Convolution estimates related to space curves

Abstract

Based on a uniform estimate of convolution operators with measures on a family of plane curves, we obtain optimal L^p -L^q boundedness of convolution operators with affine arclength measures supported on space curves satisfying a suitable condition. The result generalizes the previously known estimates.

2000 Mathematics Subject Classifications: Primary 42B15; Secondary 42B20.

1 Introduction

Let I ⊂ ℝ be an open interval and ψ : I → ℝ be a C³ function. Let γ : I → ℝ³ be the curve given by γ(t) = (t, t²/2,ψ(t)), t ∈ I. Associated to γ is the affine arclength measure dσ _γ on ℝ³ determined by

$${\int }_{{ℝ}^3}fd{\sigma }_{\gamma }={\int }_{I}f\left(\gamma \left(t\right)\right)\lambda \left(t\right)dt,f\in C_0^{\infty }\left({ℝ}^3\right)$$

with

$$\lambda \left(t\right)={\left|{\psi }^{\left(3\right)}\left(t\right)\right|}^{\frac{1}{6}},t\in I.$$

The L^p - L^q mapping properties of the corresponding convolution operator $T_{{\sigma }_{\gamma }}$ given by

$$T_{{\sigma }_{\gamma }}f\left(x\right)=f*{\sigma }_{\gamma }\left(x\right)={\int }_{I}f\left(x-\gamma \left(t\right)\right)\lambda \left(t\right)dt$$

(1.1)

have been studied by many authors [1–8]. The use of the affine arclength measure was suggested by Drury [2] to mitigate the effect of degeneracy and has been helpful to obtain uniform estimates.

We denote by Δ the closed convex hull of {(0, 0), (1, 1), (p₀^-1, q₀^-1) (p₁^-1, q₁^-1)} in the plane, where p₀ = 3/2, q₀ = 2, p₁ = 2 and q₁ = 3. The line segment joining (p₀^-1, q₀^-1) and (p₁^-1, q₁^-1) is denoted by $\mathfrak{S}$. It is well known that the typeset of $T_{{\sigma }_{\gamma }}$ is contained in Δ and that under suitable conditions $T_{{\sigma }_{\gamma }}$ is bounded from L^p (ℝ³) to L^q (ℝ³) with uniform bounds whenever $(p^{-1},q^{-1})\in \mathfrak{S}$. The most general result currently available was obtained by Oberlin [5]. In this article, we establish uniform endpoint estimates on $T_{{\sigma }_{\gamma }}$ for a wider class of curves γ.

Before we state our main result, we introduce certain conditions on functions defined on intervals. For an interval J₁ in ℝ, a locally integrable function Φ : J₁ → ℝ⁺, and a positive real number A, we let

$$\begin{array}{ll}\hfill ℭ\left(\Phi ,A\right):=\left\{\omega :J_1\to {ℝ}^{+}\mid \sqrt{\omega \left(s_1\right)\omega \left(s_2\right)}\right.& \le \frac{A}{s_2-s_1}{\int }_{s_1}^{s_2}\Phi \left(s\right)ds\\ \left.\text{whenever}{s}_1<s_2\text{and}\left[s,s_2\right]\subset J_1\right\}\end{array}$$

and

$${\mathcal{E}}_1\left(A\right):=\left\{\Phi :J\to {ℝ}^{+}\mid \Phi \in ℭ\left(\Phi ,A\right)\right\}.$$

An interesting subclass of ${\mathcal{E}}_1\left(2A\right)$ is the collection ${\mathcal{E}}_2\left(A\right)$, introduced in [9], of functions Φ : J → ℝ⁺ such that

• 1. Φ is monotone; and

• 2. whenever s ₁ < s ₂ and [s ₁, s ₂] ⊂ J,

  $$\sqrt{\Phi \left(s_1\right)\Phi \left(s_2\right)}\le A\Phi \left(\left(s_1+s_2\right)∕2\right)$$

Our main theorem is the following:

