Singular oscillatory integrals on $\mathbb{R}^{n}$

We denote by $\mathcal{P}_{d,n}$ the vector space of all real polynomials of degree at most $d$ in $\mathbb{R}^{n}$. Let $K$ be a $-n$ homogeneous function on $\mathbb{R}^{n}$, that is,

(1.1)

$$K(x)=\frac{\Omega(x/|x|)}{|x|^{n}},$$

where $\Omega$ is some function on the unit sphere $S^{n-1}$. Consider the principal value integral

$$I_{n}(P)=\bigg{|}p.v.\int_{\mathbb{R}^{n}}e^{iP(x)}K(x)dx\bigg{|}.$$

Stein has proved in [4] that if $\Omega$ has zero mean value on the unit sphere, then

(1.2)

$$|I_{n}(P)|\leq c_{d}\|\Omega\|_{L^{\infty}(S^{n-1})},$$

for some constant $c_{d}$ depending on $d$. We wish to obtain sharp estimates of the form (1.2). The one dimensional analogue, namely the estimate

(1.3)

$$\bigg{|}p.v.\int_{\mathbb{R}}e^{iP(x)}\frac{dx}{x}\bigg{|}\leq c\log d,$$

which was proved in [3], suggests that the constant $c_{d}$ in (1.2) could be replaced by $c\log d$ for some absolute positive constant $c$. The fact that this is indeed the case is the content of the following theorem.

The main ingredient of the proof of Theorem 1.1 is an estimate for the logarithmic measure of the sublevel set of a real polynomial in one dimension. This is a lemma of independent interest which we now state.

Lemma 1.3 should be compared to the following variation of a classical result of Vinogradov which can be found in [6]:

The estimates above depend on the length of the interval $[1,R]$ in all cases but the one where $r=d$. The dependence on $R$ is sharp as can be seen by a scaling argument.

When $r=d$ we get

(1.4)

$$|\{x\in[1,R]:|P(x)|\leq\alpha\}|\leq c\frac{\alpha^{\frac{1}{d}}}{|b_{d}|^{\frac{1}{d}}}.$$

The last inequality corresponds to the following more general result about sublevel sets which was proved in [1]:

Observe that inequality (1.4) can be deduced by Lemma 1.5 by taking $k=d$ derivatives of the phase function $\phi(x)=P(x)$.

In case $n=1$ the “linear” part $(\frac{a}{M})^{\frac{1}{d}}$ of the estimate of $\int_{E}\frac{1}{x}dx$ in Lemma 1.3 is enough for the proof of Theorem 1.1. In fact, the author in [3] used Lemma 1.4 in some appropriate way to prove the above ”linear” estimate of Lemma 1.3.

In case $n>1$ the “logarithmic” part of the estimate of $\int_{E}\frac{1}{x}dx$ is essential in the proof of Theorem 1.1 as can easily be seen by examining the argument therein.

The structure of the rest of this work is as follows. In section 2 we state some preliminary results. In section 3 we present the proof of Lemma 1.3 and section 3 contains the proof of Theorem 1.1. Finally in section 4 we give a proof of Theorem 1.1 in case $n=1$ which uses (the ”linear” estimate in) Lemma 1.3 and not Lemma 1.4 and which is thus simpler than the proof appearing in [3].

As is usually the case when one deals with oscillatory integrals, a key Lemma is the classical van der Corput Lemma.

The proof of Lemma 2.1 is a simple integration by parts.

We will also need a precise estimate for the Lebesgue measure of the sublevel set of a polynomial on $\mathbb{R}^{n}$.

This is a consequence of a more general Theorem of Carbery and Wright and can be found in [2].

