Topics in Harmonic Analysis Lecture 6: Pseudodifferential calculus and almost orthogonality

Transcription

1 Topics in Harmonic Analysis Lecture 6: Pseudodifferential calculus and almost orthogonality Po-Lam Yung The Chinese University of Hong Kong

2 Introduction While multiplier operators are very useful in studying constant coefficient partial differential equations, one often encounters variable coefficient partial differential equations. Thus we consider a variable coefficient generalization of multiplier operators, namely pseudodifferential operators. We study compositions and mapping properties of pseudodifferential operators. These in turn allow one to construct paramatrices to variable coefficient elliptic PDEs. We close this lecture with a beautiful almost orthogonality principle, due to Cotlar and Stein, which will play a crucial role in the proof of the T (1) theorem in the next lecture.

3 Outline Symbols of pseudodifferential operators Kernel representations Mapping properties on L 2 Compound symbols Closure under adjoints and compositions Parametrix construction Cotlar-Stein lemma (almost orthogonality)

4 Symbols of pseudodifferential operators Given a smooth function a(x, ξ) on R n R n (which we think of as the cotangent bundle of R n ), a pseudodifferential operator with symbol a is by definition T a f (x) = a(x, ξ) f (ξ)e 2πix ξ dξ. R n We will consider only symbols a that satisfy the following differential inequalities: x α β ξ a(x, ξ) (1 + ξ )m β γ+ α δ for all multiindices α and β, where m R and γ, δ [0, 1] are three fixed parameters. Following Hörmander, a symbol a is said to be of class S m γ,δ, if the above differential inequalities are satisfied for all α and β. Usually we consider only the case γ = 1, δ = 0, in which case we write S m in place of S m 1,0. m is called the order of the symbol (or the order of the associated operator).

5 Example: If p(ξ) is a polynomial of degree m, and a(x, ξ) = p(2πiξ), then a S m, and T a f (x) = p( x )f (x) is a constant coefficient differential operator of order m. More generally, if m N and a(x, ξ) = A β (x)(2πiξ) β, β m where the A β s are all Cc on R n, then we have a S m with T a f (x) = A β (x) x β f (x) β m is a variable coefficient partial differential operator of order m.

6 It is easy to see that if a Sγ,δ m for some m R, γ, δ [0, 1], then T a is a linear map from S(R n ) into itself, and the map is continuous. T a : S(R n ) S(R n ) One typical use of pseudodifferential operators is to construct paramatrices (i.e. approximate solutions) to partial differential equations. For those we usually need pseudodifferential operators of non-positive orders, which are typically integral operators. As before, let (R n R n ) be R n R n with the diagonal {(x, y) R n R n : x = y} removed.

7 Kernel estimates Theorem Let n < m 0, and a S m. Then there exists a function K 0 C ((R n R n ) ) such that T a f (x) = K 0 (x, y)f (y)dy R n for all f C c (R n ) and all x not in the support of f. Furthermore, λ x,y K 0 (x, y) for all multiindices λ and all x y. 1 x y n+m+ λ Indeed pick a smooth function η with compact support on R n with η(0) = 1. For x y and ε > 0, let K ε (x, y) := a(x, ξ)η(εξ)e 2πi(x y) ξ dξ. R n

8 We claim that K ε (x, y) C ((R n R n ) ) for all ε > 0, with λ x,y K ε (x, y) 1 x y n+m+ λ for all multiindices λ and all x y, where the constants are uniform in ε > 0. One sees this by splitting the integral depending on whether ξ x y 1 or not; when ξ > x y 1, we integrate by parts using e 2πi(x y) ξ = 1 4π 2 x y 2 ξe 2πi(x y) ξ sufficiently many times to gain enough decay in ξ. This shows 1 K ε (x, y) x y n+m when x y, and similarly one can estimate λ x,y K ε (x, y).