Theorem 1.1. Let I = (a, b) ⊂ ℝ be an open interval and let ψ : I → ℝ be a C³function such that

1. ψ⁽³⁾(t) ≥ 0, whenever t ∈ I;

2. there exists A ∈ (0, ∞) such that, for each u ∈ (0, b - a), ${\mathfrak{F}}_u:(a,b-u)\to {ℝ}^{+}$ given by${\mathfrak{F}}_u\left(s\right):=\sqrt{{\psi }^{\left(3\right)}\left(s+u\right){\psi }^{\left(3\right)}\left(s\right)}$satisfies

$${\mathfrak{F}}_u\in {\mathcal{E}}_1\left(A\right).$$

(1.2)

Then, the operator$T_{{\sigma }_{\gamma }}$defined by (1.1) is a bounded operator from L^p (ℝ³) to L^q (ℝ³) whenever $(p^{-1},q^{-1})\in \mathfrak{S}$, and the operator norm${∥T_{{\sigma }_{\gamma }}∥}_{L^p\to L^q}$is dominated by a constant that depends only on A.

The case when ${\psi }^{\left(3\right)}\in {\mathcal{E}}_2\left(A\right)$ was considered by Oberlin [5]. One can easily see that ${\psi }^{\left(3\right)}\in {\mathcal{E}}_2\left(A∕2\right)$ implies (1.2) uniformly in u ∈ (0, b - a). The theorem generalizes many results previously known for convolution estimates related to space curves, namely [1–6].

This article is organized as follows: in the following section, a uniform estimate for convolution operators with measures supported on plane curves. The proof of Theorem 1.1 based on a T*T method is given in Section 3.

2 Uniform estimates on the plane

The following theorem motivated by Oberlin [10] which is interesting in itself will be useful:

Theorem 2.1. Let J be an open interval in ℝ, and ϕ : J → ℝ be a C²function such that ϕ″ ≥ 0. Let ω : J → ℝ be a nonnegative measurable function. Suppose that there exists a positive constant A such that $\omega \in \mathfrak{G}({\varphi }^{″},A)$, i.e.

$$\omega {\left(s_1\right)}^{1∕2}\omega {\left(s_2\right)}^{1∕2}\le \frac{A}{s_2-s_1}{\int }_{s_1}^{s_2}{\varphi }^{″}\left(v\right)dv$$

holds whenever s₁< s₂and [s₁, s₂] ⊂ J. Let S be the operator given by

$$\mathcal{S}g\left(x_2,x_3\right)={\int }_{J}g\left(x_2-s,x_3-\varphi \left(s\right)\right){\omega }^{1∕3}\left(s\right)ds$$

for$g\in C_0^{\infty }\left({ℝ}^2\right)$. Then, there exists a constant C that depends only on A such that

$$\parallel \mathcal{S}g{\parallel }_{L^3\left({ℝ}^2\right)}\le C\parallel g{\parallel }_{L^{3∕2}\left({ℝ}^2\right)}$$

holds uniformly in$g\in C_0^{\infty }\left({ℝ}^2\right)$.

Proof of Theorem 2.1. Our proof is based on the method introduced by Drury and Guo [11], which was later refined by Oberlin [10].

We have

$$\begin{array}{ll}\hfill \parallel \mathcal{S}g{\parallel }_3^3& ={\int }_{ℝ}{\int }_{ℝ}{\int }_J{\int }_J{\int }_J\prod _{j=1}^3\left(g\left(x_2-s_j,x_3-\varphi \left(s_j\right)\right){\omega }^{1∕3}\left(s_j\right)\right)d{s}_{1}d{s}_{2}d{s}_{3}d{x}_{2}d{x}_3\\ ={\int }_{ℝ}{\int }_{ℝ}{\int }_{ℝ}\left[\mathcal{G}\left(g\left(z_1,\cdot \right),g\left(z_2,\cdot \right),g\left(z_3,\cdot \right)\right)\right]\left(z_1,z_2,z_3\right)d{z}_{1}d{z}_{2}d{z}_3,\end{array}$$

