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

Jordan Bell
February 16, 2015

## 1 Complex Borel measures and the Fourier transform

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

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

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

 $\hat{\mu}(\xi)=\int_{\mathbb{R}^{d}}e^{-2\pi i\xi\cdot x}d\mu(x),\qquad\xi\in% \mathbb{R}^{d}.$

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

 $|\hat{\mu}(\xi)|\leq\int_{\mathbb{R}^{d}}|e^{-2\pi i\xi\cdot x}|d|\mu|(x)=|\mu% |(\mathbb{R}^{d})=\left\|\mu\right\|.$ (1)

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

 $\Lambda_{f}=fm_{d},$

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

 $\hat{f}(\xi)=\widehat{\Lambda_{f}}(\xi)=\int_{\mathbb{R}^{d}}e^{-2\pi i\xi% \cdot x}d\Lambda_{f}(x)=\int_{\mathbb{R}^{d}}f(x)e^{-2\pi i\xi\cdot x}dm_{d}(x% ),\quad\xi\in\mathbb{R}^{d}.$

The following theorem establishes properties of the Fourier transform of a complex Borel measure with compact support.11 1 Thomas H. Wolff, Lectures on Harmonic Analysis, p. 3, Proposition 1.3.

###### Theorem 1.

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

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

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

 $\left\|D^{\alpha}\hat{\mu}\right\|_{\infty}\leq(2\pi R)^{|\alpha|_{1}}\left\|% \mu\right\|.$
###### Proof.

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

 $\Delta(h)=\frac{\hat{\mu}(\xi+he_{j})-\hat{\mu}(\xi)}{h},\qquad h\neq 0.$

We can write this as

 $\Delta(h)=\int_{\mathbb{R}^{d}}\frac{e^{-2\pi ihx_{j}}-1}{h}e^{-2\pi i\xi\cdot x% }d\mu(x).$

For any $x\in\mathbb{R}^{d}$,

 $\left|\frac{e^{-2\pi ihx_{j}}-1}{h}\right|=\frac{|e^{-2\pi ihx_{j}}-1|}{|h|}% \leq\frac{|-2\pi ihx_{j}|}{|h|}=2\pi|x_{j}|.$

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

 $\frac{e^{-2\pi ihx_{j}}-1}{h}\to-2\pi ix_{j},\qquad h\to 0.$

Therefore, the dominated convergence theorem tells us that

 $\lim_{h\to 0}\Delta(h)=\int_{\mathbb{R}^{d}}-2\pi ix_{j}e^{-2\pi i\xi\cdot x}d% \mu(x).$

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

 $(D^{\alpha}\hat{\mu})(\xi)=\lim_{h\to 0}\Delta(h),$

so

 $(D^{\alpha}\hat{\mu})(\xi)=\int_{\mathbb{R}^{d}}(-2\pi ix)^{\alpha}e^{-2\pi i% \xi\cdot x}d\mu(x)=\mathscr{F}((-2\pi ix)^{\alpha}\mu)(\xi),$

and in particular, $\hat{\mu}\in C^{1}(\mathbb{R}^{d})$. (The Fourier transform of a regular complex Borel measure on a locally compact abelian group is bounded and uniformly continuous.22 2 Walter Rudin, Fourier Analysis on Groups, p. 15, Theorem 1.3.3.) 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 $\alpha$,

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

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

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

 $(2\pi)^{|\alpha|_{1}}|x_{1}|^{\alpha_{1}}\cdots|x_{d}|^{\alpha_{d}}|\mu|,$

hence

 $\displaystyle\left\|(-2\pi ix)^{\alpha}\mu\right\|$ $\displaystyle=(2\pi)^{|\alpha|_{1}}\int_{\mathbb{R}^{d}}|x_{1}|^{\alpha_{1}}% \cdots|x_{d}|^{\alpha_{d}}d|\mu|(x)$ $\displaystyle=(2\pi)^{|\alpha|_{1}}\int_{\overline{B(0,R)}}|x_{1}|^{\alpha_{1}% }\cdots|x_{d}|^{\alpha_{d}}d|\mu|(x)$ $\displaystyle\leq(2\pi)^{|\alpha|_{1}}\int_{\overline{B(0,R)}}R^{\alpha_{1}}% \cdots R^{\alpha_{d}}d|\mu|(x)$ $\displaystyle=(2\pi R)^{|\alpha|_{1}}\int_{\overline{B(0,R)}}d|\mu|(x)$ $\displaystyle=(2\pi R)^{|\alpha|_{1}}\int_{\mathbb{R}^{d}}d|\mu|(x)$ $\displaystyle=(2\pi R)^{|\alpha|_{1}}\left\|\mu\right\|.$

