Uniform $l^2$-Decoupling in $\mathbb R^2$ for Polynomials

Abstract

For each positive integer d, we prove a uniform $l^2$-decoupling inequality for the collection of all polynomials phases of degree at most d. Our result is intimately related to Biswas et al. (Proc Am Math Soc 148(5):1987–1997, 2020), but we use a different partition that is determined by the geometry of each individual phase function.

Appendix

In the appendix, we prove Proposition 2.1 of [1] in the special case when $\nu \ge 0$.

Corollary 7.1

(Proposition 2.1 of [1]) Let $\delta \in 2^{-2\mathbb {N}}$, $d\in [3,\infty )$ and $2\le p\le 6$. For each interval $I\subseteq [0,1]$, let

$$\begin{aligned} E_I g(x,y)=\int _I g(s)e(xs+ys^d)ds. \end{aligned}$$

Let T be an axis-parallel rectangle with side lengths $\delta ^{-1}\times \delta ^{-d/2}$ (in the x and y directions, respectively). Let $\eta$ be a nonnegative Schwartz function such that $|\eta |\ge 1$ on T and $\widehat{\eta }$ is supported on the dual rectangle $T^*$ which has side lengths $\delta \times \delta ^{d/2}$.

Denote $\Delta _k=[(k-1)\delta ^{1/2},k\delta ^{1/2}]$, $1\le k\le \delta ^{-1/2}$. Then for any $g\in L^1([0,1])$, we have

$$\begin{aligned} \Vert E_{[0,1]}g \Vert _{L^p(T)}\lesssim _\varepsilon \delta ^{-\varepsilon }\left( \sum _{k=1}^{\delta ^{-1/2}}\Vert E_{\Delta _{k}}g \Vert ^2_{L^p(\eta ^p)}\right) ^{\frac{1}{2}}, \end{aligned}$$

where the implicit constant depends on p and $\varepsilon$ only.

Proof

Split the partition ${\mathcal {P}}:=\{\Delta _k:1\le k\le \delta ^{-1/2}\}$ into subpartitions ${\mathcal {P}}_n$, $1\le n\le \log _2(\delta ^{-1/2})$, defined as

$$\begin{aligned} {\mathcal {P}}_n:=\{\Delta _k:2^{n-1}\le k< 2^n\}. \end{aligned}$$

Denote $I_n$ as the union of intervals in ${\mathcal {P}}_n$. Since we can afford logarithmic losses, by triangle inequality and Cauchy–Schwarz it suffices to prove that for each n, we have

$$\begin{aligned} \Vert E_{I_n}g \Vert _{L^p(T)}\lesssim _\varepsilon \delta ^{-\varepsilon }\left( \sum _{\Delta _k\in {\mathcal {P}}_n}\Vert E_{\Delta _{k}}g \Vert ^2_{L^p(\eta ^p)}\right) ^{\frac{1}{2}}. \end{aligned}$$

(7.1)

Note that for each n, ${\mathcal {P}}_n$ is a sub-admissible partition of $I_n$ for $s^d$ at scale $a_n:=2^{dn}\delta ^{d/2}$. Therefore, by Theorem 1.6 applied to the monomial $s^d\in {\mathcal {D}}_{ d -2}$, for any $f\in L^p(\mathbb {R}^2)$ with Fourier support in ${\mathcal {N}}^{s^d}_{I_n,Ca_n}$ (where $C=C_d$ is an absolute constant), we have

$$\begin{aligned} \Vert f \Vert _{L^p(\mathbb {R}^2)}\lesssim _\varepsilon \delta ^{-\varepsilon }\left( \sum _{\Delta _k\in {\mathcal {P}}_n} \Vert f_{\Delta _k} \Vert ^2_{L^p(\mathbb {R}^2)}\right) ^{\frac{1}{2}}. \end{aligned}$$

(7.2)

The proof that (7.2) implies (7.1) is purely technical and routine. Such idea can be found in more detail in [9], but we include it here for completeness. The idea is to use a standard mollification technique that passes a global neighbourhood version of a decoupling inequality to its local extension version, with the choice of appropriate scales.

