DMat0101, Notes 8: Notes Littlewood-Paley inequalities and multipliers

In this final set of notes we will study the Littlewood-Paley decomposition and the Littlewood-Paley inequalities. These consist of very basic tools in analysis which allow us to decompose a function, on the frequency side, to pieces that have almost disjoint frequency supports. These pieces, the Littlewood-Paley pieces of the function, are almost orthogonal to each other, each piece oscillating at a different frequency.

1. The Littlewood-Paley decomposition

We start our analysis with forming a smooth Littlewood-Paley decomposition as follows. Let ${\phi}$ be a smooth real radial function supported on the closed ball ${\{\xi\in{\mathbb R}^n:0<|\xi|\leq 2\}}$ of the frequency plane, which is identically equal to ${1}$ on ${\{\xi\in{\mathbb R}^n:0\leq|\xi|\leq 1\}}$ . We then form the function ${\psi}$ as

$\displaystyle \psi(\xi):=\phi(\xi)-\phi(2\xi),\quad \xi \in {\mathbb R}^n.$

Observing that ${\phi(2\xi)=\phi(\xi)=1}$ if ${|\xi|<1/2}$ and also that ${\phi(\xi)=\phi(2\xi)=0}$ if ${|\xi|>2}$ we see that ${\psi}$ is supported on the annulus ${\{\xi\in {\mathbb R}^n:1/2 \leq |\xi|\leq 2\}}$ .

Now the sequence of functions ${\{\psi(\xi/2^k)\}_{k\in {\mathbb Z}}}$ forms a partition of unity:

$\displaystyle \sum_{k\in {\mathbb Z}}\psi(\xi/2^k)=1,\quad \xi \in {\mathbb R}^n\setminus\{0\}.$

To see this first observe that each function ${\psi(\xi/2^k)}$ is supported on the annulus ${\{2^{k-1}\leq |\xi|\leq 2^{k+1}\}}$ . Thus for each given ${\xi \in {\mathbb R}^n}$ there are only finite terms in the previous sum. In particular if ${2^\ell<|\xi_o|\leq 2^{\ell+1}}$ , then

$\displaystyle \sum_{k\in{\mathbb Z}}\psi(\xi_o/2^k)=\psi(\xi_o/2^{\ell})+\psi(\xi_o/2^{\ell+1})=\phi(\xi_o/2^{\ell+1})-\phi(\xi_o/2^{\ell-1})=1.$

Note that we miss the origin in our decomposition of the frequency space as each piece ${\psi(\xi/2^k)}$ is supported away from ${0}$ . Some attention is needed concerning this point but usually it creates no real difficulty.

Thus we partition the unity in the form ${1=\sum \psi_k}$ and each ${\psi_k}$ is smooth and has frequency support on an annulus of the form ${|\xi|\simeq 2^k}$ . Now for ${k\in {\mathbb Z}}$ let us define the multiplier operators

$\displaystyle \widehat {\Delta_k(f)}(\xi)=\psi(\xi/2^k)\hat f(\xi),$

and

$\displaystyle \widehat {S_k(f)}(\xi)=\sum_{\ell\leq k} \widehat{\Delta_\ell (f)}(\xi)=\phi(\xi/2^k)\hat f(\xi),$

initially defined for ${f\in L^2({\mathbb R}^n)}$ or ${f\in {\mathcal S(\mathbb R^n)}}$ . The operator frequency cut-off operator ${\Delta_k}$ is almost a projection to the corresponding frequency annulus ${|\xi|\simeq 2^k}$ . It is not exactly a projection since the function ${\psi(\xi/2^k)}$ is a smooth approximation of the indicator function ${\chi_{\{\xi\in{\mathbb R}^n:2^{k-1}\leq |\xi|\leq 2^k\}}}$ , introducing a small tail in the region ${\{\xi\in\mathbb R^n:2^k<|\xi|\leq 2^{k+1}}$ which is mostly harmless. Similarly, the operator ${S_k}$ is almost a projection on the ball ${|\xi|\lesssim 2^k}$ .

We have the following simple properties of the Littlewood-Paley decomposition:

Proposition 1 (i) For every ${f\in L^2({\mathbb R}^n)}$ we have ${\Delta_k(f)=S_k(f)-S_{k-1}(f)}$ that is ${\Delta_k=S_k-S_{k-1}}$ in ${L^2({\mathbb R}^n)}$ . (ii) For every ${f\in L^2({\mathbb R}^n)}$ we have ${\lim_{k\rightarrow -\infty} S_k f=0}$ and ${\lim_{k\rightarrow +\infty}S_k f=f}$ where the limits are taken in the ${L^2({\mathbb R}^n)}$ -sense.

(iii) For every ${f\in L^2({\mathbb R}^n)}$ we have that
$\displaystyle \sum_{k\in{\mathbb Z}} \Delta_k f=f$

in ${L^2({\mathbb R}^n)}$ .

Remark 1 Property (iii) above holds in a more general sense and for a wider class of functions, for example ${L^p}$ functions and more generally locally integrable functions that have some decay at infinity. The decomposition fails however if ${f}$ has no decay. Indeed, the function ${1}$ satisfies ${\Delta_k 1=0}$ for all ${k\in {\mathbb Z}}$ . Observe here that the function ${1}$ has frequency support on ${\{0\}}$ which is the point missed in our partition of unity.

Thus, with a Littlewood-Paley decomposition we managed to write any ${L^2}$ function (and thus any Schwartz function) as a sum of pieces ${\Delta_k f}$ , each piece being well localized in frequency on the annulus ${|\xi|\simeq 2^k}$ .

It is pretty obvious how the operators ${\Sigma_k,\Delta_k}$ act on the frequency variable so let us take a look on what the pieces ${\Sigma_kf, \Delta_k f}$ look in the physical space. From the general facts about the Fourier transform (see for example Exercise 2 of Notes 3) we know already that ${S_k f, \Delta_k f}$ cannot have compact spatial support. Since

$\displaystyle \widehat{S_kf}(\xi)=\phi(\xi/2^k)\hat f(\xi)=\textnormal{Dil}_{2^{-k}} ^\infty \phi \hat f(\xi)=\mathcal F (\textnormal{Dil}_{2^k} ^1 \check\phi *f)(\xi),$

and ${\check \phi =\hat{ \tilde \phi}=\hat \phi}$ , we have

$\displaystyle S_k(f)(x)=( \textnormal{Dil}_{2^k} ^1 \hat\phi*f)(x)=\int_{{\mathbb R}^n} f(x-y)2^{kn}\hat \phi(2^ky)dy=\int_{{\mathbb R}^n}f(x-2^{-k} y)\hat \phi (y)dy.$

Here note that ${\int \hat \phi =\phi(0) =1}$ . From the discussion that followed the definition of convolutions in Notes 2 we thus see that ${S_k f(x)}$ is an average of ${f}$ around the point ${x}$ at scale ${\simeq 2^{-k}}$ . Remembering that ${\widehat{S_k f }}$ is supported on the ball ${\{|\xi|\lesssim 2^k\}}$ this is also consistent with the uncertainty principle which also implies that the function ${S_k(f)}$ is essentially constant at scales ${\lesssim 2^{-k}}$ . Now since a piece ${\Delta_k f}$ has frequency support contained in ${\{|\xi|\leq 2^{k+1}\}}$ we get that

$\displaystyle \Delta_k(f)=S_{k+2}\Delta_k (f)=\int_{{\mathbb R}^n}\Delta_kf(x-2^{-(k+2)}y)\hat \phi(y)dy.$

Thus ${\Delta_k(f)}$ is almost constant on scales ${\lesssim 2^{-(k+2)}}$ . On the other hand, since ${\Delta_k f}$ has frequency support on the annulus ${\{2^{k-1}\leq|\xi|\leq 2^{k+1} \}}$ we have that

$\displaystyle S_{k-2} \Delta_k f =0.$

As before we can rewrite this as

$\displaystyle \int_{{\mathbb R}^n} \Delta_k f(x-2^{-k+2}y)\hat \phi(y)dy=0.$

The previous identity roughly says that the function ${\Delta_k(f)(x)}$ has zero mean on every ball around ${x}$ of radius ${\gtrsim 2^{k-2}}$ .