The proof of Lemma 1.3 is motivated by an argument of Vinogradov from [6], used to estimate the Lebesgue measure of the sublevel set of a polynomial in a bounded interval. We fix a polynomial $P(x)=\sum_{k=0}^{d}b_{k}x^{k}$ and look at the set $E=\{x\geq 1:|P(x)|\leq\alpha\}$. Note that by replacing $\alpha$ with $\alpha M$ in the statement of the lemma, it is enough to consider the case $M=1$. Since $E$ is a closed set we can find points $x_{0},x_{1},\ldots,x_{d}\in E$ such that $x_{0}<x_{1}<\cdots<x_{d}$ and

$$\frac{1}{d}\int_{E}\frac{dx}{x}=\int_{E\cap[x_{j},x_{j+1}]}\frac{dx}{x}\leq\log\frac{x_{j+1}}{x_{j}},\ \ \ \ \ 0\leq j\leq d-1.$$

We set $\mu=\int_{E}\frac{dx}{x}$ and $t=e^{\frac{\mu}{d}}>1$ and we have that $x_{j+1}\geq tx_{j}$, $0\leq j\leq d-1$. The Lagrange interpolation formula is

$$P(x)=\sum_{j=0}^{d}P(x_{j})\frac{(x-x_{0})\cdots(\widehat{x-x_{j}})\cdots(x-x_{d})}{(x_{j}-x_{0})\cdots(\widehat{x_{j}-x_{j}})\cdots(x_{j}-x_{d})},\ x\in\mathbb{R},$$

where $\hat{u}$ means that $u$ is omitted. Thus,

$$b_{k}=\sum_{j=0}^{d}P(x_{j})(-1)^{d-k}\frac{\sigma_{d-k}(x_{0},\ldots,\widehat{x_{j}},\ldots,x_{d})}{(x_{j}-x_{0})\cdots(\widehat{x_{j}-x_{j}})\cdots(x_{j}-x_{d})},$$

where $\sigma_{l}$ is the $l$-th elementary symmetric function of its variables. Therefore

	$\displaystyle|b_{k}|$	$\displaystyle\leq$	$\displaystyle\alpha\sum_{j=0}^{d}\frac{\sigma_{d-k}(x_{0},\ldots,\widehat{x_{j}},\ldots,x_{d})}{|x_{j}-x_{0}|\cdots|\widehat{x_{j}-x_{j}}|\cdots|x_{j}-x_{d}|}$
		$\displaystyle=$	$\displaystyle\alpha\sum_{j=0}^{d}\frac{\sigma_{k}(\frac{1}{x_{0}},\ldots,\widehat{\frac{1}{x_{j}}},\ldots,\frac{1}{x_{d}})}{(\frac{x_{j}}{x_{0}}-1)\cdots(\frac{x_{j}}{x_{j-1}}-1)(1-\frac{x_{j}}{x_{j+1}})\cdots(1-\frac{x_{j}}{x_{d}})}$
		$\displaystyle\leq$	$\displaystyle\alpha\sum_{j=0}^{d}\frac{\sigma_{k}(1,\ldots,\widehat{\frac{1}{t}},\ldots,\frac{1}{t^{d}})}{(t^{j}-1)\cdots(t-1)(1-\frac{1}{t})\cdots(1-\frac{1}{t^{d-j}})}.$

It is easy to see that there exists precisely one $j$, $0\leq j\leq\frac{d-1}{2}<d$, for which

(3.1)

$$t^{j-1}<\frac{2t^{d}}{t^{d+1}+1}\leq t^{j}.$$

It is exactly for this $j$ that $(t^{j}-1)\cdots(t-1)(1-\frac{1}{t})\cdots(1-\frac{1}{t^{d-j}})$ takes its minimum value as $j$ runs from $0$ to $d$. On the other hand we have

$$\sum_{j=0}^{d}\sigma_{k}\bigg{(}1,\ldots,\widehat{\frac{1}{t_{j}}},\ldots,\frac{1}{t^{k}}\bigg{)}=(d+1-k)\sigma_{k}\bigg{(}1,\ldots,\frac{1}{t^{d}}\bigg{)}$$

and, hence

	$\displaystyle|b_{k}|$	$\displaystyle\leq$	$\displaystyle\alpha\ (d+1-k)\sigma_{k}\bigg{(}1,\ldots,\frac{1}{t^{d}}\bigg{)}\frac{1}{(t^{j}-1)\cdots(t-1)(1-\frac{1}{t})\cdots(1-\frac{1}{t^{d-j}})}$
		$\displaystyle\leq$	$\displaystyle\frac{\alpha\ (d+1-k)\binom{d+1}{k}}{1\cdot t\cdots t^{k}}\frac{1}{(t^{j}-1)\cdots(t-1)(1-\frac{1}{t})\cdots(1-\frac{1}{t^{d-j}})}.$

