The Bernstein and Nikolsky inequalities for trigonometric polynomials

1 Introduction

Let $𝕋=ℝ/2\pi \mathbb{Z}$. For a function $f:𝕋\to ℂ$ and $\tau \in 𝕋$, we define $f_{\tau }:𝕋\to ℂ$ by $f_{\tau }(t)=f(t-\tau )$. For measurable $f:𝕋\to ℂ$ and , write

$${\parallel f\parallel }_r={\left(\frac{1}{2\pi }{\int }_{𝕋}{|f(t)|}^r𝑑t\right)}^{1/r}.$$

For $f,g\in L^1(𝕋)$, write

$$(f*g)(x)=\frac{1}{2\pi }{\int }_{𝕋}f(t)g(x-t)𝑑t,x\in 𝕋,$$

and for $f\in L^1(𝕋)$, write

$$\widehat{f}(k)=\frac{1}{2\pi }{\int }_{𝕋}f(t)e^{-ikt}𝑑t,k\in \mathbb{Z}.$$

This note works out proofs of some inequalities involving the support of $\widehat{f}$ for $f\in L^1(𝕋)$.

Let ${𝒯}_n$ be the set of trigonometric polynomials of degree $\le n$. We define the Dirichlet kernel $D_n:𝕋\to ℂ$ by

$$D_n(t)=\sum _{|j|\le n}{e}^{ijt},t\in 𝕋.$$

It is straightforward to check that if $T\in {𝒯}_n$ then

$$D_n*T=T.$$

2 Bernstein’s inequality for trigonometric polynomials

DeVore and Lorentz attribute the following inequality to Szegö.

Theorem 1.

If $T\in {𝒯}_n$ and $T$ is real valued, then for all $x\in 𝕋$,

$$T^{\prime }{(x)}^2+n^{2}T{(x)}^2\le n^2{\parallel T\parallel }_{\mathrm{\infty }}^2.$$

Proof.

If $T=0$ the result is immediate. Otherwise, take $x\in 𝕋$, and for real $c>1$ define

$$P_c(t)=\frac{T(t+x)\mathrm{sgn}{T}^{\prime }(x)}{c{\parallel T\parallel }_{\mathrm{\infty }}},t\in 𝕋.$$

$P_c\in {𝒯}_n$, and satisfies

$$P_c^{\prime }(0)=\frac{T^{\prime }(x)\mathrm{sgn}{T}^{\prime }(x)}{c{\parallel T\parallel }_{\mathrm{\infty }}}\ge 0$$

and . Since , in particular and so there is some $\alpha$, , such that $\sin n\alpha =P_c(0)$. We define $S\in {𝒯}_n$ by

$$S(t)=\sin n(t+\alpha )-P_c(t),t\in 𝕋,$$

which satisfies $S(0)=\sin n\alpha -P_c(0)=0$. For $k=-n,\mathrm{\dots },n$, let $t_k=-\alpha +\frac{(2k-1)\pi }{2n}$, for which we have

$$\sin n(t_k+\alpha )=\sin \frac{(2k-1)\pi }{2}={(-1)}^{k+1}.$$

Because ,

$$\mathrm{sgn}S(t_k)={(-1)}^{k+1},$$

so by the intermediate value theorem, for each $k=-n,\mathrm{\dots },n-1$ there is some $c_k\in (t_k,t_{k+1})$ such that $S(c_k)=0$. Because

$$t_n-t_{-n}=\frac{(2n-1)\pi }{2n}-\frac{(-2n-1)\pi }{2n}=2\pi ,$$

it follows that if $j\ne k$ then $c_j$ and $c_k$ are distinct in $𝕋$. It is a fact that a trigonometric polynomial of degree $n$ has $\le 2n$ distinct roots in $𝕋$, so if $t\in (t_k,t_{k+1})$ and $S(t)=0$, then $t=c_k$. It is the case that $t_1=-\alpha +\frac{\pi }{2n}>0$ and , so $0\in (t_0,t_1)$. But $S(0)=0$, 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 , so $S^{\prime }(0)\ge 0$. This gives

$$0\le P_c^{\prime }(0)=n\cos n\alpha -S^{\prime }(0)\le n\cos n\alpha =n\sqrt{1-{\sin }^{2}n\alpha }=n\sqrt{1-P_c{(0)}^2}.$$

