# The Bernstein and Nikolsky inequalities for trigonometric polynomials

Jordan Bell
January 28, 2015

## 1 Introduction

Let $\mathbb{T}=\mathbb{R}/2\pi\mathbb{Z}$. For a function $f:\mathbb{T}\to\mathbb{C}$ and $\tau\in\mathbb{T}$, we define $f_{\tau}:\mathbb{T}\to\mathbb{C}$ by $f_{\tau}(t)=f(t-\tau)$. For measurable $f:\mathbb{T}\to\mathbb{C}$ and $0, write

 $\left\|f\right\|_{r}=\left(\frac{1}{2\pi}\int_{\mathbb{T}}|f(t)|^{r}dt\right)^% {1/r}.$

For $f,g\in L^{1}(\mathbb{T})$, write

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

and for $f\in L^{1}(\mathbb{T})$, write

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

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

Let $\mathscr{T}_{n}$ be the set of trigonometric polynomials of degree $\leq n$. We define the Dirichlet kernel $D_{n}:\mathbb{T}\to\mathbb{C}$ by

 $D_{n}(t)=\sum_{|j|\leq n}e^{ijt},\qquad t\in\mathbb{T}.$

It is straightforward to check that if $T\in\mathscr{T}_{n}$ then

 $D_{n}*T=T.$

## 2 Bernstein’s inequality for trigonometric polynomials

DeVore and Lorentz attribute the following inequality to Szegö.11 1 Ronald A. DeVore and George G. Lorentz, Constructive Approximation, p. 97, Theorem 1.1.

###### Theorem 1.

If $T\in\mathscr{T}_{n}$ and $T$ is real valued, then for all $x\in\mathbb{T}$,

 $T^{\prime}(x)^{2}+n^{2}T(x)^{2}\leq n^{2}\left\|T\right\|_{\infty}^{2}.$
###### Proof.

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

 $P_{c}(t)=\frac{T(t+x)\mathrm{sgn}\,T^{\prime}(x)}{c\left\|T\right\|_{\infty}},% \qquad t\in\mathbb{T}.$

$P_{c}\in\mathscr{T}_{n}$, and satisfies

 $P_{c}^{\prime}(0)=\frac{T^{\prime}(x)\mathrm{sgn}\,T^{\prime}(x)}{c\left\|T% \right\|_{\infty}}\geq 0$

and $\left\|P_{c}\right\|_{\infty}\leq\frac{1}{c}<1$. Since $\left\|P_{c}\right\|_{\infty}<1$, in particular $|P_{c}(0)|<1$ and so there is some $\alpha$, $|\alpha|<\frac{\pi}{2n}$, such that $\sin n\alpha=P_{c}(0)$. We define $S\in\mathscr{T}_{n}$ by

 $S(t)=\sin n(t+\alpha)-P_{c}(t),\qquad t\in\mathbb{T},$

which satisfies $S(0)=\sin n\alpha-P_{c}(0)=0$. For $k=-n,\ldots,n$, 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 $\left\|P_{c}\right\|_{\infty}<1$,

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

so by the intermediate value theorem, for each $k=-n,\ldots,n-1$ 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 $j\neq k$ then $c_{j}$ and $c_{k}$ are distinct in $\mathbb{T}$. It is a fact that a trigonometric polynomial of degree $n$ has $\leq 2n$ distinct roots in $\mathbb{T}$, 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 $t_{0}=-\alpha-\frac{\pi}{2}<0$, 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 $S^{\prime}(0)<0$, so $S^{\prime}(0)\geq 0$. This gives

 $0\leq P_{c}^{\prime}(0)=n\cos n\alpha-S^{\prime}(0)\leq n\cos n\alpha=n\sqrt{1% -\sin^{2}n\alpha}=n\sqrt{1-P_{c}(0)^{2}}.$

Thus

 $P_{c}^{\prime}(0)\leq n\sqrt{1-P_{c}(0)^{2}},$

or

 $n^{2}P_{c}(0)+P_{c}^{\prime}(0)^{2}\leq n^{2}.$

Because

 $P_{c}(0)^{2}=\frac{T(x)^{2}}{c^{2}\left\|T\right\|_{\infty}^{2}},\qquad P_{c}^% {\prime}(0)^{2}=\frac{T^{\prime}(x)^{2}}{c^{2}\left\|T\right\|_{\infty}^{2}}$

we get

 $n^{2}T(x)^{2}+T^{\prime}(x)^{2}\leq c^{2}n^{2}\left\|T\right\|_{\infty}^{2}.$

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

 $n^{2}T(x)^{2}+T^{\prime}(x)^{2}\leq n^{2}\left\|T\right\|_{\infty}^{2},$

completing the proof. ∎

Using the above we now prove Bernstein’s inequality.22 2 Ronald A. DeVore and George G. Lorentz, Constructive Approximation, p. 98.