Remark 2 We have mentioned in passing that the operators ${\Delta_k}$ can be seen as smooth approximations of the exact projections operators
$\displaystyle \widehat {D_k (f)}(\xi)=\chi_{\{2^k\leq |\xi|\leq 2^{k+1}\}}(\xi)\hat f(\xi).$

Similarly, ${S_k}$ can be viewed as a smooth approximation of the frequency projection
$\displaystyle \widehat { { \Pi _k } (f)}(\xi)=\chi_{\{ |\xi|\leq 2^{k+1}\}}(\xi)\hat f(\xi).$

There are however important differences between the rough and smooth versions of these projections. For example, since ${\phi}$ is a Schwartz function the function ${\hat \phi}$ is also Schwartz and Young’s inequality shows that
$\displaystyle \|S_k(f)\|_p = \| \textnormal{Dil}_{2^k} ^1 \hat\phi*f \|_{L^p}\leq \|\hat \phi \|_1 \|f\|_p,$

thus ${S_k}$ is bounded on ${L^p}$ . Now, consider the rough version ${\Pi_k}$ given as
$\displaystyle \Pi_k(f)(x)= (\textnormal{Dil}_{2^k} ^1 \hat\chi_{B(0,1)}*f)(x).$

Of course ${\Pi_k}$ is still bounded on ${L^2}$ because of Plancherel’s theorem. However, the function ${\hat \chi_{B(0,1)}}$ is no longer in ${L^1}$ and Young’s inequality cannot be used. In fact, ${\Pi_k}$ is not bounded on ${L^p}$ whenever ${n\geq 2}$ and ${p\neq 2}$ . This is a deep result of C. Fefferman.

2. Littlewood-Paley Projections and derivatives

Recall the basic relation describing the interaction of derivatives with the Fourier transform:

$\displaystyle \widehat{ \partial ^\alpha f}(\xi)=(2\pi i \xi)^\alpha \hat f(\xi).$

In particular

$\displaystyle |\widehat{\nabla f} |^2=\sum_{j=1} ^n \bigg|\widehat { \frac{\partial f }{\partial x_j}}\bigg|^2=4\pi^2|\xi|^2 |\hat f|^2$

If ${f}$ has support on some annulus ${|\xi|\simeq 2^k}$ we immediately get

$\displaystyle \|\nabla f\|_{L^2({\mathbb R}^n)}\simeq _n 2^k \|f\|_{L^2({\mathbb R}^n)},$

and thus for any function ${f\in L^2({\mathbb R}^n)}$ that

$\displaystyle \|\nabla( \Delta_k f)\|_{L^2({\mathbb R}^n)}\simeq _n 2^k \|\Delta_k f\|_{L^2({\mathbb R}^n)},$

In fact the same approximate identity extends to all ${L^p}$ spaces for ${1\leq p \leq \infty}$ .

Proposition 2 For all ${1\leq p \leq \infty}$ we have that
$\displaystyle \|\nabla( \Delta_k f)\|_p \simeq_{n,p} 2^k \|\Delta_k f\|_p.$

We won’t prove this proposition here since it will be covered by a lecture in the student’s seminar.

3. The Littlewood-Paley inequalities

The Littlewood-Paley inequalities quantify the heuristic principle that the pieces ${\Delta_k(f)}$ , having well separated frequency supports, behave independently of each other, meaning that

$\displaystyle |\sum_k \Delta_k(f)|\simeq \big( \sum_k |\Delta_k(f)|^2\big)^\frac{1}{2},$

in some appropriate sense (for example in ${L^p}$ ). In ${L^2}$ this is already an easy consequence of the Plancherel identities. Indeed, note that

$\displaystyle \bigg\| \big( \sum_k |\Delta_k(f)|^2\big)^\frac{1}{2} \bigg\|^2=\int_{{\mathbb R}^n}\sum_{k \in{\mathbb Z}} |\psi(\xi/2^k)|^2 |\hat f(\xi)|^2 d\xi.$

Like before observe that for every ${\xi\in{\mathbb R}^n}$ there are only two terms ${\psi(\xi/2^\ell), \psi(\xi/2^{\ell+1})}$ which don’t vanish, and these add up to ${1}$ . Thus

$\displaystyle \begin{array}{rcl} 1 = (\psi(\xi/2^\ell)+\psi(\xi/2^{\ell+1}))^2 =|\psi(\xi/2^\ell)|^2+|\psi(\xi/2^{\ell+1})|^2+2\psi(\xi/2^\ell)\psi(\xi/2^{\ell+1}) , \end{array}$

and

$\displaystyle \sum_{k\in {\mathbb Z}}|\psi(\xi/2^k)|^2=|\psi(\xi/2^\ell)|^2+|\psi(\xi/2^{\ell+1})|^2\simeq 1.$

We can equivalently write this identity in the form

$\displaystyle \big\|\big( \sum_{k\in{\mathbb Z}}|\Delta_k(f)|^2)^\frac{1}{2}\big\|_{L^2}\simeq \|f\|_{L^2}.$

The following theorem extends this approximate identity to all ${L^p}$ spaces for ${1<p<\infty}$ .

Theorem 3 Define the Littlewood-Paley square function as
$\displaystyle S(f)(x):=\bigg( \sum_{k\in {\mathbb Z}} |\Delta_k(f)(x)|^2\bigg)^\frac{1}{2}.$

Then for all ${1<p<\infty}$ we have
$\displaystyle \|S(f)\|_{L^p({\mathbb R}^n)}\simeq_{n,p} \|f\|_{L^p({\mathbb R}^n)}.$

Proof: Consider the vector valued singular integral operator

$\displaystyle \vec S(f)(x):=\{ \Delta_kf(x)\}_{k\in Z},$

and observe that

$\displaystyle |S(f)(x)|=\|\vec S(f)(x)\|_{\ell^2({\mathbb Z})}.$

Thus the statement of the Theorem is equivalent to

$\displaystyle \|\vec S(f)\|_{L^p ({\mathbb R}^n, \ell^2({\mathbb Z}))}\simeq_{n,p}\|f\|_{L^2({\mathbb R}^n)}. \ \ \ \ \ (1)$

Observe that ${\vec S}$ is a bounded linear operator from ${L^2({\mathbb R}^n,{\mathbb C})}$ to ${L^2({\mathbb R}^n,\ell^2({\mathbb Z}))}$ . Indeed the strong ${(2,2)}$ type of ${\vec S}$ follows from the remarks before the theorem. Furthermore, defining

$\displaystyle K(x,y):=\{ 2^{nk} \hat \psi (2^k(x-y))\}_{k\in{\mathbb Z}},\quad (x,y)\in {\mathbb R}^n\times {\mathbb R}^n\setminus \Delta,$

we can verify that ${K}$ is a singular kernel:

Lemma 4 The kernel ${K}$ defined above is a singular kernel from ${{\mathbb C}}$ to ${\ell^2({\mathbb Z})}$ .

