Lemma 1. Suppose K S is a compact subset and I α is a covering of K. There is a finite subcollection {I j } such that

Transcription

1 2 Singuar Integras We start with a very usefu covering emma. Lemma. Suppose K S is a compact subset and I α is a covering of K. There is a finite subcoection {I j } such that. {I j } are disjoint. 2. The intervas {3I j } that have the same midpoints as {I j } but three times the enghth cover K. Proof. We first choose a finite subcover. From the finite subcover we pick the argest interva. In case of a tie pick any of the competing ones. Then, at any stage, of the remaining intervas from our finite subcoection we pick the argest one that is disjoint from the ones aready picked. We stop when we cannot pick any more. The coection that we end up with is ceary disjoint and finite. Let x K. This is covered by one of the intervas I from our finite subcoection covering K. IfI was picked there is nothing to prove. If I is not picked it must intersect some I j aready picked. Let us ook at the first such interva and ca it I j. I is disjoint from a the previousy picked ones and I was passed over when we picked I j. Therefore inaddition to intersecting I j, I is not arger than I j. Therefore 3I j I x. This emma is used in proving maxima inequaities. For instance, for the Hardy-Littewood maxima function we have Theorem. Let f L (S). Define M f (x) = sup 0<r< π 2 2r y x <r f(y) dy () µ[x : M f (x) >] 3 f(y) dy Proof. Let us denote by E the set (2) E = {x : M f (x) >} 6

2 and et K E be an arbitrary compact set. For each x K there is an interva I x such that I x f(y) dy µ(i x ) Ceary {I x } is a covering of K and by emma we get a finite disjoint sub coection {I j } such that {3I j } covers K. Adding them up f(y) dy µ(i j ) µ(3i j ) µ(k) 3 j j Sine K E is arbitrary we are done. There is no probem in repacing {x : M f (x) >} by {x : M f (x) }. Repace by ɛ and et ɛ 0. This theorem can be used to prove the Labesgue diffrentiabiity theorem. Theorem 2. For any f L (S), im f(y) f(x) dy =0 a.e. x (3) h 0 2h x y h Proof. It is sufficient to prove that for any δ>0 µ[x : im sup f(y) f(x) dy δ] =0 h 0 2h x y h Given ɛ>0wecanwritef = f + g with f continuous and g ɛ and µ[x : im sup f(y) f(x) dy δ] h 0 2h x y h = µ[x : im sup g(y) g(x) dy δ] h 0 2h x y h µ[x :sup g(y) g(x) dy δ] h>0 2h 3 h δ Since ɛ>0 is arbitrary we are done. 3ɛ δ 7 x y h

3 In other words the maxima inequaity is usefu to prove amost sure convergence. Typicay amost sure convergence wi be obvious for a dense set and the maxima inequaity wi be used to interchange imits in the approximation. Another summabiity method, ike the Fejer sum that is often considred is the Poisson sum S(ρ, x) = n a n ρ n e inx and the kerne corresponding to it is the Poisson kerne p(ρ, z) = 2π n ρ n e inz = ρ 2 2π ( 2ρ cos z + ρ 2 ) (4) so that P (ρ, x) = f(y)p(ρ, x y)dy It is eft as an exercise to prove that for for p<, everyf L p P (ρ, ) f( ) inl p as ρ. We wi prove a maxima inequaity for the Poisson sum, so that as a consequence we wi get the amost sure convergence of P (ρ, x) tof for every f in L. Theorem 3. For every f in L µ[x : sup P (ρ, x) ] C f 0 ρ< (5) Proof. The proof consists of estimating the Poisson maxima function interms of the Hardy-Littewood maxima function M f (x). We begin with some simpe estimates for the Poisson kerne p(ρ, z). p(ρ, z) = 2π ρ 2 ( ρ) 2 +2ρ( cos z) 2π = +ρ 2π ρ π ρ ρ 2 ( ρ) 2 The probem therefore is ony as ρ. Lets us assume that ρ 2. 8

4 For any symmetric function φ(z) the intgra π f(z)φ(z)dz = = π 0 π 0 π 0 π For the Poisson kerne 0 2M f (0) [f(z)+f( z)]φ(z)dz φ(z)[ d dz z z z f(y)dy]dz π φ (z)[ f(y)dy]dz + φ(π) f(z)dz z π 2 zφ (z) [ f(y) λ z (dy)]dz + φ(π) f(z) dz π 0 zφ (z) dz + φ(π) M f (0) z d ρ 2 p(ρ, z) = 2ρ z sin z dz 2π ( 2ρ cos z + ρ 2 ) 2 ( ρ)z 2 π ( ρ) 4 +( cos z) 2 ( ρ)z2 C ( ρ) 4 + z 4 and uniformy in ρ. π z d dz p(ρ, z) dz C π = π ρ π ρ ( ρ)z 2 ( ρ) 4 + z 4 dz z 2 +z 4 dz z 2 +z 4 dz C Interpoation theorems pay a very important roe in Harmonic Anaysis. An exampe is the foowing 9