###### Theorem 2 (Bernstein’s inequality).

If $T\in\mathscr{T}_{n}$, then

 $\left\|T^{\prime}\right\|_{\infty}\leq n\left\|T\right\|_{\infty}.$
###### Proof.

There is some $x_{0}\in\mathbb{T}$ such that $|T^{\prime}(x_{0})|=\left\|T^{\prime}\right\|_{\infty}$. Let $\alpha\in\mathbb{R}$ be such that $e^{i\alpha}T^{\prime}(x_{0})=\left\|T^{\prime}\right\|_{\infty}$. Define $S(x)=\mathrm{Re}\,(e^{i\alpha}T(x))$ for $x\in\mathbb{T}$, which satisfies $S^{\prime}(x)=\mathrm{Re}\,(e^{i\alpha}T^{\prime}(x))$ and in particular

 $S^{\prime}(x_{0})=\mathrm{Re}\,(e^{i\alpha}T^{\prime}(x_{0}))=e^{i\alpha}T^{% \prime}(x_{0})=\left\|T^{\prime}\right\|_{\infty}.$

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

 $S^{\prime}(x_{0})^{2}+n^{2}S(x_{0})^{2}\leq n^{2}\left\|S\right\|_{\infty}^{2}.$

A fortiori,

 $S^{\prime}(x_{0})^{2}\leq n^{2}\left\|S\right\|_{\infty}^{2},$

giving, because $S^{\prime}(x_{0})=\left\|T^{\prime}\right\|_{\infty}$ and $\left\|S\right\|_{\infty}\leq\left\|T\right\|_{\infty}$,

 $\left\|T^{\prime}\right\|_{\infty}^{2}\leq n^{2}\left\|T\right\|_{\infty}^{2},$

proving the claim. ∎

The following is a version of Bernstein’s inequality.33 3 Ronald A. DeVore and George G. Lorentz, Constructive Approximation, p. 101, Theorem 2.4.

###### Theorem 3.

If $T\in\mathscr{T}_{n}$ and $A\subset\mathbb{T}$ is a Borel set, there is some $x_{0}\in\mathbb{T}$ such that

 $\int_{A}|T^{\prime}(t)|dt\leq n\int_{A-x_{0}}|T(t)|dt.$
###### Proof.

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

 $Q(x)=\int_{\mathbb{T}}\chi_{A}(t)T(t+x)\mathrm{sgn}\,T^{\prime}(t)dt,\qquad x% \in\mathbb{T},$

which we can write as

 $\displaystyle Q(x)$ $\displaystyle=\int_{\mathbb{T}}\chi_{A}(t)\sum_{j}\widehat{T}(j)e^{ij(t+x)}% \mathrm{sgn}\,T^{\prime}(t)dt$ $\displaystyle=\sum_{j}\widehat{T}(j)\left(\int_{\mathbb{T}}\chi_{A}(t)e^{ijt}% \mathrm{sgn}\,T^{\prime}(t)dt\right)e^{ijx},$

showing that $Q\in\mathscr{T}_{n}$. Also,

 $Q^{\prime}(x)=\int_{\mathbb{T}}\chi_{A}(t)T^{\prime}(t+x)\mathrm{sgn}\,T^{% \prime}(t)dt,\qquad x\in\mathbb{T}.$

Let $x_{0}\in\mathbb{T}$ with $|Q(x_{0})|=\left\|Q\right\|_{\infty}$. Applying Theorem 2 we get