Postponing the proof of this lemma for now, we use the vector valued version of the Calderón-Zygmund theorem to show that ${\vec S}$ is bounded from ${L^p({\mathbb R}^n)}$ to ${L^p({\mathbb R}^n,\ell^2({\mathbb Z}))}$ :

$\displaystyle \|\vec S(f)\|_{L^p({\mathbb R}^n,\ell^2({\mathbb Z}))}\lesssim_{n,p,\psi} \|f\|_{L^p({\mathbb R}^n)},$

which is one of the estimates in (1). To prove the lower estimate we argue as follows. Let ${\vec g=\{g_j\}_{j\in{\mathbb Z}}:{\mathbb R}^n\rightarrow \ell^2({\mathbb Z})}$ . Then

$\displaystyle \begin{array}{rcl} \int_{{\mathbb R}^n}(\vec S(f)(x),\vec g(x))dx&=&\int_{{\mathbb R}^n} \sum_{k\in{\mathbb Z}} \Delta_k(f)(x)\overline{g_k(x)}dx=\int_{{\mathbb R}^n}\sum_{k\in{\mathbb Z}} \psi_k(\xi/2^k)\hat f(\xi)\overline{\widehat{g_k}(\xi)}d\xi \\ \\ &=&\int_{{\mathbb R}^n}f(x)\sum_{k\in{\mathbb Z}}\overline{\Delta_k(g_k)}(x)dx=:\int_{{\mathbb R}^n}f(x)\overline{\vec S ^*(\vec g)(x)}dx. \end{array}$

By vector valued duality and the estimate ${\|\vec S(f)\|_{L^p({\mathbb R}^n,\ell^2({\mathbb Z}))}\lesssim_{n,p,\psi}\|f\|_{L^p({\mathbb R}^n)}}$ we conclude that the adjoint operator ${\vec S^*}$ satisfies

$\displaystyle \|\sum_{k\in{\mathbb Z}} \Delta_k(g_k) \|_{L^p({\mathbb R}^n)}=\| \vec S ^* (\vec g)\|_{L^p({\mathbb R}^n)}\lesssim_{n,p,\psi}\|\vec g\|_{L^p({\mathbb R}^n,\ell^2({\mathbb Z}))},\quad 1<p<\infty.$

Now we repeat the Littlewood-Paley decomposition but starting with the function

$\displaystyle \tilde\psi (\xi)=\phi(\xi/4)-\phi(4\xi),$

and setting

$\displaystyle \widehat{\tilde \Delta_k(f)}(\xi):=\tilde\psi(\xi/(2^k))\hat f(\xi)=(\phi(\xi/{42^{k}})-\phi(4\xi/2^{k}) ) \hat f(\xi)$

or equivalently

$\displaystyle \tilde \Delta_k(f):=\Sigma_{k+2}(f)-\Sigma_{k-2}(f).$

Using exactly the same arguments as before we can show that we also have that

$\displaystyle \|\sum_{k\in{\mathbb Z}}\tilde \Delta_k(g_k)\|_{L^p({\mathbb R}^n)}\lesssim_{n,p,\psi} \| \vec g\|_{L^p({\mathbb R}^n,\ell^2({\mathbb Z}))}.$

Observe that for ${2^{k-1}\leq |\xi| \leq 2^{k+1}}$ we have that ${|\xi|/(42^k)\leq 2^{k+1}/(4^2k) =1/2}$ and ${|4\xi|/2^k\geq 42^{k+1}/2^k=8}$ thus for any function ${h}$ with ${{\mathrm{supp}}(h)\subset\{2^{k-1}<|\xi|\leq 2^{k+1}\}}$ we have that ${\tilde \Delta_k h =h}$ .

Now choose

$\displaystyle \vec g(x)=(\Delta_1(f),\Delta_2(f),\ldots,\Delta_k(f),\ldots),$

and observe that ${\tilde \Delta_k \Delta_k(f)= \Delta_k(f)}$ since we already have that ${\textnormal{supp}(\Delta_k(f))\subset\{2^{k-1}\leq |\xi| \leq 2^{k+1}\}}$ . We get

$\displaystyle \|\sum_{k\in{\mathbb Z}} \Delta_k(f)\|_{L^p({\mathbb R}^n)}\lesssim_{n,p,\psi} \|\big( \sum_{k\in {\mathbb Z}} |\Delta_kf|^2\big)^\frac{1}{2}\|_{L^p({\mathbb R}^n)}.$

However on the left hand side we have the pointwise identity ${\sum_k\Delta_k(f)(x)=f(x)}$ which shows that

$\displaystyle \|f\|_{L^p({\mathbb R}^n)}\lesssim_{n,p,\psi}\|\vec{S}(f)\|_{L^p({\mathbb R}^n,\ell^2({\mathbb Z}))},$

as we wanted to show. $\Box$

We now go back to the proof of Lemma 4.

Proof of Lemma 4: Remember that the kernel ${K}$ is given as

$\displaystyle K(x,y)=\{2^{nk}\hat \psi(2^k(x-y))\}_{k\in {\mathbb Z}}.$

Let ${\psi_k(\xi)=\psi(\xi/2^k)}$ so that

$\displaystyle 2^{nk} \hat \psi(2^k x)=\widehat{\psi_k}(x)$

First of all we prove the estimates

$\displaystyle | {\widehat{\psi_k}(x )}|\lesssim \frac{1}{|x |^n} \min((2^k|x |)^n,(2^k|x |)^{-2}), \ \ \ \ \ (2)$

and

$\displaystyle | \nabla {\widehat{\psi_k}(x )}| \lesssim \frac{1}{|x |^{n+1}} \min( (2^k|x |)^{n+1},(2^k|x |)^{-1}) , \ \ \ \ \ (3)$

For (2) we write

$\displaystyle |\widehat{\psi_k}(x)|=2^{nk}\bigg|\int_{{\mathbb R}^n}\psi(\xi) e^{-2\pi i 2^k x \cdot \xi}d\xi\bigg|.$

On the one hand we have that

$\displaystyle |\psi_k(x)|\leq 2^{nk}\int_{{\mathbb R}^n}|\psi(\xi)|d\xi \lesssim_\psi \frac{1}{|x|^n}(2^k|x|)^n .$

On the other hand for any positive integer ${N}$ we have

$\displaystyle |\psi_k(x)|=2^{nk}\bigg|\int_{{\mathbb R}^n}\psi(\xi) ( \frac{x}{2\pi i 2^k |x|^2}\cdot\nabla_\xi )^Ne^{2\pi i 2^k x\cdot \xi} dx \bigg| \lesssim_{\psi,N} 2^{nk}\frac{1}{(2^k|x|)^N},$

by integrating by parts ${N}$ times and passing the derivatives to ${\psi}$ . Applying this estimate for ${N=n+2}$ gives the second estimate in (2). The proof of (3) is very similar by observing that

$\displaystyle \partial_{x_j} {\widehat{\psi_k}(x )} =2^{nk}\int_{{\mathbb R}^n}\psi(\xi)\partial_{x_j} e^{-2\pi i 2^k x\cdot\xi }dx=2^{nk}\int_{{\mathbb R}^n}\psi(\xi)(-2\pi i 2^k\xi_j) e^{-2\pi i 2^k x\cdot\xi }dx .$

Now the same analysis as in (2) applies (with an extra ${2^k}$ factor) and gives (3). Estimates (2) and (3) now imply the size and regularity conditions for the singular kernel ${K}$ in ${\ell^2({\mathbb Z})}$ . $\Box$

3.1. A rough version for ${n}$ -dimensional dyadic intervals

So far we carried out the Littlewood-Paley decomposition based on a smooth partition of unity. The use of smooth functions to form the Littlewood-Paley decomposition has many advantages since then the projections ${\Delta_j}$ are bounded multiplier operators. On the other hand, Remark 2 shows that in dimensions ${n>1}$ , the multiplier associated with a Euclidean ball is not bounded on ${L^p}$ . This means that the Littlewood-Paley inequalities based on the projections

$\displaystyle \widehat{P_k(f)}=\chi_{\{2^{k+1}<|\xi|\leq 2^k \}}\hat f, \ \ \ \ \ (4)$

will fail in any dimension ${n\geq 2}$ .

The previous discussion leaves the one-dimensional case open. In fact we will see now that one can form the Littlewood-Paley decomposition in one dimension based on the rough partition of unity

$\displaystyle 1=\sum_{k\in{\mathbb Z}}\chi_{\{2^{k}<|\xi|\leq 2^{k+1} \}}(\xi),$

and still have the Littlewood-Paley inequalities. So let us define ${P_k}$ to be the exact frequency projection defined by (4). We have the following.

