. Definition (Heinonen–Koskela, 1996) The space X admits a (1, p)-Poincaré inequality with p ∈ [1, ∞) if

∫B ∣u − uB∣ dµ ≤ cPI diam(B)(∫λB gp dµ)

1/p

for every u ∈ L1

loc(X), its upper gradient g, and every ball B ⊂ X, where cPI, λ ≥ 1.

Lukáš Malý Poincaré inequalities that fail to be open-ended 1/17

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Self-improvement of Poincaré inequalities

Theorem (Keith–Zhong, 2008) If X admits a p-Poincaré inequality with p ∈ (1, ∞), then it admits a (p − ε)-Poincaré inequality for some ε > 0. Remark: Both completeness of X and doubling of µ are used! Example (Koskela, 1999) For every 1 < p ≤ n, there is a closed set E ⊂ Rn of zero Ln-measure such that X = Rn \ E supports a p-PI, but not q-PI for any q < p. Remark: X in this example is locally compact and µ ≔ Ln∣X is doubling. Question Does an A-PI with an averaging operator A on the RHS of PI (A is close to the Lp-average) improve to a (p − ε)-PI?

Lukáš Malý Poincaré inequalities that fail to be open-ended 2/17

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Motivation

Theorem (Durand-Cartagena, Jaramillo, Shanmugalingam, 2013) Assume that µ is Ahlfors s-regular (i.e., µ(B(x, r)) ≈ rs). Let p > s. Then, TFAE:

▶ X supports a p-PI; ▶ Every u ∈ N1,p(X) is (1 − s p)-Hölder continuous and

∣u(x) − u(y)∣ ≲ d(x, y)1−s/p∥gu∥Lp(B(x,λd(x,y)))

▶ There is C ≥ 1 such that for all x, y ∈ X

Modp({γ ∈ Γx,y ∶ len(γ) ≤ Cd(x, y)}) ≈ d(x, y)s−p

Lukáš Malý Poincaré inequalities that fail to be open-ended 3/17

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Motivation

Idea for the critical exponent Assume that µ is Ahlfors s-regular, s ≥ 1 (i.e., µ(B(x, r)) ≈ rs). Then, TFAE:

• 1. X supports a Lorentz-type Ls,1-PI
• 2. Every u ∈ N1Ls,1(X) is continuous and

∣u(x) − u(y)∣ ≲ ∥gu∥Ls,1(B(x,λd(x,y)))

• 3. There is C ≥ 1 such that for all x, y ∈ X

ModLs,1({γ ∈ Γx,y ∶ len(γ) ≤ Cd(x, y)}) ≈ 1 Works for s = 1 For s > 1, the following can be proven:

• 1. ⇐ 2. ⇔ 3.

p-PI with p < s ⇒ 2.

Lukáš Malý Poincaré inequalities that fail to be open-ended 4/17

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Lorentz spaces Lp,q

Norm in the Lebesgue Lp-spaces ∥u∥Lp(X) = (∫X ∣u(x)∣pdµ(x))

1/p

= (∑

k∈Z ∫{x∶2k<∣u(x)∣≤2k+1} ∣u(x)∣pdµ(x)) 1/p

= (∑

k∈Z

∥uχ{2k<∣u∣≤2k+1}∥

p Lp(X)) 1/p

=

• {∥uχ{2k<∣u∣≤2k+1}∥Lp(X)}

∞ k=−∞

• ℓp(Z)

Definition (functional comparable with the Lorentz norm) Let p ∈ [1, ∞) and q ∈ [1, ∞]. Then, we define ∣∣∣u∣∣∣Lp,q(X) =

• {∥uχ{2k<∣u∣≤2k+1}∥Lp(X)}

∞ k=−∞

• ℓq(Z)

Remark: Lp,1(X) ↪ Lp(X) = Lp,p(X) ↪ Lp,∞(X) for 1 ≤ p < ∞. Remark: Lp logα L(X) ↪ Lp,q(X), whenever α > (p/q) − 1 ≥ 0.

Lukáš Malý Poincaré inequalities that fail to be open-ended 5/17

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Known results on Orlicz–Poincaré inequalities

Definition (Orlicz–Poincaré inequality) Let Ψ ∶ [0, ∞) → [0, ∞) be strictly increasing, continuous, and convex with Ψ(0+) = 0 and Ψ(∞−) = ∞. The space X admits a Ψ-Poincaré inequality if

∫B ∣u − uB∣ dµ ≤ cPI diam(B)Ψ−1(∫λB Ψ(g) dµ)

for every u ∈ L1

loc(X), its upper gradient g, and every ball B ⊂ X, where cPI, λ ≥ 1.

Remark: If Ψ(t) = tp, then Ψ-PI is just a p-PI. Some self-improvement of Ψ-PI to (p − ε)-PI can be expected if Ψ(t) ∼ tpη(t), where η(t) grows (or decays) slower than any power tδ (or t−δ)

Lukáš Malý Poincaré inequalities that fail to be open-ended 6/17

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Known results on Orlicz–Poincaré inequalities

Theorem (J. Björn, 2010) The Euclidean space (R, µ) admits a Ψ-PI if and only if the Hardy–Littlewood maximal

• perator M ∶ LΨ(R, µ) → LΨ(R, µ) is bounded.

