Poincaré inequalities that fail

ي ۆ Poincaré inequalities that fail to constitute an open-ended condition Lukáš Malý Workshop on Geometric Measure Theory July 14, 2017

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 for every rectifiable curve γ [0, l γ ] X. u(γ(0)) u(γ(l γ )) γ g ds Definition (Heinonen Koskela, 1996) The space X admits a (1, p)-poincaré inequality with p [1, ) if B u u B dµ c PI diam(b)( λb g p dµ) 1/p for every u L 1 loc(x), its upper gradient g, and every ball B X, where c PI, λ 1. Lukáš Malý Poincaré inequalities that fail to be open-ended 1/17

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 R n of zero L n -measure such that X = R n \ E supports a p-pi, but not q-pi for any q < p. Remark: X in this example is locally compact and µ L n X is doubling. Question Does an A-PI with an averaging operator A on the RHS of PI (A is close to the L p -average) improve to a (p ε)-pi? Lukáš Malý Poincaré inequalities that fail to be open-ended 2/17

Motivation Theorem (Durand-Cartagena, Jaramillo, Shanmugalingam, 2013) Assume that µ is Ahlfors s-regular (i.e., µ(b(x, r)) r s ). Let p > s. Then, TFAE: X supports a p-pi; Every u N 1,p (X) is (1 s )-Hölder continuous and p There is C 1 such that for all x, y X u(x) u(y) d(x, y) 1 s/p g u L p (B(x,λd(x,y))) Mod p ({γ Γ x,y len(γ) Cd(x, y)}) d(x, y) s p Lukáš Malý Poincaré inequalities that fail to be open-ended 3/17

Motivation Idea for the critical exponent Assume that µ is Ahlfors s-regular, s 1 (i.e., µ(b(x, r)) r s ). Then, TFAE: 1. X supports a Lorentz-type L s,1 -PI 2. Every u N 1 L s,1 (X) is continuous and 3. There is C 1 such that for all x, y X u(x) u(y) g u L s,1 (B(x,λd(x,y))) Mod L s,1({γ Γ x,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

Lorentz spaces L p,q Norm in the Lebesgue L p -spaces u L p (X) = ( X u(x) p dµ(x)) 1/p = ( k Z {x 2 k < u(x) 2 k+1 } u(x) p dµ(x)) 1/p = ( uχ {2 k < u 2 k+1 } p L p (X) )1/p = { uχ {2 k < u 2 k+1 } L p (X) } k= l p (Z) k Z Definition (functional comparable with the Lorentz norm) Let p [1, ) and q [1, ]. Then, we define u L p,q (X) = { uχ {2 k < u 2 k+1 } L p (X) } k= l q (Z) Remark: L p,1 (X) L p (X) = L p,p (X) L p, (X) for 1 p <. Remark: L p log α L(X) L p,q (X), whenever α > (p/q) 1 0. Lukáš Malý Poincaré inequalities that fail to be open-ended 5/17

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 u B dµ c PI diam(b)ψ 1 ( λb Ψ(g) dµ) for every u L 1 loc(x), its upper gradient g, and every ball B X, where c PI, λ 1. Remark: If Ψ(t) = t p, then Ψ-PI is just a p-pi. Some self-improvement of Ψ-PI to (p ε)-pi can be expected if Ψ(t) t p η(t), where η(t) grows (or decays) slower than any power t δ (or t δ ) Lukáš Malý Poincaré inequalities that fail to be open-ended 6/17

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 operator M L Ψ (R, µ) L Ψ (R, µ) is bounded. Theorem (Bloom Kerman, 1994) Let Ψ(t) = t p log α (e + t) with p > 1 and α R. If M L Ψ (R, µ) L Ψ (R, µ) is bounded, then M L p ε (R, µ) L p ε (R, µ) is bounded for some ε > 0. Corollary Suppose that (R n, µ), where µ = µ 1 µ 2 µ n, admits a Ψ-PI, where Ψ(t) = t p log α (e + t) for some p > 1 and α R. Then, (R n, µ) admits a (p ε)-pi for some ε > 0. Lukáš Malý Poincaré inequalities that fail to be open-ended 7/17

Lorentz-type Poincaré inequalities Definition (Lorentz-type Poincaré inequality) Let p [1, ) and q [1, ]. The space X admits an L p,q -Poincaré inequality if B u u B dµ c PI diam(b) gχ λb L p,q (X) χ λb L p,q (X) = c PI diam(b) gχ λb L p,q (X) µ(λb) 1/p for every u L 1 loc(x), its upper gradient g, and every ball B X, where c PI, λ 1. Remark: If p = q, then L p,q -PI is just a p-pi. Observation Due to the embedding between Lorentz spaces, we have L p,q -PI p-pi L p,q -PI whenever 1 q < p < Q. Lukáš Malý Poincaré inequalities that fail to be open-ended 8/17