Then using (1),

 $\left\|\mathscr{F}((-2\pi ix)^{\alpha}\mu)\right\|_{\infty}\leq\left\|(-2\pi ix% )^{\alpha}\mu\right\|\leq(2\pi R)^{|\alpha|_{1}}\left\|\mu\right\|.$

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

## 2 Test functions

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

It is a fact that there is a test function $\phi$ satisfying: (i) $\phi(x)=1$ for $|x|\leq 1$, (ii) $\phi(x)=0$ for $|x|\geq 2$, (iii) $0\leq\phi\leq 1$, and (iv) $\phi$ is radial. We write, for $k=1,2,\ldots$,

 $\phi_{k}(x)=\phi(k^{-1}x),\qquad x\in\mathbb{R}^{d}.$

For any multi-index $\alpha$,

 $(D^{\alpha}\phi_{k})(x)=k^{-|\alpha|_{1}}(D^{\alpha}\phi)(k^{-1}x),\qquad x\in% \mathbb{R}^{d},$

hence

 $\left\|D^{\alpha}\phi_{k}\right\|_{\infty}=k^{-|\alpha|_{1}}\left\|D^{\alpha}% \phi\right\|_{\infty}.$ (2)

We use the following lemma to prove the theorem that comes after it.33 3 Thomas H. Wolff, Lectures on Harmonic Analysis, p. 4, Lemma 1.5.

###### Lemma 2.

Suppose that $f\in C^{N}(\mathbb{R}^{d})$ and $D^{\alpha}f\in L^{1}(\mathbb{R}^{d})$ for each $|\alpha|\leq N$. Then for each $|\alpha|\leq N$, $D^{\alpha}(\phi_{k}f)\to D^{\alpha}f$ in $L^{1}(\mathbb{R}^{d})$ as $k\to\infty$.

###### Proof.

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

 $\displaystyle\left\|\phi_{k}f-f\right\|_{1}$ $\displaystyle=\int_{\mathbb{R}^{d}}|\phi_{k}(x)f(x)-f(x)|dx$ $\displaystyle=\int_{|x|\geq k}|\phi_{k}(x)f(x)-f(x)|dx$ $\displaystyle\leq\int_{|x|\geq k}|f(x)|dx.$

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

Suppose that $\alpha>0$. The Leibniz rule tells us that with $c_{\beta}=\binom{\alpha}{\beta}$, we have, for each $k$,

 $D^{\alpha}(\phi_{k}f)=\phi_{k}D^{\alpha}f+\sum_{0<\beta\leq\alpha}c_{\beta}D^{% \alpha-\beta}fD^{\beta}\phi_{k}.$

For $C_{1}=\max_{\beta}|c_{\beta}|$,

 $\displaystyle\left\|D^{\alpha}(\phi_{k}f)-\phi_{k}D^{\alpha}f\right\|_{1}$ $\displaystyle\leq\sum_{0<\beta\leq\alpha}\left\|c_{\beta}D^{\alpha-\beta}fD^{% \beta}\phi_{k}\right\|_{1}$ $\displaystyle\leq C_{1}\sum_{0<\beta\leq\alpha}\left\|D^{\beta}\phi_{k}\right% \|_{\infty}\left\|D^{\alpha-\beta}f\right\|_{1}.$

Let $C_{2}=\max_{0<\beta\leq\alpha}\left\|D^{\beta}\phi\right\|_{\infty}$. By (2), for $0<\beta\leq\alpha$ we have

