Lectures on Elliptic Partial Differential Equations. Programme

Transcription

1 Lectures on Elliptic Partial Differential Equations (Method of Pseudodifferential Operators) Wien, October-December 2006 A.I.Komech 1 Faculty of Mathematics, Vienna University Nordbergstrasse 15, Vienna 1090, Austria Programme I. Tempered distributions: differentiation, multiplication by smooth functions, Fourier transform. II. Sobolev spaces, Sobolev embedding theorems. III. Pseudodifferential operators: continuity in Sobolev spaces, composition. IV. Elliptic partial differential operators: regulariser, the Fredholm properties, the Schauder a priori estimates. Methods: Operations with distributions, calculation of their Fourier transforms, construction of fundamental solutions, continuity of integral and pseudodifferential operators in Hilbert spaces, application of pseudodifferential operators to elliptic partial differential operators, construction of regulariser. Main Goals: To give an introduction to the applications of the methods of modern functional analysis to partial differential equations. References [1] A. Komech, Linear Partial Differential Equations with Constant Coefficients, p in: Yu.V. Egorov, A.I. Komech, M.A. Shubin, Elements of the Modern Theory of Partial Differential Equations. Springer, Berlin, [2] A. Komech, Book of Practical PDEs, comech/posobie [3] I.M. Gel fand, G.E. Shilov, Generalized functions. Vol. I: Properties and operations. New York and London: Academic Press, [4] W. Rudin, Functional analysis. McGraw-Hill, NY, [5] L. Schwartz, Méthodes Mathematiques pour les Sciences Physiques. Hermann, Paris, [6] M. Shubin, Pseudodifferential operators and spectral theory. Springer, Berlin, [7] M. Taylor, Pseudo differential operators. Springer, Berlin, Supported by Austrian Science Foundation (FWF) Project P19138-N13 funded by the Austrian Government.

2 2 Programme of exam I. Fourier transform of test functions and tempered distributions: Lemmas 2.2, 2.7, 2.8, and Theorem II. Sobolev spaces: Lemmas 4.3, 4.4, 6.1. III. First Sobolev s Embedding Theorem: Theorem 5.3. IV. Sobolev s Compactness Embedding Theorem: Theorem 7.2. V. Strongly elliptic PDE with constant coefficients: Theorem 8.5. VI. Schur s lemma: Lemma VII. Boundedness of multiplication operator: Lemma VIII. Boundedness of PDO: Theorem IX. Composition of PDO: Theorem X. Regulariser for strongly elliptic PDE with variable coefficients: Theorem XI. Solvability of strongly elliptic PDE with variable coefficients: Theorem 13.3.

3 Contents 1 Tempered Distributions Definitions and examples Differentiation of tempered distributions Convergence of tempered distributions Multiplication by smooth functions Fourier Transform of Tempered Distributions Fourier Transform of Test Functions Definition Parseval-Plancherel theory Generalisation to n variables Definitions and examples Differentiation Fourier transform Fourier transform of derivative The Sobolev spaces First Sobolev s embedding theorem Sobolev s spaces for integer s Sobolev s theorem on compact embedding Elliptic partial differential equations with constant coefficients Pseudodifferential operators Fourier representation for differential operators Classes of symbols and pseudodifferential operators Boundedness of pseudodifferential operators Schur s Lemma Application to operator of multiplication Boundedness of pseudodifferential operators Composition of pseudodifferential operators Composition of differential operators Composition of PDO Regulariser of elliptic equations Applications of regulariser Smoothness of solutions Solvability of elliptic equations

4 4 CONTENTS 1 Tempered Distributions 1.1 Definitions and examples Let us introduce the Schwartz space S(IR) of test functions. Definition 1.1 i) S = S(IR) is the space of functions ϕ(x) C (IR) such that (1.1) ϕ α,n := sup(1 + x ) N xϕ(x) α < x IR for any α, N = 1, 2,..., where xϕ(x) α := ϕ (α) (x). S ii) The sequence ϕ n (x) ϕ(x) iff (1.2) for any α, N = 1, 2,... ϕ n ϕ α,n 0, n Let us note that (1.1) implies the bound (1.3) α xϕ(x) ϕ α,n (1 + x ) N, x IR. for any α, N = 1, 2,... Hence, we have Corollary 1.2 For any α = 0, 1, 2,... and q IR the bound holds (1.4) α xϕ(x) ϕ α,n (1 + x ) q, x IR for any N q. Example 1.3 The function ϕ(x) = e x2 belongs to the Schwartz space S (check this!). Definition 1.4 i) S (IR) is the space of linear continuous functionals on S(IR), i.e. f S (IR) if f is the map S S, and the following two conditions hold: (1.5) (1.6) Linearity : f(αϕ + βψ) = αf(ϕ) + βf(ψ) for any α, β C and ϕ, ψ S; Continuity : f(ϕ n ) f(ϕ) if ϕ n S ϕ. ii) The functionals f S (IR) are called tempered distributions. iii) The scalar product f(x), ϕ(x) denotes the value of a functional f S (IR) at the test function ϕ S(IR), i.e. (1.7) f(x), ϕ(x) := f(ϕ), ϕ S. Example 1.5 Let us consider a continuous function f(x) C(IR) satisfying the bounds (1.8) f(x) C(1 + x ) p, x IR with some constants C, p IR. Let us define the functional (1.9) f, ϕ := f(x)ϕ(x)dx, ϕ S.

5 1. TEMPERED DISTRIBUTIONS 5 Lemma 1.6 The integral (1.9) converges, and the functional is tempered distribution. Proof i) First we note that f(x) ϕ 0,N (1 + x ) p 2 by the bound (1.4) with an q = p + 2, hence the integrand of (1.9) is bounded by C ϕ 0,N (1 + x ) 2. Therefore, the integral converges: (1.10) f, ϕ C ϕ 0,N (1 + x ) 2 dx <. ii) The linearity (1.5) of the functional follows from the properties of the integral. iii) The continuity (1.6) of the functional follows from the estimate (1.10): if ϕ n then ϕ 0,N 0, hence (1.11) f, ϕ n C ϕ n 0,N (1 + x ) 2 dx 0, n. S 0 as n, Remark 1.7 The bounds (1.9) motivate the term tempered distribution for the corresponding functionals. Exercise 1.8 Prove that functional of type (1.9) is tempered distribution if: i) f(x) L 1 (IR), ii) f(x) L 2 (IR), iii) f(x) L p (IR) with a p > 1. Hint: Use the bound (1.4) with a suitable q depending on p. For ii) use the Cauchy-Schwarz inequality, and for iii) use the Hölder inequality. Exercise 1.9 Let us check that e x S S. Hint: Construct a sequence ϕ n S such that ϕ n 0 as n while f, ϕ n 0. For example, take ϕ n (x) = φ(x n), where φ is an arbitrary nonnegative function φ D with φ(x)dx 0. Remark 1.10 We have the continuous inclusion D(IR) S(R) (check this!). Hence, each tempered distribution f S belongs also to D since the scalar product f, ϕ is linear and continuous for all test functions ϕ S, hence also for all ϕ D. Therefore, we have the map f S f D Exercise 1.11 Prove that the map D S is injective, so the map is the inclusion: D S. Hint: The space D is dense everywhere in S (check this!). Exercise 1.12 Check that δ(x) S and pv 1 x S. 1.2 Differentiation of tempered distributions Definition 1.13 For f S let us define the derivative f as the following functional on S: (1.12) f, ϕ := f, ϕ, ϕ S. Lemma 1.14 f S if f S. Exercise 1.15 Prove this lemma. Hint: ϕ n S ϕ S if ϕ n ϕ (check this!).

