Bernstein’s inequality and Nikolsky’s inequality for ${ℝ}^d$

1 Complex Borel measures and the Fourier transform

Let $ℳ({ℝ}^d)=rca({ℝ}^d)$ be the set of complex Borel measures on ${ℝ}^d$. This is a Banach algebra with the total variation norm, with convolution as multiplication; for $\mu \in ℳ({ℝ}^d)$, we denote by $|\mu |$ the total variation of $\mu$, which itself belongs to $ℳ({ℝ}^d)$, and the total variation norm of $\mu$ is $\parallel \mu \parallel =|\mu |({ℝ}^d)$.

For $\mu \in ℳ({ℝ}^d)$, it is a fact that the union $O$ of all open sets $U\subset {ℝ}^d$ such that $|\mu |(U)=0$ itself satisfies $|\mu |(O)=0$. We define $\mathrm{supp}\mu ={ℝ}^d\setminus O$, called the support of $\mu$.

For $\mu \in ℳ({ℝ}^d)$, we define $\widehat{\mu }:{ℝ}^d\to ℂ$ by

$$\widehat{\mu }(\xi )={\int }_{{ℝ}^d}{e}^{-2\pi i\xi \cdot x}𝑑\mu (x),\xi \in {ℝ}^d.$$

It is a fact that $\widehat{\mu }$ belongs to $C_u(ℝ)$, the collection of bounded uniformly continuous functions ${ℝ}^d\to ℂ$. For $\xi \in {ℝ}^d$,

$$|\widehat{\mu }(\xi )|\le {\int }_{{ℝ}^d}|e^{-2\pi i\xi \cdot x}|d|\mu |(x)=|\mu |({ℝ}^d)=\parallel \mu \parallel .$$ (1)

Let $m_d$ be Lebesgue measure on ${ℝ}^d$. For $f\in L^1({ℝ}^d)$, let

$${\mathrm{\Lambda }}_f=f{m}_d,$$

which belongs to $ℳ({ℝ}^d)$. We define $\widehat{f}:{ℝ}^d\to ℂ$ by

$$\widehat{f}(\xi )=\widehat{{\mathrm{\Lambda }}_f}(\xi )={\int }_{{ℝ}^d}{e}^{-2\pi i\xi \cdot x}𝑑{\mathrm{\Lambda }}_f(x)={\int }_{{ℝ}^d}f(x)e^{-2\pi i\xi \cdot x}𝑑m_d(x),\xi \in {ℝ}^d.$$

The following theorem establishes properties of the Fourier transform of a complex Borel measure with compact support.

Theorem 1.

If $\mu \mathrm{\in }\mathrm{M}\mathit{}\mathrm{(}{\mathrm{R}}^d\mathrm{)}$ and $\mathrm{supp}\mathit{}\mu$ is compact, then $\widehat{\mu }\mathrm{\in }{C}^{\mathrm{\infty }}\mathit{}\mathrm{(}{\mathrm{R}}^d\mathrm{)}$ and for any multi-index $\alpha$,

$$D^{\alpha }\widehat{\mu }=ℱ({(-2\pi ix)}^{\alpha }\mu ).$$

For $R\mathrm{>}\mathrm{0}$, if $\mathrm{supp}\mathit{}\mu \mathrm{\subset }\overline{B\mathit{}\mathrm{(}\mathrm{0}\mathrm{,}R\mathrm{)}}$, then

$${\parallel D^{\alpha }\widehat{\mu }\parallel }_{\mathrm{\infty }}\le {(2\pi R)}^{{|\alpha |}_1}\parallel \mu \parallel .$$

Proof.

For $j=1,\mathrm{\dots },d$, let $e_j$ be the $j$th coordinate vector in ${ℝ}^d$, with length 1. Let $\xi \in {ℝ}^d$, and define

$$\mathrm{\Delta }(h)=\frac{\widehat{\mu }(\xi +h{e}_j)-\widehat{\mu }(\xi )}{h},h\ne 0.$$

We can write this as

$$\mathrm{\Delta }(h)={\int }_{{ℝ}^d}\frac{e^{-2\pi ih{x}_j}-1}{h}{e}^{-2\pi i\xi \cdot x}𝑑\mu (x).$$

For any $x\in {ℝ}^d$,

$$\left|\frac{e^{-2\pi ih{x}_j}-1}{h}\right|=\frac{|e^{-2\pi ih{x}_j}-1|}{|h|}\le \frac{|-2\pi ih{x}_j|}{|h|}=2\pi |x_j|.$$