where for z₁, z₂, z₃ ∈ ℝ and suitable functions h₁, h₂, h₃ defined on ℝ,

$$\begin{array}{r}[\mathcal{G}(h_1\mathrm{,}{h}_2\mathrm{,}{h}_3)(z_1\mathrm{,}{z}_2\mathrm{,}{z}_3):={\displaystyle {\int }_{ℝ}{\displaystyle {\int }_{J\left(z_1\mathrm{,}{z}_2\mathrm{,}{z}_3\right)}{\displaystyle \prod _{j=1}^3[h_j(x_3-\varphi (x_2-z_j)){\omega }^{1/3}(x_2-z_j)]}}}\\ d{x}_{2}d{x}_3\mathrm{,}\end{array}$$

and

$$J\left(z_1,z_2,z_3\right):=\left(J+z_1\right)\cap \left(J+z_2\right)\cap \left(J+z_3\right).$$

We will prove that the estimate

$$\mid \left[\mathcal{G}\left(h_1,h_2,h_3\right)\right]\left(z_1,z_2,z_3\right)\mid \le \frac{C\parallel h_1{\parallel }_{L^{3∕2}\left(ℝ\right)}\parallel h_2{\parallel }_{L^{3∕2}\left(ℝ\right)}\parallel h_3{\parallel }_{L^{3∕2}\left(ℝ\right)}}{\mid \left(z_1-z_2\right)\left(z_1-z_3\right)\left(z_2-z_3\right){\mid }^{1∕3}}$$

(2.1)

holds uniformly in h₁, h₂, h₃, z₁, z₂, and z₃.

To establish (2.1) we let

$$\begin{array}{ll}\hfill \left[{\mathcal{G}}_k\left(h_1,h_2,h_3\right)\right]\left(z_1,z_2,z_3\right):=& {\int }_{ℝ}{\int }_{J\left(z_1,z_2,z_3\right)}{h}_k\left(x_3-\varphi \left(x_2-z_k\right)\right)\\ \prod _{\begin{array}{c}1\le j\le 3\\ j\ne k\end{array}}\left[h_j\left(x_3-\varphi \left(x_2-z_j\right)\right){\omega }^{1∕2}\left(x_2-z_j\right)\right]d{x}_{2}d{x}_3\end{array}$$

for k = 1, 2, 3. Then, we have

$$\begin{array}{c}\mid \left[{\mathcal{G}}_1\left(h_1,h_2,h_3\right)\right]\left(z_1,z_2,z_3\right)\mid \le \parallel h_1{\parallel }_{\infty }{\int }_{ℝ}{\int }_{J\left(z_1,z_2,z_3\right)}\prod _{j=2}^3\left(\mid h_j\left(x_3-\varphi \left(x_2-z_j\right)\right)\mid {\omega }^{1∕2}\left(x_2-z_j\right)\right)\\ d{x}_{2}d{x}_3.\end{array}$$

For z₂, z₃ ∈ ℝ and x₂ ∈ J (z₁, z₂, z₃), we have

$$\begin{array}{c}\left|{\varphi }^{\prime })(x_2-z_2)-{\varphi }^{\prime }(x_2-z_3)\right|=\left|{\displaystyle {\int }_{x_2-z_2}^{x_2-z_3}}{\varphi }^{″}(s)ds\right|\\ \ge A^{-1}\mid z_2-z_3\mid {\omega }^{1/2}(x_2-z_2){\omega }^{1/2}(x_2-z_3)\end{array}$$

by hypothesis. Hence,

$$\begin{array}{c}\mid \left[{\mathcal{G}}_1\left(h_1,h_2,h_3\right)\right]\left(z_1,z_2,z_3\right)\mid \le \frac{A\parallel h_1{\parallel }_{\infty }}{\left|z_2-z_3\right|}{\int }_{ℝ}{\int }_{J\left(z_1,z_2,z_3\right)}\prod _{j=2}^3\mid h_j\left(x_3-\varphi \left(x_2-z_j\right)\right)\mid \\ \mid {\varphi }^{\prime }\left(x_2-z_2\right)-{\varphi }^{\prime }\left(x_2-z_3\right)\mid d{x}_{2}d{x}_3.\end{array}$$