6 6 CONTENTS 1.3 Convergence of tempered distributions S Definition 1.16 f n f if (1.13) f n, ϕ f, ϕ, ϕ S. Exercise 1.17 Check that the differentiation is continuous operator in S, i.e. f n 1.4 Multiplication by smooth functions Let us consider smooth functions M(x) satisfying the bounds (1.14) M (k) (x) C(k)(1 + x ) p(k) with some constants C(k), p(k) IR for every k = 0, 1, 2... S f S if f n f. Example 1.18 i) Any polynomial M(x) = N k=0 M k x k satisfies the bounds (1.14) (find p(k)!). ii) The exponential function e x does not satisfy the bounds (1.14) (check this!). Lemma 1.19 Let M(x) satisfies the bounds (1.14). Then i) Mϕ S for ϕ S, and ii) Multiplication map M : ϕ(x) M(x)ϕ(x) is linear continuous map S S. Proof i) By definition (1.1), we have to check that (1.15) sup(1 + x ) N x(m(x)ϕ(x)) α < x IR for any α, N = 1, 2,... This follows from the Leibniz formula α (1.16) x(m(x)ϕ(x)) α = CαM k (k) (x)ϕ (α k) (x)) k=0 by the condition (1.14) since sup x IR (1+ x ) N+p(k) α xϕ(x) < by the bounds (1.4) with q = N+p(k). ii) The continuity of the map means that (1.17) sup(1 + x ) N x(m(x)ψ α n (x)) 0, n, x IR if sup x IR (1+ x ) N α xϕ n (x) 0. This follows by arguments of previous step i). Finally, the linearity of the multiplication map is obvious. Definition 1.20 For f S let us set (1.18) M(x)f(x), ϕ(x) = f(x), M(x)ϕ(x), ϕ S. Here the right hand side is defined since M(x)ϕ(x) S by previous lemma i). Lemma 1.21 i) The functional (1.18) is tempered distribution, and ii) Multiplication operator M : f(x) M(x)f(x) is linear continuous map S S if M(x) satisfies the bounds (1.14). Proof i) The functional (1.18) is obviously linear. Let us check its continuity: if ϕ n (1.19) M(x)f(x), ϕ n (x) = f(x), M(x)ϕ n (x) 0, n, S since M(x)ϕ n (x) 0 by previous lemma ii), and the functional f is continuous on S. ii) It remains to prove continuity of the map M in S S : if f n f, then (1.20) M(x)f n (x), ϕ(x) := f n (x), M(x)ϕ(x) f(x), M(x)ϕ(x), ϕ S. Hence, Mf n S Mf by definition of the convergence of tempered distributions. S 0, then

7 2. FOURIER TRANSFORM OF TEMPERED DISTRIBUTIONS 7 2 Fourier Transform of Tempered Distributions We are going to define the Fourier transform for tempered distributions. First we need to study the Fourier transform of the test functions. 2.1 Fourier Transform of Test Functions Definition 2.1 For ϕ S(IR) the Fourier transform is defined by (2.1) where e ikx := cos kx + isinkx. Fϕ(k) := ˆϕ(k) := IR e ikx ϕ(x)dx, k IR, It is well known that the inversion of the Fourier transform is given by (2.2) ϕ(x) := 1 e ikx ˆϕ(k)dk, 2π IR x IR. Lemma 2.2 i) For any test function ϕ S, its Fourier transform ˆϕ(k) also belongs to the Schwartz space S. ii) The operator F : ϕ ˆϕ is linear and continuous in S. Proof i) We have to prove the bounds (1.1) for the Fourier transform ϕ(k). i.e. (2.3) ˆϕ α,n := sup(1 + k ) N k α ˆϕ(k) <, α, N = 1, 2,... k IR Equivalently (2.4) sup k N k α ˆϕ(k) <, α, N = 1, 2,... k IR Exercise 2.3 Check that the bounds (2.4) imply (2.3). Hint: Prove the estimates (2.5) ˆϕ α,n C N N M=0 sup k M k α ˆϕ(k) k IR To prove bounds (2.4), note very important formulas of differentiation of the Fourier integral: (2.6) k ˆϕ(k) = (ix)e ikx ϕ(x)dx = F[(ix)ψ(x)], k IR. IR Another important formula of multiplication we obtain integrating by parts: (2.7) k ˆϕ(k) = ( i x )e ikx ϕ(x)dx = e ikx (i x )ϕ(x)dx = F[(i x )ϕ(x)], k IR. IR In operator form the identities (2.6) and (2.7) can be written as IR (2.8) k F = F ix, k F = F i x Applying α times the first formula and N times the second one, we obtain (2.9) k N k α ˆϕ(k) = F[(i x ) N( ) (ix) α ϕ(x) ].

8 8 CONTENTS This can be written as (2.10) k N k α ˆϕ(k) = e ikx (i x ) N( ) (ix) α ϕ(x) dx, that implies the bound (2.11) k N k α ˆϕ(k) (i x ) N( ) (ix) α ϕ(x) dx. Now we apply the Leibniz formula for differentiation of the product and obtain (2.12) (i x ) N( ) (ix) α ϕ(x) = This formula implies that (i x ) N( (ix) α (2.13) ϕ(x) ) C N M=0,...,N M=0,...,N CN N M [(i x ) M (ix) α ](i x ) N M ϕ(x). (1 + x ) α M ] ϕ (N M) (x) B α,n ϕ N,α+2 (1 + x ) 2. Substituting into (2.11), we obtain that (2.14) k N α k ˆϕ(k) C 1 ϕ N,α+2 (1 + x ) 2 dx <. that implies the desired bound (2.4). ii) The linearity of the operator F in the space S is obvious. To prove the continuity of the operator F in the space S, we note that (2.15) ϕ α,n D α,n ϕ N,α+2 by the estimates (2.5) and bounds (2.14). Now the continuity follows from definition of the convergence in S. 2.2 Definition Lemma 2.4 For any test functions f(x), ϕ(k) S the following identity holds (2.16) ˆf(k), ϕ(k) = f(x), ˆϕ(x). Proof The left hand side admits the following representation: (2.17) ˆf(k), ϕ(k) = ( 1 2π ) e ikx f(x)dx ϕ(k)dk. Applying here the Fubini theorem, we obtain (2.18) ˆf(k), ϕ(k) = that is the right hand side of (2.16). ( 1 f(x) 2π ) e ikx ϕ(k)dk dx Exercise 2.5 Check that the Fubini theorem is applicable here. Now we are going to define the Fourier transform for tempered distributions. The definition can be done using the following identity:

9 2. FOURIER TRANSFORM OF TEMPERED DISTRIBUTIONS 9 Definition 2.6 For any distribution f S let us define the functional (2.19) ˆf(k), ϕ(k) = f(x), ˆϕ(x), ϕ S, where the right hand side is well defined since ˆϕ S by Lemma 2.2 i). Lemma 2.7 i) For any tempered distribution f S, its Fourier transform ˆf is also a tempered distribution. ii) The operator F : f ˆf is linear and continuous in S. Proof i) The linearity of the functional ˆf follows easy: for any numbers α, β C and test functions ϕ, ψ S, we have (2.20) ˆf(k), αϕ(k) + βψ(k) = f(x), αˆϕ(x) + β ˆψ(x) = α f(x), ˆϕ(x) + β f(x), ˆψ(x) = α ˆf(k), ϕ(k) + β ˆf(k), ψ(k), where the first and last identities follows by definition 2.6, while the middle one follows by the linearity of the functional f. Similarly, the continuity of ˆf follows from definition 2.6 and continuity of the functional f: if S ϕ n (k) ϕ(k) as n, we have (2.21) ˆf(k), ϕ n (k) = f(x), ˆϕ n (x) f(x), ˆϕ(x) = ˆf(k), ϕ(k) since ˆϕ n (x) S ˆϕ(x) by Lemma 2.2 ii). ii) The continuity of the map F : S S follows similarly: if f n (k) S f(k) as n, we have (2.22) ˆf n (k), ϕ(k) = f n (x), ˆϕ(x) f(x), ˆϕ(x) = ˆf(k), ϕ(k), where the identities follow from definition 2.6. Finally, the linearity of the map F is easy to check (Please check!) 2.3 Parseval-Plancherel theory Now the Fourier transform ˆf(k) is defined for any tempered distribution f(x). Let us study ˆf(k) for the Lebesgue functions f(x) L p (IR) with p 1. This is possible since L p (IR) S for every p 1. I. First consider p = 1. Lemma 2.8 For f(x) L 1 (IR) the Fourier transform ˆf(k) is a bounded continuous function, i.e. ˆf C b (IR). It is given by the Fourier integral (2.23) Ff(k) := ˆf(k) := IR e ikx f(x)dx, k IR, where identities hold in the sense of distributions though the right hand side is a classical continuous function. Proof By definition 2.19 and the Fubini theorem, we have ˆf(k), ( ) (2.24) ϕ(k) = f(x), ˆϕ(x) = f(x) e ikx ϕ(k)dk dx = IR R IR ( R ) e ikx f(x)dx ϕ(k)dk for any ϕ S (check the conditions of the Fubini theorem!). The comparison of the first and last terms implies the formula (2.23) in the sense of the functionals.