Because $\mu$ has compact support, $2\pi |x_j|\in L^1(\mu )$. Furthermore, for each $x\in {ℝ}^d$ we have

$$\frac{e^{-2\pi ih{x}_j}-1}{h}\to -2\pi i{x}_j,h\to 0.$$

Therefore, the dominated convergence theorem tells us that

$$\underset{h\to 0}{lim}\mathrm{\Delta }(h)={\int }_{{ℝ}^d}-2\pi i{x}_j{e}^{-2\pi i\xi \cdot x}d\mu (x).$$

On the other hand, for ${\alpha }_k=1$ for $k=j$ and ${\alpha }_k=0$ otherwise,

$$(D^{\alpha }\widehat{\mu })(\xi )=\underset{h\to 0}{lim}\mathrm{\Delta }(h),$$

so

$$(D^{\alpha }\widehat{\mu })(\xi )={\int }_{{ℝ}^d}{(-2\pi ix)}^{\alpha }{e}^{-2\pi i\xi \cdot x}𝑑\mu (x)=ℱ({(-2\pi ix)}^{\alpha }\mu )(\xi ),$$

and in particular, $\widehat{\mu }\in C^1({ℝ}^d)$. (The Fourier transform of a regular complex Borel measure on a locally compact abelian group is bounded and uniformly continuous.) Because $\mu$ has compact support so does ${(-2\pi ix)}^{\alpha }\mu$, hence we can play the above game with ${(-2\pi ix)}^{\alpha }\mu$, and by induction it follows that for any $\alpha$,

$$D^{\alpha }\widehat{\mu }=ℱ({(-2\pi ix)}^{\alpha }\mu ),$$

and in particular, $\widehat{\mu }\in C^{\mathrm{\infty }}({ℝ}^d)$.

Suppose that $\mathrm{supp}\mu \subset \overline{B(0,R)}$. The total variation of the complex measure ${(-2\pi ix)}^{\alpha }\mu$ is the positive measure

$${(2\pi )}^{{|\alpha |}_1}{|x_1|}^{{\alpha }_1}\mathrm{\cdots }{|x_d|}^{{\alpha }_d}|\mu |,$$

hence

$\parallel {(-2\pi ix)}^{\alpha }\mu \parallel$ $={(2\pi )}^{{|\alpha |}_1}{\displaystyle {\int }_{{ℝ}^d}}{|x_1|}^{{\alpha }_1}\mathrm{\cdots }{|x_d|}^{{\alpha }_d}d|\mu |(x)$

$={(2\pi )}^{{|\alpha |}_1}{\displaystyle {\int }_{\overline{B(0,R)}}}{|x_1|}^{{\alpha }_1}\mathrm{\cdots }{|x_d|}^{{\alpha }_d}d|\mu |(x)$

$\le {(2\pi )}^{{|\alpha |}_1}{\displaystyle {\int }_{\overline{B(0,R)}}}{R}^{{\alpha }_1}\mathrm{\cdots }{R}^{{\alpha }_d}d|\mu |(x)$

$={(2\pi R)}^{{|\alpha |}_1}{\displaystyle {\int }_{\overline{B(0,R)}}}d|\mu |(x)$

$={(2\pi R)}^{{|\alpha |}_1}{\displaystyle {\int }_{{ℝ}^d}}d|\mu |(x)$

$={(2\pi R)}^{{|\alpha |}_1}\parallel \mu \parallel .$

Then using (1),

$${\parallel ℱ({(-2\pi ix)}^{\alpha }\mu )\parallel }_{\mathrm{\infty }}\le \parallel {(-2\pi ix)}^{\alpha }\mu \parallel \le {(2\pi R)}^{{|\alpha |}_1}\parallel \mu \parallel .$$

But we have already established that $D^{\alpha }\widehat{\mu }=ℱ({(-2\pi ix)}^{\alpha }\mu )$, which with the above inequality completes the proof. ∎

2 Test functions

For an open subset $\mathrm{\Omega }$ of ${ℝ}^d$, we denote by $𝒟(\mathrm{\Omega })$ the set of those $\varphi \in C^{\mathrm{\infty }}(\mathrm{\Omega })$ such that $\mathrm{supp}\varphi$ is a compact set. Elements of $𝒟(\mathrm{\Omega })$ are called test functions.