 $\left\|Q^{\prime}\right\|_{\infty}\leq n\left\|Q\right\|_{\infty}.$

Using

 $Q^{\prime}(0)=\int_{\mathbb{T}}\chi_{A}(t)T^{\prime}(t)\mathrm{sgn}\,T^{\prime% }(t)dt=\int_{\mathbb{T}}\chi_{A}(t)|T^{\prime}(t)|dt,$

this gives

 $\displaystyle\int_{\mathbb{T}}\chi_{A}(t)|T^{\prime}(t)|dt$ $\displaystyle\leq n\left\|Q\right\|_{\infty}$ $\displaystyle=n|Q(t_{0})|$ $\displaystyle=n\left|\int_{\mathbb{T}}\chi_{A}(t)T(t+x_{0})\mathrm{sgn}\,T^{% \prime}(t)dt\right|$ $\displaystyle\leq n\int_{\mathbb{T}}\chi_{A}(t)|T(t+x_{0})|dt$ $\displaystyle=n\int_{\mathbb{T}}\chi_{A-x_{0}}(t)|T(t)|dt.$

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

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

If $T\in\mathscr{T}_{n}$, then

 $\left\|T^{\prime}\right\|_{1}\leq n\left\|T\right\|_{1}.$

## 3 Nikolsky’s inequality for trigonometric polynomials

DeVore and Lorentz attribute the following inequality to Sergey Nikolsky.44 4 Ronald A. DeVore and George G. Lorentz, Constructive Approximation, p. 102, Theorem 2.6.

###### Theorem 5 (Nikolsky’s inequality).

If $T\in\mathscr{T}_{n}$ and $0, then for $r\geq\frac{q}{2}$ an integer,

 $\left\|T\right\|_{p}\leq(2nr+1)^{\frac{1}{q}-\frac{1}{p}}\left\|T\right\|_{q}.$
###### Proof.

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

 $\displaystyle|T(x)^{r}|$ $\displaystyle=\left|\frac{1}{2\pi}\int_{\mathbb{T}}T(t)^{r}D_{m}(x-t)\right|$ $\displaystyle\leq\frac{1}{2\pi}\int_{\mathbb{T}}|T(t)|^{r}|D_{m}(x-t)|dt$ $\displaystyle\leq\left\|T\right\|_{\infty}^{r-\frac{q}{2}}\cdot\frac{1}{2\pi}% \int_{\mathbb{T}}|T(t)|^{\frac{q}{2}}|D_{m}(x-t)|dt$ $\displaystyle\leq\left\|T\right\|_{\infty}^{r-\frac{q}{2}}\left\||T|^{q/2}% \right\|_{2}\left\|D_{m}\right\|_{2}$ $\displaystyle=\left\|T\right\|_{\infty}^{r-\frac{q}{2}}\left\|T\right\|_{q}^{% \frac{q}{2}}\left\|\widehat{D_{m}}\right\|_{\ell^{2}(\mathbb{Z})}$ $\displaystyle=\sqrt{2m+1}\left\|T\right\|_{\infty}^{r-\frac{q}{2}}\left\|T% \right\|_{q}^{\frac{q}{2}}.$

Hence

 $\left\|T\right\|_{\infty}^{r}\leq\sqrt{2m+1}\left\|T\right\|_{\infty}^{r-\frac% {q}{2}}\left\|T\right\|_{q}^{\frac{q}{2}},$

thus

 $\left\|T\right\|_{\infty}\leq(2m+1)^{\frac{1}{q}}\left\|T\right\|_{q}.$

Then, using $\left\|T\right\|_{p}\leq\left\|T\right\|_{\infty}^{1-\frac{q}{p}}\left\|T% \right\|_{q}^{\frac{q}{p}}$, we have

 $\left\|T\right\|_{p}\leq(2m+1)^{\frac{1}{q}-\frac{1}{p}}\left\|T\right\|_{q}^{% 1-\frac{q}{p}}\left\|T\right\|_{q}^{\frac{q}{p}}=(2m+1)^{\frac{1}{q}-\frac{1}{% p}}\left\|T\right\|_{q}.$

## 4 The complementary Bernstein inequality

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

Fejér’s kernel is, for $n\geq 0$,

 $K_{n}(t)=\sum_{|j|\leq n}\left(1-\frac{|j|}{n+1}\right)e^{ijt}=\sum_{j\in% \mathbb{Z}}\chi_{n}(j)\left(1-\frac{|j|}{n+1}\right)e^{ijt}\qquad t\in\mathbb{% T}.$

