$L^p$ norms of trigonometric polynomials

1 Introduction

A trigonometric polynomial of degree $n$ is an expression of the form

$$\sum _{k=-n}^n{c}_k{e}^{ikt},c_k\in ℂ.$$

Using the identity $e^{it}=\cos t+i\sin t$, we can write a trigonometric polynomial of degree $n$ in the form

$$a_0+\sum _{k=1}^n{a}_k\cos kt+\sum _{k=1}^n{b}_k\sin kt,a_k,b_k\in ℂ.$$

For and for a $2\pi$-periodic function $f$, we define the $L^p$ norm of $f$ by

$${\parallel f\parallel }_p={\left(\frac{1}{2\pi }{\int }_0^{2\pi }{|f(t)|}^p𝑑t\right)}^{1/p}.$$

For a continuous $2\pi$-periodic function $f$, we define the $L^{\mathrm{\infty }}$ norm of $f$ by

$${\parallel f\parallel }_{\mathrm{\infty }}=\underset{0\le t\le 2\pi }{\max }|f(t)|.$$

If $f$ is a continuous $2\pi$-periodic function, then there is a sequence of trigonometric polynomials $f_n$ such that ${\parallel f-f_n\parallel }_{\mathrm{\infty }}\to 0$ as $n\to \mathrm{\infty }$ [31, p. 54, Corollary 5.4].

If and $f$ is a continuous $2\pi$-periodic function, then

$${\parallel f\parallel }_p={\left(\frac{1}{2\pi }{\int }_0^{2\pi }{|f(t)|}^p𝑑t\right)}^{1/p}\le {\left(\frac{1}{2\pi }{\int }_0^{2\pi }{\parallel f\parallel }_{\mathrm{\infty }}^p𝑑t\right)}^{1/p}={\parallel f\parallel }_{\mathrm{\infty }}.$$

Jensen’s inequality [16, p. 44, Theorem 2.2] (cf. [30, p. 113, Problem 7.5]) tells us that if $\varphi :[0,\mathrm{\infty })\to ℝ$ is convex, then for any function $h:[0,2\pi ]\to [0,\mathrm{\infty })$ we have

$$\varphi \left(\frac{1}{2\pi }{\int }_0^{2\pi }h(t)𝑑t\right)\le \frac{1}{2\pi }{\int }_0^{2\pi }\varphi (h(t))𝑑t.$$

If , then $\varphi :[0,\mathrm{\infty })\to ℝ$ defined by $\varphi (x)=x^{q/p}$ is convex. Hence, if then for any $2\pi$-periodic function $f$,

${\parallel f\parallel }_p$ $=$ ${(\varphi ({\parallel f\parallel }_p^p))}^{1/q}$

$=$ ${\left(\varphi \left({\displaystyle \frac{1}{2\pi }}{\displaystyle {\int }_0^{2\pi }}{|f(t)|}^p𝑑t\right)\right)}^{1/q}$

$\le$ ${\left({\displaystyle \frac{1}{2\pi }}{\displaystyle {\int }_0^{2\pi }}\varphi ({|f(t)|}^p)𝑑t\right)}^{1/q}$

$=$ ${\left({\displaystyle \frac{1}{2\pi }}{\displaystyle {\int }_0^{2\pi }}{|f(t)|}^q𝑑t\right)}^{1/q}$

$=$ ${\parallel f\parallel }_q.$

The Dirichlet kernel $D_n$ is defined by

$$D_n(t)=\sum _{k=-n}^n{e}^{ikt}=1+2\sum _{k=1}^n\cos kt.$$

One can show [14, p. 71, Exercise 1.1] that

$${\parallel D_n\parallel }_1=\frac{4}{{\pi }^2}\cdot \log n+O(1).$$

(On the other hand, it can quickly be seen that ${\parallel D_n\parallel }_{\mathrm{\infty }}=2n+1$, and it follows from Parseval’s identity that ${\parallel D_n\parallel }_2=\sqrt{2n+1}$.)