It is a fact that there is a test function $\varphi$ satisfying: (i) $\varphi (x)=1$ for $|x|\le 1$, (ii) $\varphi (x)=0$ for $|x|\ge 2$, (iii) $0\le \varphi \le 1$, and (iv) $\varphi$ is radial. We write, for $k=1,2,\mathrm{\dots }$,

$${\varphi }_k(x)=\varphi (k^{-1}x),x\in {ℝ}^d.$$

For any multi-index $\alpha$,

$$(D^{\alpha }{\varphi }_k)(x)=k^{-{|\alpha |}_1}(D^{\alpha }\varphi )(k^{-1}x),x\in {ℝ}^d,$$

hence

$${\parallel D^{\alpha }{\varphi }_k\parallel }_{\mathrm{\infty }}=k^{-{|\alpha |}_1}{\parallel D^{\alpha }\varphi \parallel }_{\mathrm{\infty }}.$$ (2)

We use the following lemma to prove the theorem that comes after it.

Lemma 2.

Suppose that $f\mathrm{\in }{C}^N\mathit{}\mathrm{(}{\mathrm{R}}^d\mathrm{)}$ and $D^{\alpha }\mathit{}f\mathrm{\in }{L}^{\mathrm{1}}\mathit{}\mathrm{(}{\mathrm{R}}^d\mathrm{)}$ for each $\mathrm{|}\alpha \mathrm{|}\mathrm{\le }N$. Then for each $\mathrm{|}\alpha \mathrm{|}\mathrm{\le }N$, $D^{\alpha }\mathit{}\mathrm{(}{\varphi }_k\mathit{}f\mathrm{)}\mathrm{\to }{D}^{\alpha }\mathit{}f$ in $L^{\mathrm{1}}\mathit{}\mathrm{(}{\mathrm{R}}^d\mathrm{)}$ as $k\mathrm{\to }\mathrm{\infty }$.

Proof.

Let $|\alpha |\le N$. In the case $\alpha =0$,

${\parallel {\varphi }_{k}f-f\parallel }_1$ $={\displaystyle {\int }_{{ℝ}^d}}|{\varphi }_k(x)f(x)-f(x)|𝑑x$

$={\displaystyle {\int }_{|x|\ge k}}|{\varphi }_k(x)f(x)-f(x)|𝑑x$

$\le {\displaystyle {\int }_{|x|\ge k}}|f(x)|𝑑x.$

Because $f\in L^1({ℝ}^d)$, this tends to $0$ as $k\to \mathrm{\infty }$.

Suppose that $\alpha >0$. The Leibniz rule tells us that with $c_{\beta }=\left(\genfrac{}{}{0pt}{}{\alpha }{\beta }\right)$, we have, for each $k$,

For $C_1={\max }_{\beta }|c_{\beta }|$,

${\parallel D^{\alpha }({\varphi }_{k}f)-{\varphi }_k{D}^{\alpha }f\parallel }_1$

Let . By (2), for we have

$${\parallel D^{\beta }{\varphi }_k\parallel }_{\mathrm{\infty }}=k^{-{|\beta |}_1}{\parallel D^{\beta }\varphi \parallel }_{\mathrm{\infty }}\le C_2{k}^{-{|\beta |}_1}\le C_2{k}^{-1}.$$

Thus

which tends to $0$ as $k\to \mathrm{\infty }$. For any $k$,

${\parallel {\varphi }_k{D}^{\alpha }f-D^{\alpha }f\parallel }_1$ $={\displaystyle {\int }_{{ℝ}^d}}|{\varphi }_k(x)(D^{\alpha }f)(x)-(D^{\alpha }f)(x)|𝑑x$

$={\displaystyle {\int }_{|x|\ge k}}|{\varphi }_k(x)(D^{\alpha }f)(x)-(D^{\alpha }f)(x)|𝑑x$

$\le {\displaystyle {\int }_{|x|\ge k}}|(D^{\alpha }f)(x)|𝑑x,$

and because $D^{\alpha }f\in L^1({ℝ}^d)$, this tends to $0$ as $k\to \mathrm{\infty }$. But

$${\parallel D^{\alpha }({\varphi }_{k}f)-D^{\alpha }f\parallel }_1\le {\parallel D^{\alpha }({\varphi }_{k}f)-{\varphi }_k{D}^{\alpha }f\parallel }_1+{\parallel {\varphi }_k{D}^{\alpha }f-D^{\alpha }f\parallel }_1,$$

which completes the proof. ∎

Now we calculate the Fourier transform of the derivative of a function, and show that the smoother a function is the faster its Fourier transform decays.

Theorem 3.

If $f\mathrm{\in }{C}^N\mathit{}\mathrm{(}{\mathrm{R}}^d\mathrm{)}$ and $D^{\alpha }\mathit{}f\mathrm{\in }{L}^{\mathrm{1}}\mathit{}\mathrm{(}{\mathrm{R}}^d\mathrm{)}$ for each $\mathrm{|}\alpha \mathrm{|}\mathrm{\le }N$, then for each $\mathrm{|}\alpha \mathrm{|}\mathrm{\le }N$,

$$\widehat{D^{\alpha }f}(\xi )={(2\pi i\xi )}^{\alpha }\widehat{f}(\xi ),\xi \in {ℝ}^d.$$ (3)

There is a constant $C\mathrm{=}C\mathit{}\mathrm{(}f\mathrm{,}N\mathrm{)}$ such that

$$|\widehat{f}(\xi )|\le C{(1+|\xi |)}^{-N},\xi \in {ℝ}^d.$$

Proof.