A change of variables gives

$$\mid \left[{\mathcal{G}}_1\left(h_1,h_2,h_3\right)\right]\left(z_1,z_2,z_3\right)\mid \le \frac{A\parallel h_1{\parallel }_{\infty }}{\mid z_2-z_3\mid }{\int }_{ℝ}{\int }_{ℝ}\mid h_2\left(z_2\right)\mid \mid h_3\left(z_3\right)\mid d{z}_{2}d{z}_3.$$

Thus, we obtain

$$\mid \left[{\mathcal{G}}_1\left(h_1,h_2,h_3\right)\right]\left(z_1,z_2,z_3\right)\mid \le \frac{A\parallel h_1{\parallel }_{\infty }\parallel h_2{\parallel }_1\parallel h_3{\parallel }_1}{\mid z_2-z_3\mid }.$$

(2.2)

Similarly, we get

$$\mid \left[{\mathcal{G}}_2\left(h_1,h_2,h_3\right)\right]\left(z_1,z_2,z_3\right)\mid \le \frac{A\parallel h_1{\parallel }_1\parallel h_2{\parallel }_{\infty }\parallel h_3{\parallel }_1}{\mid z_1-z_3\mid }$$

(2.3)

and

$$\mid \left[{\mathcal{G}}_2\left(h_1,h_2,h_3\right)\right]\left(z_1,z_2,z_3\right)\mid \le \frac{A\parallel h_1{\parallel }_1\parallel h_2{\parallel }_1\parallel h_3{\parallel }_{\infty }}{\mid z_1-z_2\mid }.$$

(2.4)

Interpolating (2.2), (2.3) and (2.4) provides (2.1). Combining this with Proposition 2.2 in Christ [12] finishes the proof.

The special case in which ω = ϕ″ provides a uniform estimate for the convolution operators with affine arclength measure on plane curves.

Corollary 2.2. Let J be an open interval in ℝ, and ϕ : J → ℝ be a C²function such that ϕ″ ≥ 0. Suppose that there exists a constant A such that${\varphi }^{″}\in {\mathcal{E}}_1\left(A\right)$, i.e.

$${\varphi }^{″}{\left(s_1\right)}^{1∕2}{\varphi }^{″}{\left(s_2\right)}^{1∕2}\le \frac{A}{s_2-s_1}{\int }_{s_1}^{s_2}{\varphi }^{″}\left(v\right)dv$$

holds whenever s₁< s₂and [s₁, s₂] ⊂ J. Let$\mathcal{S}$be the operator given by

$$\mathcal{S}g\left(x_2,x_3\right)={\int }_{J}g\left(x_2-s,x_3-\varphi \left(s\right)\right){\varphi }^{″}{\left(s\right)}^{1∕3}ds$$

for$g\in C_0^{\infty }\left({ℝ}^2\right)$. Then, there exists a constant C that depends only on A such that

$$\parallel \mathcal{S}g{\parallel }_{L^3\left({ℝ}^2\right)}\le C\parallel g{\parallel }_{L^{3∕2}\left({ℝ}^2\right)}$$

holds uniformly in$g\in C_0^{\infty }\left({ℝ}^2\right)$.

3 Proof of the main theorem

Before we proceed the proof of Theorem 1.1, we note that the uniform estimate (1.2) in u ∈ (0, b - a) implies

$${\psi }^{\left(3\right)}\in {\mathcal{E}}_1\left(A\right)$$

(3.1)

by continuity of ψ⁽³⁾.

By duality and interpolation, it suffices to show that

$$\parallel T_{{\sigma }_{\gamma }}f{\parallel }_{L^2\left({ℝ}^3\right)}\le C\parallel f{\parallel }_{L^{3∕2}\left({ℝ}^3\right)}$$

(3.2)

holds uniformly for f ∈ L^3/2 (ℝ³).