Pólya and Szegő [27, Part VI] present various problems about trigonometric polynomials together with solutions to them. A result on $L^{\mathrm{\infty }}$ norms of trigonometric polynomials that Pólya and Szegő present is for the sum $A_n(t)={\sum }_{k=1}^n\frac{\sin kt}{k}$. The local maxima and local minima of $A_n$ can be explicitly determined [27, p. 74, no. 23], and it can be shown that [27, p. 74, no. 25]

$${\parallel A_n\parallel }_{\mathrm{\infty }}\sim {\int }_0^{\pi }\frac{\sin t}{t}𝑑t.$$

2 L^p norms

If , then [14, p. 123, Exercise 1.8] (cf. [7, p. 102, Theorem 2.6]) there is some $C(p,q)$ such that for any trigonometric polynomial $f$ of degree $n$, we have

$${\parallel f\parallel }_q\le C(p,q)n^{\frac{1}{p}-\frac{1}{q}}{\parallel f\parallel }_p.$$

This inequality is sharp [33, p. 230]: for there is some $C^{\prime }(p,q)$ such that if $F_n(t)=\frac{1}{n}{\sum }_{k=0}^{n-1}{D}_k(t)$ ($F_n$ is called the Fejér kernel) then

$${\parallel F_n\parallel }_q>C^{\prime }(p,q)n^{\frac{1}{p}-\frac{1}{q}}{\parallel F_n\parallel }_p.$$

Theorem 1.

Let $1\le p\le q\le \mathrm{\infty }$. If $\widehat{f}(j)=0$ for $|j|>n+1$ then

$${\parallel f\parallel }_q\le 5{(n+1)}^{\frac{1}{p}-\frac{1}{q}}{\parallel f\parallel }_p.$$

Proof.

Let $K_n(t)={\sum }_{j=-n}^n\left(1-\frac{|j|}{n+1}\right)e^{ijt}$, the Fejér kernel. From this expression we get $|K_n(t)|\le K_n(0)=n+1$. It’s straightforward to show that $K_n(t)=\frac{1}{n+1}{\left(\frac{\sin \frac{n+1}{2}t}{\sin \frac{1}{2}t}\right)}^2$. Since $\sin \frac{t}{2}>\frac{t}{\pi }$ for , we get $|K_n(t)|\le \frac{{\pi }^2}{(n+1)t^2}$, and thus we obtain

$$|K_n(t)|\le \min (n+1,\frac{{\pi }^2}{(n+1)t^2}).$$

Then, for any $r\ge 1$,

${\parallel K_n\parallel }_r^r$ $=$ ${\displaystyle \frac{1}{2\pi }}{\displaystyle {\int }_0^{2\pi }}{|K_n(t)|}^r𝑑t$

$\le$ ${\displaystyle \frac{1}{2\pi }}{\displaystyle {\int }_0^{\frac{\pi }{n+1}}}{(n+1)}^r𝑑t+{\displaystyle \frac{1}{2\pi }}{\displaystyle {\int }_{\frac{\pi }{n+1}}^{2\pi }}{\left({\displaystyle \frac{{\pi }^2}{(n+1)t^2}}\right)}^r𝑑t$

$=$ ${\displaystyle \frac{{(n+1)}^{r-1}}{2}}+{\displaystyle \frac{1}{2}}{\displaystyle \frac{1}{{(n+1)}^r}}{\displaystyle \frac{1}{2r-1}}\left({(n+1)}^{2r-1}-{\displaystyle \frac{1}{2^{2r-1}}}\right)$

$\le$ ${\displaystyle \frac{{(n+1)}^{r-1}}{2}}+{\displaystyle \frac{1}{2}}{\displaystyle \frac{1}{{(n+1)}^r}}{\displaystyle \frac{1}{2r-1}}{(n+1)}^{2r-1}$

$\le$ ${(n+1)}^{r-1}.$

Hence ${\parallel K_n\parallel }_r\le {(n+1)}^{1-\frac{1}{r}}$.

Let $V_n(t)=2K_{2n+1}(t)-K_n(t)$, the de la Vallée Poussin kernel. Then

$${\parallel V_n\parallel }_r\le 2{\parallel K_{2n+1}\parallel }_r+{\parallel K_n\parallel }_r\le 2{(2n+2)}^{1-\frac{1}{r}}+{(n+1)}^{1-\frac{1}{r}}\le 5{(n+1)}^{1-\frac{1}{r}}.$$