10 10 CONTENTS Let us check that the integral (2.23) is bounded and continuous function of k IR. The boundedness follows from the estimate (2.25) ˆf(k) f(x) dx <, k IR. The continuity means that ˆf(k n ) ˆf(k) if k n k as n, i.e. (2.26) e iknx f(x)dx e ikx f(x)dx, k n k. IR This follows from the Lebesgue dominated convergence theorem since i) The integrand converges: IR IR (2.27) e iknx f(x) e ikx f(x) for almost all x IR. ii) The integrand admits summable majorant which does not depend on n: (2.28) e iknx f(x) f(x) for almost all x IR. The lemma is proved. Exercise 2.9 Check that for f(x) L 1 (IR) we have (2.29) ˆf(k) 0, k. Hints: i) Check (2.29) for characteristic functions of intervals. ii) Approximate the function f(x), in the norm of the space L 1 (IR), by finite linear combinations of the characteristic functions. iii) Use the bound (2.25). II. Now consider p = 2. Theorem 2.10 (The Parseval Theorem) For f(x) L 2 = L 2 (IR) the Fourier transform ˆf(k) also belongs to L 2 (IR), and the Parseval identity holds: (2.30) ˆf 2 = 2π f 2. Proof Step i) First let us prove (2.30) for the functions f(x) S. For this purpose substitute ϕ(k) = ˆf(k) into (2.16): this is possible since ˆf(k) S by Lemma 2.2 i). Let us note that ˆϕ(x) = e ikx ˆf(k)dk = e ikx ˆf(k)dk = 2πf(x) according to the inversion formula (2.2). Hence, (2.16) with ϕ(k) = ˆf(k) gives that (2.31) ˆf(k), ˆf(k) = 2π f(x), f(x), i.e. (2.30) and the Parseval Theorem 2.10 are proved for the functions f(x) S. Step ii) For general functions f(x) L 2 (IR) we take an approximating sequence f n (x) S(IR) such that (2.32) f n (x) f(x) L 2 0, n. Such sequence exists since S is dense in L 2 (Please check the density!). The convergence (2.32) implies that f n is the Cauchy sequence, i.e. (2.33) f n (x) f m (x) L 2 0, n, m.

11 2. FOURIER TRANSFORM OF TEMPERED DISTRIBUTIONS 11 Applying (2.30), we obtain from (2.33) that ˆf n is also the Cauchy sequence, i.e. (2.34) ˆf n (k) ˆf m (k) L 2 0, n, m. Now we use the completeness of the Hilbert space L 2 (IR): every Cauchy sequence has limit function g(k) L 2, i.e. (2.35) ˆf n (k) g(k) L 2 0, n. Step iii) Now let us prove that g = ˆf. Namely, the convergence (2.32) implies that (Please check this!) (2.36) f n (x) S f(x), n. Similarly, the convergence (2.35) implies that (2.37) ˆf n (k) S g(k), n. However, the convergence (2.36) implies that (2.38) ˆf n (k) S ˆf(k), n. by continuity of the Fourier transform in S (Lemma 2.7 ii)). Finally, (2.37) and (2.38) imply that ˆf = g L 2. Step iv) It remains to prove the Parseval identity (2.30). We have proved the identity for the test functions from S, hence (2.39) ˆf n 2 = 2π f n 2. Let us take here the limit n. Then we get (2.40) g 2 = 2π f 2 using (2.32) in the right hand side and (2.35) in the left hand side. Finally, this implies (2.30) taking into account that g = ˆf. 1 Corollary 2.11 The Parseval identity (2.30) means that the operator 2π F : L 2 (IR) L 2 (IR) is an isometry, i.e. (2.41) 1 ˆf(k) L 2π 2 (IR) = f(x) L 2 (IR).

12 12 CONTENTS 3 Generalisation to n variables 3.1 Definitions and examples Let us introduce test functions and tempered distributions of n variables x = (x 1,..., x n ) IR n, n > 1. Let us denote x = x x2 n, and α ϕ(x) := α αn n ϕ(x) = ϕ (α) (x) for any multiindex α = (α 1,..., α n ) with α j = 0, 1, 2,... Here α 1 1 ϕ(x) := α 1 ϕ(x) x α, etc. 1 Definition 3.1 i) S = S(IR n ) is the space of functions ϕ(x) C (IR n ) such that 1 (3.1) ϕ α,n := sup x IR n (1 + x ) N α xϕ(x) < for any N = 1, 2,..., and any multiindex α = (α 1,..., α n ) with α j = 0, 1, 2,... S ii) The sequence ϕ n (x) ϕ(x) iff (3.2) ϕ n ϕ α,n 0, n for any N = 1, 2,... and α = (α 1,..., α n ) with α j = 0, 1, 2,... Example 3.2 Function ϕ(x) = e x 2 belongs to the space S(IR n ). Further, the space of tempered distributions S (IR n ) is defined similarly to the case n = 1. Example 3.3 i) Let us consider a continuous function f(x) C(IR) satisfying the bounds (3.3) f(x) C(1 + x ) p, x IR with some constants C, p IR. Let us define the functional (3.4) f, ϕ := f(x)ϕ(x)dx, ϕ S. Then the integral (3.4) converges, and the functional is tempered distribution. ii) The functional (3.4) is tempered distribution if f(x) L p (IR) with a p 1. iii) Dirac delta-function is defined as the distribution (3.5) δ(x), ϕ(x) = ϕ(0), ϕ S. It is also tempered distribution of n variables. 3.2 Differentiation Differentiation of tempered distributions is defined similarly to the case n = 1: for any k = 1,..., n, (3.6) k f(x), ϕ(x) = f(x), k ϕ(x), ϕ S. Then for any multiindex α = (α 1,..., α n ), we get general formula (3.7) α f(x), ϕ(x) = ( 1) α α n f(x), α ϕ(x).

