The Bernstein and Nikolsky inequalities for trigonometric polynomials

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1 he ernstein and Nikolsky ineualities for trigonometric polynomials Jordan ell Department of Mathematics, University of oronto January 28, Introduction Let = R/2πZ For a function f : C and τ, we define f τ : C by f τ (t) = f(t τ) For measurable f : C and 0 < r <, write For f, g L 1 (), write and for f L 1 (), write f r = (f g)(x) = 1 2π ˆf(k) = 1 2π ( 1/r 1 f(t) dt) r 2π f(t)g(x t)dt, x, f(t)e ikt dt, k Z his note works out proofs of some ineualities involving the support of ˆf for f L 1 () Let n be the set of trigonometric polynomials of degree n We define the Dirichlet kernel D n : C by D n (t) = e ijt, t j n It is straightforward to check that if n then D n = 1

2 2 ernstein s ineuality for trigonometric polynomials DeVore and Lorentz attribute the following ineuality to Szegö 1 heorem 1 If n and is real valued, then for all x, (x) 2 + n 2 (x) 2 n 2 2 Proof If = 0 the result is immediate Otherwise, take x, and for real c > 1 define P c (t) = (t + x)sgn (x) c, t P c n, and satisfies P c(0) = (x)sgn (x) c 0 and P c 1 c < 1 Since P c < 1, in particular P c (0) < 1 and so there is some α, α < π 2n, such that sin nα = P c(0) We define S n by S(t) = sin n(t + α) P c (t), t, which satisfies S(0) = sin nα P c (0) = 0 For k = n,, n, let t k = α +, for which we have (2k 1)π 2n ecause P c < 1, sin n(t k + α) = sin (2k 1)π 2 sgn S(t k ) = ( 1) k+1, = ( 1) k+1 so by the intermediate value theorem, for each k = n,, n 1 there is some c k (t k, t k+1 ) such that S(c k ) = 0 ecause t n t n = (2n 1)π 2n ( 2n 1)π 2n = 2π, it follows that if j k then c j and c k are distinct in It is a fact that a trigonometric polynomial of degree n has 2n distinct roots in, so if t (t k, t k+1 ) and S(t) = 0, then t = c k It is the case that t 1 = α + π 2n > 0 and t 0 = α π 2 < 0, so 0 (t 0, t 1 ) ut S(0) = 0, so c 0 = 0 Using S(t 1 ) = 1 > 0 and the fact that S has no zeros in (0, t 1 ) we get a contradiction from S (0) < 0, so S (0) 0 his gives 0 P c(0) = n cos nα S (0) n cos nα = n 1 sin 2 nα = n 1 P c (0) Ronald A DeVore and George G Lorentz, Constructive Approximation, p 97, heorem 2

3 hus P c(0) n 1 P c (0) 2, or n 2 P c (0) + P c(0) 2 n 2 ecause P c (0) 2 = (x)2 c 2 2, P c(0) 2 = (x) 2 c 2 2 we get n 2 (x) 2 + (x) 2 c 2 n 2 2 ecause this is true for all c > 1, n 2 (x) 2 + (x) 2 n 2 2, completing the proof Using the above we now prove ernstein s ineuality 2 heorem 2 (ernstein s ineuality) If n, then n Proof here is some x 0 such that (x 0 ) = Let α R be such that e iα (x 0 ) = Define S(x) = Re (e iα (x)) for x, which satisfies S (x) = Re (e iα (x)) and in particular S (x 0 ) = Re (e iα (x 0 )) = e iα (x 0 ) = ecause S n and S is real valued, heorem 1 yields A fortiori, S (x 0 ) 2 + n 2 S(x 0 ) 2 n 2 S 2 S (x 0 ) 2 n 2 S 2, giving, because S (x 0 ) = and S, proving the claim 24 2 n2 2, he following is a version of ernstein s ineuality 3 2 Ronald A DeVore and George G Lorentz, Constructive Approximation, p 98 3 Ronald A DeVore and George G Lorentz, Constructive Approximation, p 101, heorem 3

