Deparmens of Mahemaics Monana Sae Universiy Fall 215 Prof. Kevin Wildrick n inroducion o non-smooh analysis and geomery Lecure 1: The Poincaré Inequaliy in Euclidean space 1. Wha is he Poincaré inequaliy? In he las secion, we gave a reasonable generalizaion of he norm of he gradien of a smooh funcion on Euclidean space. This was moivaed by he fac ha our generalizaion was now a purely meric concep, and didn rely on he linear srucure of he domain. This is a prey weak effor: wha good does i do us? In order o undersand he usefulness of his generalizaion, we need o know somehing abou how he norm of he gradien of a smooh funcion on Euclidean space behaves. We ve seen ha he norm of he gradien conrols he behavior of a funcion on all recifiable curves, bu wha can we say abou is behavior on larger ses? Theorem 1.1 (The Poincaré inequaliy in R n.. Le f : R n R be a C 1 -smooh mapping, and le p 1. Then here is a consan C(n, p depending only on n and p such ha for any ball R n, ( f f dl n C(n, p diam( f p dl n 1 p. Le s examine wha his inequaliy says. a poin x, he quaniy f(x f measures he deviaion of f from is average on. We hen average his over o ge he lef hand side - so he lef hand side of he inequaliy is a measure of how much, on average, he funcion deviaes from is average value. One can hink of his as a measure of he average variaion of f on. On he righ hand side, we have a o-average of he value of he norm of he gradien: ( f p dl n 1 p. If p = 1, his is ruly jus he usual average, bu as p ges larger, he average become less spread-ou and more biased o larger values of f. In paricular, as p, he p-average ends o he essenial supremum of f divided by he measure of he ball. We hen muliply his p-average by he diameer of he ball - his is jus he good old oal change is rae imes disance. For example, consider f : R R defined by f(x = x. Then f f (,r dl 1 = r (,r 2, while ( f p dl 1 1 p = 1. So, in oal, he Poincaré inequaliy allows us o conrol he average variaion of a funcion by he average size of is gradien. The Poincaré inequaliy, and varians hereof, are very useful in regulariy heory for PDE s - if he size of he (weak gradien can be esimaed, hen bounds on he variaion of he funcion can be acheived. 1
The Poincaré inequaliy is closely relaed o geomery as well. Le us suppose, for he momen, ha f is a C 1 -smooh funcion defined no on all of R 2, bu insead only on he sli disk D = (, 1\([, 1 {}. I s easy o consruc such an f ha is idenically on he se [1/2, 3/4] (, 1/8 and idenically 1 on he se [1/2, 3/4] ( 1/8,. Then f is idenically on ((5/8, ɛ, 1/8, bu he average variaion of f on his ball is cerainly no. This is because he curves joining he boom of he sli o he op of he sli are oo long compared o he disance beween he poins hey join. We will see ha in order for a Poincaré inequaliy o be valid on some domain, he domain mus have a sufficienly large collecion of shor curves joining each poin. Le us noe ha he Poincaré inequaliy as described above makes sense in a general meric measure space if we replace f wih any upper gradien of f. 2. firs proof of he Poincaré inequaliy in Euclidean space. The Poincaré inequaliy is easy o prove on he real line. Fix a ball R, and le x and y be poins of. y he fundamenal heorem of calculus, f(x f(y f ( d = diam( f ( d. Now, noe ha since he righ hand side above is independen of x and y, f ( x f dx f(x f(y dxdy diam( f ( d, showing he 1-Poincaré inequaliy. The p-poincaré inequaliy now follows by Hölder s inequaliy. We will now give he classical proof of he Poincaré inequaliy in he case ha n 2, and along he way we will also prove some oher imporan inequaliies from classical funcion heory. The approach we will ake is based on harmonic analysis. There are wo imporan pars o his approach - he boundedness of cerain operaors on L p (which is very general, and he foliaed srucure of Euclidean space (which is no very general. 2.1. Maximal funcions. We have already me, in a way, one of he main ools from harmonic analysis ha will play a cenral role in he proof of he Poincaré inequaliy: he maximal funcion. Definiion 2.1. Le (, d, µ be a meric measure space, and le f be an inegrable funcion on. The maximal funcion Mf : R is defined by Mf(x = sup f dµ. r> (x,r Many varians of he maximal funcion are occasionally useful (for example, he resriced maximal funcion, where he supremum is only aken over radii less han some fixed R >. Noice ha by he Lebesgue differeniaion heorem, f < Mf a µ-almos every poin. Noe ha M is no a linear operaor, bu raher only sublinear: M(f + g Mf + Mg. The key propery of he maximal funcion is is boundedness on cerain funcion spaces.
