Leibniz’s Rule, Sampling and Wavelets on Mixed Lebesgue Spaces

Abstract

Several new results about Leibniz’s rule, sampling, and wavelets in the context of mixed Lebesgue spaces are presented. The results obtained rely in part on vector-valued estimates for Calderón–Zygmund operators and the use of appropriate maximal functions and Littlewood–Paley estimates.

Appendix

We sketch the proof of Corollary 2.3.

Proof

Fix $1<q<\infty$. We have $\mathbf{T}:L^q(\mathbb R^{n+1}, \mathbf{B}_1)\rightarrow L^q(\mathbb R^{n+1}, \mathbf{B}_2)$, so that for it is given by a kernel $\mathbf{K}(u,v):\mathbf{B}_1\rightarrow \mathbf{B}_2$, in the form

$$\begin{aligned} \mathbf{T}f(u) = \int _{\mathbb R^{n+1}}\mathbf{K}(u,v)(f(v)) \, dv \end{aligned}$$

for $u$ not in the support of $f$. Writing $u=(t,x)\in \mathbb R\times \mathbb R^n$ and $L^q = L^q_tL^q_x$, we can also write

$$\begin{aligned} \mathbf{T}:L^q_tL^q_x(\mathbb R\times \mathbb R^n,\mathbf{B}_1)\rightarrow L^q_tL^q_x(\mathbb R\times \mathbb R^n,\mathbf{B}_2), \end{aligned}$$

with associated kernel $\mathbf{K}((t,x),(s,y)]):\mathbf{B}_1\rightarrow \mathbf{B}_2$, so that

$$\begin{aligned} \mathbf{T}f(t,x)=\int _{R}\int _{R^n}\mathbf{K}((t,x),(s,y))(f(s,y))\,dyds \end{aligned}$$

for $(t,x)$ not in the support of $f$. We want to show that the above is true for exponents $p$ and $q$ when $p\ne q$.

With that in mind, we make the identification $L^p_tL^q_x(\mathbb R\!\times \mathbb R^n,\mathbf{B})\! =\! L^p_t(\mathbb R,L^q_x(\mathbb R^n,\mathbf{B}))$, and set $L^q(\mathbb R^n,\mathbf{B}_1) = \mathbf{B}_3$ and $L^q(\mathbb R^n,\mathbf{B}_2) = \mathbf{B}_4$. We can associate to $\mathbf{T}$, which is bounded from $L^qL^q(\mathbb R\times \mathbb R^n,\mathbf{B}_1)$ to $L^qL^q(\mathbb R\times \mathbb R^n,\mathbf{B}_2)$, an operator $\mathcal {T}$ bounded from $L^q(\mathbb R,\mathbf{B}_3)$ to $L^q(\mathbb R,\mathbf{B}_4)$ and with kernel $\mathcal {K}(t,s): L^q_x(\mathbb R^n,\mathbf{B}_1)\rightarrow L^q_x(\mathbb R^n,\mathbf{B}_2)$, $t,s\in \mathbb R$, $t\ne s$, so that

$$\begin{aligned} \mathcal {T}F(t)&= \int _{\mathbb R} \mathcal {K}(t,s) F(s)\, ds \\ \end{aligned}$$

for $t$ not in the support of $F(s)$, and so that when $F(s)= f(s,\cdot )$

$$\begin{aligned} \mathcal {T}F(t)(x)&= \int _{\mathbb R} \mathcal {K}(t,s)F(s)(x)\, ds \\&= \int _{R}\int _{R^n}\mathbf{K}((t,x),(s,y))(f(s,y))\,dyds\\&= \mathbf{T}f(t,x). \end{aligned}$$

If we show that $\mathcal {K}$ is bounded and satisfies the two estimates

$$\begin{aligned} \int _{|t-s|\ge 2|s-r|} \Vert \mathcal {K}(t,s) - \mathcal {K}(t,r)\Vert _{\mathbf{B}_3\rightarrow \mathbf{B}_4}dt\le C \end{aligned}$$

(22)

$$\begin{aligned} \int _{|t-s|\ge 2|t-t|} \Vert \mathcal {K}(t,s) - \mathcal {K}(t,s)\Vert _{\mathbf{B}_3\rightarrow \mathbf{B}_4}dt\le C \end{aligned}$$

(23)

then by Theorem 2.1 we will have the desired result (after translating appropriately).