If $g\in C_c^1({ℝ}^d)$, then for any $1\le j\le d$, integrating by parts,

$${\int }_{{ℝ}^d}({\partial }_{j}g)(x)e^{-2\pi i\xi \cdot x}𝑑x=2\pi i{\xi }_j{\int }_{{ℝ}^d}g(x)e^{-2\pi i\xi \cdot x}𝑑x.$$

It follows by induction that if $g\in C_c^N({ℝ}^d)$, then for each $|\alpha |\le N$,

$$\widehat{D^{\alpha }g}(\xi )={(2\pi i\xi )}^{\alpha }\widehat{g}(\xi ),\xi \in {ℝ}^d.$$

Let $|\alpha |\le N$. For $k=1,2,\mathrm{\dots }$, let $f_k={\varphi }_{k}f$. For each $k$ we have $f_k\in C^N({ℝ}^d)$, hence

$$\widehat{D^{\alpha }{f}_k}(\xi )={(2\pi i\xi )}^{\alpha }\widehat{f_k}(\xi ),\xi \in {ℝ}^d.$$

On the one hand,

$${\parallel \widehat{D^{\alpha }{f}_k}-\widehat{D^{\alpha }f}\parallel }_{\mathrm{\infty }}={\parallel ℱ(D^{\alpha }{f}_k-D^{\alpha }f)\parallel }_{\mathrm{\infty }}\le {\parallel D^{\alpha }{f}_k-D^{\alpha }f\parallel }_1,$$

and Lemma 2 tells us that this tends to $0$ as $k\to \mathrm{\infty }$. On the other hand, for $\xi \in {ℝ}^d$,

$|\widehat{D^{\alpha }{f}_k}(\xi )-{(2\pi i\xi )}^{\alpha }\widehat{f}(\xi )|$ $=|{(2\pi i\xi )}^{\alpha }\widehat{f_k}(\xi )-{(2\pi i\xi )}^{\alpha }\widehat{f}(\xi )|$

$=|{(2\pi i\xi )}^{\alpha }||ℱ(f_k-f)(\xi )|$

$\le |{(2\pi i\xi )}^{\alpha }|{\parallel f_k-f\parallel }_1,$

which by Lemma 2 tends to $0$ as $k\to \mathrm{\infty }$. Therefore, for $\xi \in {ℝ}^d$,

$$|\widehat{D^{\alpha }f}(\xi )-{(2\pi i\xi )}^{\alpha }\widehat{f}(\xi )|\le {\parallel \widehat{D^{\alpha }{f}_k}-\widehat{D^{\alpha }f}\parallel }_{\mathrm{\infty }}+|\widehat{D^{\alpha }{f}_k}(\xi )-{(2\pi i\xi )}^{\alpha }\widehat{f}(\xi )|,$$

and because the right-hand side tends to $0$ as $k\to \mathrm{\infty }$, we get

$$\widehat{D^{\alpha }f}(\xi )={(2\pi i\xi )}^{\alpha }\widehat{f}(\xi ).$$

If $y\in S^{d-1}$ then there is at least one $1\le j\le d$ with $y_j\ne 0$, from which we get

$$\sum _{{|\beta |}_1=N}|y^{\beta }|>0.$$

The function $y\mapsto {\sum }_{{|\beta |}_1=N}|y^{\beta }|$ is continuous $S^{d-1}\to ℝ$, so there is some $C_N>0$ such that

$$\frac{1}{C_N}\le \sum _{{|\beta |}_1=N}|y^{\beta }|,y\in S^{d-1}.$$

For nonzero $x\in {ℝ}^d$, write $x=|x|y$, with which ${\sum }_{{|\beta |}_1=N}|x^{\beta }|={|x|}^N{\sum }_{{|\beta |}_1=N}|y^{\beta }|$. Therefore

$${|x|}^N\le C_N\sum _{{|\beta |}_1=N}|x^{\beta }|,x\in {ℝ}^d.$$

For $|\alpha |\le N$, because the Fourier transform of an element of $L^1$ belongs to $C_0$, we have by (3) that $\xi \mapsto {\xi }^{\alpha }\widehat{f}(\xi )$ belongs to $C_0({ℝ}^d)$, and in particular is bounded. Then for $\xi \in {ℝ}^d$,