From (3.1) we easily see that $t^{j}<2$ and, since $\frac{\log(x-1)}{x}$ is increasing in the interval $(1,2)$, we find

			$\displaystyle\log(t-1)+\cdots+\log(t^{j}-1)$
		$\displaystyle=$	$\displaystyle\frac{t}{t-1}\bigg{(}\frac{\log(t-1)}{t}(t-1)+\cdots+\frac{\log(t^{j}-1)}{t^{j}}(t^{j}-t^{j-1})\bigg{)}$
		$\displaystyle\geq$	$\displaystyle\frac{t}{t-1}\int_{1}^{t^{j}}\frac{\log(x-1)}{x}dx=\frac{t}{t-1}\int_{0}^{t^{j}-1}\frac{\log x}{1+x}dx.$

Similarly, since $\frac{\log(1-x)}{x}$ is decreasing in the interval $(0,1)$ we get

			$\displaystyle\log\bigg{(}1-\frac{1}{t^{d-j}}\bigg{)}+\cdots+\log\bigg{(}1-\frac{1}{t}\bigg{)}$
		$\displaystyle=$	$\displaystyle\frac{1}{t-1}\bigg{(}\frac{\log(1-\frac{1}{t^{d-j}})}{\frac{1}{t^{d-j}}}\bigg{(}\frac{1}{t^{d-j-1}}-\frac{1}{t^{d-j}}\bigg{)}+\cdots+\frac{\log(1-\frac{1}{t})}{\frac{1}{t}}\bigg{(}1-\frac{1}{t}\bigg{)}\bigg{)}$
		$\displaystyle\geq$	$\displaystyle\frac{1}{t-1}\int_{\frac{1}{t^{d-j}}}^{1}\frac{\log(1-x)}{x}dx=\frac{1}{t-1}\int_{0}^{1-\frac{1}{t^{d-j}}}\frac{\log x}{1-x}dx.$

We let

$$A=\frac{t^{d}-1}{t^{d}+1},\ \ B=t^{j}-1,\ \ \Gamma=1-\frac{1}{t^{d-j}},$$

and, obviously, $0<A,B,\Gamma<1$. From (3) and (3) we have

			$\displaystyle\log(t-1)+\cdots+\log(t^{j}-1)+\log\bigg{(}1-\frac{1}{t^{d-j}}\bigg{)}+\cdots+\log\bigg{(}1-\frac{1}{t}\bigg{)}$
		$\displaystyle\geq$	$\displaystyle\frac{t}{t-1}\int_{0}^{t^{j}-1}\frac{\log x}{1+x}dx+\frac{1}{t-1}\int_{0}^{1-\frac{1}{t^{d-j}}}\frac{\log x}{1-x}dx$
		$\displaystyle=$	$\displaystyle\frac{t}{t-1}\int_{0}^{B}\frac{\log x}{1+x}dx+\frac{1}{t-1}\int_{0}^{\Gamma}\frac{\log x}{1-x}dx$
		$\displaystyle=$	$\displaystyle-\frac{t}{t-1}B\log\frac{1}{B}-\frac{1}{t-1}\Gamma\log\frac{1}{\Gamma}-O\bigg{(}\frac{t}{t-1}B\bigg{)}-O\bigg{(}\frac{1}{t-1}\Gamma\bigg{)}.$

From (3.1) we get $B,\Gamma\leq\frac{t^{d+1}-1}{t^{d+1}+1}$ and, since $\frac{t+1}{t-1}\frac{t^{d+1}-1}{t^{d+1}+1}$ is decreasing in $t\in(1,+\infty)$, we find