13 3. GENERALISATION TO N VARIABLES Fourier transform Further, the Fourier transform of test functions ϕ S(IR) is defined by (3.8) Fϕ(k) := ˆϕ(k) := eikx ϕ(x)dx, k IR n, IR n where kx = k 1 x k n x n. The inversion of the Fourier transform is given by ϕ(x) := 1 (3.9) (2π) n e ikx ˆϕ(k)dk, x IR n. IR n Similarly to the case n = 1, for any test function ϕ S(IR n ), its Fourier transform ˆϕ(k) also belongs to the Schwartz space S(IR n ), and the operator F : ϕ ˆϕ is linear and continuous in S(IR n ). Finally, the Fourier transform of tempered distribution f S (IR n ) is defined by the Parseval identity: (3.10) ˆf(k), ϕ(k) = f(x), ˆϕ(x), ϕ S(IR n ). Lemma 2.8 and Theorem 2.10 generalises to any dimension n > 1: Lemma 3.4 For f(x) L 1 (IR n ) the Fourier transform ˆf(k) is a bounded continuous function, i.e. ˆf C b (IR n ). It is given by the Fourier integral (3.11) Ff(k) := ˆf(k) := eikx f(x)dx, k IR n. IR n Theorem 3.5 (The Parseval Theorem) For f(x) L 2 = L 2 (IR n ) the Fourier transform ˆf(k) also belongs to L 2 (IR n ), and the Parseval identity holds (Cf. (2.30)): (3.12) 3.4 Fourier transform of derivative ˆf 2 = (2π) n f 2. For any tempered distribution f S (IR n ), the following generalisations of the identities (2.8) hold: (3.13) k j ˆf(k) = F[i j f(x)], j ˆf(k) = F[ixj f(x)] For the test function f S(IR n ) the identities follows similarly to (2.8). For tempered distributions f S (IR n ) the proof of the first formula is the following: (3.14) k j Ff(k), ϕ(k) = Ff(k), k j ϕ(k) = f(x), F(k j ϕ(k)) = f(x), i xj F(ϕ(k)) = i xj f(x), F(ϕ(k)) = F[i xj f(x)], ϕ(k). Here i) first identity follows by definition of distributions by smooth function k j, ii) second and last identities follow from definition of the Fourier transform for distributions, iii) third identity follows by differentiation of the Fourier integral: F(k j ϕ(k)) = eikx k j ϕ(k)dk = IR n IR n( i x j )e ikx ϕ(k)dk (3.15) = i xj IR n eikx ϕ(k)dk = i xj F(ϕ). iv) The fourth identity in (3.14) follows by definition of the derivative for distributions. Finally, second formula (3.13) follows similarly. Exercise 3.6 Prove the second formula (3.13).

14 14 CONTENTS 4 The Sobolev spaces Here we introduce the Sobolev spaces and prove simplest properties. Let s be a real number. Definition 4.1 H s = H s (IR n ) is the space of tempered distributions f(x) S (IR n ) such that (1 + k ) s ˆf(k) L 2 = L 2 (IR n ), and the corresponding Sobolev s norm is defined as (4.1) f s := (1 + k ) s ˆf(k) L 2. Remark 4.2 For s = 0 the Sobolev space H 0 (IR n ) coincides with the Hilbert space L 2 (IR n ) by the Parseval Lemma 3.5, and (4.2) f 0 := ˆf(k) L 2 = (2π) n f(x) L 2 according to (3.12). Lemma 4.3 Every Sobolev space H s is isomorphic to the Hilbert space L 2 = L 2 (IR n ). The map (1+ k ) s F : H s L 2 is continuous, and the inverse map F 1 (1+ k ) s : L 2 H s is also continuous. Proof i) By Definition 4.1, g(k) = (1 + k ) s Ff(k) = (1 + k ) s ˆf(k) L 2 if f(x) H s. Furthermore, f(x) s := g(k) L 2, hence (4.3) g(k) L 2 f(x) s that implies the continuity of the map (1 + k ) s F : f(x) g(k) from H s to L 2. ii) We have ˆf(k) = (1 + k ) s g(k), hence the inverse map is given by f(x) = F 1 ˆf = F 1 [(1 + k ) s g(k)]. It is important that this map is defined on the whole of g(k) L 2 since (1 + k ) s g(k) is a tempered distribution for every g(k) L 2 (check this!), hence f(x) = F 1 [(1 + k ) s g(k)] is also tempered distribution! Now we check that ˆf(k) = (1+ k ) s g(k), hence (1+ k ) s ˆf(k) = g(k) L 2, and therefore, by definition, f H s. Furthermore, f(x) s := g(k) L 2, hence the map F 1 (1 + k ) s : g(k) f(x) is continuous from L 2 to H s. For a multiindex α = (α 1,..., α n ) let us define monomial k α := k α kαn n of n complex variables k 1,..., k n C. Consider polynomials A(k) = a α k α of order m = 0, 1, 2,... and corresponding differential operators (4.4) α m A( )u(x) = α m a α α u(x). For any tempered distribution u(x) S, we also have A( )u(x) S since any derivative of tempered distribution also is a tempered distribution. By definition, the Sobolev spaces H s are subspaces of S, hence A( )u is a tempered distribution for any u H s. Lemma 4.4 For any s IR and u H s we have A( )u H s m, and the operator A( ) : H s H s m is linear and continuous. Proof By definition, A( )u H s m if (1 + k ) s m F[A( )u](k) L 2, and (4.5) A( )u s m = (1 + k ) s m F[A( )u](k) L 2. Let us calculate the Fourier transform: using first formulas of (3.13), we obtain that F[ α u](k) = ( i ) α û(k), hence F[A( )u](k) = A( ik)û(k), and (4.5) becomes (4.6) A( )u s m = (1 + k ) s m A( ik)û(k) L 2.

15 4. THE SOBOLEV SPACES 15 Note that the polynomial A(k) is continuous function, hence the product (1 + k ) s m A( ik)û(k) is measurable function of k IR n. Furthermore, A( ik) C(1 + k ) m for k IR n, hence (4.7) (1 + k ) s m A( ik) C(1 + k ) s, for k IR n. This implies that (1 + k ) s m A( ik)û(k) 2 L 2 = (1 + k ) 2(s m) A( ik)û(k) 2 dk (1 + k ) 2s û(k) 2 dk (4.8) Therefore (4.6) implies that (4.9) that proves the lemma. = C (1 + k ) s û(k) 2 L 2 = C u 2 s. A( )u s m C 1 u s <

16 16 CONTENTS 5 First Sobolev s embedding theorem First we state the following generalisation of Lemma 2.8 to n variables: Lemma 5.1 For f(x) L 1 (IR n ) the Fourier transform ˆf(k) is a bounded continuous function, i.e. ˆf C b (IR n ). It is given by the Fourier integral (5.1) Ff(k) := ˆf(k) := eikx f(x)dx, k IR n. IR n The proof of the lemma coincides with the proof of Lemma 2.8. The lemma can be reformulated for the inverse Fourier transform as follows: Lemma 5.2 For g(k) L 1 (IR n ) the inverse Fourier transform F 1 g(x) is a bounded continuous function, i.e. F 1 g C b (IR n ). It is given by the Fourier integral F 1 g(x) = 1 (5.2) (2π) n e ikx g(k)dk, x IR n. IR n Now we can prove first Sobolev s embedding theorem is the following. Theorem 5.3 (First Sobolev s embedding theorem) Let a tempered distribution u(x) H s (IR n ) with s > n/2. Then u(x) is a continuous function in IR n, and (5.3) sup x IR n u(x) C(s) u s, where the constant C(s) < does not depend on u(x). In other words, we have continuous embedding H s (IR n ) C b (IR n ) if s > n/2. Proof Step i) First we will prove that û(k) L 1 (IR n ). Namely, by definition, u(x) H s (IR n ) means that (1 + k ) s û(k) L 2, i.e. (5.4) (1 + k ) s û(k) 2 dk < Then by the Cauchy-Schwartz inequality, we get û(k) dk = (1 + k ) s (1 + k ) s û(k) dk (5.5) ( ) 1/2 ( ) 1/2 (1 + k ) 2s dk (1 + k ) s û(k) 2 dk < I(s) u s < since I(s) := ( ) 1/2 (1 + k ) 2s dk < for s > n/2 (Check this!). Step ii) We have proved that û(k) L 1 (IR n ). Therefore, u(x) = F 1 û(k) is bounded continuous function by Lemma 5.2. To prove the bound (5.3), let us write (5.2) with g(k) = û(k): then F 1 g(k) = u(x), hence (5.2) becomes u(x) = 1 (5.6) (2π) n e ikx û(k)dk, x IR n. IR n Therefore, (5.7) sup u(x) 1 x IR n (2π) n û(k) dk, x IR n. Substituting here (5.5), we obtain (5.3) with the constant C(s) = I(s)/(2π) n.