5 Theorem 4 (Marcinkiewicz). Let T be a subinear map defiened on L p L q that satisfies weak type inequities µ[ x : (Tf)(x) ] C i f p i p i p i (6) for i =, 2 where p <p 2 <. Then for p <p<p 2, there are constants C p such that Tf p C p f p (7) Note that T need not be inear. It need ony satisfy T (f + g) (x) Tf (x)+ Tg (x) (8) Proof. Let p (p,p 2 ) be fixed. For any function f L p and for any positive number a we deine f a = fχ { f a} and f a = χ { f >a}. Ceary f a L p2 and f a L p µ[x : Tf(x) 2] µ[x : Tf a (x) ]+µ[x : Tf a (x) ] C 2 f(x) p 2 dµ + C f(x) p dµ p 2 p f(x) a f(x) >a Take a =, mutipy by p and integrate with respect to from 0 to. Use Fubini s theorem. We get p C 2 µ[x : Tf(x) 2]d [ p 2 p + C ] f(x) p dµ (9) p p 0 Since the eft hand side is Tf p p p we are done. There is a sight variation of the argument that aows p 2 to be infinite provided T is bounded on L. If we denote the norm by C 2 we use µ[x : Tf(x) (C +)] µ[x : Tf a (x) ] and proceed as before. A different interpoation theorem for inear maps T is the foowing 0

6 Theorem 5 (Riesz-Thorin). If a inear map T is bounded from L pi into L pi with a bound C i for i =, 2 then for p p p 2 it is bounded from L p into L p with a bound C p that can be taken to be where t is determined by C p = C t C t 2 (0) p = t p +( t) p 2 () Proof. The proof uses methods from the theory of functions of a compex variabe. The starting point is the maximum moduus principe. Let us assume that u(z) is anaytic in the open strip a<rez<band bounded and continuous in the cosed strip a Rez b. LetM(x) bethemaximum moduus of the function on the ine Rez = x. Then ogm(x) is a convex function of x.this is not hard to see. Ceary the maximum principe dictates that M(x) max[m(a),m(b)] If one is worried about the maximum being attained, one can aways mutipy by e ɛz2 and et ɛ go to 0. Repacing u(z) byu(z)e tz yieds the inequaity optimizing with respect to t we get, M(x) max[m(a)e t(a x),m(b)e t(b x) ] M(x) max[[m(a)] b x b a, [M(b)] x a b a ] which is the required convexity. We note that the maximum of any coection of convex functions is again convex. The proof is competed by representing og F (p), where F (p) is the norm of T from L p to L p, as the supremum of a bunch of functions that are

7 convex in x = p. T p,p = sup f p g q g(tf)dµ = sup f p,f 0, φ = (gψ)(t (fφ))dµ g q,g 0, ψ = = sup (g x ψ)(t (f x φ))dµ f,f>0, φ = g,g>0, ψ = = sup (g z ψ)(t (f z φ))dµ f,f>0, φ = g,g>0, ψ = Rez=x = sup sup u(f,g,φ,ψ,z) f,f>0, φ = Rez=x g,g>0, ψ = In particuar for the Hardy-Littewood or Poisson maxima function the L bound is trivia and we now have a bound for the L p norm of the maxima function in terms of the L p norm of the origina function provided p>. For a convoution operator of the form (Tf)(x) = π f(y)k(x y)dy (2) we saw that for it to be bounded as an operator from L into itsef we need k to be in L. However for <p< the operator can some times be bounded even if k is not in L. This is proved by estabishing a bound from L 2 to L 2 and a weak type inequaity in L. We can then use Marcinkiewicz interpoation, foowed by Riesz-Thorin interpoation. Theorem 6. If ˆk(n) = e inz k(z)dz is bounded in absoute vaue by C, then the convoution operator given by equation (2) is bounded by C as an operator from L 2 to L 2. 2

8 Proof. Use the the orthonorma basis e inx to diagonaize T Te inx = ˆk(n)e inx (3) We now proceed to estabish weak type (, ) estimate. We sha assune that we have a kerne k in L that satisfies. 2. sup n k(y)e iny dy = C < (4) sup k(x y) k(x) dx = C 2 < (5) y x: x y >2 y Athough we have assumed that k is in L we wi prove a weak type (, ) bound. Theorem 7. The operator of convoution by k (T k f)(x) = π satisfies the weak type inequaity (,) k(x y)f(y)dy (6) µ[ x : (Tf)(x) ] C f (7) with a constant C that depends ony on C and C 2. Proof. Proof invoves severa steps. First we observe that the Hardy-Littewood maxima function given by () satisfies equation 2). The set G =[x : M f (x) ] isanopen set in [, π] and has Lebsgue measure atmost 3 f. We assume that > 3 f so that B = G c is nonempty. We write the open set G as 2π a possibe countabe union of disjoint open intervas I j of ength r j centered at x j. Note that the end points x j ± r 2 j necessariy beong to B. The maxima inequaity assures us that r j 3 f j 3