 $\left\|D^{\beta}\phi_{k}\right\|_{\infty}=k^{-|\beta|_{1}}\left\|D^{\beta}\phi% \right\|_{\infty}\leq C_{2}k^{-|\beta|_{1}}\leq C_{2}k^{-1}.$

Thus

 $\left\|D^{\alpha}(\phi_{k}f)-\phi_{k}D^{\alpha}f\right\|_{1}\leq C_{1}C_{2}k^{% -1}\sum_{0<\beta\leq\alpha}\left\|D^{\alpha-\beta}f\right\|_{1},$

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

 $\displaystyle\left\|\phi_{k}D^{\alpha}f-D^{\alpha}f\right\|_{1}$ $\displaystyle=\int_{\mathbb{R}^{d}}|\phi_{k}(x)(D^{\alpha}f)(x)-(D^{\alpha}f)(% x)|dx$ $\displaystyle=\int_{|x|\geq k}|\phi_{k}(x)(D^{\alpha}f)(x)-(D^{\alpha}f)(x)|dx$ $\displaystyle\leq\int_{|x|\geq k}|(D^{\alpha}f)(x)|dx,$

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

 $\left\|D^{\alpha}(\phi_{k}f)-D^{\alpha}f\right\|_{1}\leq\left\|D^{\alpha}(\phi% _{k}f)-\phi_{k}D^{\alpha}f\right\|_{1}+\left\|\phi_{k}D^{\alpha}f-D^{\alpha}f% \right\|_{1},$

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.44 4 Thomas H. Wolff, Lectures on Harmonic Analysis, p. 4, Proposition 1.4.

###### Theorem 3.

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

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

There is a constant $C=C(f,N)$ such that

 $|\hat{f}(\xi)|\leq C(1+|\xi|)^{-N},\qquad\xi\in\mathbb{R}^{d}.$
###### Proof.

If $g\in C^{1}_{c}(\mathbb{R}^{d})$, then for any $1\leq j\leq d$, integrating by parts,

 $\int_{\mathbb{R}^{d}}(\partial_{j}g)(x)e^{-2\pi i\xi\cdot x}dx=2\pi i\xi_{j}% \int_{\mathbb{R}^{d}}g(x)e^{-2\pi i\xi\cdot x}dx.$

It follows by induction that if $g\in C^{N}_{c}(\mathbb{R}^{d})$, then for each $|\alpha|\leq N$,

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

Let $|\alpha|\leq N$. For $k=1,2,\ldots$, let $f_{k}=\phi_{k}f$. For each $k$ we have $f_{k}\in C^{N}(\mathbb{R}^{d})$, hence

 $\widehat{D^{\alpha}f_{k}}(\xi)=(2\pi i\xi)^{\alpha}\widehat{f_{k}}(\xi),\qquad% \xi\in\mathbb{R}^{d}.$

On the one hand,

 $\left\|\widehat{D^{\alpha}f_{k}}-\widehat{D^{\alpha}f}\right\|_{\infty}=\left% \|\mathscr{F}(D^{\alpha}f_{k}-D^{\alpha}f)\right\|_{\infty}\leq\left\|D^{% \alpha}f_{k}-D^{\alpha}f\right\|_{1},$

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

 $\displaystyle|\widehat{D^{\alpha}f_{k}}(\xi)-(2\pi i\xi)^{\alpha}\hat{f}(\xi)|$ $\displaystyle=|(2\pi i\xi)^{\alpha}\widehat{f_{k}}(\xi)-(2\pi i\xi)^{\alpha}% \hat{f}(\xi)|$ $\displaystyle=|(2\pi i\xi)^{\alpha}||\mathscr{F}(f_{k}-f)(\xi)|$ $\displaystyle\leq|(2\pi i\xi)^{\alpha}|\left\|f_{k}-f\right\|_{1},$

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

 $|\widehat{D^{\alpha}f}(\xi)-(2\pi i\xi)^{\alpha}\hat{f}(\xi)|\leq\left\|% \widehat{D^{\alpha}f_{k}}-\widehat{D^{\alpha}f}\right\|_{\infty}+|\widehat{D^{% \alpha}f_{k}}(\xi)-(2\pi i\xi)^{\alpha}\hat{f}(\xi)|,$