Recall the following lemma observed by Oberlin [3]:

Lemma 3.1. Suppose there exists a constant C₁such that

$$\parallel T_{{\sigma }_{\gamma }}^{*}{T}_{{\sigma }_{\gamma }}f{\parallel }_{L^3\left({ℝ}^3\right)}\le C_1\parallel f{\parallel }_{L^{3∕2}\left({ℝ}^3\right)}$$

(3.3)

holds uniformly in f ∈ L^3/2 (R³). Then, (3.2) holds for each f ∈ L^3/2 (ℝ³).

To establish (3.3), we write

$$\begin{array}{ll}\hfill T_{{\sigma }_{\gamma }}^{*}{T}_{{\sigma }_{\gamma }}f\left(x\right)& ={\int }_I{\int }_{I}f\left(x-\gamma \left(t\right)+\gamma \left(s\right)\right)\lambda \left(t\right)\lambda \left(s\right)dtds\\ equiv{\mathcal{T}}^{\left(1\right)}f\left(x\right)+{\mathcal{T}}^{\left(2\right)}f\left(x\right),\end{array}$$

where

$$\begin{array}{c}{\mathcal{T}}^{\left(1\right)}f\left(x\right)={\iint }_{\begin{array}{c}t,s\in I\\ t>s\end{array}}f\left(x-\gamma \left(t\right)+\gamma \left(s\right)\right)\lambda \left(t\right)\lambda \left(s\right)dtds,\\ {\mathcal{T}}^{\left(2\right)}f\left(x\right)={\iint }_{\begin{array}{c}t,s\in I\\ t<s\end{array}}f\left(x-\gamma \left(t\right)+\gamma \left(s\right)\right)\lambda \left(t\right)\lambda \left(s\right)dtds.\end{array}$$

By symmetry, it suffices to prove

$$\parallel {\mathcal{T}}^{\left(1\right)}f{\parallel }_{L^3\left({ℝ}^3\right)}\le C_1\parallel f{\parallel }_{L^{3∕2}\left({ℝ}^3\right)}.$$

Next we make a change of variables, u = t - s and write for u ∈ (0, b - a)

$$\begin{array}{ll}\hfill I_u& =\left\{s\in ℝ:a<s<b-u\right\},\\ \hfill {\Psi }_u\left(s\right)& =\psi \left(s+u\right)-\psi \left(s\right).\end{array}$$

Then, we obtain:

$$\begin{array}{ll}\hfill {\mathcal{T}}^{\left(1\right)}f\left(x\right)& ={\int }_I{\int }_0^{b-s}f\left(x_1-u,x_2-u\left(s+u∕2\right),x_3-{\Psi }_u\left(s\right)\right)\lambda \left(s+u\right)\lambda \left(s\right)duds\\ ={\int }_0^{b-a}{\int }_{I_u}f\left(x_1-u,x_2-u\left(s+u∕2\right),x_3-{\Psi }_u\left(s\right)\right)\lambda \left(s+u\right)\lambda \left(s\right)dsdu,\end{array}$$

and so

$${\mathcal{T}}^{\left(1\right)}f\left(x_1,x_2,x_3\right)={\int }_0^{b-a}{\mathcal{T}}_u\left[f_u\left(x_1-u,\cdot ,\cdot \cdot \right)\right]\left(\left(x_2-u^2∕2\right)∕u,x_3\right)\frac{du}{u^{2∕3}},$$

where

$$\begin{array}{ll}\hfill f_u\left(x_1,x_2,x_3\right)& :=u^{1∕3}f\left(x_1,u{x}_2,x_3\right)\\ \hfill {\mathcal{T}}_{u}g\left(x_2,x_3\right)& :={\int }_{I_u}g\left(x_2-s,x_3-{\Psi }_u\left(s\right)\right){\Lambda }_u^{1∕3}\left(s\right)ds\\ \hfill {\Lambda }_u\left(s\right)& :=u{\lambda }^3\left(s+u\right){\lambda }^3\left(s\right)\\ =u\sqrt{{\psi }^{\left(3\right)}\left(s+u\right){\psi }^{\left(3\right)}\left(s\right)}\end{array}$$

for x₁, x₂, x₃ ∈ ℝ, u ∈ (0, b - a), s ∈ I _u .