For $|j|\le n+1$ we have $\widehat{V_n}(j)=1$, and one thus checks that $V_n*f=f$. Take $\frac{1}{q}+1=\frac{1}{p}+\frac{1}{r}$. By Young’s inequality we have

$${\parallel f\parallel }_q={\parallel V_n*f\parallel }_q\le {\parallel V_n\parallel }_r{\parallel f\parallel }_p\le 5{(n+1)}^{\frac{1}{p}-\frac{1}{q}}{\parallel f\parallel }_p.$$

∎

Let $X_n=\{a_0+{\sum }_{k=1}^n{a}_k\cos kt+b_k\sin kt:a_k,b_k\in ℝ\}$, the real vector space of real valued trigonometric polynomials of degree $n$, have norm

$${\parallel f\parallel }_{X_n}=\max \{|a_0|,|a_1|,\mathrm{\dots },|a_n|,|b_1|,\mathrm{\dots },|b_n|\}.$$

Let $Y_{n,p}$ be the same vector space with the $L^p$ norm. Ash and Ganzburg [1] give upper and lower bounds on the operator norm of the map $i:X_n\to Y_{n,p}$ defined by $i(f)=f$.

Bernstein’s inequality [14, p. 50, Exercise 7.16] states that for $1\le p\le \mathrm{\infty }$, if $f$ is a trigonometric polynomial of degree $n$, then

$${\parallel f^{\prime }\parallel }_p\le n{\parallel f\parallel }_p.$$

In the other direction, if $f\in C^1$ then

Hence

$|f(t)|$ $\le$ ${\displaystyle \frac{1}{2\pi }}{\displaystyle {\int }_0^{2\pi }}|f(s)|𝑑s+{\displaystyle \frac{1}{2\pi }}{\displaystyle {\int }_0^t}s|f^{\prime }(s)|𝑑s+{\displaystyle \frac{1}{2\pi }}{\displaystyle {\int }_t^{2\pi }}(2\pi -s)|f^{\prime }(s)|𝑑s$

$\le$ ${\displaystyle \frac{1}{2\pi }}{\displaystyle {\int }_0^{2\pi }}|f(s)|𝑑s+{\displaystyle {\int }_0^t}|f^{\prime }(s)|𝑑s+{\displaystyle {\int }_t^{2\pi }}|f^{\prime }(s)|𝑑s$

$=$ ${\parallel f\parallel }_1+2\pi {\parallel f^{\prime }\parallel }_1,$

so

$${\parallel f\parallel }_{\mathrm{\infty }}\le {\parallel f\parallel }_1+2\pi {\parallel f^{\prime }\parallel }_1.$$

This is an instance of the Sobolev inequality [26].

It turns out that for a trigonometric polynomial the mass cannot be too concentrated. More precisely, the number of nonzero terms of a trigonometric polynomial restricts how concentrated its mass can be. Let $d\mu =\frac{dt}{2\pi }$. Thus $\mu ([0,2\pi ])=1$. A result of Turán [20, p. 89, Lemma 1] states that if ${\lambda }_1,\mathrm{\dots },{\lambda }_N\in \mathbb{Z}$ and $T(t)={\sum }_{n=1}^N{b}_n{e}^{i{\lambda }_{n}t}$, $b_n\in ℂ$, then for any closed arc $I\subset [0,2\pi ]$,

$${\parallel T\parallel }_{\mathrm{\infty }}\le {\left(\frac{2e}{\mu (I)}\right)}^{N-1}\underset{t\in I}{\max }|T(t)|.$$

Nazarov [11, p. 452] shows that there is some constant $A$ such that if $E$ is a closed subset of $[0,2\pi ]$ (not necessarily an arc), then

$${\parallel \widehat{T}\parallel }_1\le {\left(\frac{A}{\mu (E)}\right)}^N\underset{t\in E}{\max }|f(T)|.$$

Nazarov [23] proves that there exists some constant $C$ such that if $0\le q\le 2$ and $\mu (E)\ge \frac{1}{3}$, then

$${\parallel T\parallel }_q\le e^{C(N-1)\left(1-\frac{\mu (E)}{2\pi }\right)}{\left(\frac{1}{2\pi }{\int }_E{|T(t)|}^q𝑑t\right)}^{1/q}.$$

These results of Turan and Nazarov are examples of the uncertainty principle [9], which is the general principle that a constrain on the support of the Fourier transform of a function constrains the support of the function itself.