4 heorem 3 If n and A is a orel set, there is some x 0 such that (t) dt n A (t) dt A x 0 Proof Let A be a orel set with indicator function χ A Define Q : C by Q(x) = χ A (t) (t + x)sgn (t)dt, x, which we can write as Q(x) = χ A (t) (j)e ij(t+x) sgn (t)dt j = ( ) (j) χ A (t)e ijt sgn (t)dt e ijx, j showing that Q n Also, Q (x) = χ A (t) (t + x)sgn (t)dt, x Let x 0 with Q(x 0 ) = Q Applying heorem 2 we get Q n Q Using this gives Q (0) = χ A (t) (t)sgn (t)dt = χ A (t) (t) dt, χ A (t) (t) dt n Q = n Q(t 0 ) = n χ A (t) (t + x 0 )sgn (t)dt n χ A (t) (t + x 0 ) dt = n χ A x0 (t) (t) dt Applying the above with A = gives the following version of ernstein s ineuality, for the L 1 norm heorem 4 (L 1 ernstein s ineuality) If n, then 1 n 1 4

5 3 Nikolsky s ineuality for trigonometric polynomials DeVore and Lorentz attribute the following ineuality to Sergey Nikolsky 4 heorem 5 (Nikolsky s ineuality) If n and 0 < p, then for r 2 an integer, p (2nr + 1) 1 1 p Proof Let m = nr hen r m, so r D m = r, and using this and the Cauchy-Schwarz ineuality we have, for x, (x) r = 1 (t) r D m (x t) 2π 1 (t) r D m (x t) dt 2π r 2 1 (t) 2 Dm (x t) dt 2π r 2 /2 2 D m 2 = r 2 2 Dm l2 (Z) = 2m + 1 r 2 2 Hence thus r 2m + 1 r 2 2, (2m + 1) 1 hen, using p 1 p p, we have p (2m + 1) 1 1 p 1 p p = (2m + 1) 1 1 p 4 he complementary ernstein ineuality We define a homogeneous anach space to be a linear subspace of L 1 () with a norm f L 1 () f with which is a anach space, such that if f and τ then f τ and f τ = f, and such that if f then f τ f in as τ Ronald A DeVore and George G Lorentz, Constructive Approximation, p 102, heorem 5

6 Fejér s kernel is, for n 0, K n (t) = ( 1 j n + 1 j n One calculates that, for t 4πZ, ) e ijt = j Z χ n (j) K n (t) = 1 n + 1 ( sin n+1 2 t sin 1 2 t ( 1 j ) e ijt t n + 1 )2 ernstein s ineuality is a statement about functions whose Fourier transform is supported only on low freuencies he following is a statement about functions whose Fourier transform is supported only on high freuencies 5 In particular, for 1 p <, L p () is a homogeneous anach space, and so is C() with the supremum norm heorem 6 Let be a homogeneous anach space and let m be a positive integer Define C m as C m = m + 1 if m is even and C m = 12m if m is odd If f(t) = a j e ijt, t, j n is m times differentiable and f (m), then f and f C m n m f (m) Proof Suppose that m is even It is a fact that if a j, j Z, is an even seuence of nonnegative real numbers such that a j 0 as j and such that for each j > 0, a j 1 + a j+1 2a j 0, then there is a nonnegative function f L 1 () such that ˆf(j) = a j for all j Z 6 Define { j m j n a j = n m + (n j )(n m (n + 1) m ) j n 1 It is apparent that a j is even and tends to 0 as j For 1 j n 2, For j = n 1, a j 1 + a j+1 2a j = 0 a j 1 + a j+1 2a j = n m + (n (n 2))(n m (n + 1) m ) + n m 2 ( n m + (n (n 1))(n m (n + 1) m ) ) = 0 5 Yitzhak Katznelson, An Introduction to Harmonic Analysis, third ed, p 55, heorem 84 6 Yitzhak Katznelson, An Introduction to Harmonic Analysis, third ed, p 24, heorem 41 6

