Regularizations of Singular Integral Operators (joint work with C. Liaw)

Transcription

1 1 Outline Regularizations of Singular Integral Operators (joint work with C. Liaw) Sergei Treil Department of Mathematics Brown University April 4, 2014

2 2 Outline 1 Examples of Calderón Zygmund operators (CZO) Definition of CZO Regularization methods of CZO 2 Setup Regularizations Two weight Muckenhoupt condition 3 Schur multipliers Splitting the supports Final remarks

3 3 Hilbert and Cauchy Transforms: Hilbert transform: T f(x) = (π) 1 (integral is defined as principal value) Examples of Calderón Zygmund operators (CZO) Definition of CZO Regularization methods of CZO R f(y) x y dy

4 Hilbert and Cauchy Transforms: Examples of Calderón Zygmund operators (CZO) Definition of CZO Regularization methods of CZO Hilbert transform: T f(x) = (π) 1 f(y) x y dµ(y) (integral is defined as principal value) Sometimes the integration is with respect to an arbitrary measure µ on R or on C. 3

5 Hilbert and Cauchy Transforms: Examples of Calderón Zygmund operators (CZO) Definition of CZO Regularization methods of CZO Hilbert transform: T f(x) = (π) 1 f(y) x y dµ(y) (integral is defined as principal value) Sometimes the integration is with respect to an arbitrary measure µ on R or on C. The case with µ on C (Cauchy Transform) appears in problems related to analytic capacity. The measure µ is one-dimensional µ(d x,r ) Cr and one is interested in boundedness of such operators in L 2 (µ). 3

6 Hilbert and Cauchy Transforms: Examples of Calderón Zygmund operators (CZO) Definition of CZO Regularization methods of CZO Hilbert transform: T f(x) = (π) 1 f(y) x y dµ(y) (integral is defined as principal value) Sometimes the integration is with respect to an arbitrary measure µ on R or on C. The case with µ on C (Cauchy Transform) appears in problems related to analytic capacity. The measure µ is one-dimensional µ(d x,r ) Cr and one is interested in boundedness of such operators in L 2 (µ). But sometimes we are interested in boundedness L 2 (µ) L 2 (ν), and there is no apriori restrictions on µ and ν. 3

7 4 An example: Examples of Calderón Zygmund operators (CZO) Definition of CZO Regularization methods of CZO Let A := M t in L 2 (µ), and let A α = A + α(, b)b, where b is cyclic for A. WLOG can assume that b 1. By the spectral theorem A α unitarily equivalent to M s in L 2 (µ α ). M s U α = U α A α, U α : L 2 (µ) L 2 (µ α ) unitary. U α b is cyclic, so WLOG can assume U α b = U α 1 = 1. Informal reasoning gives us: if U is an integral operator with Kernel K(s, t), then K(s, t) = α/(s t).

8 4 An example: Examples of Calderón Zygmund operators (CZO) Definition of CZO Regularization methods of CZO Let A := M t in L 2 (µ), and let A α = A + α(, b)b, where b is cyclic for A. WLOG can assume that b 1. By the spectral theorem A α unitarily equivalent to M s in L 2 (µ α ). M s U α = U α A α, U α : L 2 (µ) L 2 (µ α ) unitary. U α b is cyclic, so WLOG can assume U α b = U α 1 = 1. Informal reasoning gives us: if U is an integral operator with Kernel K(s, t), then K(s, t) = α/(s t). Indeed, so K(s, t) = α/(s t). sk(s, t) = U α M t + α(, b)u α b = K(s, t)t + α1

9 4 An example: Examples of Calderón Zygmund operators (CZO) Definition of CZO Regularization methods of CZO Let A := M t in L 2 (µ), and let A α = A + α(, b)b, where b is cyclic for A. WLOG can assume that b 1. By the spectral theorem A α unitarily equivalent to M s in L 2 (µ α ). M s U α = U α A α, U α : L 2 (µ) L 2 (µ α ) unitary. U α b is cyclic, so WLOG can assume U α b = U α 1 = 1. Informal reasoning gives us: if U is an integral operator with Kernel K(s, t), then K(s, t) = α/(s t). Theorem (C. Liaw, S.Treil, 2009) U α f(s) = f(s) α R f(t) f(s) dµ(t), f Cc 1. t s