In [10], Hardy and Littlewood present inequalities for norms of $2\pi$-periodic functions in terms of certain series formed from their Fourier coefficients. Let $c_k\in ℂ$, $k\in \mathbb{Z}$, be such that $c_k\to 0$ as $k\to \pm \mathrm{\infty }$, and define $c_0^{*},c_1^{*},c_{-1}^{*},c_2^{*},c_{-2}^{*},\mathrm{\dots }$ to be the absolute values of the $c_k$ ordered in decreasing magnitude. For real $r>1$, define

$$S_r^{*}(c)={\left(\sum _{k=-\mathrm{\infty }}^{\mathrm{\infty }}c_k^{*}{}^r{(|k|+1)}^{r-2}\right)}^{1/r}.$$

For instance, if $c_k=1$ for $-N\le k\le N$ and $c_k=0$ for $|k|>N$, then $S_r^{*}(c)={\left(1+2{\sum }_{k=2}^{N+1}{k}^{r-2}\right)}^{1/r}$. Hardy and Littlewood state the result [10, p. 164, Theorem 2] that if then there is some constant $A(p)$ such that for any sequence $c$, with $c_k\to 0$ as $k\to \pm \mathrm{\infty }$, if $f(t)={\sum }_{k=-\mathrm{\infty }}^{\mathrm{\infty }}{c}_k{e}^{ikt}$ and then

$$S_p^{*}(c)\le A(p){\parallel f\parallel }_p.$$

A proof of this is given in Zygmund [35, vol. II, p. 128, chap. XII, Theorem 6.3]. Asking if this inequality holds for $p=1$ suggests the following question that Hardy and Littlewood pose at the end of their paper [10, p. 168]: Is there a constant $A$ such that for all distinct positive integers $m_k,k=1,\mathrm{\dots },N$, we have

$${\parallel \sum _{k=1}^N\cos m_{k}t\parallel }_1>A\log N\mathrm{?}$$

McGehee, Pigno and Smith [18] prove that there is some $K$ such that for all $N$, if $n_1,\mathrm{\dots },n_N$ are distinct integers and $c_1,\mathrm{\dots },c_N\in ℂ$ satisfy $|c_k|\ge 1$, then

$${\parallel \sum _{k=1}^N{c}_k{e}^{i{n}_{k}t}\parallel }_1>K\log N.$$

Thus

$${\parallel \sum _{k=1}^N\cos m_{k}t\parallel }_1=\frac{1}{2}\cdot {\parallel \sum _{k=1}^N{e}^{i{m}_{k}t}+e^{-i{m}_{k}t}\parallel }_1\ge \frac{1}{2}\cdot K\log (2N).$$

For $k\ge 2$, define $T_N(t)={\sum }_{n=1}^N{e}^{i{n}^{k}t}$. Since ${\parallel T_N\parallel }_{\mathrm{\infty }}=N$, for each $p\ge 1$ we have ${\parallel T_N\parallel }_p\le N$. Hua’s lemma [22, p. 116, Theorem 4.6] states that if $ϵ>0$, then

$${\parallel T_N\parallel }_{2^k}=O\left(N^{1-\frac{k}{2^k}+ϵ}\right).$$

Hua’s lemma is used in additive number theory. The number of sets of integer solutions of the equation

$$f(x_1,\mathrm{\dots },x_n)=N,a_r\le x_r\le b_r$$

is equal to (cf. [12, p. 151])

$$\sum _{a_1\le x_1\le b_1}\mathrm{\cdots }\sum _{a_n\le x_n\le b_n}{\int }_0^1{e}^{2\pi i(f(x_1,\mathrm{\dots },x_n)-N)t}𝑑t.$$

Borwein and Lockhart [4]: what is the expected $L^p$ norm of a trigonometric polynomial of order $n$? Kahane [13, Chapter 6] also presents material on random trigonometric polynomials.

Nursultanov and Tikhonov [25]: the sup on a subset of $𝕋$ of a trigonometric polynomial $f$ of degree $n$ being lower bounded in terms of ${\parallel f\parallel }_{\mathrm{\infty }}$, $n$, and the measure of the subset.