${|\xi |}^N|\widehat{f}(\xi )|$ $\le C_N{\displaystyle \sum _{{|\beta |}_1=N}}|{\xi }^{\beta }||\widehat{f}(\xi )|$

$=C_N{\displaystyle \sum _{{|\beta |}_1=N}}|{\xi }^{\beta }\widehat{f}(\xi )|$

$\le C_N{\displaystyle \sum _{{|\beta |}_1=N}}{\parallel {\xi }^{\beta }\widehat{f}(\xi )\parallel }_{\mathrm{\infty }}$

$=C^{\prime }.$

On the one hand, for $|\xi |\ge 1$ we have

$$1+|\xi |\le 2|\xi |,$$

hence

$${|\xi |}^{-N}\le {\left(\frac{1+|\xi |}{2}\right)}^{-N}=2^N{(1+|\xi |)}^{-N},$$

giving

$$|\widehat{f}(\xi )|\le C^{\prime }{|\xi |}^{-N}\le C^{\prime }{2}^N{(1+|\xi |)}^{-N}.$$

On the other hand, for $|\xi |\le 1$ we have

$$1+|\xi |\le 2,$$

and so

$$|\widehat{f}(\xi )|\le {\parallel \widehat{f}\parallel }_{\mathrm{\infty }}{2}^N{2}^{-N}\le {\parallel \widehat{f}\parallel }_{\mathrm{\infty }}{2}^N{(1+|\xi |)}^{-N}.$$

Thus, for

$$C=\max \{2^N{C}^{\prime },2^N{\parallel \widehat{f}\parallel }_{\mathrm{\infty }}\}$$

we have

$$|\widehat{f}(\xi )|\le C{(1+|\xi |)}^{-N},\xi \in {ℝ}^d,$$

completing the proof. ∎

3 Bernstein’s inequality for L²

For a Borel measurable function $f:{ℝ}^d\to ℂ$, let $O$ be the union of those open subsets $U$ of ${ℝ}^d$ such that $f(x)=0$ for almost all $x\in U$. In other words, $O$ is the largest open set on which $f=0$ almost everywhere. The essential support of $f$ is the set

$$\mathrm{ess}\mathrm{supp}f={ℝ}^d\setminus O.$$

The following is Bernstein’s inequality for $L^{\mathrm{2}}\mathbf{}\mathrm{(}{\mathrm{R}}^d\mathrm{)}$.

Theorem 4.

If $f\mathrm{\in }{L}^{\mathrm{2}}\mathit{}\mathrm{(}{\mathrm{R}}^d\mathrm{)}$, $R\mathrm{>}\mathrm{0}$, and

$$\mathrm{ess}\mathrm{supp}\widehat{f}\subset \overline{B(0,R)},$$ (4)

then there is some $f_{\mathrm{0}}\mathrm{\in }{C}^{\mathrm{\infty }}\mathit{}\mathrm{(}{\mathrm{R}}^d\mathrm{)}$ such that $f\mathit{}\mathrm{(}x\mathrm{)}\mathrm{=}{f}_{\mathrm{0}}\mathit{}\mathrm{(}x\mathrm{)}$ for almost all $x\mathrm{\in }{\mathrm{R}}^d$, and for any multi-index $\alpha$,

$${\parallel D^{\alpha }{f}_0\parallel }_2\le {(2\pi R)}^{{|\alpha |}_1}{\parallel f\parallel }_2.$$

Proof.

Let ${\chi }_R$ be the indicator function for $\overline{B(0,R)}$. By (4), the Cauchy-Schwarz inequality, and the Parseval identity,

so $\widehat{f}\in L^1({ℝ}^d)$. The Plancherel theorem tells us that if $g\in L^2({ℝ}^d)$ and $\widehat{g}\in L^1({ℝ}^d)$, then

$$g(x)={\int }_{{ℝ}^d}\widehat{g}(\xi )e^{2\pi ix\cdot \xi }𝑑\xi$$

for almost all $x\in {ℝ}^d$. Thus, for $f_0:{ℝ}^d\to ℂ$ defined by

$$f_0(x)={\int }_{{ℝ}^d}\widehat{f}(\xi )e^{2\pi ix\cdot \xi }𝑑\xi =ℱ(\widehat{f})(-x),x\in {ℝ}^d,$$

we have $f(x)=f_0(x)$ for almost all $x\in {ℝ}^d$. Because $f=f_0$ almost everywhere,

$$\widehat{f_0}=\widehat{f}.$$

Applying Theorem 1 to $d\mu (\xi )=\widehat{f_0}(-\xi )d\xi$, we have $f_0\in C^{\mathrm{\infty }}({ℝ}^d)$ and for any multi-index $\alpha$,

$$D^{\alpha }{f}_0=ℱ({(-2\pi i\xi )}^{\alpha }\widehat{f}(-\xi )).$$

By Parseval’s identity,

${\parallel D^{\alpha }{f}_0\parallel }_2$ $={\parallel {(-2\pi i\xi )}^{\alpha }\widehat{f}(-\xi )\parallel }_2$

