Bernstein s inequality and Nikolsky s inequality for R d

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1 Bernstein s inequality and Nikolsky s inequality for d Jordan Bell jordan.bell@gmail.com Department of athematics University of Toronto February 6 25 Complex Borel measures and the Fourier transform Let ( d ) = rca( d ) be the set of complex Borel measures on d. This is a Banach algebra with the total variation norm with convolution as multiplication; for µ ( d ) we denote by µ the total variation of µ which itself belongs to ( d ) and the total variation norm of µ is µ = µ ( d ). For µ ( d ) it is a fact that the union O of all open sets U d such that µ (U) = itself satisfies µ (O) =. We define supp µ = d \ O called the support of µ. For µ ( d ) we define ˆµ : d C by ˆµ(ξ) = e 2πiξ x dµ(x) ξ d. d It is a fact that ˆµ belongs to C u () the collection of bounded uniformly continuous functions d C. For ξ d ˆµ(ξ) e 2πiξ x d µ (x) = µ ( d ) = µ. () d Let m d be Lebesgue measure on d. For f L ( d ) let Λ f = fm d which belongs to ( d ). We define ˆf : d C by ˆf(ξ) = Λ f (ξ) = e 2πiξ x dλ f (x) = d f(x)e 2πiξ x dm d (x) d ξ d. The following theorem establishes properties of the Fourier transform of a complex Borel measure with compact support. Thomas H. Wolff Lectures on Harmonic Analysis p. 3 Proposition.3.

2 Theorem. If µ ( d ) and supp µ is compact then ˆµ C ( d ) and for any multi-index α D α ˆµ = F (( 2πix) α µ). For > if supp µ B( ) then D α ˆµ (2π) α µ. Proof. For j =... d let e j be the jth coordinate vector in d with length. Let ξ d and define (h) = ˆµ(ξ + he j) ˆµ(ξ) h. h We can write this as e 2πihxj (h) = e 2πiξ x dµ(x). h d For any x d e 2πihxj h = e 2πihxj h 2πihx j h = 2π x j. Because µ has compact support 2π x j L (µ). Furthermore for each x d we have e 2πihxj 2πix j h. h Therefore the dominated convergence theorem tells us that lim (h) = 2πix j e 2πiξ x dµ(x). h d On the other hand for α k = for k = j and α k = otherwise so (D α ˆµ)(ξ) = lim h (h) (D α ˆµ)(ξ) = ( 2πix) α e 2πiξ x dµ(x) = F (( 2πix) α µ)(ξ) d and in particular ˆµ C ( d ). (The Fourier transform of a regular complex Borel measure on a locally compact abelian group is bounded and uniformly continuous. 2 ) Because µ has compact support so does ( 2πix) α µ hence we can play the above game with ( 2πix) α µ and by induction it follows that for any α D α ˆµ = F (( 2πix) α µ) 2 Walter udin Fourier Analysis on Groups p. 5 Theorem

3 and in particular ˆµ C ( d ). Suppose that supp µ B( ). The total variation of the complex measure ( 2πix) α µ is the positive measure hence Then using () (2π) α x α x d α d µ ( 2πix) α µ = (2π) α x α x d α d d µ (x) d = (2π) α x α x d α d d µ (x) (2π) α = (2π) α B() B() B() = (2π) α d d µ (x) = (2π) α µ. α α d d µ (x) d µ (x) F (( 2πix) α µ) ( 2πix) α µ (2π) α µ. But we have already established that D α ˆµ = F (( 2πix) α µ) which with the above inequality completes the proof. 2 Test functions For an open subset Ω of d we denote by D(Ω) the set of those φ C (Ω) such that supp φ is a compact set. Elements of D(Ω) are called test functions. It is a fact that there is a test function φ satisfying: (i) φ(x) = for x (ii) φ(x) = for x 2 (iii) φ and (iv) φ is radial. We write for k = 2... φ k (x) = φ(k x) x d. For any multi-index α hence (D α φ k )(x) = k α (D α φ)(k x) x d D α φ k = k α D α φ. (2) We use the following lemma to prove the theorem that comes after it. 3 3 Thomas H. Wolff Lectures on Harmonic Analysis p. 4 Lemma.5. 3

4 Lemma 2. Suppose that f C N ( d ) and D α f L ( d ) for each α N. Then for each α N D α (φ k f) D α f in L ( d ) as k. Proof. Let α N. In the case α = φ k f f = φ k (x)f(x) f(x) dx d = φ k (x)f(x) f(x) dx x k x k f(x) dx. Because f L ( d ) this tends to as k. Suppose that α >. The Leibniz rule tells us that with c β = ( α β) we have for each k D α (φ k f) = φ k D α f + c β D α β fd β φ k. For C = max β c β D α (φ k f) φ k D α f <β α C <β α cβ D α β fd β φ k <β α D β φ k D α β f. Let C 2 = max <β α D β φ. By (2) for < β α we have Thus D β φ k = k β D β φ C 2 k β C 2 k. D α (φ k f) φ k D α f C C 2 k <β α D α β f which tends to as k. For any k φ k D α f D α f = φ k (x)(d α f)(x) (D α f)(x) dx d = φ k (x)(d α f)(x) (D α f)(x) dx x k x k (D α f)(x) dx and because D α f L ( d ) this tends to as k. But D α (φ k f) D α f D α (φ k f) φ k D α f + φ k D α f D α f which completes the proof. 4