17 5. FIRST SOBOLEV S EMBEDDING THEOREM 17 Exercise 5.4 Check that C(s) as s n/2. Corollary 5.5 Let a tempered distribution u(x) H s (IR n ) with s > n/2 + k, k 0. Then for any multiindex α = (α 1,..., α n ), the derivative α u(x) is bounded continuous function in IR n if α k. In other words, we have continuous embedding H s (IR) Cb k (IR) if s > n/2 + k. Example 5.6 I. For one variable, n = 1: H 1 (IR) C b (IR) since 1 > 1/2, H 2 (IR) Cb 1 (IR) since 2 > 1/2 + 1,... II. For three variables, n = 3: H 2 (IR) C b (IR) since 2 > 3/2, H 3 (IR) Cb 1 (IR) since 3 > 1/2 + 2,...

18 18 CONTENTS 6 Sobolev s spaces for integer s 0 First we give an equivalent characterisation to the Sobolev spaces with integer s = 0, 1,... Lemma 6.1 i) For s = 0, 1,... the Sobolev space H s (IR n ) coincides with the space of functions u(x) L 2 (IR n ) such that α u(x) L 2 (IR n ) for any multiindex α = (α 1,..., α n ) with α s. ii) The Sobolev norm u s is equivalent to the norm u s defined by (6.1) u 2 s = α u 2 L 2 α s Remark 6.2 i) In this lemma the derivatives α u(x) of a function u(x) L 2 (IR n ) are understood in the sense of distributions: if one treat the derivatives in the classical sense, then the lemma would fail. ii) The equivalence of the norms means that (6.2) u s C 1 u s for u H s, for a constant C 1 <, and also (6.3) u s C 2 u s for u H s. Proof of Lemma 6.1 We have to check that i) α u(x) L 2 (IR n ) for u H s, and ii) vice versa, u H s if α u(x) L 2 (IR n ) for any α s. Step i) Lemma 4.4 implies that α u(x) L 2 (IR n ) for u H s, and the bound (6.2) holds. Step ii) Vice versa, let us consider a function u(x) L 2 (IR n ) such that α u(x) L 2 (IR n ) for any multiindex α = (α 1,..., α n ) with α s. We have to prove that u H s, and the bound (6.3) holds. By the Parseval Theorem 3.5, we have that F[ α u] L 2 (IR n ). On the other hand, F[ α u] = ( ik) α û(k) by the formula (3.13), hence ( ik) α û(k) L 2 (IR n ). Furthermore, by the Parseval identity (3.12), we have (6.4) ( ik) α û(k) 2 dk = F[ α u] 2 L 2 = (2π)n α u 2 L 2 <. Summing up the identities with α s, we obtain that (6.5) [ ( ik) α 2 ] û(k) 2 dk = C α s Let us note that the function S(k) = α s α s α u 2 L 2 = C u 2 s <. ( ik) α 2 0 for each k IR n, hence (6.5) implies that the tempered distribution û(k) is measurable Lebesgue function. Moreover, the function S(k) admits the following bound from below: (6.6) B(1 + k ) 2s S(k) := ( ik) α 2, k IR n. Exercise 6.3 Check the bound (6.6). α s The bound (6.6) implies that (6.7) B (1 + k ) 2s û(k) 2 dk [ ( ik) α 2 ] û(k) 2 dk < by (6.5). Therefore, (6.8) α s B u 2 s C u 2 s < by definitions of the norms. Hence, u H s, and the bound (6.3) holds.

19 7. SOBOLEV S THEOREM ON COMPACT EMBEDDING 19 7 Sobolev s theorem on compact embedding Definition 4.1 implies that H s 1 (IR n ) H s 2 (IR n ) if s 1 > s 2 (Check this!). We will prove that some bounded sets in H s 1 (IR n ) are precompact in H s 2 (IR n ). We will consider the case s 1 0 for the simplicity of exposition, though all statements below are valid also for s 1 < 0. Let us consider an open region Ω IR n. Definition 7.1 H o s 1 (Ω) is the subspace of tempered distributions u(x) H s 1 (IR n ) such that u(x) vanishes in the complement of Ω, i.e. u(x) = 0, x IR n \ Ω. The norm in H o s 1 (Ω) coincides with the norm in H s 1 (IR n ). By this definition, o H s 1 (Ω) H s 1 (IR n ), and hence H o s 1 (Ω) H s 2 (IR n ). Theorem 7.2 Let the region Ω be bounded in IR n, and s 1 > s 2. Then the embedding H o s 1 (Ω) H s 2 (IR n ) is compact, i.e. for any bounded sequence of the functions u j (x) H o s 1 (Ω) there exists a subsequence u j (x) converging in H s 2 (IR n ), i.e. (7.1) u j (x) u(x) s2 0 as j, where u(x) H s 2 (IR n ). Proof We will prove the theorem for the case s 1 0. The proof for general s 1 < 0 is very similar. We shall check that u j (x) is the Cauchy sequence in H s 2 (IR n ), i.e. (7.2) u j (x) u m (x) s2 0 as j, m. Then (7.1) would follow since H s 1 (IR n ) is complete Hilbert space. Step i) First we prove that the sequence u j (x) is bounded in L 1 (IR n ), i.e. (7.3) Ω u j (x) dx B 1 <, j = 1, 2,... Namely, by definition of bounded sequence in o H s 1 (Ω), (7.4) sup u j (x) s1 B <. j N Since s 1 0, we have H s 1 (IR n ) H 0 (IR n ) = L 2 (IR n ). Therefore, (7.5) Moreover, we have (7.6) u j (x) 2 dx = u j (x) 2 0 u j (x) 2 s 1 B 2. u j (x) = 0 for x IR n \ Ω since u j (x) o H s 1 (Ω). Finally, by the Cauchy-Schwartz inequality and (7.5), we obtain (7.7) Ω ( u j (x) dx Ω ) 1/2 ( 1dx Ω u j (x) 2 dx) 1/2 Ω 1/2 B.

20 20 CONTENTS Here Ω is the volume of the region Ω. The volume is finite since the region is bounded by our assumption, hence (7.7) implies (7.3) with B 1 = Ω 1/2 B. Step ii) Further let us study the Fourier transform of the functions u j (x). First, we have (7.8) û j (k) = e ikx u j (x)dx Ω by Lemma 3.4 and (7.6) since we have proved that u j (x) L 1. Let us prove that the derivatives of the functions are bounded, i.e. (7.9) sup û j (k) < B l <, j = 1, 2,... k IR n k l for any l = 1,..., n. For the proof we write definition of partial derivative: (7.10) û j (k) = lim k l ε 0 Ω e i(k+εel)x e ikx u j (x)dx, ε where e l is the unit vector e l = (0,...,1 l,...,0). The limit exists and is given by (7.11) û j (k) = e ikx u j (x)dx. k l k l This follows by the Lebesgue Dominated Convergence Theorem since I. the integrands converge for almost all points x Ω II. There exists summable majorant: (7.12) ei(k+εel)x e ikx u j (x) M(Ω) u j (x), ε since ei(k+εe l )x e ikx ε is bounded function for x Ω: (7.13) e i(k+εe l)x e ikx Exercise 7.3 Check the bound (7.13). Hints: ε Ω x Ω M(Ω) <, x Ω, ε (0, 1). i) e i(k+εe l)x e ikx ε = eiεe lx 1 ε = eiεx l 1 ε ii) e iεφ 1 ε = eiεφ/2 e iεφ/2 ε = 2 sin εφ 2 ε φ as ε 0. Step iii) Now we consider the norms in (7.2): (7.14) jm (R) := u j (x) u m (x) 2 s 2 = (1 + k ) 2s 2 û j (k) û m (k) 2 dk. Let us split the region of integration into the ball k < R and its complement k > R, where the radius R will be chosen later: jm (R) = = (1 + k ) 2s 2 û j (k) û m (k) 2 dk + (1 + k ) 2s 2 û j (k) û m (k) 2 dk (7.15) k <R = I jm (R) + J jm (R). k >R