9 Furthermore, by a similar argument, K ε (x, y) converges locally uniformly on (R n R n ) as ε 0 +, and so do λ x,y K ε (x, y) for all multiindices λ. For x y, let K 0 (x, y) := lim ε 0 + K ε(x, y) C ((R n R n ) ). Now note that if f S(R n ) and x R n, then T a f (x) = lim a(x, ξ)η(εξ) f (ξ)e 2πix ξ dξ ε 0 + R n = lim a(x, ξ)η(εξ)f (y)e 2πi(x y) ξ dydξ ε 0 + R n R n = lim f (y)k ε (x, y)dy. ε 0 + R n If in addition f Cc (R n ), and x is not in the support of f, then the last line is equal to T a f (x) = f (y)k 0 (x, y)dy R n by the dominated convergence theorem.

10 This establishes the desired kernel representation formula for T a f (x). The estimates for λ x,y K 0 (x, y) on (R n R n ) follow from the corresponding uniform estimates for λ x,y K ε (x, y). We remark that if x y 1, the above proof also shows that λ x,y K 0 (x, y) x y N for any multiindices λ and any N N (i.e. we get rapid decay as x y + ). This is closely tied to the pseudolocality of psuedodifferential operators: indeed a linear operator T : S(R n ) S (R n ) is local, if the support of Tf is contained in the support of f for every f S(R n ). This is the case if the Schwartz kernel of T is supported on the diagonal {(x, y) R n R n : x = y}. While K 0 (x, y) is not supported on the diagonal {(x, y) R n R n : x = y}, the above decay of K 0 (x, y) away from the diagonal is a close substitute for it.

11 Mapping properties on L 2 We now focus on pseudodifferential operators of order 0. Theorem Let a S 0. Then T a extends to a bounded operator on L 2 (R n ). In view of the kernel representation theorem above, and the variable coefficient singular integral theorem from Lecture 4, this establishes the following corollary. Corollary Let a S 0. Then T a extends to a bounded operator on L p (R n ) for all 1 < p <.

12 One direct proof of the theorem proceeds via pseudolocality. For j Z n, let B j be the open ball of radius 2 centered at j. Then {B j } j Z n covers R n. Let 1 = j φ2 j be a smooth partition of unity subordinate to the above cover, so that φ j Cc (B j ) for every j. Then T a f 2 L 2 = j φ j T a f 2 L 2 = j φ j T a (χ 2Bj f ) 2 L 2 + j φ j T a (χ (2Bj ) c f ) 2 L 2.

13 From our earlier kernel estimates when x y 1, we get φ j (x)t a (χ (2Bj ) c f )(x) χ B j (x) f (y) x y N dy, y / 2B j so choosing N > n and using Cauchy-Schwarz, we get φ j (x)t a (χ (2Bj ) c f )(x) 2 χ Bj (x) f (y) 2 x y N dy. y / 2B j Integrating both sides gives φ j T a (χ (2Bj ) c f ) 2 L 2 y j 1 f (y) 2 j y N dy, so summing over j gives φ j T a (χ (2Bj ) c f ) 2 L 2 j j y j 1 f (y) 2 j y N dy f 2 L 2.

14 It remains to show φ j T a (χ 2Bj f ) 2 L f 2 2 L. 2 This follows if we can show since j χ 2B j f 2 L 2 f 2 L 2. j φ j T a L 2 L 2 1 But φ j T a is a pseudodifferential operator with symbol φ j (x)a(x, ξ). The latter is just another symbol in S m, except now it has compact x-support inside some ball of radius 2. Hence it remains to prove our theorem for a S 0, under the additional assumption that a(x, ξ) has compact x-support inside some unit cube. This we obtain by expanding a(x, ξ) as Fourier series in x.

15 Without loss of generality, suppose a(x, ξ) S 0 and has compact x-support on the unit cube B centered at 0. Then a(x, ξ) = η Z n â(η, ξ)e 2πiη x where â is the Fourier transform of a in the first variable. Thus T a f (x) = η Z n e 2πiη x R n â(η, ξ) f (ξ)e 2πix ξ dξ = e 2πiη x (1 + 4π 2 η 2 ) n η Z n R n n x a(η, ξ) f (ξ)e 2πix ξ dξ. But n xa(η, ξ) is a bounded multiplier on L 2 uniformly in η. Thus triangle inequality gives T a f L 2 η Z n (1 + η ) 2n f L 2 f L 2, which finishes the proof of the Theorem.