5 Now we calculate the Fourier transform of the derivative of a function and show that the smoother a function is the faster its Fourier transform decays. 4 Theorem 3. If f C N ( d ) and D α f L ( d ) for each α N then for each α N D α f(ξ) = (2πiξ) α ˆf(ξ) ξ d. (3) There is a constant C = C(f N) such that ˆf(ξ) C( + ξ ) N ξ d. Proof. If g Cc ( d ) then for any j d integrating by parts ( j g)(x)e 2πiξ x dx = 2πiξ j g(x)e 2πiξ x dx. d d It follows by induction that if g C N c ( d ) then for each α N D α g(ξ) = (2πiξ) α ĝ(ξ) ξ d. Let α N. For k = 2... let f k = φ k f. For each k we have f k C N ( d ) hence D α f k (ξ) = (2πiξ) α fk (ξ) ξ d. On the one hand D α f k D α f = F (D α f k D α f) D α f k D α f and Lemma 2 tells us that this tends to as k. On the other hand for ξ d D α f k (ξ) (2πiξ) α ˆf(ξ) = (2πiξ) α fk (ξ) (2πiξ) α ˆf(ξ) = (2πiξ) α F (f k f)(ξ) (2πiξ) α f k f which by Lemma 2 tends to as k. Therefore for ξ d D α f(ξ) (2πiξ) α ˆf(ξ) D α f k D α f + D α f k (ξ) (2πiξ) α ˆf(ξ) and because the right-hand side tends to as k we get D α f(ξ) = (2πiξ) α ˆf(ξ). If y S d then there is at least one j d with y j from which we get y β >. β =N 4 Thomas H. Wolff Lectures on Harmonic Analysis p. 4 Proposition.4. 5

6 The function y β =N yβ is continuous S d so there is some C N > such that y β y S d. C N β =N For nonzero x d write x = x y with which β =N xβ = x N β =N yβ. Therefore x N C N x β x d. β =N For α N because the Fourier transform of an element of L belongs to C we have by (3) that ξ ξ α ˆf(ξ) belongs to C ( d ) and in particular is bounded. Then for ξ d ξ N ˆf(ξ) C N ξ β ˆf(ξ) β =N = C N C N = C. On the one hand for ξ we have hence giving β =N β =N + ξ 2 ξ ξ β ˆf(ξ) ξ β ˆf(ξ) ( ) N + ξ ξ N = 2 N ( + ξ ) N 2 On the other hand for ξ we have ˆf(ξ) C ξ N C 2 N ( + ξ ) N. + ξ 2 and so Thus for ˆf(ξ) ˆf 2 N 2 N ˆf 2 N ( + ξ ) N. { } C = max 2 N C 2 N ˆf we have completing the proof. ˆf(ξ) C( + ξ ) N ξ d 6

7 3 Bernstein s inequality for L 2 For a Borel measurable function f : d C let O be the union of those open subsets U of d such that f(x) = for almost all x U. In other words O is the largest open set on which f = almost everywhere. The essential support of f is the set ess supp f = d \ O. The following is Bernstein s inequality for L 2 ( d ). 5 Theorem 4. If f L 2 ( d ) > and ess supp ˆf B( ) (4) then there is some f C ( d ) such that f(x) = f (x) for almost all x d and for any multi-index α D α f 2 (2π) α f 2. Proof. Let χ be the indicator function for B( ). By (4) the Cauchy-Schwarz inequality and the Parseval identity ˆf = χ ˆf χ 2 ˆf = m d (B( )) /2 f 2 < 2 so ˆf L ( d ). The Plancherel theorem 6 tells us that if g L 2 ( d ) and ĝ L ( d ) then g(x) = ĝ(ξ)e 2πix ξ dξ d for almost all x d. Thus for f : d C defined by f (x) = ˆf(ξ)e 2πix ξ dξ = F ( ˆf)( x) x d d we have f(x) = f (x) for almost all x d. Because f = f almost everywhere f = ˆf. Applying Theorem to dµ(ξ) = f ( ξ)dξ we have f C ( d ) and for any multi-index α D α f = F (( 2πiξ) α ˆf( ξ)). 5 Thomas H. Wolff Lectures on Harmonic Analysis p. 3 Proposition Walter udin eal and Complex Analysis third ed. p. 87 Theorem

