8 Singular Integral Operators and L p -Regularity Theory
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1 8 Singular Integral Operators and L p -Regularity Theory 8. Motivation See hand-written notes! 8.2 Mikhlin Multiplier Theorem Recall that the Fourier transformation F and the inverse Fourier transformation F are dened by F[f](ξ) := ˆf(ξ) := e ix ξ f(x) dx, R n F [f](x) := e ix ξ f(ξ) dξ, (2π) n R n where f L (R n ). Moreover, recall that by Plancharel's theorem F : L 2 (R n ) L 2 (R n ) is an isomorphism with inverse F and that f(x)g(x) dx = ˆf(ξ)ĝ(ξ) dξ (8.) R (2π) n n R n for all f, g L 2 (R n ). THEOREM 8. (Mikhlin Multiplier Theorem) Let N = n + 2, n N. Moreover, let m: R n \ {0} C be an N-times dierentiable function such that α ξ m(ξ) C ξ α (8.2) for all ξ 0 and α N and let m(d x )f = F [m(ξ) ˆf(ξ) ] for all f S(R n ). Then m(d x ) extends to a bounded linear operator m(d x ): L p (R n ) L p (R n ) for all < p <. (8.3)
2 We note that (8.) implies that m(d x )f L 2 (R n ) m L (R n ) f L 2 (R n ) for all f L 2 (R n ). Hence m(d x ) L(L 2 (R n )). We will prove the statement in the case < p 2 with the aid of: THEOREM 8.2 (Marcinkiewisz interpolation theorem) Suppose that < r. Let T be a sub-additive mapping from L (R n ) + L r (R n ) to the space of measurable functions on R n, which is of weak type (, ) and (r, r), i.e, λ(t; T f) := {x : T f(x) > t} C q f q L q (R n ) t q (8.4) for all t > 0, f L q (R n ) and for q = and q = r if r < and T f C f if r =. Then T f p C p f p for all f L p (R n ) for all < p < r, where C p depends only on C, C r, p, and r. Remark 8.3 Note that, if g L q (R n ), q <, then g(x) q λ(t; g) = {x : g(x) > t} {x: g(x) >t} t q g q t q. (8.5) Hence, if T L(L q (R n )), then (8.4) holds. Hence (8.4) is a weaker condition than T L(L q (R n )). In order to prove Theorem 8., the main step is to show (8.4) for q =. Once this is proved, m(d x ) L(L p (R n )) for all < p 2 by the Marcinkiewicz interpolation theorem and the case 2 < p < will follow by duality. 8.3 Proof of the Mikhlin Multiplier Theorem For simplicity we will rst prove the theorem in the case H 0 = H = C and therefore m: R n \ {0} L(C) = C. How to modify the proof for the general case will be discussed below. 2
3 Note that (8.) implies that m(d x )f L 2 (R n ) m L (R n ) f L 2 (R n ) for all f L 2 (R n ). Hence m(d x ) L(L 2 (R n )). In order to prove Theorem 8., the main step is to show that m(d x ) L(L (R n ), L weak (Rn )), which is the content of the next lemma. Once this is proved, m(d x ) L(L p (R n )) for all < p 2 by the Marcinkiewicz interpolation theorem and the case 2 < p < will follow by duality. Lemma 8.4 Let M m(d x ) be as in Theorem 8.. Then {x R n : Mf(x) > t} C f L (R n ) t for all t > 0, f L (R n ) (8.6) for some C > 0 independent of t > 0. In order to prove Lemma 8.4, an essential ingredient will be the following kernel estimate: Proposition 8.5 Let m: R n \ {0} C be as in Theorem 8.. Then there is a locally integrable, continuously dierentiable function k : R n \ {0} C such that m(d x )f(x) = k(x y)f(y) dy for all x supp f, (8.7) R n for all f L 2 (R n ) with compact support which satises for all z 0. k(z) C z n k(z) C z n (8.8) We postpone the proof of Proposition 8.5 to the end of this section. Corollary 8.6 Let Q be a cube and a L (R n ) with supp a Q and a(x) dx = 0. Then there is a constant C > 0 independent of a and Q Q such that m(d x )a(x) dx C a, R n \ Q where Q = Q 2 n denotes the cube with same center as Q and 2 n times the side-length of Q. 3