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

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

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

 $\sum_{|\beta|_{1}=N}|y^{\beta}|>0.$

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

 $\frac{1}{C_{N}}\leq\sum_{|\beta|_{1}=N}|y^{\beta}|,\qquad y\in S^{d-1}.$

For nonzero $x\in\mathbb{R}^{d}$, write $x=|x|y$, with which $\sum_{|\beta|_{1}=N}|x^{\beta}|=|x|^{N}\sum_{|\beta|_{1}=N}|y^{\beta}|$. Therefore

 $|x|^{N}\leq C_{N}\sum_{|\beta|_{1}=N}|x^{\beta}|,\qquad x\in\mathbb{R}^{d}.$

For $|\alpha|\leq N$, because the Fourier transform of an element of $L^{1}$ belongs to $C_{0}$, we have by (3) that $\xi\mapsto\xi^{\alpha}\hat{f}(\xi)$ belongs to $C_{0}(\mathbb{R}^{d})$, and in particular is bounded. Then for $\xi\in\mathbb{R}^{d}$,

 $\displaystyle|\xi|^{N}|\hat{f}(\xi)|$ $\displaystyle\leq C_{N}\sum_{|\beta|_{1}=N}|\xi^{\beta}||\hat{f}(\xi)|$ $\displaystyle=C_{N}\sum_{|\beta|_{1}=N}|\xi^{\beta}\hat{f}(\xi)|$ $\displaystyle\leq C_{N}\sum_{|\beta|_{1}=N}\left\|\xi^{\beta}\hat{f}(\xi)% \right\|_{\infty}$ $\displaystyle=C^{\prime}.$

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

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

hence

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

giving

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

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

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

and so

 $|\hat{f}(\xi)|\leq\left\|\hat{f}\right\|_{\infty}2^{N}2^{-N}\leq\left\|\hat{f}% \right\|_{\infty}2^{N}(1+|\xi|)^{-N}.$

Thus, for

 $C=\max\left\{2^{N}C^{\prime},2^{N}\left\|\hat{f}\right\|_{\infty}\right\}$

we have

 $|\hat{f}(\xi)|\leq C(1+|\xi|)^{-N},\qquad\xi\in\mathbb{R}^{d},$

completing the proof. ∎

## 3 Bernstein’s inequality for L2

For a Borel measurable function $f:\mathbb{R}^{d}\to\mathbb{C}$, let $O$ be the union of those open subsets $U$ of $\mathbb{R}^{d}$ 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

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

The following is Bernstein’s inequality for $L^{2}(\mathbb{R}^{d})$.55 5 Thomas H. Wolff, Lectures on Harmonic Analysis, p. 31, Proposition 5.1.

###### Theorem 4.

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

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

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

 $\left\|D^{\alpha}f_{0}\right\|_{2}\leq(2\pi R)^{|\alpha|_{1}}\left\|f\right\|_% {2}.$
###### Proof.

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

 $\left\|\hat{f}\right\|_{1}=\left\|\chi_{R}\hat{f}\right\|_{1}\leq\left\|\chi_{% R}\right\|_{2}\left\|\hat{f}\right\|_{2}=m_{d}(\overline{B(0,R)})^{1/2}\left\|% f\right\|_{2}<\infty,$

so $\hat{f}\in L^{1}(\mathbb{R}^{d})$. The Plancherel theorem66 6 Walter Rudin, Real and Complex Analysis, third ed., p. 187, Theorem 9.14. tells us that if $g\in L^{2}(\mathbb{R}^{d})$ and $\hat{g}\in L^{1}(\mathbb{R}^{d})$, then

 $g(x)=\int_{\mathbb{R}^{d}}\hat{g}(\xi)e^{2\pi ix\cdot\xi}d\xi$

for almost all $x\in\mathbb{R}^{d}$. Thus, for $f_{0}:\mathbb{R}^{d}\to\mathbb{C}$ defined by