21 7. SOBOLEV S THEOREM ON COMPACT EMBEDDING 21 First we estimate the second integral: J jm (R) = (7.16) (1 + R ) 2s 2 2s 1 since s 2 s 1 < 0. Therefore, (7.17) according to (7.3). k >R k >R J jm (R) (1 + R ) 2s 2 2s 1 (1 + R ) 2s 2 2s 1 (1 + k ) 2s 1 (1 + k ) 2s 2 2s 1 û j (k) û m (k) 2 dk (1 + k ) 2s 1 û j (k) û m (k) 2 dk k >R (1 + k ) 2s 1 2( û j (k) 2 + û m (k) 2 )dk IR n (1 + k ) 2s1 2( û j (k) 2 + û m (k) 2 )dk 2B 2 (1 + R ) 2(s 1 s 2 ) Step iv) Finally we can prove (7.2). First we use the bounds (7.9): they imply that the functions û j (k) are equicontinuous in IR n. On the other hand, the bound (7.3) implies that the functions are uniformly bounded in IR n. Hence, by the Arzela-Ascoli Theorem, there exists a subsequence û j (k) converging at every point of IR n. Therefore, (7.18) û j (k) û m (k) 0 as j, m, and this convergence is uniform in every compact set of k. In particular, for any fixed R > 0, (7.19) max k R û j (k) û m (k) 0 as j, m. This implies that first integral in (7.15) converges to zero: for any fixed R > 0, (7.20) I j m (R) 0 as j, m. It remains to note that J j m (R ε) is small for large R by (7.17) uniformly in j and m. This proves (7.2). With detail: for any ε > 0 a) (7.17) implies that there exists R ε > 0 such that (7.21) J j m (R ε) < ε/2 for all j, m since s 1 s 2 > 0. b) (7.20) implies that for the fixed R ε there exists a number N ε such that (7.22) I j m (R ε) < ε/2 for j, m N ε. On the other hand, j m = I j m (R ε) + J j m (R ε) by (7.15). Therefore, (7.21) and (7.22) imply that (7.23) j m I j m (R ε) + J j m (R ε) < ε for j, m N ε.

22 22 CONTENTS 8 Elliptic partial differential equations with constant coefficients We will consider partial differential equations in IR n with variable coefficients (8.1) a α (x) α u(x) = f(x), x IR n. α m We plan to study in the future the questions on existence, uniqueness and smoothness of solutions u(x). Example 8.1 For example, we will consider the stationary Schrödinger equation (8.2) (H E)u(x) := ( + V (x) E)u(x) = f(x) = f(x), x IR n, where = n 2 j=1 x 2 j is the Laplace operator, H = + V (x) is the Schrödinger operator, V (x) is the potential energy, and E C is a complex (energy) parameter. The investigation of the inverse operator (H E) 1 = ( + V (x) E) 1 is one of main goals of quantum scattering theory. Here we will study the equations with constant coefficients (8.3) Au(x) := a α α u(x) = f(x), x IR n. α m We assume that the right hand side f(x) is tempered distribution, f(x) S (IR n ), and seek for the solution u(x) also in the space of tempered distributions, u(x) S (IR n ). Definition 8.2 The polynomial a(k) = α m In the Fourier transform, the equation (8.3) becomes a α ( ik) α is the symbol of differential operator A. (8.4) a(k)û(k) = ˆf(k), k IR n. due to formulas (3.13). Definition 8.3 i) Differential operator A( ) is elliptic of order m if (8.5) a(k) > c(1 + k ) m, k > R, k IR n for some R > 0 and c > 0. ii) Differential operator A( ) is strongly elliptic of order m if (8.6) a(k) > c(1 + k ) m, k IR n, where c > 0. Example 8.4 i) The Laplace operator (and ) is elliptic (check this!). Hint: Symbol of the Laplace operator is n kj 2 = k 2. j=1 ii) The Schrödinger operator E is elliptic for every E C (check this!). iii) The Schrödinger operator E is strongly elliptic a) for every E < 0 (check this!); b) for every E C \ IR + where IR + = {E IR : E 0} (check this!).

23 8. ELLIPTIC PARTIAL DIFFERENTIAL EQUATIONS WITH CONSTANT COEFFICIENTS 23 Theorem 8.5 Let us consider equation with constant coefficients (8.3). Let us assume that A is strongly elliptic operator of order m, and f(x) H s m (IR n ). Then for any s IR, we have i) Equation (8.3) admits unique solution u(x) H s (IR n ). ii) The bound holds, (8.7) u s C f s m. Proof In Fourier transform, equation (8.3) becomes (8.3) where û(k) and ˆf(k) are measurable Lebesgue functions in IR n. Therefore, the solution is given by (8.8) û(k) = ˆf(k) a(k) for almost all k IR n since a(k) 0 for k IR n. It remains to set u(x) := F 1 û(k) and check that u(x) belongs to the Sobolev space H s (IR n ). By definition, (8.9) u 2 s = (1 + k ) 2s û(k) 2 dk = (1 + k ) 2s ˆf(k) 2 (1 + k ) 2s dk = a(k) a(k) 2 ˆf(k) 2 dk. Note that (8.10) by (8.6). Hence, (8.9) implies that (8.11) since f H s m. This proves the bound (8.7). (1 + k ) 2s a(k) 2 C(1 + k ) 2(s m), k IR n u 2 s C (1 + k ) 2(s m) ˆf(k) 2 dk = C f 2 s m < Corollary 8.6 Strongly elliptic operator A of order m is isomorphism H s H s m. The continuity of the direct operator A : H s H s m follows from Lemma4.4, and the continuity of the inverse operator A 1 : H s m H s follows from the bound (8.7). Exercise 8.7 Check the inequality (8.10). Hints: Due to (8.6), i) The function (8.12) is continuous in IR n, and ii) (8.13) R(k) := (1 + k ) 2s a(k) 2 (1 + k ) 2(s m) lim R(k) <. k In next lectures we will extend Theorem 8.5 and Corollary 8.6 to strongly elliptic partial differential equations with variable coefficients.

