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Suppose $\zeta\in\mathcal{S}(\mathbb{R}^{n})$ is a Schwartz function such that $\widehat{\zeta}$ is compactly supported away from the origin (say in the annulus $2^{-l_{0}}<\left|\xi\right|<2^{l_{0}}$ for some $l_{0}>0$). Let $\Delta_{j}^{\zeta}$ be the Littlewood-Paley projection defined by

$$\widehat{\Delta_{j}^{\zeta}f}(\xi)=\widehat{\zeta}\left(\dfrac{\xi}{2^{j}}\right)\widehat{f}(\xi),\quad{f\in \mathcal{S}(\mathbb{R}^{n})}$$

and consider the associated Littlewood-Paley decomposition $\sum_{j}\Delta_{j}^{\zeta}f$. It is a well-known theorem, the Littlewood-Paley inequality (L. Grafakos, Classical Fourier Analysis, Theorem 5.1.2) that the square function associated to the $\Delta_{j}^{\zeta}$ satisfies the inequality

$$\left\|\left(\sum_{j\in\mathbb{Z}}\left|\Delta_{j}^{\zeta}f\right|^{2}\right)^{1/2}\right\|_{L^{p}(\mathbb{R}^{n})}\lesssim_{n,p}\left\|f\right\|_{L^{p}(\mathbb{R}^{n})} \tag{1}$$ for $1<p<\infty$. Using the Littlewood-Paley inequality, we can show that the operator $f\mapsto\sum_{j}\Delta_{j}^{\zeta}f$ is bounded $L^{p}\rightarrow L^{p}$, for $1<p<\infty$. See below for details.

Question: Is there a simpler way (i.e. without using the Littlewood-Paley inequality) to show that the operator $$f\mapsto > \sum_{j\in\mathbb{Z}}\Delta_{j}^{\zeta}f \quad{f\in\mathcal{S}(\mathbb{R}^{n})}\tag{2}$$ bounded $L^{p}\rightarrow L^{p}$, for $1<p<\infty$?

By duality and the density of $\mathcal{S}(\mathbb{R}^{n})$ in $L^{p}$ spaces, it suffices to show that

$$\left|\langle{\sum_{j\in\mathbb{Z}}\Delta_{j}^{\zeta}f,g}\rangle\right|\lesssim_{n,p}\left\|f\right\|_{L^{p}(\mathbb{R}^{n})}\left\|g\right\|_{L^{p'}(\mathbb{R}^{n})},\quad{\forall f,g\in\mathcal{S}(\mathbb{R}^{n})}\tag{3}$$

Given $f\in\mathcal{S}(\mathbb{R}^{n})$, the decomposition $\sum_{j}\Delta_{j}^{\zeta}f$ converges in $L^{2}(\mathbb{R}^{n})$, whence in $\mathcal{S}'(\mathbb{R}^{n})$. Now let $\varphi$ be a Schwartz function such that $\widehat{\varphi}$ is compactly supported in the annulus $1/2<\left|\xi\right|<2$ and $\sum_{j}\widehat{\varphi}(2^{-j}\xi)=1$ for all $\xi\in\mathbb{R}^{n}\setminus\left\{0\right\}$. For $g\in\mathcal{S}(\mathbb{R}^{n})$, one can verify that $\sum_{k}\Delta_{k}^{\varphi}g$ converges in $L^{2}$. Whence,

$$\begin{array}{lcl}\displaystyle\left|\langle{\sum_{j\in\mathbb{Z}}\Delta_{j}^{\zeta}f,g}\rangle\right|=\left|\sum_{j\in\mathbb{Z}}\sum_{k\in\mathbb{Z}}\langle{\Delta_{j}^{\zeta}f,\Delta_{k}^{\varphi}g}\rangle\right|\end{array}\tag{4}$$