Thus

$$P_c^{\prime }(0)\le n\sqrt{1-P_c{(0)}^2},$$

or

$$n^2{P}_c(0)+P_c^{\prime }{(0)}^2\le n^2.$$

Because

$$P_c{(0)}^2=\frac{T{(x)}^2}{c^2{\parallel T\parallel }_{\mathrm{\infty }}^2},P_c^{\prime }{(0)}^2=\frac{T^{\prime }{(x)}^2}{c^2{\parallel T\parallel }_{\mathrm{\infty }}^2}$$

we get

$$n^{2}T{(x)}^2+T^{\prime }{(x)}^2\le c^2{n}^2{\parallel T\parallel }_{\mathrm{\infty }}^2.$$

Because this is true for all $c>1$,

$$n^{2}T{(x)}^2+T^{\prime }{(x)}^2\le n^2{\parallel T\parallel }_{\mathrm{\infty }}^2,$$

completing the proof. ∎

Using the above we now prove Bernstein’s inequality.

Theorem 2 (Bernstein’s inequality).

If $T\in {𝒯}_n$, then

$${\parallel T^{\prime }\parallel }_{\mathrm{\infty }}\le n{\parallel T\parallel }_{\mathrm{\infty }}.$$

Proof.

There is some $x_0\in 𝕋$ such that $|T^{\prime }(x_0)|={\parallel T^{\prime }\parallel }_{\mathrm{\infty }}$. Let $\alpha \in ℝ$ be such that $e^{i\alpha }{T}^{\prime }(x_0)={\parallel T^{\prime }\parallel }_{\mathrm{\infty }}$. Define $S(x)=\mathrm{Re}(e^{i\alpha }T(x))$ for $x\in 𝕋$, which satisfies $S^{\prime }(x)=\mathrm{Re}(e^{i\alpha }{T}^{\prime }(x))$ and in particular

$$S^{\prime }(x_0)=\mathrm{Re}(e^{i\alpha }{T}^{\prime }(x_0))=e^{i\alpha }{T}^{\prime }(x_0)={\parallel T^{\prime }\parallel }_{\mathrm{\infty }}.$$

Because $S\in {𝒯}_n$ and $S$ is real valued, Theorem 1 yields

$$S^{\prime }{(x_0)}^2+n^{2}S{(x_0)}^2\le n^2{\parallel S\parallel }_{\mathrm{\infty }}^2.$$

A fortiori,

$$S^{\prime }{(x_0)}^2\le n^2{\parallel S\parallel }_{\mathrm{\infty }}^2,$$

giving, because $S^{\prime }(x_0)={\parallel T^{\prime }\parallel }_{\mathrm{\infty }}$ and ${\parallel S\parallel }_{\mathrm{\infty }}\le {\parallel T\parallel }_{\mathrm{\infty }}$,

$${\parallel T^{\prime }\parallel }_{\mathrm{\infty }}^2\le n^2{\parallel T\parallel }_{\mathrm{\infty }}^2,$$

proving the claim. ∎

The following is a version of Bernstein’s inequality.

Theorem 3.

If $T\in {𝒯}_n$ and $A\subset 𝕋$ is a Borel set, there is some $x_0\in 𝕋$ such that

$${\int }_A|T^{\prime }(t)|𝑑t\le n{\int }_{A-x_0}|T(t)|𝑑t.$$

Proof.

Let $A\subset 𝕋$ be a Borel set with indicator function ${\chi }_A$. Define $Q:𝕋\to ℂ$ by

$$Q(x)={\int }_{𝕋}{\chi }_A(t)T(t+x)\mathrm{sgn}{T}^{\prime }(t)𝑑t,x\in 𝕋,$$

which we can write as

$Q(x)$ $={\displaystyle {\int }_{𝕋}}{\chi }_A(t){\displaystyle \sum _j}\widehat{T}(j)e^{ij(t+x)}\mathrm{sgn}{T}^{\prime }(t)dt$