8 By Parseval s identity D α f 2 = ( 2πiξ) α 2 ˆf( ξ) = (2πiξ) α χ (ξ) ˆf(ξ) 2 (2πiξ) α χ (ξ) ˆf (2π) α ˆf 2 = (2π) α f 2 proving the claim. 4 Nikolsky s inequality Nikolsky s inequality tells us that if the Fourier transform of a function is supported on a ball centered at the origin then for p q the L q norm of the function is bounded above in terms of its L p norm. 7 Theorem 5. There is a constant C d such that if f S ( d ) > and p q then Proof. Let g = f i.e. supp ˆf B( ) f q C d d( p q ) f p. g(x) = d f( x) x d. Then for ξ d ĝ(ξ) = g(x)e 2πiξ x dx = d d f( x)e 2πiξ x dx = ˆf(ξ) d showing that supp ĝ = supp ˆf B( ). Let χ D( d ) with χ(ξ) = for ξ with which ĝ = χĝ. Then g = (F χ) g and using Young s inequality with + q = r + p g q F χ r g q = ˆχ r g q. (5) 7 Camil uscalu and Wilhelm Schlag Classical and ultilinear Harmonic Analysis volume I p. 83 Lemma

9 oreover ( g a = d f( x) a dx ( d = da+d f(y) a dy d = d( a ) f a ) /a ) /a so (5) tells us d( ) q f q ˆχ r d( ) p f p i.e. f q ˆχ r d( p q ) f p. Now r = + q p so r because p q namely r. By the log-convexity of L r norms for r = θ we have Thus with we have proved the claim. ˆχ r ˆχ θ ˆχ θ. C d = max{ ˆχ ˆχ } 5 The Dirichlet kernel and Fejér kernel for The function D C () defined by D (x) = sin 2πx x πx and D () = 2 is called the Dirichlet kernel. Let χ function for the set [ ]. We have for x χ (x) = χ (ξ)e 2πixξ dξ = = e 2πixξ 2πix e 2πixξ dξ = e 2πix 2πix + e2πix 2πix = e 2πix e 2πix πx 2i sin 2πx =. πx be the indicator 9

10 For x = χ () = 2 = D (). Thus D = χ. For f L () and > we define (S f)(x) = ˆf(ξ)e 2πiξx dξ x. It is straightforward to check that sin 2πt (S f)(x) = f(x t)dt = (D f)(x) x. πt For f L () > and x (S m f)(x)dm = = = = = = = ( m m ( m ( f(y) m ( We define the Fejér kernel K C () by K (x) = ) ˆf(ξ)e 2πiξx dξ dm f(y)e 2πiξy dy ( m ) e 2πiξ(y x) dξ m ) ( f(y) D m (y x)dm ) e 2πiξx dξ dm ) ) dy dm ( ) sin 2πm(y x) f(y) dm dy π(y x) ( ) cos 2πm(y x) f(y) 2π 2 (y x) 2 dy ( f(y) cos 2π(y x) 2π 2 (y x) 2 2π 2 (y x) 2 cos 2πx 2π 2 x 2 x and K () =. Thus because K is an even function (S m f)(x)dm = (K f)(x). One proves that K is an approximate identity: K K (x)dx = dy ) dy.

11 and for any δ > lim K (x)dx =. x >δ The fact that K is an approximate identity implies that for any f L () K f f in L () as. We shall use the Fejér kernel to prove Bernstein s inequality for. 8 Theorem 6. If µ () > and supp µ [ ] then ˆµ 4π ˆµ. Proof. For x let dµ x (t) = e 2πixt dµ(t). µ x has the same support has µ and µ x (x) = e 2πixt dµ x (t) = e 2πixt e 2πixt dµ(t) = ˆµ(x + x ). It follows that to prove the claim it suffices to prove that ˆµ () 4π ˆµ. Write f = ˆµ C u (). Define C c () by { t t < (t) = t. t We calculate for x so (t)e 2πixt dt = e 2πix ( + e 2πix ) 2 4π 2 x 2 (sin πx)2 = π 2 x 2 cos 2πx = 2π 2 x 2. (x) = K (x). Then for t [ ] (e 2πiξ e 2πiξ )K (ξ)e 2πiξt dξ = K (t ) K (t + ) = ( t + ) ( t ) = t. 8 ark A. Pinsky Introduction to Fourier Analysis and Wavelets p. 22 Theorem

12 On the one hand the integral of the left-hand side with respect to µ is (e 2πiξ e 2πiξ )K (ξ)e 2πiξt dξdµ(t) = (e 2πiξ e 2πiξ )K (ξ)f(ξ)dξ. On the other hand the integral of the right-hand side with respect to µ is t dµ(t) = 2πitdµ(t) 2πi = 2πi F (( 2πit)µ)() = 2πi f (). Hence giving proving the claim. 2πi f () = (e 2πiξ e 2πiξ )K (ξ)f(ξ)dξ f () 4π f K = 4π f 2