16 Compound symbols We will turn soon to the adjoints and compositions of pseudodifferential operators whose symbols are in S m for some m R. A convenient tool is the concept of compound symbols. Let m R, γ [0, 1], δ [0, 1). A compound symbol of class CS m γ,δ is a smooth function c(x, y, ξ) on Rn R n R n where x,y α β ξ c(x, y, ξ) (1 + ξ )m β γ+ α δ for all multiindices α and β. To every c CS m γ,δ, we associate an operator T [c] on S(R n ) by T [c] f (x) = lim T [c],εf (x), where ε 0 + T [c],ε f (x) := c(x, y, ξ)η(εξ)f (y)e 2πi(x y) ξ dξdy R n R n and η is a fixed function in Cc (R n ) with η(0) = 1.

17 Since δ < 1, one can show that for f S(R n ), T [c],ε f defines a Schwartz function on R n for every ε > 0, and that T [c],ε f converges in the topology of S(R n ) as ε 0 +. Indeed this follows from multiple integrating by parts via e 2πi(x y) ξ = (I y )e 2πi(x y) ξ 1 + 4π 2 ξ 2. Thus T [c] defines a linear mapping T [c] : S(R n ) S(R n ); it is easy to check that this map is also continuous. We are mainly interested in CSγ,δ m when γ = 1 and δ = 0. We write CS m for CS1,0 m if m R. The main theorem about compound symbols is the following:

18 Theorem If c CS m for some m R, then there exists a S m such that T [c] f = T a f for all f S(R n ). Also, we have the asymptotic expansion a(x, ξ) γ (2πi) γ y γ γ γ! ξ c(x, y, ξ) y=x, in the sense that the sum of those terms with γ < N on the right hand side differ from a(x, ξ) by a symbol in S m N for all N N. A proof is outlined in Homework 6. We note that different compound symbols c may give rise to the same symbol a in the above theorem. In particular, the map c T [c] is not injective. We will use this non-injectivity to our advantage in what follows.

19 Closure under adjoints and compositions We will prove two theorems using compound symbols. Theorem Let m R. If a S m, then there exists a symbol a S m, such that the formal adjoint of T a is T a, in the sense that T a f (x)g(x)dx = R n f (x)t a g(x)dx R n for all f, g S(R n ). Also, we have the asymptotic expansion a (x, ξ) γ (2πi) γ y γ γ γ! ξ a(y, ξ) y=x. In particular, a (x, ξ) = a(x, ξ) (mod S m 1 ).

20 Theorem Let m 1, m 2 R. If a 1 S m 1 and a 2 S m 2, then there exists a symbol a S m 1+m 2, such that T a1 T a2 f = T a f for all f S(R n ). Also, we have the asymptotic expansion a(x, ξ) γ (2πi) γ γ γ! ξ a 1(x, ξ) x γ a 2 (x, ξ). In particular, a(x, ξ) = a 1 (x, ξ)a 2 (x, ξ) (mod S m 1+m 2 1 ).

21 Indeed, let m R, a S m and f, g S(R n ). Let η Cc on R n with η(0) = 1. Then by dominated convergence, T a f (x)g(x)dx = lim a(x, ξ)η(εξ) f (ξ)g(x)e 2πix ξ dξdx. R n ε 0 + R n R n Writing c(x, y, ξ) := a(y, ξ), the above limit is equal to lim ε 0 + R n R n a(x, ξ)η(εξ)f (y)g(x)e 2πi(x y) ξ dydξdx R n = lim ε 0 + f (y)t [c],ε g(y)dy. R n Since T [c],ε g converges to T [c] g in S(R n ) as ε 0 +, the above limit is just R n f (y)t [c] g(y)dy. It remains to write T [c] as T a for some a S m, using our previous theorem about compound symbols.