$={\displaystyle \sum _j}\widehat{T}(j)\left({\displaystyle {\int }_{𝕋}}{\chi }_A(t)e^{ijt}\mathrm{sgn}{T}^{\prime }(t)𝑑t\right)e^{ijx},$

showing that $Q\in {𝒯}_n$. Also,

$$Q^{\prime }(x)={\int }_{𝕋}{\chi }_A(t)T^{\prime }(t+x)\mathrm{sgn}{T}^{\prime }(t)𝑑t,x\in 𝕋.$$

Let $x_0\in 𝕋$ with $|Q(x_0)|={\parallel Q\parallel }_{\mathrm{\infty }}$. Applying Theorem 2 we get

$${\parallel Q^{\prime }\parallel }_{\mathrm{\infty }}\le n{\parallel Q\parallel }_{\mathrm{\infty }}.$$

Using

$$Q^{\prime }(0)={\int }_{𝕋}{\chi }_A(t)T^{\prime }(t)\mathrm{sgn}{T}^{\prime }(t)𝑑t={\int }_{𝕋}{\chi }_A(t)|T^{\prime }(t)|𝑑t,$$

this gives

${\displaystyle {\int }_{𝕋}}{\chi }_A(t)|T^{\prime }(t)|𝑑t$ $\le n{\parallel Q\parallel }_{\mathrm{\infty }}$

$=n|Q(t_0)|$

$=n\left|{\displaystyle {\int }_{𝕋}}{\chi }_A(t)T(t+x_0)\mathrm{sgn}{T}^{\prime }(t)𝑑t\right|$

$\le n{\displaystyle {\int }_{𝕋}}{\chi }_A(t)|T(t+x_0)|𝑑t$

$=n{\displaystyle {\int }_{𝕋}}{\chi }_{A-x_0}(t)|T(t)|𝑑t.$

∎

Applying the above with $A=𝕋$ gives the following version of Bernstein’s inequality, for the $L^1$ norm.

Theorem 4 ($L^1$ Bernstein’s inequality).

If $T\in {𝒯}_n$, then

$${\parallel T^{\prime }\parallel }_1\le n{\parallel T\parallel }_1.$$

3 Nikolsky’s inequality for trigonometric polynomials

DeVore and Lorentz attribute the following inequality to Sergey Nikolsky.

Theorem 5 (Nikolsky’s inequality).

If $T\in {𝒯}_n$ and , then for $r\ge \frac{q}{2}$ an integer,

$${\parallel T\parallel }_p\le {(2nr+1)}^{\frac{1}{q}-\frac{1}{p}}{\parallel T\parallel }_q.$$

Proof.

Let $m=nr$. Then $T^r\in {𝒯}_m$, so $T^r*D_m=T^r$, and using this and the Cauchy-Schwarz inequality we have, for $x\in 𝕋$,

$|T{(x)}^r|$ $=\left|{\displaystyle \frac{1}{2\pi }}{\displaystyle {\int }_{𝕋}}T{(t)}^r{D}_m(x-t)\right|$

$\le {\displaystyle \frac{1}{2\pi }}{\displaystyle {\int }_{𝕋}}{|T(t)|}^r|D_m(x-t)|𝑑t$

$\le {\parallel T\parallel }_{\mathrm{\infty }}^{r-\frac{q}{2}}\cdot {\displaystyle \frac{1}{2\pi }}{\displaystyle {\int }_{𝕋}}{|T(t)|}^{\frac{q}{2}}|D_m(x-t)|𝑑t$

$\le {\parallel T\parallel }_{\mathrm{\infty }}^{r-\frac{q}{2}}{\parallel {|T|}^{q/2}\parallel }_2{\parallel D_m\parallel }_2$

$={\parallel T\parallel }_{\mathrm{\infty }}^{r-\frac{q}{2}}{\parallel T\parallel }_q^{\frac{q}{2}}{\parallel \widehat{D_m}\parallel }_{{\mathrm{\ell }}^2(\mathbb{Z})}$

$=\sqrt{2m+1}{\parallel T\parallel }_{\mathrm{\infty }}^{r-\frac{q}{2}}{\parallel T\parallel }_q^{\frac{q}{2}}.$