One calculates that, for $t\not\in 4\pi\mathbb{Z}$,

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

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.55 5 Yitzhak Katznelson, An Introduction to Harmonic Analysis, third ed., p. 55, Theorem 8.4. In particular, for $1\leq p<\infty$, $L^{p}(\mathbb{T})$ is a homogeneous Banach space, and so is $C(\mathbb{T})$ 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

 $f(t)=\sum_{|j|\geq n}a_{j}e^{ijt},\qquad t\in\mathbb{T},$

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

 $\left\|f\right\|_{B}\leq C_{m}n^{-m}\left\|f^{(m)}\right\|_{B}.$
###### 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 $|j|\to\infty$ and such that for each $j>0$,

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

then there is a nonnegative function $f\in L^{1}(\mathbb{T})$ such that $\hat{f}(j)=a_{j}$ for all $j\in\mathbb{Z}$.66 6 Yitzhak Katznelson, An Introduction to Harmonic Analysis, third ed., p. 24, Theorem 4.1. Define

 $a_{j}=\begin{cases}j^{-m}&|j|\geq n\\ n^{-m}+(n-|j|)(n^{-m}-(n+1)^{-m})&|j|\leq n-1.\end{cases}$

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

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

For $j=n-1$,

 $\displaystyle a_{j-1}+a_{j+1}-2a_{j}$ $\displaystyle=n^{-m}+(n-(n-2))(n^{-m}-(n+1)^{-m})+n^{-m}$ $\displaystyle-2\left(n^{-m}+(n-(n-1))(n^{-m}-(n+1)^{-m})\right)$ $\displaystyle=0.$

The function $j\mapsto j^{-m}$ is convex on $\{n,n+1,\ldots\}$, as $m\geq 1$, so for $j\geq n$ we have $a_{j-1}+a_{j+1}-2a_{j}\geq 0$. Therefore, there is some nonnegative $\phi_{m,n}\in L^{1}(\mathbb{T})$ such that

 $\widehat{\phi_{m,n}}(j)=a_{j},\qquad j\in\mathbb{Z}.$

Because $\phi_{m,n}$ is nonnegative, and using $n^{-m}-(n+1)^{-m}<\frac{m}{n}n^{-m}$,

 $\left\|\phi_{m,n}\right\|_{1}=\widehat{\phi_{m,n}}(0)=n^{-m}+n(n^{-m}-(n+1)^{-% m})<(m+1)n^{-m}.$

Define $d\mu_{m,n}(t)=\frac{1}{2\pi}\phi_{m,n}(t)dt$. For $|j|\geq n$,

 $\displaystyle\widehat{f^{(m)}*\mu_{m,n}}(j)$ $\displaystyle=\widehat{f^{(m)}}(j)\widehat{\mu_{m,n}}(j)$ $\displaystyle=(ij)^{m}\hat{f}(j)\widehat{\phi_{m,n}}(j)$ $\displaystyle=(ij)^{m}\hat{f}(j)\cdot|j|^{-m}$ $\displaystyle=i^{m}\hat{f}(j).$

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

 $\widehat{f^{(m)}*\mu_{m,n}}(j)=(ij)^{m}\hat{f}(j)\widehat{\phi_{m,n}}(j)=0=i^{% m}\hat{f}(j),$

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

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

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

 $\displaystyle\left\|f\right\|_{B}$ $\displaystyle=\left\|i^{m}f\right\|_{B}$ $\displaystyle=\left\|f^{(m)}*\mu_{m,n}\right\|_{B}$ $\displaystyle\leq\left\|f^{(m)}\right\|_{B}\left\|\mu_{m,n}\right\|_{M(\mathbb% {T})}$ $\displaystyle=\left\|\phi_{m,n}\right\|_{1}\left\|f^{(m)}\right\|_{B}$ $\displaystyle\leq(m+1)n^{-m}\left\|f^{(m)}\right\|_{B}.$

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 $\psi_{l}:\mathbb{T}\to\mathbb{C}$ by

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

There is a unique $l_{n}$ such that $n\in\{2l_{n},2l_{n}+1\}$. For $k\geq 0$ an integer, define $\Psi_{n,k}:\mathbb{T}\to\mathbb{C}$ by