22 Next, let m 1, m 2 R, a 1 S m 1, a 2 S m 2. By the previous theorem, there exists a2 S m 2, such that the formal adjoint of T a 2 is T a2, which in view of the computation on the previous page implies T a2 f (x) = lim a ε (y, ξ)η(εξ)f (y)e2πi(x y) ξ dydξ. R n R n for all f S(R n ) and all x R n (limit taken in S(R n )). Hence T a1 T a2 f (x) is given by lim ε 0 + R n R n The latter is lim ε 0 + T [c],ε f (x) if R n a 1 (x, ξ)a 2 (y, ξ)η(εξ)f (y)e2πi(x y) ξ dydξ. c(x, y, ξ) := a 1 (x, ξ)a 2 (y, ξ). Since such c CS m 1+m 2, by the previous theorem about compound symbols, there exists a S m 1+m 2, such that the above limit is equal to T a f (x).

23 Parametrix construction Let m N. Let P(D) = α m p α (x) α x be a differential operator of order m with C coefficients. P(D) is a pseudodifferential operator with symbol p(x, ξ) := α m p α (x)(2πiξ) α. It is said to be elliptic, if there exists a constant C > 0, such that p(x, ξ) C ξ m for all x R n and all ξ with ξ 1. We use the following theorem to construct parametrices of such elliptic partial differential operators.

24 Theorem Let m R. Given a sequence of symbols a 0, a 1,..., with a k S m k for every k 0, then there exists a symbol a S m, such that for every N N, there exists e N S m N with a(x, ξ) = N 1 k=0 a k (x, ξ) + e N (x, ξ). See Homework 6 for its proof.

25 Let P(D) be an elliptic partial differential operator of order m with C coefficients. Let p(x, ξ) be its symbol. Let ϕ(ξ) be a smooth function that is identically 0 on B(0, 1), and identically 1 outside B(0, 2). Let T a0 be the pseudodifferential operator of order m with symbol a 0 (x, ξ) := ϕ(ξ) p(x, ξ). Then the composition theorem shows that P(D)T a0 = I E 1 for some pseudodifferential operator of order 1. We compose both sides on the right by E 1 k (where k N), and get P(D)T a0 E 1 k = E 1 k E 1 k+1. Summing over k and telescoping, we get P(D)[T a0 + T a0 E T a0 E N 1] = I E N+1 1 for any N N.

26 Using the composition theorem again, for any k N, there exists a symbol a k S m k, such that T a0 E k 1 = T ak. From the previous theorem, there exists a symbol a S m, such that for every N N, there exists e N S m N such that T a = N 1 k=0 T ak + T en. Let T e be the pseudodifferential operator defined by P(D)T a = I + T e. The calculation on the previous slide shows that e S m N for any N N. In this sense T a is an approximate solution to P(D), aka a parametrix for P(D).

27 Cotlar-Stein lemma (almost orthogonality) Earlier we proved L 2 boundedness of psuedodifferential operators with symbols in S 0 by using the Fourier transform. We now describe another important tool about establishing L 2 boundedness of linear operators, namely Cotlar-Stein lemma. This can be used to prove a theorem of Calderón and Vaillaincourt, namely T a is bounded on L 2 whenever a S 0 γ,γ for all γ [0, 1) (see Homework 6); in particular, this recovers the L 2 boundedenss of psuedodifferential operators with symbols in S 0. The Cotlar-Stein lemma also plays a key role in the proof of many celebrated theorems. We will prove a proposition this time, which will play a crucial role in the proof of the T (1) theorem in the next lecture.