Hence

$${\parallel T\parallel }_{\mathrm{\infty }}^r\le \sqrt{2m+1}{\parallel T\parallel }_{\mathrm{\infty }}^{r-\frac{q}{2}}{\parallel T\parallel }_q^{\frac{q}{2}},$$

thus

$${\parallel T\parallel }_{\mathrm{\infty }}\le {(2m+1)}^{\frac{1}{q}}{\parallel T\parallel }_q.$$

Then, using ${\parallel T\parallel }_p\le {\parallel T\parallel }_{\mathrm{\infty }}^{1-\frac{q}{p}}{\parallel T\parallel }_q^{\frac{q}{p}}$, we have

$${\parallel T\parallel }_p\le {(2m+1)}^{\frac{1}{q}-\frac{1}{p}}{\parallel T\parallel }_q^{1-\frac{q}{p}}{\parallel T\parallel }_q^{\frac{q}{p}}={(2m+1)}^{\frac{1}{q}-\frac{1}{p}}{\parallel T\parallel }_q.$$

∎

4 The complementary Bernstein inequality

We define a homogeneous Banach space to be a linear subspace $B$ of $L^1(𝕋)$ with a norm ${\parallel f\parallel }_{L^1(𝕋)}\le {\parallel f\parallel }_B$ with which $B$ is a Banach space, such that if $f\in B$ and $\tau \in 𝕋$ then $f_{\tau }\in B$ and ${\parallel f_{\tau }\parallel }_B={\parallel f\parallel }_B$, and such that if $f\in B$ then $f_{\tau }\to f$ in $B$ as $\tau \to 0$.

Fejér’s kernel is, for $n\ge 0$,

$$K_n(t)=\sum _{|j|\le n}\left(1-\frac{|j|}{n+1}\right)e^{ijt}=\sum _{j\in \mathbb{Z}}{\chi }_n(j)\left(1-\frac{|j|}{n+1}\right)e^{ijt}\mathit{\hspace{1em}\hspace{1em}}t\in 𝕋.$$

One calculates that, for $t\notin 4\pi \mathbb{Z}$,

$$K_n(t)=\frac{1}{n+1}{\left(\frac{\sin \frac{n+1}{2}t}{\sin \frac{1}{2}t}\right)}^2.$$

Bernstein’s inequality is a statement about functions whose Fourier transform is supported only on low frequencies. The following is a statement about functions whose Fourier transform is supported only on high frequencies. In particular, for , $L^p(𝕋)$ is a homogeneous Banach space, and so is $C(𝕋)$ with the supremum norm.

Theorem 6.

Let $B$ be a homogeneous Banach 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

$$f(t)=\sum _{|j|\ge n}{a}_j{e}^{ijt},t\in 𝕋,$$

is $m$ times differentiable and $f^{(m)}\in B$, then $f\in B$ and

$${\parallel f\parallel }_B\le C_m{n}^{-m}{\parallel f^{(m)}\parallel }_B.$$

Proof.

Suppose that $m$ is even. It is a fact that if $a_j,j\in \mathbb{Z}$, is an even sequence of nonnegative real numbers such that $a_j\to 0$ as $|j|\to \mathrm{\infty }$ and such that for each $j>0$,

$$a_{j-1}+a_{j+1}-2a_j\ge 0,$$

then there is a nonnegative function $f\in L^1(𝕋)$ such that $\widehat{f}(j)=a_j$ for all $j\in \mathbb{Z}$. Define

It is apparent that $a_j$ is even and tends to $0$ as $|j|\to \mathrm{\infty }$. For $1\le j\le n-2$,

$$a_{j-1}+a_{j+1}-2a_j=0.$$

For $j=n-1$,

$a_{j-1}+a_{j+1}-2a_j$ $=n^{-m}+(n-(n-2))(n^{-m}-{(n+1)}^{-m})+n^{-m}$

$-2\left(n^{-m}+(n-(n-1))(n^{-m}-{(n+1)}^{-m})\right)$

$=0.$

The function $j\mapsto j^{-m}$ is convex on $\{n,n+1,\mathrm{\dots }\}$, as $m\ge 1$, so for $j\ge n$ we have $a_{j-1}+a_{j+1}-2a_j\ge 0$. Therefore, there is some nonnegative ${\varphi }_{m,n}\in L^1(𝕋)$ such that

$$\widehat{{\varphi }_{m,n}}(j)=a_j,j\in \mathbb{Z}.$$

Because ${\varphi }_{m,n}$ is nonnegative, and using ,

Define $d{\mu }_{m,n}(t)=\frac{1}{2\pi }{\varphi }_{m,n}(t)dt$. For $|j|\ge n$,

$\widehat{f^{(m)}*{\mu }_{m,n}}(j)$ $=\widehat{f^{(m)}}(j)\widehat{{\mu }_{m,n}}(j)$