$={\parallel {(2\pi i\xi )}^{\alpha }{\chi }_R(\xi )\widehat{f}(\xi )\parallel }_2$

$\le {\parallel {(2\pi i\xi )}^{\alpha }{\chi }_R(\xi )\parallel }_{\mathrm{\infty }}{\parallel \widehat{f}\parallel }_2$

$\le {(2\pi R)}^{{|\alpha |}_1}{\parallel \widehat{f}\parallel }_2$

$={(2\pi R)}^{{|\alpha |}_1}{\parallel f\parallel }_2,$

proving the claim. ∎

4 Nikolsky’s inequality

Nikolsky’s inequality tells us that if the Fourier transform of a function is supported on a ball centered at the origin, then for $1\le p\le q\le \mathrm{\infty }$, the $L^q$ norm of the function is bounded above in terms of its $L^p$ norm.

Theorem 5.

There is a constant $C_d$ such that if $f\mathrm{\in }\mathrm{S}\mathit{}\mathrm{(}{\mathrm{R}}^d\mathrm{)}$, $R\mathrm{>}\mathrm{0}$,

$$\mathrm{supp}\widehat{f}\subset \overline{B(0,R)},$$

and $\mathrm{1}\mathrm{\le }p\mathrm{\le }q\mathrm{\le }\mathrm{\infty }$, then

$${\parallel f\parallel }_q\le C_d{R}^{d\left(\frac{1}{p}-\frac{1}{q}\right)}{\parallel f\parallel }_p.$$

Proof.

Let $g=f_R$, i.e.

$$g(x)=R^{-d}f(R^{-1}x),x\in {ℝ}^d.$$

Then for $\xi \in {ℝ}^d$,

$$\widehat{g}(\xi )={\int }_{{ℝ}^d}g(x)e^{-2\pi i\xi \cdot x}𝑑x={\int }_{{ℝ}^d}{R}^{-d}f(R^{-1}x)e^{-2\pi i\xi \cdot x}𝑑x=\widehat{f}(R\xi ),$$

showing that $\mathrm{supp}\widehat{g}=R^{-1}\mathrm{supp}\widehat{f}\subset \overline{B(0,1)}$. Let $\chi \in 𝒟({ℝ}^d)$ with $\chi (\xi )=1$ for $|\xi |\le 1$, with which

$$\widehat{g}=\chi \widehat{g}.$$

Then $g=({ℱ}^{-1}\chi )*g$, and using Young’s inequality, with $1+\frac{1}{q}=\frac{1}{r}+\frac{1}{p}$,

$${\parallel g\parallel }_q\le {\parallel {ℱ}^{-1}\chi \parallel }_r{\parallel g\parallel }_q={\parallel \widehat{\chi }\parallel }_r{\parallel g\parallel }_q.$$ (5)

Moreover,

${\parallel g\parallel }_a$ $={\left({\displaystyle {\int }_{{ℝ}^d}}{|R^{-d}f(R^{-1}x)|}^a𝑑x\right)}^{1/a}$

$={\left({\displaystyle {\int }_{{ℝ}^d}}{R}^{-da+d}{|f(y)|}^a𝑑y\right)}^{1/a}$

$=R^{d\left(\frac{1}{a}-1\right)}{\parallel f\parallel }_a,$

so (5) tells us

$$R^{d\left(\frac{1}{q}-1\right)}{\parallel f\parallel }_q\le {\parallel \widehat{\chi }\parallel }_r{R}^{d\left(\frac{1}{p}-1\right)}{\parallel f\parallel }_p,$$

i.e.

$${\parallel f\parallel }_q\le {\parallel \widehat{\chi }\parallel }_r{R}^{d\left(\frac{1}{p}-\frac{1}{q}\right)}{\parallel f\parallel }_p.$$

Now, $\frac{1}{r}=1+\frac{1}{q}-\frac{1}{p}$, so $0\le \frac{1}{r}\le 1$ because $1\le p\le q\le \mathrm{\infty }$, namely, $1\le r\le \mathrm{\infty }$. By the log-convexity of $L^r$ norms, for $\frac{1}{r}=1-\theta$ we have

$${\parallel \widehat{\chi }\parallel }_r\le {\parallel \widehat{\chi }\parallel }_1^{1-\theta }{\parallel \widehat{\chi }\parallel }_{\mathrm{\infty }}^{\theta }.$$