$$\frac{t}{t-1}B\leq\frac{t+1}{t-1}\frac{t^{d+1}-1}{t^{d+1}+1}\leq d+1$$

and, similarly,

$$\frac{1}{t-1}\Gamma\leq\frac{t+1}{t-1}\frac{t^{d+1}-1}{t^{d+1}+1}\leq d+1.$$

Therefore

			$\displaystyle\log(t-1)+\cdots+\log(t^{j}-1)+\log\bigg{(}1-\frac{1}{t^{d-j}}\bigg{)}+\cdots+\log\bigg{(}1-\frac{1}{t}\bigg{)}$
		$\displaystyle\geq$	$\displaystyle-\frac{t}{t-1}B\log\frac{1}{B}-\frac{1}{t-1}\Gamma\log\frac{1}{\Gamma}-cd$
		$\displaystyle\geq$	$\displaystyle-\frac{2}{t-1}A\log\frac{1}{A}-\frac{1}{t-1}\bigg{(}B\log\frac{1}{B}+\Gamma\log\frac{1}{\Gamma}-2A\log\frac{1}{A}\bigg{)}-cd.$

Now

	$\displaystyle B\log\frac{1}{B}+\Gamma\log\frac{1}{\Gamma}-2A\log\frac{1}{A}$	$\displaystyle=$	$\displaystyle(B+\Gamma-2A)\log\frac{1}{A}+A\frac{B}{A}\log\frac{A}{B}+A\frac{\Gamma}{A}\log\frac{A}{\Gamma}$
		$\displaystyle\leq$	$\displaystyle\ \bigg{(}\frac{B+\Gamma}{A}-2\bigg{)}A\log\frac{1}{A}+cA.$

Using (3.1)

$$\frac{B+\Gamma}{A}-1\leq\frac{2(t-1)}{t^{d+1}+1}$$

and we conclude that

	$\displaystyle\frac{1}{t-1}\bigg{(}B\log\frac{1}{B}+\Gamma\log\frac{1}{\Gamma}-2A\log\frac{1}{A}\bigg{)}$	$\displaystyle\leq$	$\displaystyle\frac{2}{t^{d+1}+1}A\log\frac{1}{A}+\frac{c}{t-1}A$
		$\displaystyle\leq$	$\displaystyle c+c\frac{t+1}{t-1}\frac{t^{d}-1}{t^{d}+1}\leq cd.$

Therefore

			$\displaystyle\log(t-1)+\cdots+\log(t^{j}-1)+\log\bigg{(}1-\frac{1}{t^{d-j}}\bigg{)}+\cdots+\log\bigg{(}1-\frac{1}{t}\bigg{)}$
		$\displaystyle\geq$	$\displaystyle-\frac{2}{t-1}A\log\frac{1}{A}-cd$

and, finally, (3) implies that for some $k>\frac{d}{2}$

$$1\leq\frac{c_{o}^{d}\alpha}{t^{\frac{k(k-1)}{2}}}\bigg{(}\frac{1}{A}\bigg{)}^{\frac{2A}{t-1}},$$

where $c_{o}$ is an absolute positive constant.

Let $\Omega$ be a function with zero mean value on the unit sphere $S^{n-1}$ belonging to the class $L\log L(S^{n-1})$, that is

$$\|\Omega\|_{L\log L(S^{n-1})}=\int_{S^{n-1}}|\Omega(x^{\prime})|(1+\log^{+}|\Omega(x^{\prime})|)d\sigma_{n-1}(x^{\prime})<\infty.$$

Set $K(x)=\Omega(x/|x|)/|x|^{n}$ and let $P\in\mathcal{P}_{d,n}$. We will show the theorem for $d=2^{m}$, for some $m\geq 0$. The general case is then an immediate consequence.

We set

$$C_{d}=\sup_{\begin{subarray}{c}0<\epsilon<R\\ P\in\mathcal{P}_{d,n}\end{subarray}}\left|\int_{\epsilon\leq|x|\leq R}e^{iP(x)}K(x)dx\right|,$$

where $C_{d}$ is a constant depending on $d$, $\Omega$ and $n$. For $0<\epsilon<R$ and $P\in\mathcal{P}_{d,n}$ we write,

$$I_{\epsilon,R}(P)=\int_{\epsilon\leq|x|\leq R}e^{iP(x)}K(x)dx=\int_{S^{n-1}}\int_{\epsilon}^{R}e^{iP(rx^{\prime})}\frac{dr}{r}\Omega(x^{\prime})d\sigma_{n-1}(x^{\prime}).$$