$={(ij)}^m\widehat{f}(j)\widehat{{\varphi }_{m,n}}(j)$

$={(ij)}^m\widehat{f}(j)\cdot {|j|}^{-m}$

$=i^m\widehat{f}(j).$

For , since $\widehat{f}(j)=0$ we have

$$\widehat{f^{(m)}*{\mu }_{m,n}}(j)={(ij)}^m\widehat{f}(j)\widehat{{\varphi }_{m,n}}(j)=0=i^m\widehat{f}(j),$$

so for all $j\in \mathbb{Z}$,

$$\widehat{f^{(m)}*{\mu }_{m,n}}(j)=i^m\widehat{f}(j).$$

This implies that $f^{(m)}*{\mu }_{m,n}=i^{m}f$, which in particular tells us that $f\in B$. Then,

${\parallel f\parallel }_B$ $={\parallel i^{m}f\parallel }_B$

$={\parallel f^{(m)}*{\mu }_{m,n}\parallel }_B$

$\le {\parallel f^{(m)}\parallel }_B{\parallel {\mu }_{m,n}\parallel }_{M(𝕋)}$

$={\parallel {\varphi }_{m,n}\parallel }_1{\parallel f^{(m)}\parallel }_B$

$\le (m+1)n^{-m}{\parallel f^{(m)}\parallel }_B.$

This 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 ${\psi }_l:𝕋\to ℂ$ by

$${\psi }_l(t)=\left(e^{2lit}+\frac{1}{2}{e}^{3lit}\right)K_{l-1}(t),t\in 𝕋.$$

There is a unique $l_n$ such that $n\in \{2l_n,2l_n+1\}$. For $k\ge 0$ an integer, define ${\mathrm{\Psi }}_{n,k}:𝕋\to ℂ$ by

$${\mathrm{\Psi }}_{n,k}(t)={\psi }_{l_n{2}^k}(t),t\in 𝕋.$$

${\mathrm{\Psi }}_{n,k}$ satisfies

$${\parallel {\mathrm{\Psi }}_{n,k}\parallel }_1\le \frac{3}{2}{\parallel K_{k-1}\parallel }_1=\frac{3}{2}\cdot \frac{1}{2\pi }{\int }_{𝕋}|K_{k-1}(t)|𝑑t=\frac{3}{2}\cdot \frac{1}{2\pi }{\int }_{𝕋}{K}_{k-1}(t)𝑑t=\frac{3}{2}.$$

On the one hand, for $j\le 0$, from the definition of ${\psi }_l$ we have $\widehat{{\mathrm{\Psi }}_{n,k}}(j)=0$, hence ${\sum }_{k=0}^{\mathrm{\infty }}\widehat{{\mathrm{\Psi }}_{n,k}}(j)=0$. On the other hand, for $j\ge n$ we assert that

$$\sum _{k=0}^{\mathrm{\infty }}\widehat{{\mathrm{\Psi }}_{n,k}}(j)=1.$$

We define ${\mathrm{\Phi }}_n:𝕋\to ℂ$ by

$${\mathrm{\Phi }}_n(t)=\sum _{k=0}^{\mathrm{\infty }}({\mathrm{\Psi }}_{n,k}*{\varphi }_{1,n{2}^k})(t),t\in 𝕋.$$

We calculate the Fourier coefficients of ${\mathrm{\Phi }}_n$. For $j\ge n$,

$$\widehat{{\mathrm{\Phi }}_n}(j)=\sum _{k=0}^{\mathrm{\infty }}\widehat{{\mathrm{\Phi }}_{n,k}}(j)\widehat{{\varphi }_{1,n{2}^k}}(j)=\frac{1}{j}\sum _{k=0}^{\mathrm{\infty }}\widehat{{\mathrm{\Phi }}_{n,k}}(j)=\frac{1}{j}.$$