 $\Psi_{n,k}(t)=\psi_{l_{n}2^{k}}(t),\qquad t\in\mathbb{T}.$

$\Psi_{n,k}$ satisfies

 $\left\|\Psi_{n,k}\right\|_{1}\leq\frac{3}{2}\left\|K_{k-1}\right\|_{1}=\frac{3% }{2}\cdot\frac{1}{2\pi}\int_{\mathbb{T}}|K_{k-1}(t)|dt=\frac{3}{2}\cdot\frac{1% }{2\pi}\int_{\mathbb{T}}K_{k-1}(t)dt=\frac{3}{2}.$

On the one hand, for $j\leq 0$, from the definition of $\psi_{l}$ we have $\widehat{\Psi_{n,k}}(j)=0$, hence $\sum_{k=0}^{\infty}\widehat{\Psi_{n,k}}(j)=0$. On the other hand, for $j\geq n$ we assert that

 $\sum_{k=0}^{\infty}\widehat{\Psi_{n,k}}(j)=1.$

We define $\Phi_{n}:\mathbb{T}\to\mathbb{C}$ by

 $\Phi_{n}(t)=\sum_{k=0}^{\infty}(\Psi_{n,k}*\phi_{1,n2^{k}})(t),\qquad t\in% \mathbb{T}.$

We calculate the Fourier coefficients of $\Phi_{n}$. For $j\geq n$,

 $\widehat{\Phi_{n}}(j)=\sum_{k=0}^{\infty}\widehat{\Phi_{n,k}}(j)\widehat{\phi_% {1,n2^{k}}}(j)=\frac{1}{j}\sum_{k=0}^{\infty}\widehat{\Phi_{n,k}}(j)=\frac{1}{% j}.$

As well,

 $\left\|\Phi_{n}\right\|_{1}\leq\sum_{k=0}^{\infty}\left\|\Psi_{n,k}*\phi_{1,n2% ^{k}}\right\|_{1}\leq\sum_{k=0}^{\infty}\left\|\Psi_{n,k}\right\|_{1}\left\|% \phi_{1,n2^{k}}\right\|_{1}\leq\frac{3}{2}\sum_{k=0}^{\infty}2(n2^{k})^{-1}=% \frac{6}{n}$

We now define

 $d\mu_{1,n}(t)=\frac{1}{2\pi}(\Phi_{n}(t)-\Phi_{n}(-t))dt,$

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

 $\widehat{\mu_{1,n}}(j)=\widehat{\Phi_{n}}(j)-\widehat{\Phi_{n}}(-j)=\frac{1}{j}$

and hence

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

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

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

and therefore,

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

Then

 $\left\|f\right\|_{B}=\left\|if\right\|_{B}=\left\|f^{\prime}*\mu_{1,n}\right\|% _{B}\leq\left\|\mu_{1,n}\right\|_{M(\mathbb{T})}\left\|f^{\prime}\right\|_{B}% \leq 2\left\|\Phi_{n}\right\|_{1}\left\|f^{\prime}\right\|_{B}\leq\frac{12}{n}% \left\|f^{\prime}\right\|_{B}.$

That is, with $C_{1}=12$ we have

 $\left\|f\right\|_{B}\leq 12n^{-1}\left\|f^{\prime}\right\|_{B}.$

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

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

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

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

It follows that

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

whence

 $\displaystyle\left\|f\right\|_{B}$ $\displaystyle=\left\|i^{m}f\right\|_{B}$ $\displaystyle=\left\|f^{(m)}*\mu_{m,n}\right\|_{B}$ $\displaystyle\leq\left\|\mu_{m,n}\right\|_{M(\mathbb{T})}\left\|f^{(m)}\right% \|_{B}$ $\displaystyle\leq\left\|\mu_{1,n}\right\|_{M(\mathbb{T})}\left\|\mu_{2\nu,n}% \right\|_{M(\mathbb{T})}\left\|f^{(m)}\right\|_{B}$ $\displaystyle\leq\frac{12}{n}\cdot(2\nu+1)n^{-2\nu}\left\|f^{(m)}\right\|_{B}$ $\displaystyle=12mn^{-m}\left\|f^{(m)}\right\|_{B}.$

That is, with $C_{m}=12m$, we have

 $\left\|f\right\|_{B}\leq C_{m}n^{-m}\left\|f^{(m)}\right\|_{B},$

completing the proof. ∎