Lorentz-type Poincaré inequalities Comparison to Muckenhoupt weights on R (R, µ) admits an L p,q -PI, p (1, ), q [1, ], if and only if the HL maximal operator M L p,q (R, µ) L p,q (R, µ) is bounded. M L p,q (R, µ) L p,q (R, µ) is bounded if and only if M L p (R, µ) L p (R, µ) is bounded, p (1, ), q (1, ]. (Chung Hunt Kurtz, 1982) If (R n, µ), where µ = µ 1 µ n, admits an L p,q -PI with p (1, ) and q (1, ], then it admits a (p ε)-pi for some ε > 0. Question Does an L p,1 -PI undergo self-improvement? Lukáš Malý Poincaré inequalities that fail to be open-ended 9/17

General Poincaré inequalities Let A L 0 +(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 g 1 g 2 a.e. in B, then A(g 1, B) A(g 2, B). Let M A be the associated (non-centered) maximal operator M A f(x) = sup A( f, B), x X. B x Definition (Generalized Poincaré inequality) The space X admits an A-Poincaré inequality if B u u B dµ c PI diam(b) A(g, λb) for every u L 1 loc(x), its upper gradient g, and every ball B X, where c PI, λ 1. Lukáš Malý Poincaré inequalities that fail to be open-ended 10/17

General Poincaré inequalities Lemma (M.) Assume that X supports an A-PI. Suppose that A(g, B) ( B g p A dµ) 1/p A for all g, B. Then, X admits a (p A ε)-pi for some ε > 0. Corollary Let p > 1. Suppose that X supports an L p,q -PI with q [p, ], OR a Ψ-PI with Ψ(t) t p log α (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

General Poincaré inequalities Theorem (M.) Assume that X supports an A-PI and that M A RI 1 (X) w-ri 2 (X) is bounded, where RI 1 and RI 2 are rearrangement invariant spaces, such that: RI 1 and RI 2 are close to each other, viz., their fundamental functions satisfy t φ 1 (φ 1 2 (t)) t (1 + log t) β, β 0; RI 1 satisfies an upper q-estimate, i.e., f χ k E k q RI 1 C f χ Ek q RI 1, q 1; RI 1 lies close to L p A, specifically, φ 1 (t p A ) t(log t) 1 (β+1/q) 0 as t 0. k Then, X admits a (p A ε)-pi for some ε > 0. Lukáš Malý Poincaré inequalities that fail to be open-ended 12/17

General Poincaré inequalities Corollary (M.) Assume that X supports an L p,q -PI with p (1, ) and q (1, ]. Then, X admits a (p ε)-pi. Proof: Choose A(g, B) = gχ B L p,q µ(b) 1/p. (a) q p. Then, M A L p,q (X) L p, (X) is bounded [Chung Hunt Kurtz, 1982]. Hence, φ 1 (t) = φ 2 (t) = t 1/p. L p,q satisfies an upper q-estimate [CHK 82]. (b) q > p. Then, M A L p (X) L p, (X) is bounded. Hence, φ 1 (t) = φ 2 (t) = t 1/p. L p,q satisfies an upper p-estimate [Maz ya]. Beware L p,1 -PI is NOT included in the self-improvement result. Lukáš Malý Poincaré inequalities that fail to be open-ended 13/17

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 L p,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) = g χ B RI(X), where RI(X) is a rearrangement-invariant Banach function χ B RI(X) space s.t. RI loc (X) L p,1 loc (X); OR A(g, B) = g RI(B, µb ), where µ B (E) = µ(e B)/µ(B) and RI loc (X) L p,1 loc (X); OR A(g, B) = Φ 1 ( B Ψ(g) dµ) provided that inf κ>0 κa( g κ, B) g RI(B, µ B ). Lukáš Malý Poincaré inequalities that fail to be open-ended 14/17

Lack of self-improvement Let X = X + X R n be endowed with the Euclidean distance and the n-dimensional Lebesgue measure, where X + = {(x 1,..., x n ) R n x j 0 for all j} X = {(x 1,..., x n ) R n x j 0 for all j}. and Both X + and X admit a 1-PI. If u N 1 L n,1 (X), then u X + N 1 L n,1 (X + ) and u X N 1 L n,1 (X ) Then, u(x) u(y) g u L n,1 (B x y X + ) whenever x, y X + ; analogously in X. Thus, u(x) u(y) g u L n,1 ( B x y ) whenever x, y X. Direct calculation yields for a ball B that B u(x) u B dx B B u(x) u(y) dx dy g u L n,1 (λb) diam(b) µ(b) 1/n g u L n,1 (λb). Lukáš Malý Poincaré inequalities that fail to be open-ended 15/17

Lack of self-improvement Let h RI \ L n,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, g k g/k, k > 0, is an upper gradient of u χ X. For the ball B = B(0, 1), we have 1 2 = u(x) u B dx? c PI A(g k, λb) = c PI A( g, λb) 0 as k. B k Lukáš Malý Poincaré inequalities that fail to be open-ended 16/17

Final remarks The motivating problem (characterization of L s,1 -PI by the estimates for modulus of continuity of N 1 L s,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 (R n, µ) with product measure supports a Ψ-PI (where the conjugate of Ψ is doubling), then there is p < such that (R n, µ) admits a p-pi. This applies in particular to Ψ(t) = exp(t α ), 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 L 1 µ (Durand-Cartagena Jaramillo Shanmugalingam, 2013) Ψ-PI for (R, µ) requires µ L 1 (J. Björn, 2010) Lukáš Malý Poincaré inequalities that fail to be open-ended 17/17

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