Theorem 5 Let ${f\in L^p({\mathbb R})}$ , ${1<p<\infty}$ . Then we have the one dimensional Littlewood-Paley inequalities for the rough projections ${P_k}$ :
$\displaystyle \bigg\|\big( \sum_{k\in {\mathbb Z}}|P_k(f)|^2\big)^\frac{1}{2}\bigg\|_{L^p({\mathbb R})}\simeq_p \|f\|_{L^p({\mathbb R})}.$

Proof: Writing ${P_k(f)}$ in the form

$\displaystyle \widehat{P_k f}=\chi_{[-2^{k+1},2^k)}\hat f+\chi_{(2^k,2^{k+1} ]}\hat f,$

we have the following representation in terms of the Hilbert transform

$\displaystyle P_k(f)=\frac{i}{2}\big(\textnormal{Mod}_{2^k}H\textnormal{Mod}_{-2^{k}}f -\textnormal{Mod}_{2^{k+1}} H\textnormal{Mod}_{-2^{k+1}}f\big). \ \ \ \ \ (5)$

For ${\vec g =(g_1,g_2,\ldots,g_k,\ldots)\in L^p({\mathbb R},\ell^2({\mathbb Z}) )}$ let us define the vector valued analogue

$\displaystyle \vec P(\vec g):=\{P_k(g_k)\}_{k\in{\mathbb Z}}.$

Using the fact that ${H}$ is a CZO and the representation (5) of ${P_k}$ in terms of ${H}$ we can see that ${\vec P}$ is a vector valued Calderón-Zygmund operator, thus ${\vec P}$ is bounded from ${L^p({\mathbb R},\ell^2({\mathbb Z}))}$ to ${L^p({\mathbb R},\ell^2({\mathbb Z}))}$ . Applying this property to the function

$\displaystyle \vec g =\{\Delta_k f\}_{k\in {\mathbb Z}}$

we get

$\displaystyle \bigg\|\big( \sum_{k\in {\mathbb Z}}|P_k\Delta_k (f)|^2\big)^\frac{1}{2}\bigg\|_{L^p({\mathbb R})}\lesssim_p \bigg\|\big( \sum_{k\in {\mathbb Z}}|\Delta_k (f)|^2\big)^\frac{1}{2}\bigg\|_{L^p({\mathbb R})} .$

Now observe that since ${{\mathrm{supp}}(\widehat{\Delta_k(f)})=\{2^{k-1}<|\xi|\leq 2^{k+1}\}}$ we have the identity ${P_k\Delta_k (f) = P_k (f)}$ . Thus the previous estimate implies that

$\displaystyle \bigg\|\big( \sum_{k\in {\mathbb Z}}|P_k (f)|^2\big)^\frac{1}{2}\bigg\|_{L^p({\mathbb R})}\lesssim_p \bigg\|\big( \sum_{k\in {\mathbb Z}}|\Delta_k (f)|^2\big)^\frac{1}{2}\bigg\|_{L^p({\mathbb R})} .$

By Theorem 3 we get one of the inequalities in the statement of the theorem:

$\displaystyle \bigg\|\big( \sum_{k\in {\mathbb Z}}|P_k (f)|^2\big)^\frac{1}{2}\bigg\|_{L^p({\mathbb R})}\lesssim_p \|f\|_{L^p({\mathbb R})} .$

To prove the opposite inequality, we write the dual estimate that was obtained in proof of Theorem 3:

$\displaystyle \|\sum_{k\in{\mathbb Z}} \tilde \Delta_k(\vec g)\|_{L^p({\mathbb R})}\lesssim_{n,p,\psi} \|\vec g\|_{L^p({\mathbb R},\ell^2({\mathbb Z}))}.$

Now take ${\vec g :=\{P_kf\}_{k\in {\mathbb Z}}}$ and observe that

$\displaystyle \tilde \Delta_k P_k f=P_k(f)$

so the previous estimate implies

$\displaystyle \|f\|_{L^p({\mathbb R})}=\|\sum P_kf\|_{L^p({\mathbb R})}\lesssim_p \|\vec g\|_{L^p({\mathbb R},\ell^2({\mathbb Z}))}=\bigg\|\big(\sum_{k\in{\mathbb Z}}|P_k(f)|^2\big)^\frac{1}{2}\bigg\|_{L^p({\mathbb R})},$

which gives the other inequality in the theorem. $\Box$

Exercise 1 Let ${T}$ be a scalar valued CZO and ${\vec f\in L^p({\mathbb R}^n,\ell^r({\mathbb Z}))}$ . Show that
$\displaystyle \bigg\|\big(\sum_{k\in{\mathbb Z}} |T(f_k)|^r \big)^\frac{1}{r}\bigg\|_{L^p({\mathbb R}^n)}\lesssim_{n,p,r,T} \bigg\|\big(\sum_{k\in{\mathbb Z}}|f_j|^r\big)^\frac{1}{r}\bigg\|_{L^p({\mathbb R})}.$

Hint: Consider the vector valued operator
$\displaystyle \vec T(\vec f)=\{Tf_k\}_{k\in{\mathbb Z}}.$

The problem reduces to showing that ${\vec T}$ is bounded from ${L^p({\mathbb R}^n,\ell^r({\mathbb Z}))}$ to ${L^p({\mathbb R}^n,\ell^r({\mathbb Z}))}$ . Observe that ${\vec T}$ is associated with the kernel
$\displaystyle \vec K(x,y)=K(x,y)\textnormal{id}_{\ell^r},$

where ${\textnormal{id}_{\ell^r}}$ is the identity from ${\ell^r({\mathbb Z})}$ to ${\ell^r({\mathbb Z})}$ and ${K}$ is the (scalar) kernel associated with ${T}$ . You can assume a Banach space version of the vector valued Calderón-Zygmund theorem.

Exercise 2 Let ${\{I_k\}_{k\in\Lambda}}$ be a sequence of bounded or unbounded intervals on the real line, where ${\Lambda}$ is a finite or countably infinite index set. Define the frequency projections
$\displaystyle \widehat{P_{I_k}f}=\chi_{I_j}\hat f.$

Show that
$\displaystyle \bigg\|\big( \sum_{k\in\Lambda} |P_{I_k}f|^r \big)^\frac{1}{r}\bigg\|_{L^p({\mathbb R})}\lesssim_{p,r} \big\|\big( \sum_{k\in\Lambda} |f|^r \big)^\frac{1}{r}\big\|_{L^p({\mathbb R})}.$

Hint: Like in the proof of Theorem 5 use the representation of the projections ${P_{I_k}}$ in terms of the Hilbert transform and Exercise 1.

We have already remarked (see remark 2) that Theorem 5 does not generalize to annuli in the ${n}$ -dimensional Euclidean space if we insist on using the rough projections ${\chi_{\{2^k<|\xi|\leq 2^{k+1}\}}\hat f}$ . However, there is a generalization of the `rough’ Littlewood-Paley theorem to dimensions ${n>1}$ . This is based on decomposing the frequency space ${{\mathbb R}^n}$ to a union of disjoint dyadic `intervals’, that is, ${n}$ -dimensional rectangles with axes parallel to the coordinate axes, where every side of the rectangle is an interval of the form ${( 2^k, 2^{k+1}]}$ or ${[-2^{k+1},-2^k]}$ . This allows for `tensoring’ Theorem 5 to several dimensions without great difficulty. This is done as follows. For ${k=(k_1,\ldots,k_n)\in {\mathbb N}^n}$ we set

$\displaystyle P^{(k)}=P_{k_1}P_{k_2}\cdots P_{k_n}$

where each ${P_{k_j}}$ is the one-dimensional projection previously defined acting only on the ${j}$ -th variable. For ${f:{\mathbb R}^n\rightarrow {\mathbb C}}$ we have

$\displaystyle \widehat{P_{k_j} f }(\xi)= \chi_{\{2^{k_j}<|\xi_j|\leq 2^{k_j+1}\}}(\xi_j) \hat f(\xi),\quad \xi \in {\mathbb R}^n.$

The corresponding square function is defined as

$\displaystyle S_\square(f)(\xi):=\bigg( \sum_{k=(k_1,\ldots,k_n)\in{\mathbb N}^n}|P^{(k)}f(\xi)|^2\bigg)^\frac{1}{2}$

This leads to

Theorem 6 For ${1<p<\infty}$ we have
$\displaystyle \|S_\square(f)\|_{L^p({\mathbb R}^n)}\simeq_{p,n}\|f\|_{L^p({\mathbb R}^n)}.$

We omit the proof of this theorem as it is mostly technical, based on induction and starting from the one dimensional version of the theorem already proved. You can find the proof for example in [D] or [S].