As well,

$${\parallel {\mathrm{\Phi }}_n\parallel }_1\le \sum _{k=0}^{\mathrm{\infty }}{\parallel {\mathrm{\Psi }}_{n,k}*{\varphi }_{1,n{2}^k}\parallel }_1\le \sum _{k=0}^{\mathrm{\infty }}{\parallel {\mathrm{\Psi }}_{n,k}\parallel }_1{\parallel {\varphi }_{1,n{2}^k}\parallel }_1\le \frac{3}{2}\sum _{k=0}^{\mathrm{\infty }}2{(n{2}^k)}^{-1}=\frac{6}{n}$$

We now define

$$d{\mu }_{1,n}(t)=\frac{1}{2\pi }({\mathrm{\Phi }}_n(t)-{\mathrm{\Phi }}_n(-t))dt,$$

which satisfies for $|j|\ge n$,

$$\widehat{{\mu }_{1,n}}(j)=\widehat{{\mathrm{\Phi }}_n}(j)-\widehat{{\mathrm{\Phi }}_n}(-j)=\frac{1}{j}$$

and hence

$$\widehat{f^{\prime }*{\mu }_{1,n}}(j)=\widehat{f^{\prime }}(j)\widehat{{\mu }_{1,n}}(j)=ij\widehat{f}(j)\cdot \frac{1}{j}=i\widehat{f}(j).$$

Because $\widehat{f}(j)=0$ for , $\widehat{f^{\prime }*{\mu }_{1,n}}(j)=0$ for , it follows that for any $j\in \mathbb{Z}$,

$$\widehat{f^{\prime }*{\mu }_{1,n}}(j)=i\widehat{f}(j),$$

and therefore,

$$f^{\prime }*{\mu }_{1,n}=if.$$

Then

$${\parallel f\parallel }_B={\parallel if\parallel }_B={\parallel f^{\prime }*{\mu }_{1,n}\parallel }_B\le {\parallel {\mu }_{1,n}\parallel }_{M(𝕋)}{\parallel f^{\prime }\parallel }_B\le 2{\parallel {\mathrm{\Phi }}_n\parallel }_1{\parallel f^{\prime }\parallel }_B\le \frac{12}{n}{\parallel f^{\prime }\parallel }_B.$$

That is, with $C_1=12$ we have

$${\parallel f\parallel }_B\le 12n^{-1}{\parallel f^{\prime }\parallel }_B.$$

For $m=2\nu +1$, we define

$${\mu }_{m,n}={\mu }_{1,n}*{\mu }_{2\nu ,n},$$

for which we have, for $|j|\ge n$,

$$\widehat{f^{(m)}*{\mu }_{m,n}}(j)={(ij)}^m\widehat{f}(j)\widehat{{\mu }_{1,n}}(j)\widehat{{\mu }_{2\nu ,n}}(j)={(ij)}^m\widehat{f}(j)\cdot \frac{1}{j}\cdot j^{-2\nu }=i^m\widehat{f}(j).$$

It follows that

$$f^{(m)}*{\mu }_{m,n}=i^{m}f,$$

whence

${\parallel f\parallel }_B$ $={\parallel i^{m}f\parallel }_B$

$={\parallel f^{(m)}*{\mu }_{m,n}\parallel }_B$

$\le {\parallel {\mu }_{m,n}\parallel }_{M(𝕋)}{\parallel f^{(m)}\parallel }_B$

$\le {\parallel {\mu }_{1,n}\parallel }_{M(𝕋)}{\parallel {\mu }_{2\nu ,n}\parallel }_{M(𝕋)}{\parallel f^{(m)}\parallel }_B$

$\le {\displaystyle \frac{12}{n}}\cdot (2\nu +1)n^{-2\nu }{\parallel f^{(m)}\parallel }_B$

$=12m{n}^{-m}{\parallel f^{(m)}\parallel }_B.$

That is, with $C_m=12m$, we have

$${\parallel f\parallel }_B\le C_m{n}^{-m}{\parallel f^{(m)}\parallel }_B,$$

completing the proof. ∎