For $x^{\prime}\in S^{n-1}$, we have that $P(rx^{\prime})=\sum_{j=1}^{d}P_{j}(x^{\prime})r^{j}$ where $P_{j}$ is a homogeneous polynomial of degree $j$. Observe that we can omit the constant term, without loss of generality. Set also $m_{j}=\|P_{j}\|_{L^{\infty}(S^{n-1})}$. Since $\epsilon$ and $R$ are arbitrary positive numbers, by a dilation in $r$ we can assume that $\max_{\frac{d}{2}<j\leq d}m_{j}=1$ and, in particular, that $m_{j_{o}}=1$ for some $\frac{d}{2}<j_{o}\leq d$. We also write $Q(x)=\sum_{j=1}^{\frac{d}{2}}P_{j}(x)$. We split the integral in two parts as follows

	$\displaystyle|I_{\epsilon,R}(P)|$	$\displaystyle\leq$	$\displaystyle\left|\int_{S^{n-1}}\int_{\epsilon}^{1}e^{iP(rx^{\prime})}\frac{dr}{r}\Omega(x^{\prime})d\sigma_{n-1}(x^{\prime})\right|$
		$\displaystyle+$	$\displaystyle\left|\int_{S^{n-1}}\int_{1}^{R}e^{iP(rx^{\prime})}\frac{dr}{r}\Omega(x^{\prime})d\sigma_{n-1}(x^{\prime})\right|=I_{1}+I_{2}.$

For $I_{1}$ we have that

	$\displaystyle I_{1}$	$\displaystyle\leq$	$\displaystyle\int_{S^{n-1}}\int_{0}^{1}\left|e^{iP(rx^{\prime})}-e^{iQ(rx^{\prime})}\right|\frac{dr}{r}|\Omega(x^{\prime})|d\sigma_{n-1}(x^{\prime})$
		$\displaystyle+$	$\displaystyle\left|\int_{S^{n-1}}\int_{\epsilon}^{1}e^{iQ(rx^{\prime})}\frac{dr}{r}\Omega(x^{\prime})d\sigma_{n-1}(x^{\prime})\right|$
		$\displaystyle\leq$	$\displaystyle\sum_{\frac{d}{2}<j\leq d}\frac{m_{j}}{j}\|\Omega\|_{L^{1}(S^{n-1})}+C_{\frac{d}{2}}\leq c\|\Omega\|_{L^{1}(S^{n-1})}+C_{\frac{d}{2}}.$

For $I_{2}$ we write

	$\displaystyle I_{2}$	$\displaystyle\leq$	$\displaystyle\int_{S^{n-1}}\left|\int_{\{r\in[1,R]:|\frac{\partial P(rx^{\prime})}{\partial r}|>d\}}e^{iP(rx^{\prime})}\frac{dr}{r}\right||\Omega(x^{\prime})|d\sigma_{n-1}(x^{\prime})$
		$\displaystyle+$	$\displaystyle\int_{S^{n-1}}\int_{\{r\in[1,R]:|\frac{\partial P(rx^{\prime})}{\partial r}|\leq d\}}\frac{dr}{r}|\Omega(x^{\prime})|d\sigma_{n-1}(x^{\prime}).$

Since $\{r\in[1,R]:|\frac{\partial P(rx^{\prime})}{\partial r}|>d\}$ consists of at most $O(d)$ intervals where $\frac{\partial P(rx^{\prime})}{\partial r}$ is monotonic, a simple corollary to van der Corput’s lemma for the first derivative [5, corollary on p. 334] gives the bound

$$\int_{S^{n-1}}\left|\int_{\{r\in[1,R]:|\frac{\partial P(rx^{\prime})}{\partial r}|>d\}}e^{iP(rx^{\prime})}\frac{dr}{r}\right||\Omega(x^{\prime})|d\sigma_{n-1}(x^{\prime})\leq c\|\Omega\|_{L^{1}(S^{n-1})}.$$