7 he function j j m is convex on {n, n + 1, }, as m 1, so for j n we have a j 1 + a j+1 2a j 0 herefore, there is some nonnegative φ m,n L 1 () such that φ m,n (j) = a j, j Z ecause φ m,n is nonnegative, and using n m (n + 1) m < m n n m, φ m,n 1 = φ m,n (0) = n m + n(n m (n + 1) m ) < (m + 1)n m Define dµ m,n (t) = 1 2π φ m,n(t)dt For j n, For j < n, since ˆf(j) = 0 we have so for all j Z, f (m) µ m,n (j) = f (m) (j) µ m,n (j) = (ij) m ˆf(j) φ m,n (j) = (ij) m ˆf(j) j m = i m ˆf(j) f (m) µ m,n (j) = (ij) m ˆf(j) φ m,n (j) = 0 = i m ˆf(j), f (m) µ m,n (j) = i m ˆf(j) his implies that f (m) µ m,n = i m f, which in particular tells us that f hen, f = i m f = f (m) µ m,n f (m) µ m,n M() f (m) = φ m,n 1 (m + 1)n m f (m) his shows what we want in the case that m is even, with C m = m + 1 Suppose that m is odd For l a positive integer, define ψ l : C by ψ l (t) = (e 2lit + 12 ) e3lit K l 1 (t), t here is a uniue l n such that n {2l n, 2l n + 1} For k 0 an integer, define Ψ n,k : C by Ψ n,k (t) = ψ ln2k(t), t Ψ n,k satisfies Ψ n,k K k 1 1 = K k 1 (t) dt = 3 2π 2 1 K k 1 (t)dt = 3 2π 2 7

8 On the one hand, for j 0, from the definition of ψ l we have Ψ n,k (j) = 0, hence Ψ n,k (j) = 0 On the other hand, for j n we assert that Ψ n,k (j) = 1 We define Φ n : C by Φ n (t) = (Ψ n,k φ 1,n2 k)(t), t We calculate the Fourier coefficients of Φ n For j n, Φ n (j) = Φ n,k (j) φ 1,n2 k(j) = 1 j Φ n,k (j) = 1 j As well, Φ n 1 Ψn,k φ 1,n2 k 1 Ψ n,k φ1,n2 1 k We now define dµ 1,n (t) = 1 2π (Φ n(t) Φ n ( t))dt, which satisfies for j n, 2(n2 k ) 1 = 6 n and hence ecause ˆf(j) = 0 for j < n, j Z, and therefore, hen µ 1,n (j) = Φ n (j) Φ n ( j) = 1 j f µ 1,n (j) = f (j) µ 1,n (j) = ij ˆf(j) 1 j = i ˆf(j) f µ 1,n (j) = 0 for j < n, it follows that for any f µ 1,n (j) = i ˆf(j), f µ 1,n = if f = if = f µ 1,n µ 1,n M() f 2 Φ n 1 f 12 n f hat is, with C 1 = 12 we have f 12n 1 f 8

9 For m = 2ν + 1, we define for which we have, for j n, µ m,n = µ 1,n µ 2ν,n, f (m) µ m,n (j) = (ij) m ˆf(j) µ1,n (j) µ 2ν,n (j) = (ij) m 1 ˆf(j) j j 2ν = i m ˆf(j) It follows that whence f (m) µ m,n = i m f, f = i m f = f (m) µ m,n f (m) µ m,n M() f (m) µ 1,n M() µ 2ν,n M() 12 f n (2ν + 1)n 2ν (m) = 12mn m f (m) hat is, with C m = 12m, we have f C m n m f (m), completing the proof 9