4. Two theorems on multipliers

We now go back to multiplier operators and reconsider them from the point of view of Calderón-Zygmund theory. We have already seen that a multiplier operator is the linear operator ${T_m}$ with ${\widehat{T_mf}=m(\xi)\hat f(\xi)}$ for some ${m\in L^\infty({\mathbb R}^n)}$ . This definition automatically implies that ${T_m}$ is bounded on ${L^2}$ with norm ${\|T_m\|_{L^2\rightarrow L^2}=\|m\|_{L^\infty}}$ . Alternatively, the discussion from Paragraph 8.1 of Notes 4 reveals that these are all the bounded linear operators on ${L^2}$ that commute with translations and can be realized in the form

$\displaystyle T_m(f)(x)=(K*f)(x),\quad f\in {\mathcal S(\mathbb R^n)},$

where ${K\in{\mathcal S'(\mathbb R^n)}}$ is the unique tempered distribution such that ${\hat K=m}$ .

If the operator ${T_m}$ extends to a bounded linear operator on ${L^p({\mathbb R}^n)}$ we say that ${m}$ is an ${L^p}$ -multiplier and write ${m\in \mathcal M^p}$ . We set

$\displaystyle \|m\|_{\mathcal M ^p}:=\|T_m\|_{L^p\rightarrow L^p}.$

The previous remarks then show that ${\|m\|_{\mathcal M^2}=\|m\|_{L^\infty}}$ . It turns out that the space ${(\mathcal M^p,\|\cdot\|_{\mathcal M^p})}$ is a Banach space but we will not dwell on this issue here. We also have the following easy proposition:

Proposition 7 (i) Let ${1\leq p \leq \infty}$ and ${p'}$ be the conjugate exponent of ${p}$ . Then
$\displaystyle m\in \mathcal M^p\iff m \in\mathcal M^{p'}$

and in this case we have that
$\displaystyle \|m\|_{\mathcal M^p}=\|m\|_{\mathcal M^{p'}}.$

(ii) For all ${1\leq p \leq \infty}$ we have
$\displaystyle \|m\|_{L^\infty({\mathbb R}^n)}\leq \|m\|_{\mathcal M^p}.$

Proof: This is a consequence of the following obvious identity; for ${f,g\in{\mathcal S(\mathbb R^n)}}$ we have

$\displaystyle \begin{array}{rcl} \int_{{\mathbb R}^n} T_m(f)(x)\overline{ g(x)} =\int_{{\mathbb R}^n} m(\xi)\hat f(\xi)\overline{\hat g(\xi)}=\int_{{\mathbb R}^n} f(x) T_{\overline m}g(x)dx, \end{array}$

That is, ${T_{\overline m} }$ is the adjoint of ${T_m}$ . Thus

$\displaystyle \|m\|_{\mathcal M^p}=\|T_m\|_{L^p\rightarrow L^p}=\| T_{\overline m}\|_{L^{p'}\rightarrow L^{p'}}=\|\overline m\|_{\mathcal M^{p'}}=\| m\|_{\mathcal M^{p'}},$

since ${m}$ and ${\overline m}$ have the same norm. To prove the second assertion assume that ${\|m\|_{\mathcal M^p}<+\infty}$ otherwise there is nothing to prove. By (i), the linear operator ${T_m}$ is of strong type ${(p,p)}$ and ${(p',p')}$ with the same operator norm. By the Riesz-Thorin interpolation theorem we get that

$\displaystyle \|m\|_{L^\infty({\mathbb R}^n)}=\|m\|_{L^2({\mathbb R}^n)}\leq \|m\|_{\mathcal M^p} ^\theta \|m\|_{\mathcal M^{p'}} ^{1-\theta}=\|m\|_{\mathcal M^p},$

which proves ${(ii)}$ . $\Box$

Remark 3 Observation (ii) above shows that ${L^p}$ multipliers are necessarily bounded functions. The opposite however is not true. Another easy consequence of the discussion above is the following. We always have
$\displaystyle T_m(f)(x)=(K*f)(x)=( \check m*f)(x),$

where ${K\in{\mathcal S'(\mathbb R^n)}}$ and ${K=\check m}$ as observed above. The problem with this representation is that we don’t know whether ${K=\check m}$ is actually a function that can give meaning to the formula
$\displaystyle T_m(f)(x)=\int_{{\mathbb R}^n}K(x-y)f(y)dy.$

If however it happens that ${K=\check m\in L^1({\mathbb R}^n)}$ then Young’s inequality readily applies to yield that
$\displaystyle \|T_m(f)\|_{L^p({\mathbb R}^n)}\leq \|K\|_{L^1({\mathbb R}^n)}\|f\|_{L^p({\mathbb R}^n)},$

so that
$\displaystyle \|m\|_{\mathcal M^p}=\|T_m\|_{L^p\rightarrow L^p}\leq \|\check m \|_{L^1({\mathbb R}^n)}.$

The main problem in the theory of multipliers is to get away from the case ${p=2}$ and place suitable conditions on ${m}$ so that we can conclude that ${m\in\mathcal M^p}$ . The previous generalities easily imply that if ${m\in {\mathcal S(\mathbb R^n)}}$ then ${m\in \mathcal M^p}$ since ${\check m\in L^1({\mathbb R}^n)}$ in this case. A similar result with weaker hypothesis is the following.

Proposition 8 For ${m:{\mathbb R}^n\rightarrow {\mathbb C}}$ we define the Sobolev space ${L^2 _s}$ to be the space of tempered distributions ${f}$ such that ${\hat f}$ agrees with a function that satisfies
$\displaystyle \|f\|_{L^2 _s}:=\int_{{\mathbb R}^n}|\hat f(\xi)|^2(1+4\pi^2|\xi|^2)^s d\xi <+\infty.$

Suppose that ${m\in L^2 _s}$ for some ${s>n/2}$ . Then ${m\in \mathcal M^p }$ and ${\|m\|_{\mathcal M^p}\le \|m\|_{L^2 _s}}$ .

Remark 4 Observe that for any tempered distribution ${f}$ we have that
$\displaystyle \widehat{ (-\Delta f)}(\xi)=4\pi^2|\xi|^2 \hat f(\xi).$

If ${k}$ is an even integer we can write
$\displaystyle \mathcal F ( (I-\Delta)^\frac{k}{2} f ) (\xi)=(1+4\pi^2|\xi|^2)^\frac{k}{2} \hat f(\xi).$

Thus, at least when ${k}$ is an even integer, the Sobolev space ${L^2 _k}$ is the space of tempered distributions such that
$\displaystyle (I-\Delta)^\frac{k}{2} f \in L^2({\mathbb R}^n),$

where ${(I-\Delta)^\frac{k}{2}}$ makes sense as a partial differentiable operator since ${k/2}$ is an integer. Similarly one can define the Sobolev spaces ${L^p _k}$ to be the space of tempered distributions ${f}$ such that
$\displaystyle (I-\Delta)^\frac{k}{2} f \in L^p({\mathbb R}^n).$

In fact one can take one step further and define the space ${L^p _s}$ for any real number ${s}$ . In the case ${p=2}$ this presents no difficulty since one has a direct interpretation of ${(I-\Delta)^\frac{s}{2}}$ as a Fourier integral operator. In particular, ${(I-\Delta)^\frac{s}{2}}$ is a pseudo-differential operator. Although this sounds a bit cryptic at the moment, we want to make the point here that ${f\in L^2 _s}$ for example is a condition that imposes decay on ${s}$ derivatives of ${f}$ .

Exercise 3 Prove Proposition 8 above.

The general flavor of the previous results is that if a function ${m}$ has no local singularities and, together with its derivatives, decays fast enough at infinity, then ${m}$ is an ${\mathcal M^p}$ multiplier for all ${1\leq p\leq \infty}$ . Besides a (controllable) singularity at infinity, one can also allow for a singularity at the origin.

We present two instances of this principle, usually referred to as the Hörmander multiplier theorem. We start with an `easy’ version where the function ${m}$ is bounded, to assure the ${(2,2)}$ hypothesis is satisfied, ${C^\infty}$ away from the origin and its derivatives decay at least as fast as their order.