Let $I'_n=\cup _{k=2^{n-1}+1}^{2^n-2}\Delta _k$ be the union of intervals in ${\mathcal {P}}_k$ excluding the first and the last one. By triangle inequality and Cauchy–Schwarz, it suffices to show

$$\begin{aligned} \Vert E_{I'_n}g \Vert _{L^p(T)}\lesssim _\varepsilon \delta ^{-\varepsilon }\left( \sum _{k=2^{n-1}}^{2^n-1}\Vert E_{\Delta _k}g \Vert ^2_{L^p(\eta ^p)}\right) ^{\frac{1}{2}}. \end{aligned}$$

(7.3)

On the left-hand side of (7.3), we have

$$\begin{aligned} \Vert E_{I'_n}g \Vert _{L^p(T)}\le \Vert \eta E_{I'_n}g \Vert _{L^p(\mathbb {R}^2)}. \end{aligned}$$

Observe that $\widehat{E_{I'_n}g}*\widehat{\eta }$ is a smooth function supported on the Minkowski sum of $\Gamma _n$ and $T^*$ where $\Gamma _n$ is the graph of $s^d$ over $I'_n$. Since each $\Delta _k$ has length $\delta ^{1/2}$ and $T^*$ has length $\delta$ in the horizontal direction, we see that $\Gamma _n+T^*$ does not exceed [0, 1] in the horizontal axis. In addition, for each $2^{n-1}\le k\le 2^n-1$, $2^{n-2}\le k'\le 2^n-1$, we have $(\eta E_{\Delta _{k'}} g)_{\Delta _k}\ne 0$ only if $k'=k-1$, k or $k+1$.

In the vertical axis, for $(s,s^d)\in \Gamma _n$ and $(u,v)\in T^*$, we have

$$\begin{aligned} |s^d+v-(s+u)^d|&\le |v|+|u|\sum _{j=0}^{d-1}(s+u)^j s^{d-1-j}\\&\lesssim _d \delta ^{\frac{d}{2}}+\delta \cdot 2^{n(d-1)}\delta ^{\frac{d-1}{2}}\\&\lesssim _d 2^{dn}\delta ^{\frac{d}{2}}, \end{aligned}$$

since $2^n\le \delta ^{-1/2}$ and for $s\in I'_n$ we have $2^{n-1}\delta ^{1/2}\le s\le 2^n \delta ^{1/2}$. Thus we have

$$\begin{aligned} \Gamma _n+T^*\subseteq {\mathcal {N}}^{s^3}_{I_n,Ca_n}. \end{aligned}$$

Thus, for $2^{n-1}\le k\le 2^n-1$,

$$\begin{aligned} \Vert (\eta E_{I_n'}g)_{\Delta _k} \Vert _{L^p(\mathbb {R}^2)}&\le \Vert (\eta E_{\Delta _{k-1}}g)_{\Delta _k} \Vert _{L^p(\mathbb {R}^2)}+\Vert (\eta E_{\Delta _k}g)_{\Delta _k} \Vert _{L^p(\mathbb {R}^2)}+\Vert (\eta E_{\Delta _{k+1}}g)_{\Delta _k} \Vert _{L^p(\mathbb {R}^2)}\\&\lesssim \Vert \eta E_{\Delta _{k-1}}g \Vert _{L^p(\mathbb {R}^2)}+\Vert \eta E_{\Delta _k}g \Vert _{L^p(\mathbb {R}^2)}+\Vert \eta E_{\Delta _{k+1}}g \Vert _{L^p(\mathbb {R}^2)}, \end{aligned}$$

since the Fourier multiplier $(s,t)\mapsto 1_I(s)$ for any interval I has $L^p(\mathbb {R}^2)\rightarrow L^p(\mathbb {R}^2)$ operator norm bounded by an absolute constant, whenever $1<p<\infty$.

Thus, by (7.2) with $f=\eta E_{I'_n}g$,

$$\begin{aligned} \Vert \eta E_{I'_n}g \Vert _{L^p(\mathbb {R}^2)}&\lesssim _\varepsilon \delta ^{-\varepsilon }\left( \sum _{k=2^{n-1}}^{2^n-1}\Vert (\eta E_{I'_n}g)_{\Delta _k} \Vert ^2_{L^p(\mathbb {R}^2)}\right) ^{\frac{1}{2}}\\&\lesssim \delta ^{-\varepsilon }\left( \sum _{k=2^{n-1}}^{2^n-1}\left( \Vert \eta E_{\Delta _{k-1}}g \Vert _{L^p(\mathbb {R}^2)}+\Vert \eta E_{\Delta _k}g \Vert _{L^p(\mathbb {R}^2)}+\Vert \eta E_{\Delta _{k+1}}g \Vert _{L^p(\mathbb {R}^2)}\right) ^2\right) ^{\frac{1}{2}}\\&\lesssim \delta ^{-\varepsilon }\left( \sum _{k=2^{n-1}}^{2^n-1}\Vert \eta E_{\Delta _k}g \Vert ^2_{L^p(\mathbb {R}^2)}\right) ^{\frac{1}{2}}, \end{aligned}$$

where in the last line we have used Cauchy–Schwarz. $\square$