Fixing $t,s,\in \mathbb R$, $t\ne s$ we have

$$\begin{aligned} (\mathcal {K}(t,s)H)(x)&= \int _{\mathbb R^n} \mathbf{K}((t,x),(s,y))H(y) \, dy, \end{aligned}$$

with

$$\begin{aligned} \Vert \mathbf{K}((t,x),(s,y)) \Vert _{\mathbf{B}_1\rightarrow \mathbf{B}_2}&\lesssim {\frac{1}{|(t,x)-(s,y)|^{n+1}}}\\&\lesssim {\frac{1}{(|t-s|^2+|x-y|^2)^{(n+1)/2}}}. \end{aligned}$$

If we integrate in $x$ we have,

$$\begin{aligned} \int _{\mathbb R^n}\Vert \mathbf{K}((t,x),(s,y)) \Vert _{\mathbf{B}_1\rightarrow \mathbf{B}_2} \, dx&\lesssim \int _{\mathbb R^n} \frac{dx}{(|t-s|^2+|x-y|^2)^{(n+1)/2}}\\&\lesssim \frac{1}{|t-s|^{n+1}} \int _{\mathbb R^n} \frac{dx}{\left( 1+\left( \frac{|x|}{|t-s|}\right) ^2\right) ^{(n+1)/2}}\\&\lesssim \frac{1}{|t-s|}. \end{aligned}$$

Similarly, integrating in $y$ yields

$$\begin{aligned} \int _{\mathbb R^n}\Vert \mathbf{K}((t,x),(s,y)) \Vert _{\mathbf{B}_1\rightarrow \mathbf{B}_2} \, dy \lesssim \frac{1}{|t-s|}. \end{aligned}$$

By the Schur test (which also holds in the vector-valued setting; see e.g. [27] for details), we get that $\mathcal {K}$ is a bounded operator and

$$\begin{aligned} \Vert \mathcal {K}(t,s)\Vert _{\mathbf{B}_3\rightarrow \mathbf{B}_4} \lesssim \frac{1}{|t-s|}. \end{aligned}$$

The proofs of (22) and (23) are completely analogous. For example, now fix $t,s,r\in \mathbb R$ with $|t-s|\ge 2|s-r|$ and define an operator $G$ as follows:

$$\begin{aligned} G(t,s,r)H(x)&= (\mathcal {K}(t,s)-\mathcal {K}(t,r))H(x)\\&= \int _{\mathbb R^n}\left( \mathbf{K}((t,x),(s,y))- \mathbf{K}((t,x),(r,y))\right) H(y)dy\\&= \int _{\mathbb R^n}g(t,s,r)(x,y)H(y)dy. \end{aligned}$$

Now consider the kernel $g(t,s,r)(x,y)$. Using (10)

$$\begin{aligned} \Vert g(t,s,r)(x,y)\Vert _{\mathbf{B}_1\rightarrow \mathbf{B}_2}&= \Vert \mathbf{K}((t,x),(s,y)) - \mathbf{K}((t,x),(r,y))\Vert _{\mathbf{B}_1\rightarrow \mathbf{B}_2}\\&\lesssim {\frac{|(s,y)-(r,y)|^{\delta }}{|(t,x)-(s,y)|^{n+1+\delta }}}\\&\lesssim {\frac{|s-r|^{\delta }}{(|t-s|^2+|x-y|^2)^{(n+1+\delta )/2}}}, \end{aligned}$$

and integrating in $x$ we have,

$$\begin{aligned} \int _{\mathbb R^n}\Vert g(t,s,r)(x,y)\Vert _{\mathbf{B}_1\rightarrow \mathbf{B}_2} \,dx&\lesssim \int _{\mathbb R^n} \frac{|s-r|^{\delta }}{(|t-s|^2+|x-y|^2)^{(n+1+\delta )/2}} \, dx\\&\lesssim {\frac{|s-r|^{\delta }}{|t-s|^{1+\delta }}}. \end{aligned}$$

Likewise, if we integrate in $y$ we get

$$\begin{aligned} \int _{\mathbb R^n}\Vert g(t,s,r)(x,y)\Vert _{\mathbf{B}_1\rightarrow \mathbf{B}_2}dy \le C\frac{|s-r|^{\delta }}{|t-s|^{1+\delta }}. \end{aligned}$$

Again by the Schur test, $G(t,s,r)$ maps $\mathbf{B}_3$ into $\mathbf{B}_4$ with norm

$$\begin{aligned} \Vert G(t,s,r)\Vert _{\mathbf{B}_3\rightarrow \mathbf{B}_4} = \Vert \mathcal {K}(t,s)-\mathcal {K}(t,r)\Vert _{\mathbf{B}_3\rightarrow \mathbf{B}_4} \le C\frac{|s-r|^{\delta }}{|t-s|^{1+\delta }} \end{aligned}$$

and therefore $\mathcal {K}$ satisfies (22).

We leave the rest of the details to the interested reader. $\square$