Theorem 9 (Hörmander-Mikhlin multiplier theorem version I) Let ${m:{\mathbb R}^n\rightarrow {\mathbb C}}$ be a bounded function which belongs to the class ${C^\infty({\mathbb R}^n\setminus\{0\})}$ and satisfies
$\displaystyle |\partial^\alpha _\xi m(\xi)|\lesssim_{n,\alpha} |\xi|^{-|\alpha|},\quad \xi\in {\mathbb R}^n\setminus\{0\},$

for all multi-indices ${\alpha}$ . Then ${K=\check m}$ agrees with a ${C^\infty}$ function away from the origin and satisfies
$\displaystyle |\partial^\alpha K(x)|\lesssim_\alpha |x|^{-n-|\alpha|},\quad x\in {\mathbb R}^n\setminus\{0\},$

for all multi-indices ${\alpha}$ . In particular, ${m}$ is an ${\mathcal M^p}$ multiplier for all ${1< p < \infty}$ with ${\|m\|_{\mathcal M^p}\lesssim_{p,n} 1}$ .

Proof: Using the Littlewood-Paley decomposition we can write

$\displaystyle m(\xi)=\sum_{j\in {\mathbb Z}} \psi(\xi/2^k) m(\xi)=:\sum_{j\in{\mathbb Z}} m_j(\xi),$

whenever ${\xi\neq 0}$ . Each piece ${m_j}$ is supported on the annulus ${2^{j-1}\leq|\xi|\leq 2^{j+1}}$ and ${C^\infty}$ as a product of smooth functions so it makes sense to define

$\displaystyle K_j(x)=\int_{{\mathbb R}^n}m_j(\xi)e^{2\pi i x\cdot \xi} d\xi=\check{m}_j (x).$

Furthermore, from our hypotheses on ${m}$ we can get some good estimates on each ${K_j}$ together with its derivatives. Indeed since ${\|m_j\|_{L^\infty}\leq \|m\|_{L^\infty}\lesssim_n 1}$ by our hypothesis (with the zero multi-index ${\alpha}$ ) we have

$\displaystyle |K_j(x)|\leq \int_{|\xi|\simeq 2^j }|m_j(\xi)|d\xi \lesssim_{n} 2^{jn}.$

Likewise

$\displaystyle |\partial^\alpha K_j(x)|\leq \int_{ |\xi|\simeq 2^j } |(2\pi i \xi)^\alpha m_j(\xi)| d\xi \lesssim_{n,\alpha} \int_{|\xi|\simeq 2^j}d\xi \leq 2^{j(n+ |\alpha|)}.$

On the other hand for every multi-index ${\alpha }$ we have

$\displaystyle |\partial^\alpha K_j(x)|=|\int_{|\xi|\simeq 2^j}(2\pi i \xi)^\alpha m_j(\xi)\big( \frac{x\cdot \nabla_\xi}{2\pi i |x|^2}\big)^M e^{2\pi i x\cdot \xi}d\xi|,$

for every non-negative integer ${M}$ . Integrating by parts ${M}$ times to pass the derivatives to the term ${(2\pi i \xi)^\alpha m_j(\xi)}$ , using Leibniz’s rule and the hypothesis on the derivatives ${\partial ^\alpha m_j}$ we get the estimate

$\displaystyle |\partial^\alpha K_j(x)|\lesssim_{n,\alpha,M} |x|^{-M}2^{j(n+|\alpha|-M)},$

for all multi-indices ${\alpha}$ and non-negative integers ${M}$ . We summarize these estimates in the form

$\displaystyle |\partial^\alpha K_j(x)|\lesssim_{n,\alpha,M} \min(2^{j(n+|\alpha|)},|x|^{-M}2^{j(n+|\alpha|-M)}) \ \ \ \ \ (6)$

for all multi-indices ${\alpha}$ and non-negative integers ${M}$ . Using (6) for ${M=0}$ we have

$\displaystyle \sum_{2^j\leq |x|^{-1}}|\partial^\alpha K_j(x)|\leq \sum_{2^j\leq|x|^{-1}}2^{j(n+|\alpha|)}\lesssim_{n,\alpha}|x|^{-(n+\alpha)}$

On the other hand, using (6) for ${M>n+|\alpha|}$ we get

$\displaystyle \sum_{2^j> |x|^{-1}}|\partial^\alpha K_j(x)|\leq|x|^{-M} \sum_{2^j> |x|^{-1}} 2^{j(n+|\alpha|-M)}\simeq_{n,\alpha} |x|^{-M}|x|^{-(n+|\alpha|-M)}=|x|^{-(n+|\alpha|)}.$

Now since the series ${\sum_j \partial^\alpha K_j(x)}$ converges absolutely and uniformly in ${x}$ (when ${x\neq0}$ ) for every multi-index ${\alpha}$ we conclude that the series ${\sum_j K_j(x)}$ converges in ${C^\infty({\mathbb R}^n\setminus\{0\})}$ to some function ${\tilde K\in C^\infty({\mathbb R}^n\setminus\{0\})}$ which also satisfies the estimate

$\displaystyle |\partial^\alpha \tilde K(x)|\lesssim_{n,\alpha}|x|^{-(n+|\alpha|)}, \quad x\in {\mathbb R}^n\setminus\{0\},$

for all multi-indices ${\alpha}$ . On the other hand ${\sum_j m_j=\sum_j \widehat{K_j}}$ converges to ${m=\hat K}$ in ${L^2({\mathbb R}^n)}$ we conclude that ${K(x)=\tilde K(x)}$ when ${x\neq 0}$ . In particular,

$\displaystyle T_m(f)=(K*f)(x)=\int \tilde K(x-y)f(y)dy,$

whenever ${f\in L^2({\mathbb R}^n)}$ has compact support and ${x\notin {\mathrm{supp}}(f)}$ since then ${x-y\neq 0}$ . However, ${\tilde K}$ satisfies

$\displaystyle |\tilde K(x-y)| \lesssim_n |x-y|^{-n},\quad x\neq y,$

by taking the zero multi-index and furthermore

$\displaystyle |\nabla_y K(x-y)|,|\nabla_x K(x-y)|\lesssim_n |x-y|^{-(n+1)}, \quad x\neq y,$

by considering multi-indices ${\alpha}$ with ${|\alpha|=1}$ . These estimates are enough to assure that ${\tilde K}$ and thus ${K}$ is a singular kernel so ${T_m}$ is a CZO associated with ${K}$ . However this means that ${m\in\mathcal M^p}$ and we are done. $\Box$

Observe that what we really used in order to show that ${m\in\mathcal M^p}$ is the estimates with ${|\alpha|=1}$ of the derivatives of ${K}$ which in turn required a control of the derivatives of ${m}$ up to order ${M>n+|\alpha|=n+1 }$ . Thus we have the following corollary.

Corollary 10 Let ${m:{\mathbb R}^d\rightarrow {\mathbb C}}$ be a function such that
$\displaystyle |\partial_\xi ^\alpha m(\xi)|\lesssim_{\alpha} |\xi|^{-|\alpha|},$

for all multi-indices ${\alpha}$ with ${0\leq |\alpha|\leq n+2}$ . Then ${m\in \mathcal M^p}$ for all ${1<p<\infty}$ .

Remark 5 The hypothesis of the previous theorem is not optimal as one can get away with less derivatives of ${m}$ . However it already applies to many practical case. For example for any multi-index ${\beta}$ of order ${|\beta|=2}$ , consider the operator ${T_m}$ with symbol
$\displaystyle m_\beta (\xi)=\frac{ \xi ^\beta }{|\xi|^2} .$

Observe that ${m_\beta}$ falls into the scope of Theorem 9 since
$\displaystyle |\partial^\alpha m_\beta (\xi)|\lesssim \frac{1}{|\xi|^{|\alpha|}},$

for all multi-indices ${\alpha}$ . So ${m_\beta\in \mathcal M^p}$ for all ${1<p<\infty}$ . Now observe that for ${f\in\mathcal S({\mathbb R}^n)}$ (say) we have

$\displaystyle \widehat{(\partial^\beta f)}(\xi)=(-2\pi i \xi)^\beta \hat f(\xi)=\frac{(-2\pi i \xi)^\beta}{4\pi^2 |\xi|^2}4\pi^2|\xi|^2 \hat f(\xi)=m_\beta(\xi)\widehat{\Delta f}(\xi).$

which shows in particular that
$\displaystyle \|\partial^\beta f\|_{L^p}\lesssim \|\Delta f\|_{L^p}$

for all multi-indices of order ${2}$ , whenever ${\Delta f\in L^p({\mathbb R}^n)}$ . Thus all partial derivatives of order ${2}$ are control by the Laplacian in ${L^p}$ .