Theorem 2.2. Le (, d, µ be a doubling meric measure space. Then here is a consan C 1, depending only on he doubling consan of µ, such ha if f L 1 (, µ, hen for all >, µ({x : Mf(x > } < C 1 f dµ. Moreover, for each p > 1, hen here is a consan C p, depending only on he doubling consan of µ, such ha if f L p (, µ, hen Mf p C p f p. In oherwords, M is a bounded sublinear operaor L 1 (, µ weak L 1 (, µ and L p (, µ L p (, µ. Lemma 2.3. Le f be a measurable funcion on (, µ. Then for p 1, f p dµ = p p 1 µ({x : f(x > } d. Proof. Use he fac ha and Fubini s Theorem. f(x p = p f(x Proof. For R >, le us denoe M R f = sup <r<r (x,r p 1 d f dµ. We will prove he resuls for M R f, and hen le R. We sar wih he weak-ype esimae. For each x such ha M R f(x >, choose a ball (x, r wih r < R such ha f dµ >. (x,r Using he 5 covering lemma, exrac a counable collecion G of such balls so ha 5-imes dialed balls cover. Then µ({x : M R f(x > } µ(5 C µ( G G C f dµ C f dµ. Leing R now shows he firs saemen. The second saemen will follow only from he firs inequaliy, he sublineariy of M, and he fac ha Mf(x f. This is an example of an inerpolaion argumen; we have shown ha M is bounded from L 1 o weak-l 1, and i is clearly bounded from L o L, and now we everyhing in beween for free. To his end, le f L p (, µ, and fix >. Le s wrie f as he sum of a good par g and a bad par b: f = fχ { f /2} + fχ { f >/2} = g + b. Since Mg g, he sublineariy of M yields G Mf 2 + Mb, implying ha {x : Mf(x > } {x : Mb(x > /2}.
Moreover, if we apply Lemma 2.3 o b, we see ha b dµ = µ({x : fχ f >/2 (x > s} ds = 2 µ({x : f > /2} + µ({x : f(x > s} ds /2 Now, by Lemma 2.3, he weak ype esimae proved above applied o b, he above esimae, and Fubini s heorem Mf p p = p p 2Cp 2Cp p 1 µ({x : Mf(x > } d p 1 µ({x : Mb(x > /2} d p 2 b dµ d ( p 1 µ({x : f > /2} + p 2 2 2 p C f p p + 2Cp 2 p C f p p + 2p C p 1 f p p. This complees he proof. We now move owards he Poincaré inequaliy. /2 µ({x : f(x > s} ds p 2 µ({x : f(x > s}χ {<2s} d ds Theorem 2.4 (The Riesz Poenial Esimae. For each n N, here is a consan C = C(n such ha if f : R n R is a C 1 -smooh funcion, is a ball in R n, and x, hen f(z f(x f C z x n 1 dln (z. Proof. We prove he heorem only when n 2; he case ha n = 1 is mosly he same, bu easier. Think of fixing x and leing y vary, and consider he parameerized line l : R R n beween hem, defined by l y ( = x + (y x. Then, by he fundamenal heorem of calculus and he chain rule f(x f(y = 1 (f l y ( d = 1 f(l y ( (y x d Now, fix a ball in R n of radius R, and le x and y be poins of. Inegraing wih respec o y and applying Fubini s heorem yields L n ( f(x f f(l y ( (y x dl 1 ( dl n (y. [,1] [,1] f(l y ( (y x dl n (y dl 1 (. We now condsider he change of variables ψ : ψ ( defined by ψ (y = l y (. d