 $f_{0}(x)=\int_{\mathbb{R}^{d}}\hat{f}(\xi)e^{2\pi ix\cdot\xi}d\xi=\mathscr{F}(% \hat{f})(-x),\qquad x\in\mathbb{R}^{d},$

we have $f(x)=f_{0}(x)$ for almost all $x\in\mathbb{R}^{d}$. Because $f=f_{0}$ almost everywhere,

 $\widehat{f_{0}}=\hat{f}.$

Applying Theorem 1 to $d\mu(\xi)=\widehat{f_{0}}(-\xi)d\xi$, we have $f_{0}\in C^{\infty}(\mathbb{R}^{d})$ and for any multi-index $\alpha$,

 $D^{\alpha}f_{0}=\mathscr{F}((-2\pi i\xi)^{\alpha}\hat{f}(-\xi)).$

By Parseval’s identity,

 $\displaystyle\left\|D^{\alpha}f_{0}\right\|_{2}$ $\displaystyle=\left\|(-2\pi i\xi)^{\alpha}\hat{f}(-\xi)\right\|_{2}$ $\displaystyle=\left\|(2\pi i\xi)^{\alpha}\chi_{R}(\xi)\hat{f}(\xi)\right\|_{2}$ $\displaystyle\leq\left\|(2\pi i\xi)^{\alpha}\chi_{R}(\xi)\right\|_{\infty}% \left\|\hat{f}\right\|_{2}$ $\displaystyle\leq(2\pi R)^{|\alpha|_{1}}\left\|\hat{f}\right\|_{2}$ $\displaystyle=(2\pi R)^{|\alpha|_{1}}\left\|f\right\|_{2},$

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 $1\leq p\leq q\leq\infty$, the $L^{q}$ norm of the function is bounded above in terms of its $L^{p}$ norm.77 7 Camil Muscalu and Wilhelm Schlag, Classical and Multilinear Harmonic Analysis, volume I, p. 83, Lemma 4.13.

###### Theorem 5.

There is a constant $C_{d}$ such that if $f\in\mathscr{S}(\mathbb{R}^{d})$, $R>0$,

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

and $1\leq p\leq q\leq\infty$, then

 $\left\|f\right\|_{q}\leq C_{d}R^{d\left(\frac{1}{p}-\frac{1}{q}\right)}\left\|% f\right\|_{p}.$
###### Proof.

Let $g=f_{R}$, i.e.

 $g(x)=R^{-d}f(R^{-1}x),\qquad x\in\mathbb{R}^{d}.$

Then for $\xi\in\mathbb{R}^{d}$,

 $\hat{g}(\xi)=\int_{\mathbb{R}^{d}}g(x)e^{-2\pi i\xi\cdot x}dx=\int_{\mathbb{R}% ^{d}}R^{-d}f(R^{-1}x)e^{-2\pi i\xi\cdot x}dx=\hat{f}(R\xi),$

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

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

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

 $\left\|g\right\|_{q}\leq\left\|\mathscr{F}^{-1}\chi\right\|_{r}\left\|g\right% \|_{q}=\left\|\hat{\chi}\right\|_{r}\left\|g\right\|_{q}.$ (5)

Moreover,

 $\displaystyle\left\|g\right\|_{a}$ $\displaystyle=\left(\int_{\mathbb{R}^{d}}|R^{-d}f(R^{-1}x)|^{a}dx\right)^{1/a}$ $\displaystyle=\left(\int_{\mathbb{R}^{d}}R^{-da+d}|f(y)|^{a}dy\right)^{1/a}$ $\displaystyle=R^{d\left(\frac{1}{a}-1\right)}\left\|f\right\|_{a},$

so (5) tells us

 $R^{d\left(\frac{1}{q}-1\right)}\left\|f\right\|_{q}\leq\left\|\hat{\chi}\right% \|_{r}R^{d\left(\frac{1}{p}-1\right)}\left\|f\right\|_{p},$

i.e.