Theorem (Bloom–Kerman, 1994) Let Ψ(t) = tp logα(e + t) with p > 1 and α ∈ R. If M ∶ LΨ(R, µ) → LΨ(R, µ) is bounded, then M ∶ Lp−ε(R, µ) → Lp−ε(R, µ) is bounded for some ε > 0. Corollary Suppose that (Rn, µ), where µ = µ1 × µ2 × ⋯ × µn, admits a Ψ-PI, where Ψ(t) = tp logα(e + t) for some p > 1 and α ∈ R. Then, (Rn, µ) admits a (p − ε)-PI for some ε > 0.

Lukáš Malý Poincaré inequalities that fail to be open-ended 7/17

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Lorentz-type Poincaré inequalities

Definition (Lorentz-type Poincaré inequality) Let p ∈ [1, ∞) and q ∈ [1, ∞]. The space X admits an Lp,q-Poincaré inequality if

∫B ∣u − uB∣ dµ ≤ cPI diam(B)

∥gχλB∥Lp,q(X) ∥χλB∥Lp,q(X) = cPI diam(B) ∥gχλB∥Lp,q(X) µ(λB)1/p for every u ∈ L1

loc(X), its upper gradient g, and every ball B ⊂ X, where cPI, λ ≥ 1.

Remark: If p = q, then Lp,q-PI is just a p-PI. Observation Due to the embedding between Lorentz spaces, we have Lp,Q-PI ⇒ p-PI ⇒ Lp,q-PI whenever 1 ≤ q < p < Q ≤ ∞.

Lukáš Malý Poincaré inequalities that fail to be open-ended 8/17

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Lorentz-type Poincaré inequalities

Comparison to Muckenhoupt weights on R

▶ (R, µ) admits an Lp,q-PI, p ∈ (1, ∞), q ∈ [1, ∞], if and only if the HL maximal

• perator M ∶ Lp,q(R, µ) → Lp,q(R, µ) is bounded.

▶ M ∶ Lp,q(R, µ) → Lp,q(R, µ) is bounded if and only if M ∶ Lp(R, µ) → Lp(R, µ) is

bounded, p ∈ (1, ∞), q ∈ (1, ∞]. (Chung–Hunt–Kurtz, 1982)

▶ If (Rn, µ), where µ = µ1 × ⋯ × µn, admits an Lp,q-PI with p ∈ (1, ∞) and

q ∈ (1, ∞], then it admits a (p − ε)-PI for some ε > 0. Question Does an Lp,1-PI undergo self-improvement?

Lukáš Malý Poincaré inequalities that fail to be open-ended 9/17

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General Poincaré inequalities

Let A ∶ L0

+(X) × B(X) → [0, ∞] be an averaging operator, i.e., ▶ A(c, B) ≈ c for every constant c ≥ 0 and every ball B ⊂ X; ▶ if 0 ≤ g1 ≤ g2 a.e. in B, then A(g1, B) ≤ A(g2, B).

Let MA be the associated (non-centered) maximal operator MAf(x) = sup

B∋x

A(∣f∣, B), x ∈ X. Definition (Generalized Poincaré inequality) The space X admits an A-Poincaré inequality if

∫B ∣u − uB∣ dµ ≤ cPI diam(B) A(g, λB)

for every u ∈ L1

loc(X), its upper gradient g, and every ball B ⊂ X, where cPI, λ ≥ 1.

Lukáš Malý Poincaré inequalities that fail to be open-ended 10/17

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General Poincaré inequalities

Lemma (M.) Assume that X supports an A-PI. Suppose that A(g, B) ≲ (∫B gpA dµ)

1/pA

for all g, B. Then, X admits a (pA − ε)-PI for some ε > 0. Corollary Let p > 1. Suppose that X supports

▶ an Lp,q-PI with q ∈ [p, ∞],

OR

▶ a Ψ-PI with Ψ(t) ∼ tp logα(e + t), where α ∈ R.

Then, X admits a (p − ε)-PI for some ε > 0.

Lukáš Malý Poincaré inequalities that fail to be open-ended 11/17

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General Poincaré inequalities

Theorem (M.) Assume that X supports an A-PI and that MA ∶ RI1(X) → w-RI2(X) is bounded, where RI1 and RI2 are rearrangement invariant spaces, such that:

▶ RI1 and RI2 are close to each other, viz., their fundamental functions satisfy

t ≤ φ1(φ−1

2 (t)) ≤ t (1 + log− t)β,

β ≥ 0;

▶ RI1 satisfies an upper q-estimate, i.e.,

∥f χ⋃k Ek∥

q RI1 ≤ C ∑ k

∥f χEk∥q

RI1,

q ≥ 1;

▶ RI1 lies close to LpA, specifically,

φ1(tpA) t(log− t)1−(β+1/q) → 0 as t → 0. Then, X admits a (pA − ε)-PI for some ε > 0.