4 Proof: Let x 0 denote the center of Q. Then x Q = Q 2 n and y Q implies x x 0 > 2 y x 0. Therefore m(d x )a(x) dx = k a(x) dx R n \ Q R n \ Q k(x y) k(x x 0 ) dx a(y) dy Q x x 0 >2 y x 0 C a(y) dy, Q since a(x) dx = 0 provided that Q k(x y) k(x) dx C for all y R n. (8.9) x >2 y In order to prove the latter estimate, we use If x > 2 y, then k(x y) k(x) = 0 y k(x ty) dt. k(x y) k(x) sup x k(x ty) y C x n y t [0,] since x ty x for all t [0, ]. Hence 2 k(x y) k(x) dx C x >2 y uniformly in y 0. x >2 y x n dx y C Proposition 8.7 Let f L (R n ) be continuous and let t > 0. Then there are cubes Q j, j N N 0, with disjoint interior and parallel to the axis such that. t < f(y) dy 2 n t for all j N. (8.0) Q j Q j 2. f(x) t for (almost) all x j N Q j. 4
5 Proof: In the following D k, k Z, denotes the set of all dyadic cubes with side length 2 k meaning the collection of all (closed) cubes Q with corners on neighboring points of the lattice 2 k Z n. Moreover, we set D = k Z D k. Let C t for given t > 0 be the set of all Q D satisfying the condition t < Q Q f(x) dx and let C t be the subset of all Q C t that are maximal with respect to inclusion in C t. Every Q C t is contained in some Q C t since Q t f for all Q C t. Next, if Q C t D k and Q Q D k, then by the maximality of Q we have Q C t, i.e., Q Q f(x) dx t. Moreover, since Q = 2 n Q, we get t < f(x) dx 2n f(x) dx 2 n t for all Q C Q Q Q t. Q Hence C t = {Q j : j N}, where Q j, j N N 0, are non-overlapping and (8.0) is satised for all j N. Now let F := R n \ j N Q j. If x F, then f(y) dy t for every Q Q Q D such that x Q. Hence, choosing a sequence of cubes Q k D k with x Q k, we obtain f(x) = lim k f(y) dy Q k t for all x F. Q k Remark 8.8 The latter proposition holds for a general f L (R n ) if one applies Lebesgue's dierentiation theorem in the last step of the proof. Proof of Lemma 8.4: First of all, since C0 (R n ) is dense in L (R n ), it is enough to consider f C0 (R n ). Moreover, let t > 0 be xed and let Q j, j N N 0 be the cubes due to Proposition 8.7, Ω = j N Q j and F = R n \ Ω. We dene g, b L (R n ) L 2 (R n ) by { f(x) if x F g(x) = Q j Q j f(y) dy if x Q j 5
6 and b(x) = f(x) g(x). Note that this implies Then. g(x) 2 n t almost every in R n, 2. b(x) = 0 for every x F and Q j b(x) dx = 0 for each j N. {x : Mf(x) > t} {x : Mg(x) > t/2} + {x : Mb(x) > t/2} and it is sucient to estimate each term separately. In order to estimate Mg, we use that g(x) 2 n t for almost every x R n, f(x) = g(x) for x F, t Ω f, and that M L(L 2 (R n )). More precisely, {x : Mg(x) > t/2} 4 Mg(x) 2 dx 4 t 2 t m 2 g(x) 2 dx ( ) C m t 2 t f(x) dx + t 2 Ω Ct f In order to estimate T b, we apply Corollary 8.6 to b j (x) := b(x)χ Qj (x) and conclude Mb j (x) dx C b j 2C R n \ Q j f(x) dx Q j where Q j = Q 2 n j. On the other hand, since b L 2 (R n ), j N b j and therefore j N Mb j converge in L 2 (R n ) to b and T b, respectively. Hence F Mb(x) j N Mb j (x) almost everywhere and Mb(x) dx 2C f(x) dx 2C f, R n \ Ω j N Q j where Ω = Q j N j. Finally, {x : Mb(x) > t/2} Ω + 2 t 6 R n \ Ω Mb(x) dx C t f,