24 24 CONTENTS 9 Pseudodifferential operators 9.1 Fourier representation for differential operators Let us consider the Fourier representation for partial differential operators in IR n with variable coefficients (9.1) Au(x) = a α (x) α u(x), x IR n, α m for test functions u(x) S(IR n ). First let us consider the case of constant coefficients: a α (x) = a α. Then Au(x) = α m a α α u(x), and in the Fourier representation we have FAu(k) = a(k)û(k), where a(k) is the symbol of the operator A: (9.2) a(k) := a α ( ik) α. α m Since u(x) S(IR n ), we have also û(k) S(IR n ), and hence a( ik)û(k) L 1 (IR n ) (Check this!). Then the inverse Fourier transform Au(x) = F 1 a(k)û(k) can be written as the standard Fourier integral Au(x) = 1 (9.3) (2π) n e ikx a(k)û(k)dk. Now consider the case of variable coefficients (8.1). Applying (9.3) to the operator α, we write α u(x) = 1 (9.4) (2π) n e ikx ( ik) α û(k)dk. Multiplying by a α (x) and summing up, we obtain that Au(x) = a α (x) α u(x) = 1 (9.5) (2π) n e ikx This can be written as (9.6) where the polynomial (9.7) α m Au(x) = 1 (2π) n a(x, k) := α m α m e ikx a(x, k)û(k)dk, a α (x)( ik) α a α (x)( ik) α û(k)dk. is the symbol of the differential operator A (Cf. Definition 8.2). For general, not necessarily polynomial, functions a(x, k), the operators of type (9.6) are called pseudodifferential operators if a(x, k) satisfies appropriate estimates that we are going to discuss. 9.2 Classes of symbols and pseudodifferential operators Definition 9.1 For a real number m IR, class of symbols S m consists of the functions a(x, k) C (IR n ) such that a(x, k) = a 0 (k) + a (x, k), where the terms a 0 (k) and a (x, k) satisfy the following estimates I. For any multiindex α, the bound holds, (9.8) α k a 0 (k) C(α)(1 + k ) m α, k IR n. II. For any multiindexes α, β, and number N > 0, the bound holds, (9.9) (1 + x ) N α k β xa (x, k) C(α, β, N)(1 + k ) m α, x, k IR n.

25 9. PSEUDODIFFERENTIAL OPERATORS 25 Exercise 9.2 i) Check that a(k) = (1 + k 2 ) m/2 S m for any m IR. ii) Check that the symbol a(x, k) = a α (x)( ik) α belongs to the class S m if for any multiindex α α m we have a α (x) = a 0 α(k) + a α(x), where a α(x) S(IR n ). Exercise 9.3 Check that a 1 (x, k)a 2 (x, k) S m 1+m 2 if a 1 (x, k) S m 1 and a 2 (x, k) S m 2, i.e. the union of the classes m IR S m is an algebra. Now consider an operator A of type (9.6 with the symbol a(x, k) from the class S m with an m IR. The operator is defined for the test functions u(x) S(IR n ) since the integral (9.6) converges then (Check this!). Definition 9.4 An operator A of type (9.6) is called pseudodifferential operator of class A m if the corresponding symbol a(x, k) belongs to the class S m with m IR.

26 26 CONTENTS 10 Boundedness of pseudodifferential operators Here we discuss boundedness of pseudodifferential operators in the Sobolev spaces Schur s Lemma Let us consider an integral operator defined for test functions v(k) S(IR n ) by the following formula (10.1) Sv(k) = S(k, k )v(k )dk, k IR n. IR n Lemma 10.1 Let the integral kernel S(k, k ) satisfies the estimates (10.2) S(k, k ) dk C <, k IR n, IR n (10.3) Then Sv(k) L 2 (IR n ) for v(k) S(IR n ), and (10.4) IR n S(k, k ) dk C <, k IR n. Sv L 2 C v L 2. Proof Definition (10.1) implies the inequality (10.5) Sv(k) S(k, k )v(k ) dk = S(k, k ) 1/2 S(k, k ) 1/2 v(k ) dk, k IR n. IR n IR n Applying here the Cauchy-Schwartz inequality, we obtain the inequality Sv(k) 2 S(k, k )v(k ) dk = S(k, k ) dk S(k, k ) v(k ) 2 dk IR n IR n IR n (10.6) C S(k, k ) v(k ) 2 dk, k IR n IR n by condition (10.2). Integrating (10.6) over k IR n, we obtain that ( ) (10.7) Sv(k) 2 dk C S(k, k ) v(k ) 2 dk dk. IR n IR n This implies by the Fubini theorem, that ( ) (10.8) Sv(k) 2 dk C S(k, k ) dk v(k ) 2 dk. IR n Now we use second condition (10.3) and obtain (10.9) Sv(k) 2 dk C 2 v(k ) 2 dk IR n that implies (10.4). IR n Corollary 10.2 Under the conditions (10.2), (10.3), the integral operator S admits unique extension from test functions v(k) S(IR n ) to the functions v(k) L 2 (IR n ) as bounded operator S : L 2 (IR n ) L 2 (IR n ). Exercise 10.3 Justify the application of the Fubini theorem in (10.8). Hint: Use the condition (10.3).

27 10. BOUNDEDNESS OF PSEUDODIFFERENTIAL OPERATORS Application to operator of multiplication Let us consider the operator of multiplication, M : u(x) a(x)u(x), by a test function a(x) S(IR n ). Lemma 10.4 Let a(x) S(IR n ). Then the operator M is continuous H s (IR n ) H s (IR n ) for any s IR. Proof We have to prove the bound (10.10) By definition, (10.11) a(x)u(x) s C(s) u(x) s, u H s. a(x)u(x) 2 s = (1 + k ) s [F(au)](k) L 2. Let us calculate the Fourier transform F(au): (10.12) F(au)(k) = e ikx a(x)u(x)dx. Substituting the Fourier representation u(x) = 1 (2π) n e ik xû(k )dk, we obtain by the Fubini theorem that ( 1 F(au)(k) = e ikx a(x) (2π) n e ik xû(k )dk ) dx (10.13) = Therefore, (10.11) becomes (10.14) 1 (2π) n ( e i(k k )x a(x)dx )û(k )dk = 1 (2π) n a(x)u(x) 2 s = 1 (1 + k )s (2π) n â(k k )û(k )dk L 2 Hence, the bound (10.10) can be written in the form (10.15) (1 + k ) s â(k k )û(k )dk L 2 C(s) (1 + k ) s û(k ) L 2 Finally, denoting v(k ) := (1 + k ) s û(k ), rewrite (10.15) as (1 + k ) s â(k k v(k ) (10.16) ) (1 + k ) sdk L 2 C(s) ˆv(k ) L 2 â(k k )û(k )dk. This bound is equivalent to boundedness in L 2 of integral operator with integral kernel (10.17) S(k, k ) = (1 + k ) s â(k k ) (1 + k ) s = â(k (1 + k k )s ) (1 + k ) s It remains to prove the estimates (10.2) and (10.3) for the kernel (10.17). To prove the estimates, we first note that (1 + k ) s (10.18) (1 + k ) s C(s)(1 + k k ) s, k, k IR n. (this is known as the Peetre inequality). Further, â(k k ) C(N)(1 + k k ) N for any N > 0. Therefore, we obtain the bound (10.19) Taking s N > n, we obtain (10.2) and (10.3). S(k, k ) C(s, N))(1 + k k ) s N Exercise 10.5 Prove the Peetre inequality (10.18). Hints: i) Consider first s = 1, then s > 0. ii) Reduce the case s < 0 to s > 0. Exercise 10.6 Check that (10.19) with s N > n implies (10.2) and (10.3).