Lukáš Malý Poincaré inequalities that fail to be open-ended 12/17

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General Poincaré inequalities

Corollary (M.) Assume that X supports an Lp,q-PI with p ∈ (1, ∞) and q ∈ (1, ∞]. Then, X admits a (p − ε)-PI. Proof: Choose A(g, B) = ∥gχB∥Lp,q µ(B)1/p . (a) q ≤ p. Then, MA ∶ Lp,q(X) → Lp,∞(X) is bounded [Chung–Hunt–Kurtz, 1982]. Hence, φ1(t) = φ2(t) = t1/p. Lp,q satisfies an upper q-estimate [CHK ’82]. (b) q > p. Then, MA ∶ Lp(X) → Lp,∞(X) is bounded. Hence, φ1(t) = φ2(t) = t1/p. Lp,q satisfies an upper p-estimate [Maz′ya]. Beware Lp,1-PI is NOT included in the self-improvement result.

Lukáš Malý Poincaré inequalities that fail to be open-ended 13/17

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Lack of self-improvement

Theorem (M.) For every p ∈ (1, ∞), there is a complete metric space X endowed with a doubling measure such that it supports an Lp,1-PI but no better. If in addition p ∈ N, then µ can be chosen Ahlfors n-regular. How to interpret “no better” X does NOT admit an A-PI, whenever

▶ A(g, B) = ∥gχB∥RI(X) ∥χB∥RI(X) , where RI(X) is a rearrangement-invariant Banach function

space s.t. RIloc(X) ⊋ Lp,1

loc(X);

OR

▶ A(g, B) = ∥g∥RI(B, ̃ µB), where ̃

µB(E) = µ(E ∩ B)/µ(B) and RIloc(X) ⊋ Lp,1

loc(X);

OR

▶ A(g, B) = Φ−1(∫B Ψ(g) dµ) provided that infκ>0 κA( g κ , B) ≲ ∥g∥RI(B, ̃ µB).

Lukáš Malý Poincaré inequalities that fail to be open-ended 14/17

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Lack of self-improvement

Let X = X+ ∪ X− ⊂ Rn be endowed with the Euclidean distance and the n-dimensional Lebesgue measure, where X+ = {(x1, . . . , xn) ∈ Rn ∶ x j ≥ 0 for all j} and X− = {(x1, . . . , xn) ∈ Rn ∶ x j ≤ 0 for all j}.

▶ Both X+ and X− admit a 1-PI. ▶ If u ∈ N1Ln,1(X), then u∣X+ ∈ N1Ln,1(X+) and u∣X− ∈ N1Ln,1(X−) ▶ Then, ∣u(x) − u(y)∣ ≲ ∥gu∥Ln,1(Bx y∩X+) whenever x, y ∈ X+; analogously in X−. ▶ Thus, ∣u(x) − u(y)∣ ≲ ∥gu∥Ln,1( ˜ Bx y) whenever x, y ∈ X. ▶ Direct calculation yields for a ball B that

∫B ∣u(x)−uB∣ dx ≈ ∫B ∫B ∣u(x)−u(y)∣ dx dy ≲ ∥gu∥Ln,1(λB) ≈ diam(B)

µ(B)1/n ∥gu∥Ln,1(λB) .

Lukáš Malý Poincaré inequalities that fail to be open-ended 15/17

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Lack of self-improvement

Let h ∈ RI \ Ln,1(X). Let g be the radially decreasing rearrangement of h with peak in 0, i.e.,

▶ 0 ≤ g(x) ≤ g(y) whenever ∣x∣ ≥ ∣y∣ > 0; ▶ ∣{x ∈ X ∶ g(x) > τ}∣ = ∣{x ∈ X ∶ ∣h(x)∣ > τ}∣ for every τ > 0.

Then, gk ≔ g/k, k > 0, is an upper gradient of u ≔ χX−. For the ball B = B(0, 1), we have 1 2 = ∫B ∣u(x) − uB∣ dx

?

≤ cPIA(gk, λB) = cPI A(g k , λB) → 0 as k → ∞.

Lukáš Malý Poincaré inequalities that fail to be open-ended 16/17

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Final remarks

▶ The motivating problem (characterization of Ls,1-PI by the estimates for

modulus of continuity of N1Ls,1(X) functions) is still unsolved.

▶ The theorem does not say anything about the Orlicz-type Ψ-Poincaré

inequalities if Ψ is non-doubling.

▶ The Muckenhoupt analogue gives that if (Rn, µ) with product measure supports

a Ψ-PI (where the conjugate of Ψ is doubling), then there is p < ∞ such that (Rn, µ) admits a p-PI. This applies in particular to Ψ(t) = exp(tα), α > 0.

▶ It is possible to find a doubling measure µ on R such that (R, µ) supports an

∞-PI, but no Ψ-PI. (Not even for non-doubling Ψ).

▶ ∞-PI for (R, µ) is equivalent to L1 ≪ µ

(Durand-Cartagena–Jaramillo–Shanmugalingam, 2013)

▶ Ψ-PI for (R, µ) requires µ ≪ L1 (J. Björn, 2010) Lukáš Malý Poincaré inequalities that fail to be open-ended 17/17

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Thank you for your attention!