7 where we have used that Ω Q j (2 n) n Q j C t j N j N j N Q j f(x) dx C t f This nishes the proof of (8.6). Proof of Theorem 8.: Since m(d x ) L(L 2 (R n )) m and because of (8.6), we can apply the Marcinkiewicz interpolation theorem to conclude m(d x ) L(L p (R n )) for all < p 2. The statement for 2 < p < follows by duality since m(d x )f(x)g(x) dx = m(ξ) R (2π) ˆf(ξ)ĝ(ξ) dξ n n R n = ˆf(ξ)m(ξ)ĝ(ξ) dξ (2π) n R n = f(x)m(d x )g(x) dx R n for all f, g C 0 (R n ) due to (8.), where m(ξ) = m(ξ) satises the same estimates as m(ξ). Therefore m(d x ) = m(d x ) L(L p (R n )) if 2 < p < by the rst part. It remains to prove Proposition 8.5. To this end we will use the so-called Littlewood-Paley partition of unity, which is nowadays a standard tool in the theory of function spaces and harmonic analysis. It is frequently used to analyze mapping properties of certain operators. Usually a Littlewood-Paley or dyadic partition of unity on R n \ {0} is a decomposition of unity ϕ j (ξ), j Z of R n \ {0} such that supp ϕ j (ξ) {ξ R n : c2 j ξ C2 j } for all j Z. Here c, C > 0 are some suitable xed numbers often chosen to be c = 2, C = 2, which we will assume in the following. Such a partition can be easily constructed by choosing some non-negative ψ C 0 (R n ) such that ψ(ξ) > 0 if and only if 2 < ξ < 2. Then dening ψ j(ξ) := ψ(2 j ξ), j Z we have Φ(ξ) = j Z ψ j (ξ) > 0 for all ξ 0, 7
8 where we note that for each ξ 0 the sum above contains at most two non-vanishing terms. Hence ϕ j (ξ) = Φ(ξ) ψ j (ξ) denes a decomposition of unity with the desired properties. Moreover, in this case ϕ j (ξ) = ϕ 0 (2 j ξ) since Φ(2 j ξ) = Φ(ξ), which implies that α ξ ϕ j (ξ) C α ξ ϕ 0 L (R n )2 α j (8.) for all α N n 0 and j Z. The idea of the proof of Proposition 8.5 is to decompose m(ξ) = j Z m j (ξ), ξ 0, where m j (ξ) = ϕ j (ξ)m(ξ). Then ξ α m j (ξ) ( ) α α β ξ ϕ j (ξ) β ξ β m(ξ) C2 j α (8.2) 0 β α because of (8.2), (8.), and since 2 j ξ 2 j+ on supp ϕ j. For each part m j (ξ), we have m j (D x )f = F [m j (ξ) ˆf(ξ) ] = k j (x y)f(y) dy, R n where k j (x) = F [m j ](x) C 0 (R n ) since m j (ξ) L (R n ). Here we have used that Hence formally F[f g](ξ) = ˆf(ξ)ĝ(ξ) for all ξ R n, f, g L (R n ). m(d x )f = j Z R n k j (x y)f(y) dy, where it remains to show that the sum on the right-hand side converges for x supp f and that k(z) = j Z k j (z), z 0, 8
9 converges to a function satisfying (8.8). To this end, we need some suitable uniform estimates of k j (z). These are a consequence of the following simple lemma: Lemma 8.9 Let N N 0 and let g : R n C be an N-times dierentiable function, N N 0, with compact support. Then F [g](x) C N supp g x N sup β ξ g L (R n ) (8.3) β =N uniformly in x 0 and g, where C N depends only on n, N. Proof: Let β N n 0 with β = N. Then and therefore ( ix) β F [g] = F [ β ξ g] x β F [g](x) C β ξ g L (R n ) C supp g sup β ξ g L (R n ). β =N Since β N n 0 with β = N was arbitrary, x N F [g](x) C N supp g sup β ξ g L (R n ) β =N for all x R n, which completes the proof. Corollary 8.0 Let m be as in the assumptions of Theorem 8. and let m j (ξ) = m(ξ)ϕ j (ξ), j Z, where ϕ j, j Z, is the dyadic decomposition of unity of R n \ {0} as above. Then k j (x) := F ξ x [m j] satises α z k j (z) C α,n,m 2 j(n+ α M) z M (8.4) for all z 0, M = 0,..., n+2, α N n 0 for some constant C α,n,m independent of j Z, z 0. Proof: First of all, α z k j (z) = F [(iξ) α m j (ξ)]. 9