Thus with

$$C_d=\max \{{\parallel \widehat{\chi }\parallel }_1,{\parallel \widehat{\chi }\parallel }_{\mathrm{\infty }}\}$$

we have proved the claim. ∎

5 The Dirichlet kernel and Fejér kernel for R

The function $D_M\in C_0(ℝ)$ defined by

$$D_M(x)=\frac{\sin 2\pi Mx}{\pi x},x\ne 0$$

and $D_M(0)=2M$, is called the Dirichlet kernel. Let ${\chi }_M$ be the indicator function for the set $[-M,M]$. We have, for $x\ne 0$,

$\widehat{{\chi }_R}(x)$ $={\displaystyle {\int }_{ℝ}}{\chi }_R(\xi )e^{-2\pi ix\xi }𝑑\xi$

$={\displaystyle {\int }_{-M}^M}{e}^{-2\pi ix\xi }𝑑\xi$

$={{\displaystyle \frac{e^{-2\pi ix\xi }}{-2\pi ix}}|}_{-M}^M$

$={\displaystyle \frac{e^{-2\pi iMx}}{-2\pi ix}}+{\displaystyle \frac{e^{2\pi iMx}}{2\pi ix}}$

$={\displaystyle \frac{1}{\pi x}}{\displaystyle \frac{e^{2\pi iMx}-e^{-2\pi iMx}}{2i}}$

$={\displaystyle \frac{\sin 2\pi Mx}{\pi x}}.$

For $x=0$, $\widehat{{\chi }_R}(0)=2M=D_M(0)$. Thus,

$$D_M=\widehat{{\chi }_R}.$$

For $f\in L^1(ℝ)$ and $M>0$, we define

$$(S_{M}f)(x)={\int }_{-M}^M\widehat{f}(\xi )e^{2\pi i\xi x}𝑑\xi ,x\in ℝ.$$

It is straightforward to check that

$$(S_{M}f)(x)={\int }_{ℝ}\frac{\sin 2\pi Mt}{\pi t}f(x-t)𝑑t=(D_M*f)(x),x\in ℝ.$$

For $f\in L^1(ℝ)$, $M>0$, and $x\in ℝ$,

${\displaystyle \frac{1}{M}}{\displaystyle {\int }_0^M}(S_{m}f)(x)𝑑m$ $={\displaystyle \frac{1}{M}}{\displaystyle {\int }_0^M}\left({\displaystyle {\int }_{-m}^m}\widehat{f}(\xi )e^{2\pi i\xi x}𝑑\xi \right)𝑑m$

$={\displaystyle \frac{1}{M}}{\displaystyle {\int }_0^M}\left({\displaystyle {\int }_{-m}^m}\left({\displaystyle {\int }_{ℝ}}f(y)e^{-2\pi i\xi y}𝑑y\right)e^{2\pi i\xi x}𝑑\xi \right)𝑑m$

$={\displaystyle \frac{1}{M}}{\displaystyle {\int }_{ℝ}}f(y)\left({\displaystyle {\int }_0^M}\left({\displaystyle {\int }_{-m}^m}{e}^{-2\pi i\xi (y-x)}𝑑\xi \right)𝑑m\right)𝑑y$

$={\displaystyle \frac{1}{M}}{\displaystyle {\int }_{ℝ}}f(y)\left({\displaystyle {\int }_0^M}{D}_m(y-x)𝑑m\right)𝑑y$

$={\displaystyle \frac{1}{M}}{\displaystyle {\int }_{ℝ}}f(y)\left({\displaystyle {\int }_0^M}{\displaystyle \frac{\sin 2\pi m(y-x)}{\pi (y-x)}}𝑑m\right)𝑑y$

$={\displaystyle \frac{1}{M}}{\displaystyle {\int }_{ℝ}}f(y)\left(-{{\displaystyle \frac{\cos 2\pi m(y-x)}{2{\pi }^2{(y-x)}^2}}|}_0^M\right)𝑑y$

$={\displaystyle \frac{1}{M}}{\displaystyle {\int }_{ℝ}}f(y)\left({\displaystyle \frac{1}{2{\pi }^2{(y-x)}^2}}-{\displaystyle \frac{\cos 2\pi M(y-x)}{2{\pi }^2{(y-x)}^2}}\right)𝑑y.$