3 ℓᵖ norms

For a $2\pi$-periodic function $f$, we define $\widehat{f}:\mathbb{Z}\to ℂ$ by

$$\widehat{f}(k)=\frac{1}{2\pi }{\int }_0^{2\pi }{e}^{-ikt}f(t)𝑑t.$$

For , we define the ${\mathrm{\ell }}^p$ norm of $\widehat{f}$ by

$${\parallel \widehat{f}\parallel }_p={\left(\sum _{k=-\mathrm{\infty }}^{\mathrm{\infty }}{|\widehat{f}(k)|}^p\right)}^{1/p},$$

and we define the ${\mathrm{\ell }}^{\mathrm{\infty }}$ norm of $\widehat{f}$ by

$${\parallel \widehat{f}\parallel }_{\mathrm{\infty }}=\underset{k\in \mathbb{Z}}{\max }|\widehat{f}(k)|.$$

Parseval’s identity [31, p. 80, Theorem 1.3] states that ${\parallel f\parallel }_2={\parallel \widehat{f}\parallel }_2$.

If , then

$${\parallel \widehat{f}\parallel }_{\mathrm{\infty }}\le {\left(\mathrm{\cdots }+{\parallel \widehat{f}\parallel }_{\mathrm{\infty }}^p+\mathrm{\cdots }\right)}^{1/p}={\parallel \widehat{f}\parallel }_p.$$

If , then, since for each $k$, $\frac{|\widehat{f}(k)|}{{\parallel \widehat{f}\parallel }_q}\le 1$,

$$1={\left(\sum _{k=-\mathrm{\infty }}^{\mathrm{\infty }}{\left(\frac{|\widehat{f}(k)|}{{\parallel \widehat{f}\parallel }_q}\right)}^q\right)}^{1/q}\le {\left(\sum _{k=-\mathrm{\infty }}^{\mathrm{\infty }}{\left(\frac{|\widehat{f}(k)|}{{\parallel \widehat{f}\parallel }_q}\right)}^p\right)}^{1/q}=\frac{{\parallel \widehat{f}\parallel }_p^{p/q}}{{\parallel \widehat{f}\parallel }_q^{p/q}}.$$

Hence for ,

$${\parallel \widehat{f}\parallel }_q\le {\parallel \widehat{f}\parallel }_p.$$

For , if $f$ is a trigonometric polynomial of degree $n$ then

$${\parallel \widehat{f}\parallel }_p={\left(\sum _{k=-n}^n{|\widehat{f}(k)|}^p\right)}^{1/p}\le {\left(\sum _{k=-n}^n{\parallel \widehat{f}\parallel }_{\mathrm{\infty }}^p\right)}^{1/p}={(2n+1)}^{1/p}{\parallel \widehat{f}\parallel }_{\mathrm{\infty }}.$$

For , we have [30, p. 123, Problem 8.3] (this is Jensen’s inequality for sums)

$${\left(\sum _{k=-n}^n\frac{1}{2n+1}{|\widehat{f}(k)|}^p\right)}^{1/p}\le {\left(\sum _{k=-n}^n\frac{1}{2n+1}{|\widehat{f}(k)|}^q\right)}^{1/q},$$

i.e.

$${(2n+1)}^{-1/p}{\parallel \widehat{f}\parallel }_p\le {(2n+1)}^{-1/q}{\parallel \widehat{f}\parallel }_q.$$

Hence for ,

$${\parallel \widehat{f}\parallel }_p\le {(2n+1)}^{\frac{1}{p}-\frac{1}{q}}{\parallel \widehat{f}\parallel }_q.$$

For any $t$,

$$|f(t)|=\left|\sum _{k=-\mathrm{\infty }}^{\mathrm{\infty }}\widehat{f}(k)e^{ikt}\right|\le \sum _{k=-\mathrm{\infty }}^{\mathrm{\infty }}|\widehat{f}(k)e^{ikt}|=\sum _{k=-\mathrm{\infty }}^{\mathrm{\infty }}|\widehat{f}(k)|={\parallel \widehat{f}\parallel }_1.$$

Hence

$${\parallel f\parallel }_{\mathrm{\infty }}\le {\parallel \widehat{f}\parallel }_1.$$

For any $k\in \mathbb{Z}$,

$$|\widehat{f}(k)|=\left|\frac{1}{2\pi }{\int }_0^{2\pi }{e}^{-ikt}f(t)𝑑t\right|\le \frac{1}{2\pi }{\int }_0^{2\pi }|f(t)|𝑑t={\parallel f\parallel }_1.$$

Hence

$${\parallel \widehat{f}\parallel }_{\mathrm{\infty }}\le {\parallel f\parallel }_1.$$

