Inequalities for trigonometric polynomials

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(a)

Prove that for any trigonometric polynomial ∑∣k∣⩽nckeikx\sum_{\left|k\right| \leqslant n} c_{k} e^{i k x} one can choose numbers εk=±1(∣k∣⩽n)\varepsilon_{k}= \pm 1 \quad (\left|k\right| \leqslant n) such that

∫−ππ∣∑∣k∣⩽nεkckeikx∣dx⩾c(∑∣k∣⩽n∣ck∣2)1/2, \int _{-\pi }^{\pi }\left|\sum _{\left|k\right| \leqslant n} \varepsilon _{k} c_{k} e^{i k x}\right| d x \geqslant c\left(\sum _{\left|k\right| \leqslant n}\left|c_{k}\right|^{2}\right)^{1 / 2},

where c>0c>0 is an absolute constant.

(b)

Consider the trigonometric polynomial Tn(φ)=∑k=1nakeikφT_{n}(\varphi )=\sum_{k=1}^{n} a_{k} e^{i k \varphi }, whose coefficients a1,…,ana_{1}, \ldots , a_{n} form a permutation of the numbers 1,2,…,n1,2, \ldots , n. Prove that the trivial estimate

∥Tn∥1=∫−ππ∣Tn(φ)∣dφ⩽2π∥Tn∥2=2π12+…+n2⩽const⁡n3/2, \left\| T_{n}\right\| _{1}=\int _{-\pi }^{\pi }\left|T_{n}(\varphi )\right| d \varphi \leqslant \sqrt{2 \pi }\left\| T_{n}\right\| _{2}=2 \pi \sqrt{1^{2}+\ldots +n^{2}} \leqslant \operatorname {const} n^{3 / 2},

cannot, in general, be improved in order: for every n∈Nn \in \mathbb {N} there exists a permutation a1,…,ana_{1}, \ldots , a_{n} of the numbers 1,…,n1, \ldots , n such that ∥Tn∥1⩾cn3/2(c>0\left\| T_{n}\right\|_{1} \geqslant c n^{3 / 2}(c>0 and independent of n)n).

Prove that for any trigonometric polynomial ∑∣k∣⩽nckeikx(n⩾2)\sum_{\left|k\right| \leqslant n} c_{k} e^{i k x}(n \geqslant 2) one can choose numbers εk=±1(∣k∣⩽n)\varepsilon_{k}= \pm 1(\left|k\right| \leqslant n) such that

sup⁡x∈R∣∑∣k∣⩽nεkckeikx∣⩽10ln⁡n(∑∣k∣⩽n∣ck∣2)1/2. \sup _{x \in \mathbb {R}}\left|\sum _{\left|k\right| \leqslant n} \varepsilon _{k} c_{k} e^{i k x}\right| \leqslant 10 \sqrt{\ln n}\left(\sum _{\left|k\right| \leqslant n}\left|c_{k}\right|^{2}\right)^{1 / 2} .
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The Cesàro-Fejér sum of a function f∈L1(−π,π)f \in \mathscr {L}^{1}(-\pi , \pi ) is the trigonometric polynomial

σn(f,x)=σn(x)=S0(x)+S1(x)+…+Sn−1(x)n(x∈R), \sigma _{n}(f, x)=\sigma _{n}(x)=\frac{S_{0}(x)+S_{1}(x)+\ldots +S_{n-1}(x)}{n} \quad (x \in \mathbb {R}),

where S0,S1,…,SnS_{0}, S_{1}, \ldots , S_{n} are the partial sums of the Fourier series of ff. Cesàro-Fejér sums can be represented as the convolution of the function ff and the Fejér kernel Φn\Phi_{n}:

σn(f,x)=∫−ππΦn(x−y)f(y)dy, where Φn(x)=12πnsin⁡2(nx/2)sin⁡2(x/2); \sigma _{n}(f, x)=\int _{-\pi }^{\pi } \Phi _{n}(x-y) f(y) d y, \quad \text{ where } \quad \Phi _{n}(x)=\frac{1}{2 \pi n} \frac{\sin ^{2}(n x / 2)}{\sin ^{2}(x / 2)};

moreover, σn(f,x)=∑∣k∣<n(1−∣k∣n)f^(k)eikx\sigma_{n}(f, x)=\sum_{\left|k\right|<n}\left(1-\frac{\left|k\right|}{n}\right) \widehat{f}(k) e^{i k x}.