9 Let us define the averages and write f in the form m j = r j I j f(y)dy f(x) =[f(x) B (x)+ j m j Ij (x)] + j [f(x) m j ] Ij (x) = g(x)+ j h j (x) We have the bounds m j r j f(y) dy f(y) dy I j r j Ĩ j 2 f(y) dy 2M f (x j ± r j ) 2 2r j Ĩ j Here Ĩj is the interva centered around x j ± r j 2 of ength 2r j. In particuar g 2. On the other hand h j = j j We therefore have I j f(y) m j dy 2 j I j f(y) dy 2 f g 3 f Let us write the corre- Note that the decomposition depends on. sponding sum u = T k f = T k g + j T k h j = v + j w j = v + w We estimate the L 2 norm of v and the L norm of w on arge enough set. Then use Tchebychev s inequaity. µ[x : v(x) 2 ] v 2 2 C g 2 2 2C g = 6C f

10 Let us denote by Îj the interva centered around x j of ength 3r j and by U = j Î j. We begin by estimating w. U c. w. U c k(x y)[f(y) m j ]dy dx U c j I j = [k(x y) k(x x j )][f(y) m j ]dy dx U c j I j k(x y) k(x x j ) f(y) m j dydx U c j I j = f(y) m j dy k(x y) k(x x j ) dx j I j U c f(y) m j dy k(x y) k(x x j ) dx j I j Îj c f(y) m j dy k(x y) k(x x j ) dx j I j x: x y 2 y x j C 2 f(y) m j dy I j j 2C 2 f We have used here two facts. f(y) m j has mean zero on I j.ify I j and x Ĩc j,then y x r j 2 y x j. On the other hand µ(u) µ(ĩj) 3 µ(i j )=3 j r j 9 f Finay we can put the pieces together. µ(x : u(x) 2) µ(x : v(x) )+µ(x : w(x) ) or 6C f + 9 f + 2C 2 f µ(x : u(x) ) (2C +8+4C 2 ) f = C f 5

11 There is one point that we shoud note. For the interva doubing construction on the circe we shoud be sure that we do not see for instance any interva of enghth arger than π in G. This can be ensured if we take > 6 f.the 2 π inequaity is however satisfied for a because C 2. We want to ook at the specia kerne k(y) = which is not in L y. We consider its truncation k δ (y) = y { y δ}(y) First we estimate the Fourier transform e iny π sin ny dy =2 dy y δ y δ y =2 nπ sin y dy 4 sup y nδ 0<a< a sin y 0 y dy C Next in order to verify the condition (5) we need to estimate the foowing quantity uniformy in y and δ. x: x y >2 y k δ (x y) k δ (x) dx There are three sets over which the integra does not vanish. F = {x : x y > 2 y, x y δ, x δ} F 2 = {x : x y > 2 y, x y δ, x δ} F 3 = {x : x y > 2 y, x y δ, x δ} We consider F x y x dx x: x y 2 y z 2 x y x dx z z dz = C 3 6

12 It is cear that F 2 [ 2δ, 2δ]. Therefore F 2 x dx 2 δ 2δ dx x = C 4 Finay F 3 [x : x y 2δ] and works simiary. With C 2 = C 3 +2C 4 we are done. We are now ready to prove Theorem 8. For any f L p the partia sums s N (f,x) converge to f in L p provided <p<. Proof. We need ony prove, for <p<, a bound from L p to L p, for the partia sum operators (T N f)(x) = f(x y)k N (y)dy with k N (z) = sin(n + )z 2 2π sin z 2 that is uniform in N. In terms of mutipiers we are ooking at a uniform L p bound for the operators defined by ˆk N (n) = { n N} (n) Let us define the operators M k as mutipication by e ikx which are isometries in every L p. P 0 is the operator of projection to constants, i.e. the operator with mutipier {0} (n) which is ceary bounded in every L p. Finay the Hibert transform S is the one with mutipier signum n. Itiseasytoverify that T N = M N 2 [(S + I)+P 0]M N M (N+) ] 2 [(S + I)+P 0]M (N+) ] This reduces the probem to proving that a singe operator S is bounded on L p. The kerne is cacuated to be s(z) = 2π cot z 2 This can be repaced by the modified kerne and we are done. k(z) = πz 7