 $\left\|f\right\|_{q}\leq\left\|\hat{\chi}\right\|_{r}R^{d\left(\frac{1}{p}-% \frac{1}{q}\right)}\left\|f\right\|_{p}.$

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

 $\left\|\hat{\chi}\right\|_{r}\leq\left\|\hat{\chi}\right\|_{1}^{1-\theta}\left% \|\hat{\chi}\right\|_{\infty}^{\theta}.$

Thus with

 $C_{d}=\max\{\left\|\hat{\chi}\right\|_{1},\left\|\hat{\chi}\right\|_{\infty}\}$

we have proved the claim. ∎

## 5 The Dirichlet kernel and Fejér kernel for R

The function $D_{M}\in C_{0}(\mathbb{R})$ defined by

 $D_{M}(x)=\frac{\sin 2\pi Mx}{\pi x},\qquad x\neq 0$

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 $x\neq 0$,

 $\displaystyle\widehat{\chi_{R}}(x)$ $\displaystyle=\int_{\mathbb{R}}\chi_{R}(\xi)e^{-2\pi ix\xi}d\xi$ $\displaystyle=\int_{-M}^{M}e^{-2\pi ix\xi}d\xi$ $\displaystyle=\frac{e^{-2\pi ix\xi}}{-2\pi ix}\bigg{|}_{-M}^{M}$ $\displaystyle=\frac{e^{-2\pi iMx}}{-2\pi ix}+\frac{e^{2\pi iMx}}{2\pi ix}$ $\displaystyle=\frac{1}{\pi x}\frac{e^{2\pi iMx}-e^{-2\pi iMx}}{2i}$ $\displaystyle=\frac{\sin 2\pi Mx}{\pi x}.$

For $x=0$, $\widehat{\chi_{R}}(0)=2M=D_{M}(0)$. Thus,

 $D_{M}=\widehat{\chi_{R}}.$

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

 $(S_{M}f)(x)=\int_{-M}^{M}\hat{f}(\xi)e^{2\pi i\xi x}d\xi,\qquad x\in\mathbb{R}.$

It is straightforward to check that

 $(S_{M}f)(x)=\int_{\mathbb{R}}\frac{\sin 2\pi Mt}{\pi t}f(x-t)dt=(D_{M}*f)(x),% \qquad x\in\mathbb{R}.$

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

 $\displaystyle\frac{1}{M}\int_{0}^{M}(S_{m}f)(x)dm$ $\displaystyle=\frac{1}{M}\int_{0}^{M}\left(\int_{-m}^{m}\hat{f}(\xi)e^{2\pi i% \xi x}d\xi\right)dm$ $\displaystyle=\frac{1}{M}\int_{0}^{M}\left(\int_{-m}^{m}\left(\int_{\mathbb{R}% }f(y)e^{-2\pi i\xi y}dy\right)e^{2\pi i\xi x}d\xi\right)dm$ $\displaystyle=\frac{1}{M}\int_{\mathbb{R}}f(y)\left(\int_{0}^{M}\left(\int_{-m% }^{m}e^{-2\pi i\xi(y-x)}d\xi\right)dm\right)dy$ $\displaystyle=\frac{1}{M}\int_{\mathbb{R}}f(y)\left(\int_{0}^{M}D_{m}(y-x)dm% \right)dy$ $\displaystyle=\frac{1}{M}\int_{\mathbb{R}}f(y)\left(\int_{0}^{M}\frac{\sin 2% \pi m(y-x)}{\pi(y-x)}dm\right)dy$ $\displaystyle=\frac{1}{M}\int_{\mathbb{R}}f(y)\left(-\frac{\cos 2\pi m(y-x)}{2% \pi^{2}(y-x)^{2}}\bigg{|}_{0}^{M}\right)dy$ $\displaystyle=\frac{1}{M}\int_{\mathbb{R}}f(y)\left(\frac{1}{2\pi^{2}(y-x)^{2}% }-\frac{\cos 2\pi M(y-x)}{2\pi^{2}(y-x)^{2}}\right)dy.$

We define the Fejér kernel $K_{M}\in C_{0}(\mathbb{R})$ by

 $K_{M}(x)=\frac{1-\cos 2\pi Mx}{2M\pi^{2}x^{2}},\qquad x\neq 0,$

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: $K_{M}\geq 0$,

 $\int_{\mathbb{R}}K_{M}(x)dx=1,$

and for any $\delta>0$,

 $\lim_{M\to\infty}\int_{|x|>\delta}K_{M}(x)dx=0.$

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

We shall use the Fejér kernel to prove Bernstein’s inequality for $\mathbb{R}$.88 8 Mark A. Pinsky, Introduction to Fourier Analysis and Wavelets, p. 122, Theorem 2.3.17.