Let TT be a trigonometric polynomial of order at most nn. Prove that

T′=−2nT∗Ψn, where Ψn(x)=Φn(x)sin⁡nx. T^{\prime }=-2 n T * \Psi _{n}, \quad \text{ where } \Psi _{n}(x)=\Phi _{n}(x) \sin n x .

Deduce from this the weakened Bernstein inequality:

max⁡x∣T′(x)∣⩽2nmax⁡x∣T(x)∣ \max _{x}\left|T^{\prime }(x)\right| \leqslant 2 n \max _{x}\left|T(x)\right|

and analogous estimates for LpL^{p}-norms (regarding the Bernstein inequality see problem I.2.22).

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Consider two sequences of algebraic polynomials {Pn}n⩾0\left\{ P_{n}\right\}_{n \geqslant 0} and {Qn}n⩾0\left\{ Q_{n}\right\}_{n \geqslant 0} (the Rudin-Shapiro polynomials), which are defined recursively as follows: P0=Q0=1P_{0}=Q_{0}=1 and for n=0,1,2,…n=0,1,2, \ldots

Pn+1(z)=Pn(z)+z2nQn(z),Qn+1(z)=Pn(z)−z2nQn(z) P_{n+1}(z)=P_{n}(z)+z^{2^{n}} Q_{n}(z), \quad Q_{n+1}(z)=P_{n}(z)-z^{2^{n}} Q_{n}(z)

Prove that

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(a)

the degree of the polynomials PnP_{n} and QnQ_{n} equals 2n−12^{n}-1;

(b)

all coefficients of the polynomials PnP_{n} and QnQ_{n} equal ±1\pm 1;

(c)

if z∈C,∣z∣=1z \in \mathbb {C},\left|z\right|=1, then ∣Pn(z)∣2+∣Qn(z)∣2=2n+1\left|P_{n}(z)\right|^{2}+\left|Q_{n}(z)\right|^{2}=2^{n+1}.

Let RN(N=0,1,2,…)R_{N}(N=0,1,2, \ldots ) be the natural projection of the set of all algebraic polynomials onto the set of polynomials of degree not exceeding NN (RNR_{N} annihilates monomials whose degree is greater than NN, and leaves unchanged monomials whose degree does not exceed NN).

Prove that for z∈C,∣z∣=1z \in \mathbb {C},\left|z\right|=1, and any N,n∈NN, n \in \mathbb {N}, for the Rudin-Shapiro polynomials PnP_{n} and QnQ_{n} (see problem 4.25) the following inequalities hold:

∣RN(Pn)(z)∣⩽10N,∣RN(Qn)(z)∣⩽10N. \left|R_{N}\left(P_{n}\right)(z)\right| \leqslant 10 \sqrt{N}, \quad \left|R_{N}\left(Q_{n}\right)(z)\right| \leqslant 10 \sqrt{N} .
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Prove that there exists a series ∑k⩾0εkeikθ\sum_{k \geqslant 0} \varepsilon_{k} e^{i k \theta }, where εk=±1\varepsilon_{k}= \pm 1, for all of whose partial sums SnS_{n} the following inequalities hold:

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(a)

max⁡0⩽θ⩽2π∣Sn(θ)∣⩽10n\max_{0 \leqslant \theta \leqslant 2 \pi }\left|S_{n}(\theta )\right| \leqslant 10 \sqrt{n};

(b)

∫02π∣Sn(θ)∣dθ⩾n3\int_{0}^{2 \pi }\left|S_{n}(\theta )\right| d \theta \geqslant \sqrt{\frac{n}{3}}.