10 Rigidity theorem Examples of Calderón Zygmund operators (CZO) Definition of CZO Regularization methods of CZO Theorem (C. Liaw S. Treil, 2009) Let an operator V defined on C 1 c by V f(s) = f(s) α R f(t) f(s) dµ(s) t s extends to a bounded operator L 2 (µ) L 2 (ν), and let ker V = {0}. Then there exists h 0 such that 1/h L (ν), and M h V is a unitary operator from L 2 (µ) L 2 (ν). Moreover, the unitary operator U := M h V gives the spectral representation of the operator A α := M t + α(, b)b, b 1, in L 2 (µ), namely UA α = M s U, where M s is the multiplication by the independent variable s in L 2 (ν). In other words, all operators of such type appear from rank one perturbations. 5

11 6 Calderón Zygmund operators Examples of Calderón Zygmund operators (CZO) Definition of CZO Regularization methods of CZO A Calderón Zygmund operator of dimension d in R N, d N, is an integral operator, bounded in L 2 and with kernel K satisfying the following growth and smoothness conditions 1 K(x, y) C cz x y d for all x, y RN, x y. 2 There exists α > 0 such that K(x, y) K(x, y) + K(y, x) K(y, x ) C cz x x α for all x, x, y R N such that x x < x y /2. Has singularity at the diagonal x = y. The measure is at best d-dimensional, µ(b x,r ) Cr d, so integral is not well defined. How to interpret? x y d+α

12 7 Interpreting CZO: regularization Examples of Calderón Zygmund operators (CZO) Definition of CZO Regularization methods of CZO Let m be a function in R N, m 0 in a neighborhood of 0, and m(x) 1 in a neighborhood of. Kernels K ε (x, y) = K(x y)m((x y)/ε) do not have singularity near the diagonal x = y, so the integral T ε f(x) = K ε (x, y)f(y)dµ(y) is well defined for f L c. We say that T is bounded if T ε are uniformly (in ε bounded). If m(x) = 1 [1, ) ( x ) we get the classical truncation. Smooth truncations are also used (and are more natural from my point of view). Does boundedness depend on regularization?

13 8 Interpreting CZO: axiomatic approach Examples of Calderón Zygmund operators (CZO) Definition of CZO Regularization methods of CZO We assume that bilinear form T f, g is defined on a dense set; in classical situations when one works with Lebesgue measure in R N or on a smooth manifold it is often Cc Cc. The fact that T is an integral operators means simply that T f, g = K(x, y)f(y)g(x)dµ(y)dν(x) for compactly supported f and g with separated supports. A multiplication operator M ϕ, M ϕ f = ϕf is an operator with kernel K(x, y) 0. If µ = ν, µ(b x,r ) Cr d, the generalization of Cotlar inequality implies that the truncations T ε are uniformly bounded for all reasonable truncating functions m. Nothing was known in the two-weight case!

14 Setup Setup Regularizations Two weight Muckenhoupt condition Radon measures µ and ν in R n are fixed. µ and ν do not have common atoms; no other restriction is assumed. A singular kernel K(x, y) is a function in L 2 loc (µ ν) off the diagonal x = y (can blow-up at the diagonal). Definition (Restricted boundedness) A kernel K is called L p restrictedly bounded if T f, g = K(x, y)f(y)g(x)dµ(y)dν(x) C f L p (µ) g L p (ν) for all f, g L c, supp f supp g =. Here 1/p + 1/p = 1. The best constant C is called the restricted norm. 9

15 10 Setup Regularizations Two weight Muckenhoupt condition Recall: m is a cut-off function, m 0 in a neighborhood of 0 and m 1 in a neighborhood of. T ε are truncated operators with kernels K ε, K ε (x, y) = K(x, y)m((x y)/ε); Measures µ and ν do not have common atoms.

16 10 Setup Regularizations Two weight Muckenhoupt condition Recall: m is a cut-off function, m 0 in a neighborhood of 0 and m 1 in a neighborhood of. T ε are truncated operators with kernels K ε, K ε (x, y) = K(x, y)m((x y)/ε); Measures µ and ν do not have common atoms. Life is not as simple as it seems.