Now consider the space ${L^p _2({\mathbb R}^n)}$ to be the space of ${L^p}$ functions ${f}$ such that all the partial derivatives of order up to ${2}$ are in ${L^p}$ and equip this space with the norm
$\displaystyle \|f\|_{L^p _2}:=\sum_{0\leq |\alpha|\leq 2} \|\partial^\alpha f\|_{L^p({\mathbb R}^n)}.$

By the remarks above this norm is equivalent to
$\displaystyle \|f\|_{L^p _2}\simeq_{n,p}\|(I-\Delta)f \|_{L^p({\mathbb R}^n)}.$

Similar conclusions hold for any even integer ${k}$ and the space ${L^p _k}$ . Thus the two definitions of the Sobolev space ${L^p _k}$ , the one given here and then one given in Remark 3 coincide whenever ${k}$ is an even integer:
$\displaystyle \|f\|_{L^p _k}=\sum_{0\leq |\alpha|\leq k}\|\partial^\alpha f\|_{L^p({\mathbb R}^n)} \simeq _{n,p} \|(I-\Delta)^\frac{k}{2}f\|_{L^p({\mathbb R}^n)},\quad k\in 2{\mathbb N}.$

We now give a sharper form of the multiplier theorem which requires control only on ${\sim n/2}$ derivatives of ${m}$ .

Theorem 11 (Hörmander-Mikhlin multiplier theorem version II) (i) Let ${k}$ be the smallest integer ${>n/2}$ and suppose that the multiplier ${m: {\mathbb R}^n\rightarrow {\mathbb C}}$ is of class ${C^k({\mathbb R}^n\setminus \{0\})}$ with
$\displaystyle |\partial^\alpha m(\xi)|\lesssim_\alpha |\xi|^{-|\alpha|},$

for all multi-indices ${\alpha}$ with ${|\alpha|\leq k}$ . Then ${\check m}$ agrees with a function ${K(x)}$ away from the origin which is locally integrable away from the origin and satisfies
$\displaystyle \int_{|x|>2|y|}|K(x-y)-K(x)|dx\lesssim_n 1,$

for all ${y\neq 0}$ .

(ii) Under the assumptions of (i) we have that ${m\in \mathcal M^p}$ .

Proof: As in the proof of Theorem 9 it will be enough to control the pieces ${K_j}$ . For this, let ${\beta}$ be a multi-index. We have

$\displaystyle \int_{{\mathbb R}^n}|(-2\pi i x)^\beta K_j(x)dx|^2dx =\int_{{\mathbb R}^n} |\partial^\beta _\xi m_j(\xi)|^2 d\xi.$

For ${M \leq k }$ this implies that

$\displaystyle \int_{{\mathbb R}^n}( |x|^M)^2 |K_j(x)|^2dx =\int_{{\mathbb R}^n}(x_1 ^2+\cdots+x_n^2)^M |K_j(x)|^2dx \lesssim_{n,M} 2^{nj} 2^{-2Mj}.$

Now for any ${R>0}$ we have

$\displaystyle \int_{|x|\leq R}|K_j(x)|dx\leq \bigg( \int_{|x|\leq R}|K_j(x)|^2dx\bigg)^\frac{1}{2} R^\frac{n}{2}\lesssim_{n,M} 2^\frac{nj}{2} R^\frac{n}{2}.$

On the other hand

$\displaystyle \int_{|x|>R}|K_j(x)|dx=\int_{|x|>R}|K_j(x)||x|^k |x|^{-k}dx\leq 2^\frac{nj}{2}2^{-kj} R^ {n/2-k}, \ \ \ \ \ (7)$

where ${n/2-k<0}$ . Choosing ${M=0}$ and ${R=2^{-j}}$ these estimates imply that

$\displaystyle \int_{{\mathbb R}^n}|K_j(x)|dx\lesssim_{n}1+2^{j(n-2k)}\lesssim_n 1.$

We will now prove a similar estimate for the derivatives of ${K_j}$ using a very similar approach. Indeed, we start from the identity

$\displaystyle \int_{{\mathbb R}^n}|(-2\pi i x)^\beta \partial^\alpha K_j(x)dx|^2dx =\int_{{\mathbb R}^n} |\partial^\beta _\xi [ (2\pi i \xi)^\alpha m_j(\xi)]|^2 d\xi.$

Now for ${M\leq k}$ and using the Leibniz rule we get

$\displaystyle \int_{{\mathbb R}^n}(|x|^M)^2| \partial^\alpha K_j(x)|^2dx\lesssim_{M,\alpha} 2^{nj} 2^{-2Mj} 2^{2j |\alpha| }.$

Thus we have

$\displaystyle \int_{|x|\leq R} |\partial^\alpha K_j(x)|dx\leq \bigg(\int_{{\mathbb R}^n}|\partial^\alpha K_j(x)|^2 \bigg)^\frac{1}{2} R^\frac{n}{2}\lesssim_{\alpha,n}2^\frac{nj}{2} 2^{j|\alpha|}R^\frac{n}{2}.$

Also

$\displaystyle \int_{|x|>R}|\partial^\alpha K_j(x)|dx =\int_{|x|>R}|x|^{-k}|x|^k|\partial^\alpha K_j(x)|dx\lesssim_{n,\alpha,k} 2^\frac{nj}{2} 2^{j|\alpha|} 2^{-kj} R^{\frac{n}{2}-k}.$

Choosing ${R=2^{-kj}}$ and combining the last two estimates we conclude

$\displaystyle \int_{{\mathbb R}^n}|\partial^\alpha K_j(x)|dx \lesssim_{n,\alpha} 2^{j|\alpha|}.$

This estimate for ${|\alpha|=1}$ together with the mean value theorem implies that

$\displaystyle \int_{{\mathbb R}^n}|K_j(x+h)-K_j(x)|dx\lesssim_{n} 2^j|h|.$

We now have for all ${y\neq 0}$

$\displaystyle \sum_{2^j>|y|^{-1}}\int_{|x| \geq 2|y| }|K_j(x-y)-K(x)|dx \lesssim_{n}\sum_{2^j>|y|^{-1}}2^j|y|\lesssim_{n,k}1.$

On the other hand

$\displaystyle \begin{array}{rcl} \sum_{2^j<|y|^{-1}} \int_{|x| \geq 2|y| }|K_j(x-y)-K(x)|dx &\lesssim&\sum_{2^j<|y|^{-1}}\int_{|x|\geq |y|}|K_j(x)|dx \\ \\ &\lesssim_n& \sum_{2^j<|y|^{-1}} 2^\frac{nj}{2}2^{-kj}|y|^{\frac{n}{2}-k}\lesssim_{n,k}1, \end{array}$

by (7). Using now that ${\sum_j K_j(x)}$ converges in ${L^1(V)}$ to some locally integrable function for every compact set ${V}$ that doesn’t contain ${0}$ we conclude that ${K}$ coincides with a locally integrable function away from ${0}$ and satisfies

$\displaystyle \int_{|x|\geq 2|y|}|K(x-y)-K(x)|dx\lesssim_{n}1, \ \ \ \ \ (8)$

for ${y\neq 0}$ .