28 Theorem (Cotlar-Stein) Suppose {T j } is a sequence of bounded linear operators between two Hilbert spaces. If there exist constants A and B such that sup T j Ti 1/2 A and sup Ti T j 1/2 B j i for all i, j, then j T j converges strongly to a bounded linear operator T between the two Hilbert spaces, with T AB. This is called an almost orthogonality lemma, because one situation where the hypothesis are fulfilled are when all T j B, the images of the different T j s are orthogonal, and the images of the different Ti s are orthogonal. (Indeed then Ti T j = 0 and T j Ti = 0 whenever j i.) The proof of Cotlar-Stein involves a classic application of the tensor power trick. i j

29 To prove Cotlar-Stein, suppose first {T j } is a finite sequence (say J terms). Then for any positive integer N, T 2N = (TT ) N = T j1 Tj 2 T j3 Tj 4... T j2n 1 Tj 2N j 1,...,j 2N But the summand is bounded by both T j1 Tj 2 T j3 Tj 4... T j2n 1 Tj 2N and T j1 T j2 T j3... T j2n 2 T j2n 1 T j2n. So taking geometric average, the sum is bounded by j 1,...,j 2N T j1 1/2 Tj1 T j 2 1/2 T j2 T j3 1/2... Tj2N 1 T j 2N 1/2 T j2n 1/2 J max T j1 1/2 A N B N 1 max T j2n 1/2 j 1 j 2N Taking 2N-th root and letting N + yields T AB.

30 The general case when we have infinitely many operators T j follows once we have the following lemma: Lemma Let {f j } be a sequence in a Hilbert space, and A R. Suppose for any sequence {ε j } that has only finitely many non-zero terms, and that satisfies ε j 1 for all j, we have ε j f j A. Then j f j converges in the Hilbert space. See Homework 6 for the proof of this lemma. j We close this lecture by the following application of Cotlar-Stein.

31 Proposition Suppose δ > 0. Let {k j (x, y)} j Z be a sequence of C kernels on R n R n that satisfies k j (x, y) α x β y k j (x, y) 2 jn (1 + 2 j x y ) n+δ (1) 2 j(n+ α + β ) (1 + 2 j x y ) n+δ for all α, β (2) R n k j (x, y)dy = 0 for all x R n (3) R n k j (x, y)dx = 0 for all y R n. (4) For each j Z, let T j f (x) = R n f (y)k j (x, y)dy. Then T j T i L 2 L 2 + T i T j L 2 L 2 2 δ i j, so j T j converges strongly to a bounded linear operator T on L 2.

32 One can use this to prove, for instance, that the Hilbert transform is bounded on L 2 ; see Homework 6. Indeed, morally speaking, (1) to (4) says that j T j is almost like a translation-invariant singular integral we studied in Lecture 4. Next time we use this proposition to prove the T (1) theorem. We prove this proposition as follows. We will only show that T j Ti L 2 L2 2 δ i j since the bound for Ti T j L 2 L 2 is similar. The kernel for T j Ti is K j,i (x, y) := k j (x, z)k i (y, z)dz, R n and if i j, then we rewrite this using (3) as K j,i (x, y) = (k j (x, z) k j (x, y))k i (y, z)dz. R n

33 From (1) and (2), we get k j (x, z) k j (x, y) ( (2 j z y ) δ 2 2 jn (1 + 2 j x z ) n+δ + 2 jn (1 + 2 j x y ) n+δ indeed if 2 j y z 1, this follows by estimating k j (x, z) k j (x, y) k j (x, z) + k j (x, y) and using (1), whereas if 2 j y z 1, then we use (2) and the mean-value theorem, noting that if u is on the straight line segment connecting y and z, then which implies min{ x y, x z } 2 j + x u, ) ; 2 jn (1 + 2 j x u ) n+δ 2 jn (1 + 2 j x z ) n+δ + 2 jn (1 + 2 j x y ) n+δ.

34 Now we invoke an elementary inequality, which says Rn 2 jn 2 in dz (1 + 2 j x z ) n+ δ 2 (1 + 2 i z ) n+ δ 2 (1 + 2 j x ) n+ δ 2 if i j (this can be proved by rescaling to j = 0, and then dividing the domain of integration into 2 parts, depending on whether z x x x 2 or z x 2 ). Hence if i j, then 2 jn K j,i (x, y) 2 (i j) δ 2 ; (1 + 2 j x y ) n+ δ 2 similarly if j i, then 2 in K j,i (x, y) 2 (j i) δ 2. (1 + 2 i x y ) n+ δ 2 From Young s inequality, it follows that T j T i L 2 L 2 2 i j δ 2. 2 jn