17 10 Setup Regularizations Two weight Muckenhoupt condition Recall: m is a cut-off function, m 0 in a neighborhood of 0 and m 1 in a neighborhood of. T ε are truncated operators with kernels K ε, K ε (x, y) = K(x, y)m((x y)/ε); Measures µ and ν do not have common atoms. Life is not as simple as it seems. It is much simpler.

18 10 Setup Regularizations Two weight Muckenhoupt condition Recall: m is a cut-off function, m 0 in a neighborhood of 0 and m 1 in a neighborhood of. T ε are truncated operators with kernels K ε, K ε (x, y) = K(x, y)m((x y)/ε); Measures µ and ν do not have common atoms. Life is not as simple as it seems. It is much simpler. Theorem (C. Liaw, S. Treil) Let K be an L p restrictedly bounded singular kernel, and let m C. Then the resularized operators T ε are uniformly (in ε) bounded.

19 Setup Regularizations Two weight Muckenhoupt condition Recall: m is a cut-off function, m 0 in a neighborhood of 0 and m 1 in a neighborhood of. T ε are truncated operators with kernels K ε, K ε (x, y) = K(x, y)m((x y)/ε); Measures µ and ν do not have common atoms. Life is not as simple as it seems. It is much simpler. Theorem (C. Liaw, S. Treil) Let K be an L p restrictedly bounded singular kernel, and let m C. Then the resularized operators T ε are uniformly (in ε) bounded. In fact it is sufficient to assume that 1 m is in the Sobolev space H s, s > N/2. Recall: f H s R N (1 + x ) 2s f(x) 2 dx <. 10

20 11 Classical truncations Setup Regularizations Two weight Muckenhoupt condition For interesting operators the result holds with classical truncations (m(x) = 1 [1, ) ( x )) as well! Interesting operator: convolution kernel, K(x, y) = K(y x), K(x) = A( x )B(x/ x ), where A(r) 0 for all r > 0 and B is a function (with values in some R m ) in the Sobolev space H k, k > N/2 on the unit sphere S N 1 in R N. Examples K(z) = 1/z, z C Cauchy Transform. K(z) = 1/z 2, z C Beurling Ahlfors Transform; K(x) = x/ x d+1, x R N Riesz Transform of order d in R N.

21 Setup Regularizations Two weight Muckenhoupt condition 12 Recall: Riesz Transform of order d is convolution integral operator with kernel K(x, y) = (x y)/ x y d+1. Theorem (C. Liaw, S. Treil) For Riesz Transform of order d the restricted L p boundedness, 1 < p <, implies the following generalized two-weight Muckenhoupt A d p condition of order d; sup (diam B) d µ(b) 1/p ν(b) 1/p < ; B here the supremum is taken over all balls in R N. In particular, if µ = ν then µ is d-dimensional, µ(b x,r ) Cr d. It was known before, but the proofs were ugly. If d = N then µ s ν s.

22 A simple idea: Schur multipliers Schur multipliers Splitting the supports Final remarks Replacing K(x, y) by K(x, y)e iax e iay, a R does not change its restricted norm. So if ρ L 1 (R N ), then for the kernel K(x, y)e iax e iay ρ(a)da = K(x, y) ρ(x y) R its restricted norm increases at most ρ 1 times. Functions m in the Wiener Algebra W = F(L 1 ) + c are Schur multipliers (multipliers with respect to restricted norm). If m = ρ, ρ L 1, then m(s/ε) = F{x ε N ρ(εx)}(s), and ε N ρ(εx) dx = ρ(x) dx = ρ 1. R N R N so the Schur multipliers m((x y)/ε) are uniformly (in ε) bounded. 13

23 Schur multipliers, continued Schur multipliers Splitting the supports Final remarks If ϕ H k, k > N/2, then ϕ F(L 1 ) (easy exercise in Cauchy Schwartz). Note that the rescaling ϕ ϕ ε, m ε (x) = m(x/ε) does change Sobolev norm, but the norm in the Wiener Algebra W is preserved. So, if 1 m C c W k, k > N/2, then the kernels K ε, K ε (x, y)m((x y)/ε) are uniformly (in ε) restrictedly bounded. 14