The Hausdorff-Young inequality [32, p. 57, Corollary 2.4] states that for $1\le p\le 2$ and $\frac{1}{p}+\frac{1}{q}=1$, if $f\in L^p$ then

$${\parallel \widehat{f}\parallel }_q\le {\parallel f\parallel }_p.$$

The dual Hausdorff-Young inequality [32, p. 58, Corollary 2.5] states that for $1\le p\le 2$ and $\frac{1}{p}+\frac{1}{q}=1$, if $f\in L^q$ then

$${\parallel f\parallel }_q\le {\parallel \widehat{f}\parallel }_q.$$

A survey on the Hausdorff-Young inequality is given in [6])

For $M+1\le k\le M+N$, let $a_k\in ℂ$ and let $S(t)={\sum }_{k=M+1}^{N+1}{a}_k{e}^{ikt}$. Let $t_1,\mathrm{\dots },t_R\in ℝ$, and let $\delta$ be such that if $r\ne s$ then

$$\parallel t_r-t_s\parallel \ge \delta ,$$

where $\parallel t\parallel ={\min }_k|t-k|$ is the distance from $t$ to a nearest integer. The large sieve [19] is an inequality of the form

$$\sum _{r=1}^R{|S(2\pi t_r)|}^2\le \mathrm{\Delta }(N,\delta )\sum _{k=M+1}^{M+N}{|a_k|}^2.$$

A result of Selberg [19, p. 559, Theorem 3] shows that the large sieve is valid for $\mathrm{\Delta }=N-1+{\delta }^{-1}$.

Kristiansen [15]

Boas [2]

For $F:\mathbb{Z}/n\to ℂ$, its Fourier transform $\widehat{F}:\mathbb{Z}/n\to ℂ$ (called the discrete Fourier transform) is defined by

$$\widehat{F}(k)=\frac{1}{n}\sum _{j=0}^{n-1}F(j)e^{-2\pi ijk/n},0\le k\le n-1,$$

and one can prove [31, p. 223, Theorem 1.2] that

$$F(j)=\sum _{k=0}^{n-1}\widehat{F}(k)e^{2\pi ikj/N},0\le j\le n-1.$$

One can also prove Parseval’s identity for the Fourier transform on $\mathbb{Z}/n$ [31, p. 223, Theorem  1.2]. It states

$$\sum _{k=0}^{n-1}{|\widehat{F}(k)|}^2=\frac{1}{n}\sum _{j=0}^{n-1}{|F(j)|}^2.$$

Let $P(t)={\sum }_{k=0}^{n-1}{a}_k{e}^{ikt}$. Define $F:\mathbb{Z}/n\to ℂ$ by

$$F(j)=\sum _{k=0}^{n-1}{a}_k{e}^{2\pi ikj/n},0\le j\le n-1.$$

(That is, $\widehat{F}(k)=a_k$.) We then have

$$\sum _{k=0}^{n-1}{|a_k|}^2=\frac{1}{n}\sum _{j=0}^{n-1}{|F(j)|}^2=\frac{1}{n}\sum _{j=0}^{n-1}{\left|P\left(\frac{2\pi j}{n}\right)\right|}^2.$$

Thus

$${\parallel P\parallel }_2={\left(\frac{1}{n}\sum _{j=0}^{n-1}{\left|P\left(\frac{2\pi j}{n}\right)\right|}^2\right)}^{1/2}.$$

The Marcinkiewicz-Zygmund inequalities [35, vol. II, p. 28, chap. X, Theorem 7.5] state that there is a constant $A$ such that for $1\le p\le \mathrm{\infty }$, if $f$ is a trigonometric polynomial of degree $n$ then

$${\left(\frac{1}{2n+1}\sum _{k=0}^{2n}{\left|f\left(\frac{2\pi k}{2n+1}\right)\right|}^p\right)}^{1/p}\le A{(2\pi )}^{1/p}{\parallel f\parallel }_p,$$

and for each there exists some $A_p$ such that if $f$ is a trigonometric polynomial of degree $n$ then

$${\parallel f\parallel }_p\le A_p{\left(\frac{1}{2n+1}\sum _{k=0}^{2n}{\left|f\left(\frac{2\pi k}{2n+1}\right)\right|}^p\right)}^{1/p}.$$