Now since ${K=\check m}$ away from the origin we have that ${T_m}$

$\displaystyle T_m(f)(x)=\int K(x-y) f(y)dy$

whenever ${f}$ in ${L^2({\mathbb R}^n)}$ and has compact support and ${x\notin {\mathrm{supp}}(f)}$ . Furthermore, by the assumption ${m\in L^\infty({\mathbb R}^n)}$ we automatically get that ${T_m}$ is bounded on ${L^2({\mathbb R}^n)}$ . Here condition (8) is enough to substitute the conditions given in the definition of a singular kernel and show that ${T_m}$ is a CZO with ${K}$ playing the role of the kernel. Indeed, the ${(2,2)}$ type of ${T_m}$ can be used to treat the bad part in the Calderón-Zygmund decomposition of a function ${f}$ . On the other hand, if ${b_Q}$ is a bad piece supported on a dyadic cube ${Q}$ with center ${w_Q}$ and ${Q^*}$ is the cube with the same center and twice the side-length, we have

$\displaystyle \begin{array}{rcl} \int_{{\mathbb R}^n\setminus Q^*}|\int_Q K(x-y)b_Q(y)dy | dx &\leq& \int_{{\mathbb R}^n\setminus Q^*}\int_Q |K(x-y)-K(x-w_Q)||b_Q(y)|dy dx \\ \\ &\leq & \int_Q |b_Q(y)|\int_{{\mathbb R}^n\setminus Q^*} |K(x-y)-K(x-w_Q)|dx dy. \end{array}$

Now if ${y\in Q}$ and ${x\notin Q^*}$ we have that ${|x-w_Q|\geq \textnormal{side}(Q)\geq 2 |y-w_Q| }$ . Thus for ${y\in Q}$ we have from (8) that

$\displaystyle \int_{{\mathbb R}^n\setminus Q^*}|K(x-y)-K(x-w_Q)|dx = \int_{{\mathbb R}^n\setminus Q^*}|K(x-w_Q-(y-w_Q))-K(x-w_Q)|dx\lesssim_{n} 1,$

so that

$\displaystyle \int_{{\mathbb R}^n\setminus Q^*}|\int_Q K(x-y)b_Q(y)dy | dx\lesssim_n \|b_Q\|_{L^1(Q)}.$

This treats the bad part of the Calderón-Zygmund decomposition of ${f}$ so we conclude the proof that ${T_m}$ is of weak type ${(1,1)}$ as in the general case of a CZO. Interpolating between this bound and the strong ${(2,2)}$ bound we get that ${m\in\mathcal M^p}$ for ${1<p<2}$ . By Proposition 7 or using the symmetry of ${K(x-y)}$ in ${x}$ and ${y}$ , we also get the range ${2<p'<\infty}$ with ${\|m\|_{\mathcal M^p}=\|m\|_{\mathcal M^{p'}}=\|T_m\|_{L^p\rightarrow L^p}}$ . $\Box$

Exercise 4 The purpose of this exercise is to clear out some of the calculation in the proofs of the two versions of Hörmander’s theorem. (i) Prove the identity
$\displaystyle \bigg(\frac{x\cdot \nabla_\xi}{2\pi i |x|^2}) \bigg )^N e^{2\pi i x\cdot \xi}=e^{2\pi i x\cdot \xi},$

for any positive integer ${N}$ . Here the meaning of the symbol ${x\cdot \nabla_\xi}$ is
$\displaystyle x\cdot \nabla_\xi :=x_1\partial_{\xi_1}+\cdots+x_m\partial_{\xi_m}.$

Let ${\alpha=(\alpha_1,\ldots,\alpha_n)}$ and ${\beta=(\beta_1,\ldots , \beta_n)}$ be two multi-indices in ${\mathbb N^n}$ . We write ${\beta\leq \alpha}$ of ${\beta_j\leq \alpha_j}$ for all ${j\in\{1,2,\ldots,n\}}$ . With this notation the Leibniz rule says that for any multi-index ${\alpha}$ and functions ${f,g}$ which are say smooth, we have
$\displaystyle \partial^\alpha(fg)=\sum_{\beta\leq \alpha}\binom {\alpha} {\beta}(\partial ^{\alpha-\beta}f)( \partial ^\beta g).$

Here the generalized binomial coefficients ${\binom{\alpha}{\beta}}$ are defined as
$\displaystyle \binom{\alpha}{\beta}:=\binom{\alpha_1}{ \beta_1} \binom {\alpha_2}{\beta_2}\cdots \binom{\alpha_n}{\beta_n}.$

Alternatively we use the notation
$\displaystyle \alpha!:=\alpha_1 ! \cdots \alpha_n!\quad\mbox{so that}\quad \binom{\alpha}{\beta}=\frac{\alpha!}{\beta!(\alpha-\beta)!},\quad \beta\leq \alpha.$

(ii) For any two multi-indices ${\alpha,\beta\in{\mathbb N}^n _o}$ show that
$\displaystyle \partial_x ^\alpha (x^\beta)= \begin{cases} \binom{\beta}{\alpha} x^{\beta-\alpha},\quad \alpha\leq \beta,\\ 0\quad\mbox{otherwise}.\end{cases}$

(iii) Let ${m:{\mathbb R}^n\rightarrow {\mathbb C}}$ satisfy the estimate
$\displaystyle |\partial^\alpha m(\xi)|\lesssim_{n,\alpha} |\xi|^{-|\alpha|},$

for some ${\xi \in {\mathbb R}^n}$ and ${\psi_j=\psi(\xi/2^j),\ j\in{\mathbb Z}}$ , be as in the Littlewood-Paley decomposition. Show that ${m_j=m\psi_j}$ satisfies the same estimates, that is,
$\displaystyle |\partial^\alpha m_j(\xi)|\lesssim_{n,\alpha} |\xi|^{-|\alpha|},$

with different implied constants of course. Remember that ${\psi_j}$ and thus ${m_j}$ is supported on ${|\xi|\simeq 2^j}$ .

(iv) Let ${m:{\mathbb R}^n\rightarrow {\mathbb C}}$ satisfy the estimate
$\displaystyle |\partial^\alpha m(\xi)|\lesssim_{n,\alpha} |\xi|^{-|\alpha|},$

for some ${\xi \in {\mathbb R}^n}$ and ${\psi_j=\psi(\xi/2^j),\ j\in{\mathbb Z}}$ , be as in the Littlewood-Paley decomposition. Set ${m_j=m\psi_j}$ . Show that for any multi-index ${\gamma}$ of order ${|\gamma|=M}$ we have
$\displaystyle |\partial^\gamma( (2\pi i \xi)^\alpha m_j(\xi))|\lesssim_{n,\alpha,M} |\xi|^{|\alpha|-M }.$

(v) Let ${h}$ be a smooth function which is supported on ${A_k:=\{2^{k-1}<|\xi|\leq 2^{k+1}\}}$ . Show that
$\displaystyle \int_{A_k} h(\xi) \bigg(\frac{x\cdot \nabla_\xi}{2\pi i |x|^2}\bigg) e^{2\pi i x\cdot\xi}d\xi = \int_{A_k}\bigg[ \bigg(-\frac{x\cdot \nabla_\xi}{2\pi i |x|^2}\bigg) h(\xi) \bigg] e^{2\pi i x\cdot\xi}d\xi,$

and by iterating that
$\displaystyle \int_{A_k} h(\xi) \bigg(\frac{x\cdot \nabla_\xi}{2\pi i |x|^2}\bigg)^N e^{2\pi i x\cdot\xi}d\xi = \int_{A_k} \bigg[ \bigg(-\frac{x\cdot \nabla_\xi}{2\pi i |x|^2}\bigg)^N h(\xi) \bigg]e^{2\pi i x\cdot\xi}d\xi,$

for all positive integers ${N}$ .

Exercise 5 Let ${K\in L^2( {\mathbb R}^n) }$ be such that ${m:=\hat K\in L^\infty ({\mathbb R}^n)}$ . Furthermore suppose that ${K}$ satisfies the mean regularity condition
$\displaystyle \int_{|x|>2|y|}|K(x-y)-K(x)|dx \lesssim_{n} 1,\quad y\neq 0.$

Show that ${m\in \mathcal M^p({\mathbb R}^n)}$ .

Hint: Briefly describe the key elements of the proof showing that ${T_m(f)=K*f}$ is of weak type ${(1,1)}$ . Argue why this implies that ${m\in\mathcal M^p}$ for ${1<p<2}$ . You get the complementary interval ${2<p<\infty}$ for free (why?).

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