Using the sequence {εk}k⩾0\left\{ \varepsilon_{k}\right\}_{k \geqslant 0} from problem 4.27, verify that there exists a uniformly convergent series on R\mathbb {R}, f(θ)=∑k⩾0akeikθf(\theta )=\sum_{k \geqslant 0} a_{k} e^{i k \theta }, such that ∑∣ak∣p=+∞\sum \left|a_{k}\right|^{p}=+\infty for any p<2p<2 (Carleman's singularity).

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Let the sequence {εk}⊂C\left\{ \varepsilon_{k}\right\} \subset \mathbb {C} satisfy the condition max⁡θ∈R∣∑k=1nεkeikθ∣=O(n)\max_{\theta \in \mathbb {R}}\left|\sum_{k=1}^{n} \varepsilon_{k} e^{i k \theta }\right|=O(\sqrt{n}) (cf. problem 4.27). Prove that for α>0\alpha >0

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(a)

max⁡θ∈R∣∑1⩽k⩽n!εkk1/2−αeikθ∣=O(nα)\max_{\theta \in \mathbb {R}}\left|\sum_{1 \leqslant k \leqslant n!} \frac{\varepsilon_{k}}{k^{1 / 2-\alpha }} e^{i k \theta }\right|=O\left(n^{\alpha }\right);

(b)

max⁡θ∈R∣∑k⩾nεkk1/2+αeikθ∣=O(n−α)\max_{\theta \in \mathbb {R}}\left|\sum_{k \geqslant n} \frac{\varepsilon_{k}}{k^{1 / 2+\alpha }} e^{i k \theta }\right|=O\left(n^{-\alpha }\right);

(c)

the function f(θ)=∑εkk1/2+αeikθf(\theta )=\sum \frac{\varepsilon_{k}}{k^{1 / 2+\alpha }} e^{i k \theta } belongs to the class Lip⁡α\operatorname {Lip}_{\alpha } for α∈(0,1)\alpha \in (0,1).

Let 1=n0<n1<n2<…<nm1=n_{0}<n_{1}<n_{2}<\ldots <n_{m} be natural numbers, and let a1a_{1}, a2,…,ama_{2}, \ldots , a_{m} be arbitrary complex numbers, and let

fk(x)=∑2k−1⩽j<2kajeinjx(x∈R) f_{k}(x)=\sum _{2^{k-1} \leqslant j<2^{k}} a_{j} e^{i n_{j} x} \quad (x \in \mathbb {R})

(we assume that k⩾1k \geqslant 1 and aj=0a_{j}=0 for j>mj>m). Further, let

gk=∣fk∣,hk=gk−ig~k, and Fε=∑k⩾1fke−εΣj⩾khj, g_{k}=\left|f_{k}\right|, \quad h_{k}=g_{k}-i \tilde{g}_{k}, \quad \text{ and } \quad F_{\varepsilon }=\sum _{k \geqslant 1} f_{k} e^{-\varepsilon \Sigma _{j \geqslant k} h_{j}},

where ε\varepsilon is an arbitrary positive number, g~k\widetilde{g}_{k} is the function conjugate to gkg_{k} (see the definition in problem 4.32). Prove that

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(a)

∣Fε(x)∣⩽ε−1\left|F_{\varepsilon }(x)\right| \leqslant \varepsilon^{-1} for any x∈(−π,π)x \in (-\pi , \pi );

(b)

if 2p−1⩽q<2p2^{p-1} \leqslant q<2^{p}, then

∣F^ε(nq)−aq∣⩽ε2∑k⩾p∥fk∥2∑j⩾k∥fj∥2. \left|\widehat{F}_{\varepsilon }\left(n_{q}\right)-a_{q}\right| \leqslant \frac{\varepsilon }{2} \sum _{k \geqslant p}\left\| f_{k}\right\| _{2} \sum _{j \geqslant k}\left\| f_{j}\right\| _{2} .
Remark.
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The function f~\widetilde{f} defined in problem 4.32 is called conjugate to ff. The reader can become acquainted with one of the numerous applications of this important notion in problems 4.33 and 4.34, borrowed from [MGPS].