Notice that for u ∈ (0, b - a) and [s₁, s₂] ⊂ I _u , we have

$$\begin{array}{ll}\hfill {\Lambda }_u^{1∕2}\left(s_1\right){\Lambda }_u^{1∕2}\left(s_2\right)& \le \frac{Au}{s_2-s_1}{\int }_{s_1}^{s_2}\sqrt{{\psi }^{\left(3\right)}\left(s+u\right){\psi }^{\left(3\right)}\left(s\right)}ds\\ \le \frac{A^{2}u}{s_2-s_1}{\int }_{s_1}^{s_2}\frac{1}{u}{\int }_s^{s+u}{\psi }^{\left(3\right)}\left(v\right)dvds\\ =\frac{A^2}{s_2-s_1}{\int }_{s_1}^{s_2}\left({\psi }^{″}\left(s+u\right)-{\psi }^{″}\left(s\right)\right)ds\\ =\frac{A^2}{s_2-s_1}{\int }_{s_1}^{s_2}{{\Psi }^{″}}_u\left(s\right)ds\end{array}$$

by (1.2) and (3.1). By Theorem 2.1, $\parallel {\mathcal{T}}_u{\parallel }_{L^{3∕2}\left({ℝ}^2\right)\to L^3\left({ℝ}^2\right)}$ is uniformly bounded. Hence, we obtain

$$\begin{array}{ll}\hfill \parallel {\mathcal{T}}^{\left(1\right)}f{\parallel }_3& \le {\left({\int }_{ℝ}{\left[{\iint }_{{ℝ}^2}{\left({\int }_0^{b-a}\left|{\mathcal{T}}_u{f}_u\left(x_1-u,\cdot ,\cdot \cdot \right)\left(\frac{x_2-u^2∕2}{u},x_3\right)\right|\frac{du}{u^{2∕3}}\right)}^{3}d{x}_{2}d{x}_3\right]}^{\frac{1}{3}\cdot 3}d{x}_1\right)}^{\frac{1}{3}}\\ \le {\left({\int }_{ℝ}{\left[{\int }_0^{b-a}{\left({\iint }_{{ℝ}^2}{\left|{\mathcal{T}}_u{f}_u\left(x_1-u,\cdot ,\cdot \cdot \right)\left(\frac{x_2-u^2∕2}{u},x_3\right)\right|}^{3}d{x}_{2}d{x}_3\right)}^{\frac{1}{3}}\frac{du}{u^{2∕3}}\right]}^{3}d{x}_1\right)}^{\frac{1}{3}}\\ \le C\left(A\right){\left({\int }_{ℝ}{\left[{\int }_0^{b-a}{u}^{\frac{1}{3}}\parallel f_u\left(x_1-u,\cdot ,\cdot \cdot \right){\parallel }_{L^{3∕2}\left({ℝ}^2\right)}\frac{du}{u^{2∕3}}\right]}^{3}d{x}_1\right)}^{\frac{1}{3}}\\ \le C\left(A\right){\left({\int }_{ℝ}{\left[{\int }_0^{b-a}\parallel f\left(x_1-u,\cdot ,\cdot \cdot \right){\parallel }_{L^{3∕2}\left({ℝ}^2\right)}\frac{du}{u^{2∕3}}\right]}^{3}d{x}_1\right)}^{\frac{1}{3}}.\end{array}$$

By Hardy-Littlewood-Sobolev theorem on fractional integration, we obtain

$$\parallel {\mathcal{T}}^{\left(1\right)}f{\parallel }_3\le C_1\left(A\right)\parallel f{\parallel }_{3∕2}$$

This finishes the proof of Theorem 1.1.