24 Schur multipliers: an example Schur multipliers Splitting the supports Final remarks Let ρ(x) = e x 1 [0, ), and let m(s) := 1 ρ(s) = s s i. 15

25 Schur multipliers: an example Schur multipliers Splitting the supports Final remarks Let ρ(x) = e x 1 [0, ), and let Then m(s/ε) = s s iε. m(s) := 1 ρ(s) = s s i. 15

26 15 Schur multipliers: an example Schur multipliers Splitting the supports Final remarks Let ρ(x) = e x 1 [0, ), and let m(s) := 1 ρ(s) = Then m(s/ε) = s s iε. Then for K(x, y) = 1/(x y) we have s s i. K ε (x, y) = 1 x y x y x y iε = 1 x + iε y

27 15 Schur multipliers: an example Schur multipliers Splitting the supports Final remarks Let ρ(x) = e x 1 [0, ), and let m(s) := 1 ρ(s) = Then m(s/ε) = s s iε. Then for K(x, y) = 1/(x y) we have s s i. K ε (x, y) = 1 x y x y x y iε = 1 x + iε y We used Schur multipliers before in complex analysis, probably without noticing it!

28 Schur multipliers Splitting the supports Final remarks From restricted boundedness to boundedness Proposition Let K L 2 loc (µ ν) (everywhere, not just off the diagonal x = y), and let K be restrictedly bounded with restricted norm C. Then the integral operator T with kernel K is bounded, T 2C. Note that T is well defined on f L c. Idea of the proof: Restricting everything to a compact set assume WLOG that T is compact L 2 (µ) L 2 (ν). Take f and g. Considering f only on almost half of the support and g on almost the other half, we get the sequences f n, g n with separated supports such that f n 1 2 f, g n 1 2 g weakly. T f n, g n can be estimated because of restricted boundedness, and T f n, g n 1 4 T f, g 16

29 Schur multipliers Splitting the supports Final remarks 17 Lemma (One weight version) Let σ be a Radon measure without atoms in R N. There exist Borel sets E 1 n, E 2 n, n N such that dist(e 1 n, E 2 n) > 0 n. 1 The operators Pn k, Pn k f := 1 E k n f, k = 1, 2 converge to 1 2I in weak operator topology in L 2 (σ). 2 For any p [1, ) and for k = 1, 2 lim n 1 E k n f L p (σ) = 2 1/p f L p (σ), f L p (σ). Trivially to do for Lebesgue measure More work needed for general case. One can get a two weigh version from this Lemma

30 Schur multipliers Splitting the supports Final remarks Lemma (Two weight version) Let µ and ν be Radon measures in R N without common atoms. here exist Borel sets E 1 n, E 2 n, n N such that dist(e 1 n, E 2 n) > 0 n. 1 The operators Pn 1 and Pn, 2 given by Pnf 1 ( := 1 E 1 fµc n f µa), Png 2 ( := 1 E 2 gνc n g νa) converge to 1 2I in weak operator topology of L 2 (µ) and L 2 (ν) respectively. 2 For any p [1, ) and for any f L p (µ), g L p (ν) lim n 1 E 1 n f L p (µ) 2 1/p f L p (µ), lim n 1 E 2 n g L p (ν) 2 1/p g L p (ν). If µ and ν do not have atoms, apply one weight lemma to σ = µ + ν: the sets En 1,2 from this lemma would do a trick. Atoms: add one atom to each set at each step, and remove small neighborhoods of added atoms from the other set. 18

31 Proof of the one-weight lemma Schur multipliers Splitting the supports Final remarks Lemma (One weight version) Let σ be a Radon measure without atoms in R N. There exist Borel sets E 1 n, E 2 n, n N such that dist(e 1 n, E 2 n) > 0 n. 1 The operators Pn k, Pn k f := 1 E k n f, k = 1, 2 converge to 1 2I in weak operator topology in L 2 (σ). 2 For any p [1, ) and for k = 1, 2 lim n 1 E k n f L p (σ) = 2 1/p f L p (σ), f L p (σ). 19

32 19 Proof of the one-weight lemma Schur multipliers Splitting the supports Final remarks Since P n are always contractions, sufficiently to check everything on dense sets. Sufficient to check everything only on 1 Q for standard dyadic cubes in R N, i.e. on cubes of form Q = 2 k ([0, 1) N + j), k Z, j Z N. Sufficient to show that given ε > 0 and the size 2 k we can split all the cubes of this size almost in half with relative error ε: σ(q E1,2 ) 1 2 σ(q) εσ(q). Note that the estimate for size 2 k implies the same estimate for bigger dyadic cubes.