Máté and Nevai [17, p. 148, Theorem 6] prove that for $p>0$, if $S_n$ is a trigonometric polynomial of degree $n$ then

$${\parallel S_n\parallel }_{\mathrm{\infty }}\le {\left(\frac{(1+np)e}{2}\right)}^{1/p}{\parallel S_n\parallel }_p.$$

Máté and Nevai [17] prove a version of Bernstein’s inequality for , and their result can be sharpened to the following [34]: For , if $T_n$ is a trigonometric polynomial of order $n$ then

$${\parallel T_n^{\prime }\parallel }_p\le n{\parallel T_n\parallel }_p.$$

Let $supp\widehat{f}=\{k\in \mathbb{Z}:\widehat{f}(k)\ne 0\}$. A subset $\mathrm{\Lambda }$ of $\mathbb{Z}$ is called a Sidon set [28, p. 121, §5.7.2] if there exists a constant $B$ such that for every trigonometric polynomial $f$ with $supp\widehat{f}\subseteq \mathrm{\Lambda }$ we have

$${\parallel \widehat{f}\parallel }_1\le B{\parallel f\parallel }_{\mathrm{\infty }}.$$

Let $B(\mathrm{\Lambda })$ be the least such $B$. A sequence of positive integers ${\lambda }_k$ is said to be lacunary if there is a constant $\rho$ such that ${\lambda }_{k+1}>\rho {\lambda }_k$ for all $k$. If ${\lambda }_k$ is a lacunary sequence, then $\{{\lambda }_k\}$ is a Sidon set [21, p. 154, Corollary 6.17]. If $\mathrm{\Lambda }\subset \mathbb{Z}$ is a Sidon set, then [28, p. 128, Theorem 5.7.7] (cf. [21, p. 157, Corollary 6.19]) for any , for every trigonometric polynomial $f$ with $supp\widehat{f}\subseteq \mathrm{\Lambda }$ we have

$${\parallel f\parallel }_p\le B(\mathrm{\Lambda })\sqrt{p}{\parallel f\parallel }_2,$$

and

$${\parallel f\parallel }_2\le 2B(\mathrm{\Lambda }){\parallel f\parallel }_1.$$

Let . A subset $E$ of $\mathbb{Z}$ is called a $\mathrm{\Lambda }(p)$-set if for every there is some $A(E,p)$ such that for all trigonometric polynomials $f$ with $supp\widehat{f}\subset E$ we have

$${\parallel f\parallel }_p\le A(E,p){\parallel f\parallel }_2.$$

$\mathrm{\Lambda }(p)$ sets were introduced by Rudin, and he discusses them in his autobiography [29, Chapter 28]. A modern survey of $\mathrm{\Lambda }(p)$-sets is given by Bourgain [5].

Bochkarev [3] proves various lower bounds on the $L^1$ norms of certain trigonometric polynomials. Let $c_k\in ℂ$, $k\ge 1$. If there are constants $A$ and $B$ such that

$$A\frac{{(\log k)}^s}{\sqrt{k}}\le |c_k|\le B\frac{{(\log k)}^s}{\sqrt{k}},k\ge 1,$$

then [3, p. 58, Theorem 19]

If $P(t)={\sum }_{k=0}^n{a}_k{e}^{ikt}$ with $a_k\in \{-1,1\}$, then by the Cauchy-Schwarz inequality and Parseval’s identity we have

$${\parallel P\parallel }_1=\frac{1}{2\pi }{\int }_0^{2\pi }1\cdot |P(t)|𝑑t\le {\parallel 1\parallel }_2\cdot {\parallel P\parallel }_2=1\cdot {\parallel \widehat{P}\parallel }_2=\sqrt{n+1}.$$

Newman [24] shows that in fact we can do better than what we get using the Cauchy-Schwarz inequality and Parseval’s identity:

A Fekete polynomial is a polynomial of the form ${\sum }_{k=1}^{l-1}\left(\frac{k}{l}\right)z^k$, $l$ prime, where $\left(\frac{k}{l}\right)$ is the Legendre symbol. Let $P_l(t)={\sum }_{k=1}^{l-1}\left(\frac{k}{l}\right)e^{ikt}$. Erdélyi [8] proves upper and lower bounds on ${\left(\frac{1}{|I|}{\int }_I{|P_l(t)|}^q𝑑t\right)}^{1/q}$, $q>0$, where $I$ is an arc in $[0,2\pi ]$.