###### Theorem 6.

If $\mu\in\mathcal{M}(\mathbb{R})$, $M>0$, and

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

then

 $\left\|\hat{\mu}^{\prime}\right\|_{\infty}\leq 4\pi M\left\|\hat{\mu}\right\|_% {\infty}.$
###### Proof.

For $x_{0}\in\mathbb{R}$, let $d\mu_{x_{0}}(t)=e^{-2\pi ix_{0}t}d\mu(t)$. $\mu_{x_{0}}$ has the same support has $\mu$, and

 $\widehat{\mu_{x_{0}}}(x)=\int_{\mathbb{R}}e^{-2\pi ixt}d\mu_{x_{0}}(t)=\int_{% \mathbb{R}}e^{-2\pi ixt}e^{-2\pi ix_{0}t}d\mu(t)=\hat{\mu}(x+x_{0}).$

It follows that to prove the claim it suffices to prove that $|\hat{\mu}^{\prime}(0)|\leq 4\pi M\left\|\hat{\mu}\right\|_{\infty}$.

Write $f=\hat{\mu}\in C_{u}(\mathbb{R})$. Define $\Delta_{M}\in C_{c}(\mathbb{R})$ by

 $\Delta_{M}(t)=\begin{cases}M-|t|&|t|

We calculate, for $x\neq 0$,

 $\displaystyle\int_{\mathbb{R}}\Delta_{M}(t)e^{-2\pi ixt}dt$ $\displaystyle=-\frac{e^{-2\pi iMx}(-1+e^{2\pi iMx})^{2}}{4\pi^{2}x^{2}}$ $\displaystyle=\frac{(\sin\pi Mx)^{2}}{\pi^{2}x^{2}}$ $\displaystyle=\frac{1-\cos 2\pi Mx}{2\pi^{2}x^{2}}.$

so

 $\widehat{\Delta_{M}}(x)=MK_{M}(x).$

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

 $\displaystyle\int_{\mathbb{R}}(e^{2\pi iM\xi}-e^{-2\pi iM\xi})K_{M}(\xi)e^{-2% \pi i\xi t}d\xi$ $\displaystyle=\widehat{K_{M}}(t-M)-\widehat{K_{M}}(t+M)$ $\displaystyle=\frac{\Delta_{M}(-t+M)-\Delta_{M}(-t-M)}{M}$ $\displaystyle=\frac{t}{M}.$

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

 $\begin{split}&\displaystyle\int_{\mathbb{R}}\int_{\mathbb{R}}(e^{2\pi iM\xi}-e% ^{-2\pi iM\xi})K_{M}(\xi)e^{-2\pi i\xi t}d\xi d\mu(t)\\ \displaystyle=&\displaystyle\int_{\mathbb{R}}(e^{2\pi iM\xi}-e^{-2\pi iM\xi})K% _{M}(\xi)f(\xi)d\xi.\end{split}$

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

 $\displaystyle\int_{\mathbb{R}}\frac{t}{M}d\mu(t)$ $\displaystyle=\frac{1}{-2\pi iM}\int_{\mathbb{R}}-2\pi itd\mu(t)$ $\displaystyle=\frac{1}{-2\pi iM}\mathscr{F}((-2\pi it)\mu)(0)$ $\displaystyle=\frac{1}{-2\pi iM}f^{\prime}(0).$

Hence

 $\frac{1}{-2\pi iM}f^{\prime}(0)=\int_{\mathbb{R}}(e^{2\pi iM\xi}-e^{-2\pi iM% \xi})K_{M}(\xi)f(\xi)d\xi,$

giving

 $|f^{\prime}(0)|\leq 4\pi M\left\|f\right\|_{\infty}\left\|K_{M}\right\|_{1}=4% \pi M\left\|f\right\|_{\infty},$

proving the claim. ∎