33 20 Schur multipliers Splitting the supports Final remarks Proof of the one-weight lemma: splitting a cube Want to split cubes with relative error ε, σ(q E1,2 ) 1 2 σ(q) εσ(q). Let us construct the disjoint sets E 1,2, we then shrink them to make separated. Pick a size δ = 2 m such that σ(r) 1 2εσ(Q) for every dyadic cube R Q of this size. Follows from continuity of the measure. Split cubes Q into dyadic cubes R of this size, and order these cubes.

34 20 Schur multipliers Splitting the supports Final remarks Proof of the one-weight lemma: splitting a cube Want to split cubes with relative error ε, σ(q E1,2 ) 1 2 σ(q) εσ(q). Let us construct the disjoint sets E 1,2, we then shrink them to make separated. Pick a size δ = 2 m such that σ(r) 1 2εσ(Q) for every dyadic cube R Q of this size. Follows from continuity of the measure. Split cubes Q into dyadic cubes R of this size, and order these cubes. Put R 1 to E 1 and R 2 to E 2.

35 20 Schur multipliers Splitting the supports Final remarks Proof of the one-weight lemma: splitting a cube Want to split cubes with relative error ε, σ(q E1,2 ) 1 2 σ(q) εσ(q). Let us construct the disjoint sets E 1,2, we then shrink them to make separated. Pick a size δ = 2 m such that σ(r) 1 2εσ(Q) for every dyadic cube R Q of this size. Follows from continuity of the measure. Split cubes Q into dyadic cubes R of this size, and order these cubes. Put R 1 to E 1 and R 2 to E 2. Then always put the next cube to the set of smaller measure σ.

36 20 Schur multipliers Splitting the supports Final remarks Proof of the one-weight lemma: splitting a cube Want to split cubes with relative error ε, σ(q E1,2 ) 1 2 σ(q) εσ(q). Let us construct the disjoint sets E 1,2, we then shrink them to make separated. Pick a size δ = 2 m such that σ(r) 1 2εσ(Q) for every dyadic cube R Q of this size. Follows from continuity of the measure. Split cubes Q into dyadic cubes R of this size, and order these cubes. Put R 1 to E 1 and R 2 to E 2. Then always put the next cube to the set of smaller measure σ. We end up with error 1 2 εσ(q).

37 21 Schur multipliers Splitting the supports Final remarks Proof of the one-weight lemma: shrinking the cubes We constructed disjoint sets E 1,2, now we want to make them separated. Replace each small cube R by rr where r < 1 and close to 1; rr means dilation with respect to the the corner, not the center: r[0, 1) N = [0, r) N, i.e. we dilate this cube with respect to the origin. With such dilation, R = r (0,1) rr, and thus it follows from countable additivity that σ(r) = lim r 1 σ(rr).

38 22 Remarks Schur multipliers Splitting the supports Final remarks Assumption about no common atoms is not really restriction: Interaction between µ c and ν and the interaction between µ and ν c is given by the main result. Interaction between µ a and ν a is described by a matrix. The result about two weight Muckenhoupt condition is obtained by multiplying K by an appropriate Schur multiplier to get he kernel satisfying K(x, y) r d for x y r. The modified kernel is positive in some direction. The result about classical truncations is obtained similarly: we use an appropriate Schur multiplier to get the positive kernel dominating the difference between the classical and smooth truncation.

39 23 Bibliography Schur multipliers Splitting the supports Final remarks C. Liaw, S. Treil, Rank one perturbations and singular integral operators, arxiv: v1 [math.fa], J. Funct. Anal. 257(2009), no. 6, C. Liaw, S. Treil, Regularizations of general singular integral operators, arxiv: v2 [math.ca], 2010, 23 pp.