We define the Fejér kernel $K_M\in C_0(ℝ)$ by

$$K_M(x)=\frac{1-\cos 2\pi Mx}{2M{\pi }^2{x}^2},x\ne 0,$$

and $K_M(0)=M$. Thus, because $K_M$ is an even function,

$$\frac{1}{M}{\int }_0^M(S_{m}f)(x)𝑑m=(K_M*f)(x).$$

One proves that $K_M$ is an approximate identity: $K_M\ge 0$,

$${\int }_{ℝ}{K}_M(x)𝑑x=1,$$

and for any $\delta >0$,

$$\underset{M\to \mathrm{\infty }}{lim}{\int }_{|x|>\delta }{K}_M(x)𝑑x=0.$$

The fact that $K_M$ is an approximate identity implies that for any $f\in L^1(ℝ)$, $K_M*f\to f$ in $L^1(ℝ)$ as $M\to \mathrm{\infty }$.

We shall use the Fejér kernel to prove Bernstein’s inequality for $ℝ$.

Theorem 6.

If $\mu \mathrm{\in }\mathrm{M}\mathit{}\mathrm{(}\mathrm{R}\mathrm{)}$, $M\mathrm{>}\mathrm{0}$, and

$$\mathrm{supp}\mu \subset [-M,M],$$

then

$${\parallel {\widehat{\mu }}^{\prime }\parallel }_{\mathrm{\infty }}\le 4\pi M{\parallel \widehat{\mu }\parallel }_{\mathrm{\infty }}.$$

Proof.

For $x_0\in ℝ$, let $d{\mu }_{x_0}(t)=e^{-2\pi i{x}_{0}t}d\mu (t)$. ${\mu }_{x_0}$ has the same support has $\mu$, and

$$\widehat{{\mu }_{x_0}}(x)={\int }_{ℝ}{e}^{-2\pi ixt}𝑑{\mu }_{x_0}(t)={\int }_{ℝ}{e}^{-2\pi ixt}{e}^{-2\pi i{x}_{0}t}𝑑\mu (t)=\widehat{\mu }(x+x_0).$$

It follows that to prove the claim it suffices to prove that $|{\widehat{\mu }}^{\prime }(0)|\le 4\pi M{\parallel \widehat{\mu }\parallel }_{\mathrm{\infty }}$.

Write $f=\widehat{\mu }\in C_u(ℝ)$. Define ${\mathrm{\Delta }}_M\in C_c(ℝ)$ by

We calculate, for $x\ne 0$,

${\displaystyle {\int }_{ℝ}}{\mathrm{\Delta }}_M(t)e^{-2\pi ixt}𝑑t$ $=-{\displaystyle \frac{e^{-2\pi iMx}{(-1+e^{2\pi iMx})}^2}{4{\pi }^2{x}^2}}$

$={\displaystyle \frac{{(\sin \pi Mx)}^2}{{\pi }^2{x}^2}}$

$={\displaystyle \frac{1-\cos 2\pi Mx}{2{\pi }^2{x}^2}}.$

so

$$\widehat{{\mathrm{\Delta }}_M}(x)=M{K}_M(x).$$

Then for $t\in [-M,M]$,

${\displaystyle {\int }_{ℝ}}(e^{2\pi iM\xi }-e^{-2\pi iM\xi })K_M(\xi )e^{-2\pi i\xi t}𝑑\xi$ $=\widehat{K_M}(t-M)-\widehat{K_M}(t+M)$

$={\displaystyle \frac{{\mathrm{\Delta }}_M(-t+M)-{\mathrm{\Delta }}_M(-t-M)}{M}}$

$={\displaystyle \frac{t}{M}}.$

On the one hand, the integral of the left-hand side with respect to $\mu$ is

On the other hand, the integral of the right-hand side with respect to $\mu$ is

${\displaystyle {\int }_{ℝ}}{\displaystyle \frac{t}{M}}𝑑\mu (t)$ $={\displaystyle \frac{1}{-2\pi iM}}{\displaystyle {\int }_{ℝ}}-2\pi itd\mu (t)$

$={\displaystyle \frac{1}{-2\pi iM}}ℱ((-2\pi it)\mu )(0)$

$={\displaystyle \frac{1}{-2\pi iM}}{f}^{\prime }(0).$

Hence

$$\frac{1}{-2\pi iM}{f}^{\prime }(0)={\int }_{ℝ}(e^{2\pi iM\xi }-e^{-2\pi iM\xi })K_M(\xi )f(\xi )𝑑\xi ,$$

giving

$$|f^{\prime }(0)|\le 4\pi M{\parallel f\parallel }_{\mathrm{\infty }}{\parallel K_M\parallel }_1=4\pi M{\parallel f\parallel }_{\mathrm{\infty }},$$

proving the claim. ∎