28 28 CONTENTS 10.3 Boundedness of pseudodifferential operators Theorem 10.7 Let A be pseudodifferential operator of class A m with an m IR. Then Au(x) is defined by formula (9.6) for test functions u S(IR n ), and for any s IR the bound holds (10.20) Au(x) s m C(s) u(x) s where C(s) < does not depend on u(x). Example 10.8 For example, let us prove the theorem for particular case of differential operators of class A m from Exercise 9.2 ii) with integer m = 0, 1, 2. I. First we consider integer s = m, m+1,... In this case the Sobolev norms Au(x) s m and u(x) s admit equivalent characterisation (6.1), hence (10.20) is equivalent to the estimate (10.21) Au(x) s m C(s) u(x) s. By (9.1) and triangle inequality for the norm, we have (10.22) Au(x) s m a α (x) α u(x) s m. α m Hence, it suffices to prove the estimate for every summand in (10.22), i.e. (10.23) a α (x) α u(x) s m C(s) u(x) s for α m: then summing up in α, we obtain (10.21). Further, (10.23) means by definition that (10.24) β[ a α (x) α u(x) ] L 2 C(s) γ u(x) L 2. β s m Finally, this inequality holds since every derivative β[ ] a α (x) α u(x) is the sum of products of derivatives of the coefficients a α (x) and derivatives of γ u(x) with γ s since β s m and α m. It remains to note that every derivative of a α (x) is bounded function in IR n since a α (x) = a 0 α(k)+a α(x) with a α(x) S(IR n ). II. Now let us prove the bound (10.20) for the same class of differential operators and for all s IR. Similarly to (10.22), we have by triangle inequality for the norm that (10.25) Au(x) s m a α (x) α u(x) s m. α m γ s Hence, again it suffices to prove the estimate for every summand in (10.25), i.e. (10.26) a α (x) α u(x) s m C(s) u(x) s for α m. Since a α (x) = a 0 α(k) + a α(x), we have (10.27) a α (x) α u(x) s m a 0 α α u(x) s m + a α(x) α u(x) s m. The first term in the right hand side is bounded by C(s) u(x) s by Lemma 4.4. Finally, for the second term we obtain from Lemma 10.4 that (10.28) a α(x) α u(x) s m C(s m) α u(x) s m since a α(x) S(IR n ).

29 10. BOUNDEDNESS OF PSEUDODIFFERENTIAL OPERATORS 29 Proof of Theorem 10.7 for general case Step i) By definition 9.1, the formula (9.6) can be rewritten as Au(x) = A 0 u(x) + A u(x) = 1 (2π) n e ikx a 0 (k)û(k)dk + 1 (10.29) (2π) n e ikx a (x, k)û(k)dk. It suffices to prove the bound of type (10.20) for each of two terms. Step ii) For the first term the bound follows easily. Namely, by definition of the Sobolev norm, (10.30) A 0 u(x) s m = (1 + k ) s mâ0 u(k) L 2. [ ] Further, definition of A 0 u(x) in (10.29) implies that F A 0 u (k) = a 0 (k)û(k), hence (10.31) (1 + k ) s mâ0 u(k) L 2 = (1 + k ) s m a 0 (k)û(k) L 2 C (1 + k ) s û(k) L 2 since a 0 (k) C(1 + k ) m by condition (9.8) with α = 0. Combining the bounds (10.30) and (10.31), we obtain the desired estimate (10.32) A 0 u(x) s m C u(x) s. Step iii) It remains to prove similar bound for second term, A u(x) = 1 (10.33) (2π) n e ikx a (x, k)û(k)dk. As in (10.30) (10.34) A u(x) s m = (1 + k ) s mâ u(k) L 2. Let us calculate the Fourier transform  u(k). First let us prove that A u(x) L 1 (IR n ). Indeed, condition (9.8) with α = β = 0 implies that a (x, k) C(N)(1 + x ) N (1 + k ) m for any N > 0. On the other hand, û(k) C(M)(1 + k ) M for any M > 0 since u(x) S and also û(k) S. Therefore, (10.35) a (x, k)û(k) C(N, M)(1 + x ) N (1 + k ) m M Therefore, (10.33) implies that (10.36) A u(x) C (N, M)(1 + x ) N (1 + k ) m M dk C (N, M)(1 + x ) N if we take M sufficiently large, so that m M < n. Finally, this inequality implies that [ A u(x) ] L 1 (IR n ) if we take N > n. Therefore, Lemma 5.1 implies that the Fourier transform F A u (k) is given by standard Fourier integral:  u(k) = e ikx A u(x)dx = e ikx( 1 (10.37) (2π) n e ik x a (x, k )û(k )dk ) dx. Applying here the Fubini theorem (it is possible by (10.35) with m M < n and N > n), we obtain  u(k) = 1 ( (2π) n e i(k k )x a (x, k )dx )û(k )dk = 1 (10.38) (2π) n â (k k, k )û(k )dk. Step iv) Now we substitute (10.38) into (10.39) and obtain that A u(x) s m = C (1 + k ) s m â (k k, k )û(k )dk L 2 (10.39) = C = C (1 + k ) s m (1 + k ) s â (k k, k )(1 + k ) s û(k )dk L 2 S(k, k )v(k )dk L 2,

30 30 CONTENTS where S(k, k ) = (1 + k )s m (1 + k ) s â (k k, k ) and v(k ) = (1 + k ) s û(k ). Lemma 10.9 The kernel S(k, k ) satisfies conditions (10.2) and (10.3) of Schur s lemma. Proof The crucial observation is that condition (9.9) with α = 0 implies the bound (10.40) â (k k, k ) C(N)(1 + k k ) N (1 + k ) m for any N > 0. This follows by arguments similar to the proof of Lemma 2.2. The bound implies that S(k, k ) C (1 + k )s m (1 + k ) s (1 + k k ) N (1 + k ) m (10.41) = C (1 + k )s m (1 + k ) s m(1 + k k ) N C(1 + k k ) N+ s m. by the Peetre inequality (10.18) with s m instead of s. Now conditions (10.2) and (10.3) follow if we take N sufficiently large so that N + s m < n. Finally, applying Schur s lemma in (10.39), we obtain that (10.42) A u(x) s m = S(k, k )v(k )dk L 2 C v(k) L 2 = C u(x) s, that proves bound of type (10.20) for operator (10.39).

31 11. COMPOSITION OF PSEUDODIFFERENTIAL OPERATORS Composition of pseudodifferential operators Here we consider composition of two pseudodifferential operators A 1 and A 2 classes A m 1 and A m 2 respectively. Theorem 11.1 Let the symbols a 1 (x, k) and a 2 (x, k) of pseudodifferential operators A 1 and A 2, belong to the classes S m 1 and S m 2 respectively. Then the operator A 1 A 2 belongs to the class A m 1+m 2, and its symbol a(x, k) admits asymptotic expansion (generalised Leibniz Formula) (11.1) a(x, k) γ 0 1 γ! (i k) γ a 1 (x, k) γ xa 2 (x, k). First let us discuss the meaning of asymptotic expansion (11.1): by definition, it means that for any N = 0, 1, 2,... the finite expansion holds (11.2) a(x, k) = N 1 γ 0 1 γ! (i k) γ a 1 (x, k) γ xa 2 (x, k) + R N (x, k), where the remainder R N (x, k) belongs to class S m 1+m 2 N. Next let us consider the properties of all terms in the asymptotic expansion. Definition 9.1 implies that i) the symbol (i k ) α a 1 (x, k) belongs to class S m 1 α (Exercise: Check this!), and ii) the symbol α xa 2 (x, k) belongs to class S m 2 (Exercise: Check this!). Therefore, their product (i k ) α a 1 (x, k) α xa 2 (x, k) belongs to class S m 1+m 2 α by Exercise Composition of differential operators I. First consider simplest example of differential operators A 1 = d dx and A 2 = a(x) in dimension n = 1. Their composition is (11.3) Au(x) := A 1 A 2 u(x) = d [ ] a(x)u(x) = a (x)u(x) + a(x)u (x). dx This means that A = A 1 A 2 = a(x) d dx + a (x), hence the symbol of the composition is (11.4) a(x, k) = a(x)( ik) + a (x). This formula coincides with (11.2) since the symbols a 1 (x, k) = ik and a 1 (x, k) = a(x), hence i) the term in (11.2) with γ = 0 is the product ( ik)a(x), ii) next term with γ = 1 is a(x), and iii) all terms with γ > 1 vanish. II. Second, let us consider general differential operators A 1 and A 2 in dimension n = 1: (11.5) A l u(x) = a lα (x) α u(x), l = 1, 2. α m l Their composition is also a differential operator (11.6) Au(x) := A 1 A 2 u(x) = α 1 m 1 α 2 m 2 [ ] a 1α1 (x) α 1 a 2α2 (x) α 2 u(x).