I claim that $\Delta_{j}^{\zeta}\Delta_{k}^{\varphi}=0$ if $\left|j-k\right|>c_{0}$, where $c_{0}$ is a fixed constant depending only on $\zeta$ and $\varphi$. Indeed,

$$\left(\Delta_{j}^{\zeta}\Delta_{k}^{\varphi}f\right)^{\widehat{}}(\xi)=\widehat{\zeta}_{j}(\xi)\widehat{\varphi}_{k}(\xi)=\widehat{\zeta}(2^{-j}\xi)\widehat{\varphi}(2^{-k}\xi)\widehat{f}(\xi)$$

and $\text{supp}(\widehat{\zeta}_{j})\subset(2^{j-l_{0}},2^{j+l_{0}}), \text{supp}(\widehat{\varphi}_{k})\subset(2^{k-1},2^{k+1})$. The claim follows immediately from considering $2^{j+l_{0}}>2^{k-1}$ and $2^{k+1}>2^{j-l_{0}}$.

Using this result together with the triangle inequality, we obtain that (4) is $\leq$

$$\begin{array}{lcl}\displaystyle\sum_{j\in\mathbb{Z}}\sum_{k=j-c_{0}}^{j+c_{0}}\int_{\mathbb{R}^{n}}\left|\Delta_{j}^{\zeta}f\overline{\Delta_{k}^{\varphi}g}\right|&=&\displaystyle\sum_{k=-c_{0}}^{c_{0}}\sum_{j\in\mathbb{Z}}\int_{\mathbb{R}^{n}}\left|\Delta_{j}^{\zeta}f\Delta_{j-k}^{\varphi}g\right|\\[2 em]&\leq&\displaystyle\sum_{k=-c_{0}}^{c_{0}}\int_{\mathbb{R}^{n}}\left(\sum_{j\in\mathbb{Z}}(\Delta_{j}^{\zeta}f)^{2}\right)^{1/2}\left(\sum_{j\in\mathbb{Z}}(\Delta_{j-k}^{\varphi}g)^{2}\right)^{1/2}\\[2 em]&\leq&\displaystyle(2c_{0}+1)\left\|\left(\sum_{j\in\mathbb{Z}}(\Delta_{j}^{\zeta}f)^{2}\right)^{1/2}\right\|_{L^{p}}\left\|\left(\sum_{j\in\mathbb{Z}}(\Delta_{j}^{\varphi}g)^{2}\right)^{1/2}\right\|_{L^{p'}},\end{array}$$

where we use the $\ell^{2}$ Cauchy-Schwarz inequality in the penultimate step and Holder's inequality in the ultimate step. Applying (1) completes the proof.

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1 Answer 1

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I think it will be difficult to find a simple proof of this.

Let $\alpha(x)$ be a smooth cutoff function on $\mathbb R$ with $\sum_{j \in \mathbb Z} \alpha(2^j x) = 1$. Then if you let your $\hat{\zeta_j}(\xi)$ be ${\displaystyle \alpha\bigg( {|\xi| \over 2^j}\bigg) {\xi_1 \over |\xi|}}$, the operator you are trying to show is bounded on $L^p$ becomes a Riesz transform. And proofs that Riesz transforms are bounded on L^p are not that easy. Probably Littlewood-Paley theory is the easiest way.

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  • $\begingroup$ The hypothesis on $\zeta$ is a bit prohibitive in this case. My intuition tells me you are right about Littlewood-Paley being the easiest way. My motivation for asking the question was to make sure that Littlewood-Paley wasn't overkill, since it's a nontrivial result. I believe one can also obtain the $L^{p}$ boundedness using the Hormander-Mihlin multiplier theorem applied to $m(\xi)=\sum_{j}\widehat{\zeta}(2^{-j}\xi)$, but that it is a nontrivial result, as well. $\endgroup$ May 2, 2015 at 22:16
  • $\begingroup$ $L^p$ boundness results for singular integrals or multiplier operators are rarely very simple when $p \neq 2$.. as far as I know even $L^p$ boundedness for the Hilbert transform in one dimension doesn't have that easy a proof. $\endgroup$
    – Zarrax
    May 2, 2015 at 23:28

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