Let 1⩽n1<n2<…<nm(nj∈N)1 \leqslant n_{1}<n_{2}<\ldots <n_{m}\left(n_{j} \in \mathbb {N}\right). Using the result of the previous problem, prove that for any complex numbers c1,c2,…,cmc_{1}, c_{2}, \ldots , c_{m} the inequality

∫−ππ∣∑1⩽j⩽mcjeinjx∣dx⩾δ∑1⩽j⩽m∣cj∣j, \int _{-\pi }^{\pi }\left|\sum _{1 \leqslant j \leqslant m} c_{j} e^{i n_{j} x}\right| d x \geqslant \delta \sum _{1 \leqslant j \leqslant m} \frac{\left|c_{j}\right|}{j},

holds, where δ\delta is a positive absolute constant (δ⩾10−2\delta \geqslant 10^{-2}).

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Remark.
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The function f~\widetilde{f} defined in problem 4.32 is called conjugate to ff. The reader can become acquainted with one of the numerous applications of this important notion in problems 4.33 and 4.34, borrowed from [MGPS].

For a set Ω⊂R\Omega \subset \mathbb {R} the projector PΩP_{\Omega } retains in a trigonometric polynomial T(x)=∑ncneinxT(x)=\sum_{n} c_{n} e^{i n x} only the terms with frequencies n∈Ωn \in \Omega: PΩ(T;x)=∑n∈ΩcneinxP_{\Omega }(T ; x)=\sum_{n \in \Omega } c_{n} e^{i n x} (similarly for polynomials T(x,y)=∑cm,nei(mx+ny)T(x, y)=\sum c_{m, n} e^{i(m x+n y)} in two variables and Ω⊂R2\Omega \subset \mathbb {R}^{2}). The projector is called bounded if max⁡∣PΩ(T)∣⩽Cmax⁡∣T∣\max \left|P_{\Omega }(T)\right| \leqslant C \max \left|T\right| for all TT; the smallest such CC is denoted ∥PΩ∥\left\| P_{\Omega }\right\|. For bounded Ω\Omega we have ∥PΩ∥=∥DΩ∥1\left\| P_{\Omega }\right\| =\left\| D_{\Omega }\right\|_{1}, where DΩ(u)=12π∑n∈Ωe−inuD_{\Omega }(u)=\frac{1}{2 \pi } \sum_{n \in \Omega } e^{-i n u} is the Dirichlet kernel (in the two-dimensional case DΩ(u,v)=14π2∑(m,n)∈Ωe−i(mu+nv)D_{\Omega }(u, v)=\frac{1}{4 \pi^{2}} \sum_{(m, n) \in \Omega } e^{-i(m u+n v)}, with the norm in L1([−π,π]2)\mathscr {L}^{1}\left(\left[-\pi , \pi \right]^{2}\right)).

Using the result of problem 4.34, prove that the norms of the projectors onto the intervals ΩN=[−N,N]\Omega_{N}=\left[-N, N\right] have the minimal order of growth: for any set Ω⊂R,card⁡(Ω∩Z)=N\Omega \subset \mathbb {R}, \operatorname {card}(\Omega \cap \mathbb {Z})=N, the inequality ∥PΩ∥⩾c∥PΩN∥\left\| P_{\Omega }\right\| \geqslant c\left\| P_{\Omega_{N}}\right\| holds with an absolute constant c>0c>0.

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References

[МГЛ+04] Макаров Б.М., Голузина М.Г., Лодкин А.А., Подкорытов А.Н. Избранные задачи по вещественному анализу. 2-е изд., Невский Диалект; БХВ-Петербург, 2004.