On the other hand, the logarithmic measure lemma implies that

			$\displaystyle\int_{S^{n-1}}\int_{\{r\in[1,R]:|\frac{\partial P(rx^{\prime})}{\partial r}|\leq d\}}\frac{dr}{r}|\Omega(x^{\prime})|d\sigma_{n-1}(x^{\prime})$
		$\displaystyle\leq$	$\displaystyle c\|\Omega\|_{L^{1}(S^{n-1})}+c\frac{1}{d}\int_{S^{n-1}}\log\frac{d}{\max_{\frac{d}{2}<j\leq d}\{j|P_{j}(x^{\prime})|\}}|\Omega(x^{\prime})|d\sigma_{n-1}(x^{\prime}).$

Combining the estimates we get

$$C_{d}\leq c\|\Omega\|_{L^{1}(S^{n-1})}+C_{\frac{d}{2}}+c\frac{2j_{o}}{d}\int_{S^{n-1}}\ \log\frac{\|P_{j_{o}}\|^{\frac{1}{2j_{o}}}_{L^{\infty}(S^{n-1})}}{|P_{j_{o}}(x^{\prime})|^{\frac{1}{2j_{o}}}}|\Omega(x^{\prime})|d\sigma_{n-1}(x^{\prime})$$

and, from Young’s inequality,

	$\displaystyle C_{d}\leq c\|\Omega\|_{L^{1}(S^{n-1})}+C_{\frac{d}{2}}$	$\displaystyle+$	$\displaystyle c\int_{S^{n-1}}\ \frac{\|P_{j_{o}}\|^{\frac{1}{2j_{o}}}_{L^{\infty}(S^{n-1})}}{|P_{j_{o}}(x^{\prime})|^{\frac{1}{2j_{o}}}}d\sigma_{n-1}(x^{\prime})$
		$\displaystyle+$	$\displaystyle c\int_{S^{n-1}}|\Omega(x^{\prime})|(1+\log^{+}|\Omega(x^{\prime})|)d\sigma_{n-1}(x^{\prime}).$

Now, using corollary 2.3 we get

$$C_{d}\leq C_{\frac{d}{2}}+c(\|\Omega\|_{L\log L(S^{n-1})}+1).$$

Since $d=2^{m}$, this means that

$$C_{2^{m}}\leq C_{2^{m-1}}+c(\|\Omega\|_{L\log L(S^{n-1})}+1).$$

Using induction on $m$ we get that $C_{2^{m}}\leq C_{1}+cm(\|\Omega\|_{L\log L(S^{n-1})}+1)$. Observe that $C_{1}$ corresponds to some polynomial $P(x)=b_{1}x_{1}+\cdots+b_{n}x_{n}$. We write

	$\displaystyle\left|\int_{\epsilon<|x|<R}e^{iP(x)}K(x)dx\right|=$
	$\displaystyle=\left|\int_{S^{n-1}}\int_{\epsilon}^{R}\{e^{irP(x^{\prime})}-e^{ir\|P\|_{L^{\infty}(S^{n-1})}}\}\frac{dr}{r}\Omega(x^{\prime})d\sigma_{n-1}(x^{\prime})\right|.$

Using the simple estimate

$$\left|\int_{\epsilon}^{R}\{e^{iar}-e^{ibr}\}\frac{dr}{r}\right|\leq c+c\left|\log\left|\frac{b}{a}\right|\right|$$

we get

	$\displaystyle\left|\int_{\epsilon<|x|<R}e^{iP(x)}K(x)dx\right|$	$\displaystyle\leq$	$\displaystyle c\|\Omega\|_{L^{1}(S^{n-1})}+$
			$\displaystyle+c\int_{S^{n-1}}\log\frac{\|P\|^{\frac{1}{2}}_{L^{\infty}(S^{n-1})}}{|P(x^{\prime})|^{\frac{1}{2}}}|\Omega(x^{\prime})|d\sigma_{n-1}(x^{\prime}).$

Hence, $C_{1}\leq c\|\Omega\|_{L^{1}(S^{n-1})}+c+\|\Omega\|_{L\log L(S^{n-1})}$ and

$$C_{2^{m}}\leq cm(\|\Omega\|_{L\log L(S^{n-1})}+1).$$

The case of general d is now trivial. If $2^{m-1}<d\leq 2^{m}$ then

$$C_{d}\leq C_{2^{m}}\leq cm(\|\Omega\|_{L\log L(S^{n-1})}+1)\leq c\log d(\|\Omega\|_{L\log L(S^{n-1})}+1).$$