10 Using (8.2) and 2 j ξ 2 j+ on supp ϕ j, one easily veries β ξ ((iξ)α m j (ξ)) C α,β 2 j( α β ) for all β n + 2. Hence, applying Lemma 8.9 to (iξ) α m j with N = M = 0,..., n + 2, we obtain α z k j (z) C n,m supp ϕ j 2 j( α M) z M C n,m2 j(n+ α M) z M, which nishes the proof. Proof of Proposition 8.5: Firstly, we will show that j Z α z k j (z), α converges absolutely and uniformly on every compact subset of R n \ {0} to a function α z k(z) satisfying (8.8). The main idea of the proof is to split for given z 0 the sum j Z k j(z) into the two parts 2 j z α z k j (z) and 2 j > z α z k j (z) and to show convergence and the estimate (8.8) separately. For the rst sum we use (8.4) with α and M = 0. Then z α k j (z) C 2 j(n+ α ) C z n α j ld z j ld z where ld denotes the logarithm with respect to basis 2. For the second sum we apply (8.4) with α and M = n + α + and obtain z α k j (z) 2 2 j z n α C z n α. ld z <j ld z <j Hence j Z α z k j (z) converges absolutely and uniformly on every closed subset of R n \ {0} to a function k(z) that satises (8.8) for all α. Finally, it remains to show that k(z) satises (8.7). First of all, m(ξ) ˆf(ξ) = j Z m j (ξ) ˆf(ξ), f C 0 (R n ) since j Z ϕ j(ξ) is locally nite. Moreover, since m j (ξ) is uniformly bounded w.r.t. ξ and ˆf(ξ) L 2 (R n ), the sum on the right-hand side converges in 0
11 L 2 (R n ) to the left-hand side by Lebesgue's theorem on dominated convergence. Hence m(d x )f = lim m j (D x )f = lim k j (x y)f(y) dy, k k j N R n k j N k in L 2 (R n ) and almost everywhere for every f C 0 (R n ) and some N k as k since Fourier transformation is bounded operator from L 2 (R n ) to L 2 (R n ). Therefore it only remains to interchange the summation and integration in last term above provided that x supp f. But this can be done since j Z k j(z) converges absolutely and uniformly on every closed subset of R n \ {0} as shown above. Hence (8.7) follows. 8.4 Proof of the Marcinkiewicz Interpolation Theorem Proof of Theorem 8.2: First we consider the case r <. Let f L p (R n ) and consider the distribution function λ(t; T f), t > 0, dened as above. For given t > 0 we dene f = f + f 2 by { f(x) if f(x) > t, f (x) = 0 else. Then f L (R n ) and f 2 L r (R n ) since f (x) dx = f (x) p f (x) p dx C p (t) f p p and similarly f 2 (x) r dx C r,p (t) f p p. Now, since T f(x) T f (x) + T f 2 (x), we have Therefore {x : T f(x) > t} {x : T f (x) > t/2} {x : T f 2 (x) > t/2}. λ(t; T f) λ(t/2; T f ) + λ(t/2; T f 2 ) C f (x) dx + Cr r t/2 (t/2) r = 2C t f(x) >t f(x) dx + 2r C r r t r f 2 (x) r dx f(x) r dx f(x) t
12 where we have used the weak type (, ) and (r, r) estimate and (8.4). Next we use that T f(x) p dx = p t p λ(t; T f) dt, cf. e.g. [, Theorem 8.6]. We combine this with the estimate before. To this end we calculate ( ) f t p t f dx dt = f t p 2 dt dx 0 f >t R n 0 = f f p dx p R n since p > and similarly ( ) t p t r f r dx dt = f r t p r dt dx 0 f t R n f = f r f p r dx r p R n since p < r. Altogether T f p C p f p for all f L p (R n ). Finally, if r =, we assume for simplicity that C =. Otherwise replace T by C T. Then we use the same splitting of f as before, but cut at height t/2 instead of t. Hence T f 2 (x) t since T 2 L(L (R n )). Therefore {x : T f(x) > t} {x : T f (x) > t/2} and λ(t, T f) λ(t/2, T f ). The rest of the proof in this case is done as before having only the rst term Applications See handwritten notes! References [] W. Rudin. Real and complex analysis. McGraw-Hill Book Co., New York, third edition,
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