Noe ha Thus, L n ( f(x f de ψ = n ψ ( (x, 2R y x = ψ (y x. [,1] 2R [,1] (x,2r (x,2r f(z z x n dl n (z dl 1 ( f(z n dl n (z dl 1 ( Noe ha if [, 1] and z saisfies z x < 2R, hen saisfies z x 2R < < 1. Hence, Fubini s heorem and an inegraion imply 1 L n ( f(x f 2R f(z n dl n (z dl 1 ( z x (2R 1 2Rn f(z n 1 z x n 1 dln (z. Given his resul, we make he following definiion. Definiion 2.5 (The Riesz Transform. Le g L 1 loc (Rn and le be a ball. The Riesz ransform of g on is defined by g(z I 1 (g(x = z x n 1 dln (z. Hence, we have shown ha for a C 1 -smooh funcion f, a ball R n, and x, f(x f C(nI 1 ( f(x. Theorem 2.6 (The boundedness of he Riesz ransform. Le be a ball in R n, n 2. Then here are consans C 1 and C p, 1 < p < n, such ha if g L 1 (, hen ( n L n g 1 n 1 ({I 1 (g > } C 1, and if g L p (, 1 < p < n, hen I 1 (g np (n p C p g p. We will no prove his heorem, bu menion ha i is a corollary of he Hardy- Lilewood maximal funcion heorem. s a relaively quick corollary, we achieve Corollary 2.7 (The Sobolev-Poincaré Inequaliy. Le 1 p < n, and le f : R n R be a C 1 -smooh funcion. Then for any ball, f f np/(n p C(n, p f p.
This is an exraordinary fac - i says ha he p-norm of he gradien of f conrols is he np/(n p-norm of is variaion. This fac has profound impacs in parial differenial equaions. Tracing hrough he argumen, we see ha he following is rue: le f : R n R be any funcion ha has an upper gradien in L p (R n, for 1 < p < n. Then f f is in L np/(n p for any ball. The Sobolev-Poincaré inequaliy, afer wo applicaions of Hölder s inequaliy, leads o a proof of he Poincaré inequaliy. However, we can proceed direcly from he Riesz poenial esimae o give a quick proof. Proof of he Poincaré inequaliy. y he Riesz poenial esimae and Fubini s inequaliy f(x f dl n f(z (x C z x n 1 dln (z dl n (x 1 C f(z z x n 1 dln (x dl n (z. Polar coordinaes shows ha z x 1 n dl n (x (z,diam z x 1 n dl n (x C(n diam. This shows he 1-Poincaré inequaliy. The general case now follows from Hölder s inequaliy. If f : R n R is a C 1 -smooh funcion, is a ball in R n, and x and y are in, hen he riangle inequaliy and he Riesz poenial esimae proven before imply he symmeric esimae ( f(x f(y C f(z z x n 1 dln (z + f(z z y n 1 dln (z. This inequaliy can also be used o prove he Poincaré inequaliy, since f(x f dl n (x = f(x f(y dl n (y dln (x f(x f(y dl n (y dl n (x. Plugging in he symmeric esimae and arguing as before now yields he desired resul. 3. Pencils of Curves We now give a geomeric way of verifying he Poincaré inequaliy - one ha does no rely so heavily on he srucure of R n. Le f : R n R be a C 1 -funcion, and le x and y be poins of a ball in R n. Suppose ha here is a family of recifiable curves and a probabiliy measure P on. Then we may form anoher measure µ on by µ( := ds dp (γ = lengh(γ dp (γ. γ If one can show ha for each orel se, ( f(z µ( C z x n 1 dln (z + hen a similar resul follows for orel funcions: ( ρ(z ρ ds dp (γ C z x n 1 dln (z + γ γ f(z z y n 1 dln (z, ρ(z z y n 1 dln (z.
Thus inegraing he upper gradien inequaliy yields f(x f(y = f(x f(y dp f ds dp γ ( C f(z z x n 1 dln (z + f(z z y n 1 dln (z, and he Poincaré inequaliy follows. Thus we have reduced he Poincaré inequaliy o a quesion abou finding sufficienly large pah families.