A fractional Hörmander type multiplier theorem


— 1. A fractional Hörmander type multiplier theorem —

In a few months ago we deal with translation invariant operators {T} in {L^{p}(\mathbb{R}^{d})}, see previous post. More precisely, was shown that there exists a tempered distribution {K\in\mathcal{S}'(\mathbb{R}^{d})} such that {Tu=u\ast K}. However, what does happens with {\widehat{K}} to get {L^{p}(\mathbb{R}^{d})}-boundedness of {T}? Such questions belongs to a fruitiful area of Fourier analysis which many personages work in that, such as Marcinkiewicz, Minhklin, Hörmander, Lizorkin and more recentily Fefferman’s work about ball multiplier conjecture. In this post will be initiated the saga related to {L^{p}(\mathbb{R}^{d})} spaces so far. Before, we recall some important definitions. Let {K\in\mathcal{S}'(\mathbb{R}^{d})}, we set

\displaystyle \widehat{K}(\varphi)=K(\widehat{\varphi}), \; \; \widehat{\varphi}(\xi)=(\mathcal{F}\varphi)(\xi):=(2\pi)^{-d}\int_{\mathbb{R}^{d}}e^{-ix\cdot\xi}\varphi(x)dx \ \ \ \ \ (1)

for any {\varphi\in\mathcal{S}(\mathbb{R}^{d})}. Also, recall that

\displaystyle (u\ast K)(\varphi)=K(Ru\ast\varphi). \ \ \ \ \ (2)

Then via Theorem 3 the operator {T} may be written in {\mathcal{S}'(\mathbb{R}^{d})} as

\displaystyle Tu(x)=\mathcal{F}^{-1}[\sigma(\xi)\mathcal{F}u(\xi)](x) \ \ \ \ \ (3)

where {\sigma=\widehat{K}(\varphi)}. In fact, by observing that

\displaystyle \mathcal{F}(ab)(\xi)=\mathcal{F}a(\xi)\mathcal{F}b(\xi)

we have from (2) that

\displaystyle \begin{array}{rcl} \mathcal{F}(u\ast K)(\varphi)=u\ast K(\mathcal{F}\varphi)=K(Ru\ast \mathcal{F}\varphi)=K(\mathcal{F}(\check{Ru}\,\varphi))=\widehat{K}(\varphi)\mathcal{F}u \end{array}

being {\check{a}} the inverse Fourier transform of {a}. We call the operator {T} in (3) of multiplier operator associated to symbol {\sigma} and we will written it as {T_{\sigma}}. Also, we say that {T_{\sigma}} is a multiplier on {L^{p}(\mathbb{R}^{d})}, if for all {u\in L^{2}\cap L^{p}} the map {u\mapsto T_{\sigma}u} is bounded on {L^{p}(\mathbb{R}^{d})}.

Example 1 The Fourier transform in classical sense of

\displaystyle \begin{array}{rcl} f(x)=\frac{1}{\vert x\vert^{2}}, \text{ for } x\in\mathbb{R}^{3}\backslash\{0\} \end{array}

is not possible, because {f} not belong to any {L^{p}(\mathbb{R}^{3})} space. However, Fourier transform of {f} in distribution sense is given by

\displaystyle \begin{array}{rcl} \widehat{\frac{1}{\vert x\vert^{2}}}(\xi)=\frac{\pi}{\vert \xi\vert}, \end{array}

for more details, see notes of Iannis Parissis.

Let {s\in\mathbb{R}}, the Bessel space {L_{s}^{p}(\mathbb{R}^{d})} is the space of {g} functions such that {(1+\vert \xi\vert^{2})^{s/2}\widehat{g}(\xi)\in L^{p}(\mathbb{R}^{d})} which endowed with norm

\displaystyle \Vert g\Vert_{L_{s}^{p}(\mathbb{R}^{d})}=\Vert (1+\vert \cdot\vert^{2})^{s/2}\widehat{g}\Vert_{L^{p}(\mathbb{R}^{d})}, \text{ for } 1\leq p\leq\infty

is a Banach space. The following proposition give a condition on symbol {\sigma} for which {T_{\sigma}} is a multiplier on {L^{p}(\mathbb{R}^{d})}.

Proposition 1 Let {s>d/2} and {\sigma\in L^{2}_{s}(\mathbb{R}^{d})}. Then {T_{\sigma}} is a multiplier on {L^{p}(\mathbb{R}^{d})}, for {1\leq p\leq \infty}.

Proof: In view of {T_{\sigma}f(x)=(K\ast f)(x)}, where {\widehat{K}(\xi)=\sigma(\xi)}. From Young’s inequality, just we need to show {K\in L^{1}(\mathbb{R}^{d})}. To do this, let {\widehat{h}(\xi)=(1+\vert \xi\vert^{2})^{s/2}\sigma(\xi)\in L^2(\mathbb{R}^{d})}. As {\Vert h\Vert_{L^2}=\Vert \widehat{h}\Vert_{L^2}=\Vert \sigma\Vert_{L_{s}^{2}}}, from Cauchy-Schwartz inequality one gets

\displaystyle \begin{array}{rcl} \int_{\mathbb{R}^{d}}\vert K(x)\vert dx&=&\int_{\mathbb{R}^{d}}\vert \check{\sigma}(x)\vert dx\\ \\ &=& \left (\int_{\mathbb{R}^{d}}(1+\vert x\vert^{2})^{-s} dx\right)^{\frac{1}{2}} \left(\int_{\mathbb{R}^{d}} \vert h(x)\vert^{2} dx\right)^{\frac{1}{2}}\\ \\ &\leq& C \Vert \sigma\Vert_{L_{s}^{2}}. \end{array}

\Box

Example 2 Let {\sigma} be bounded, that is, {\sigma\in L^{\infty}(\mathbb{R}^{d})}. We claim that

\displaystyle \Vert\sigma\Vert_{L^{\infty}(\mathbb{R}^{d})}=\Vert T_{\sigma}\Vert_{L^2(\mathbb{R}^{d})\rightarrow L^2(\mathbb{R}^{d})}.

It’s quite easy to get {\Vert T_{\sigma}\Vert_{L^2(\mathbb{R}^{d})\rightarrow L^2(\mathbb{R}^{d})}\leq \Vert\sigma\Vert_{L^{\infty}(\mathbb{R}^{d})}} via Plancherel’s theorem. It remains to get the other inequality. Let {D_{r}(x)} be a ball in {\mathbb{R}^{d}} with radius {r}. Let {\varphi_r\in C^{\infty}_c(\mathbb{R}^{d})} such that {supp\,(\varphi_r)\subset D_{2r}(x)} and {\varphi_r\equiv1} on {D_r(x)}. Note that

\displaystyle \begin{array}{rcl} \sigma(\xi)\varphi_r=\mathcal{F}(T_{\sigma}\ast \mathcal{F}^{-1}\varphi_r) &\Rightarrow& \Vert \sigma\varphi_r\Vert_{L^2(\mathbb{R}^{d})}=\\ \\ &=&\Vert T_{\sigma}\ast\mathcal{F}^{-1}\varphi_r\Vert_{L^2(\mathbb{R}^{d})}=\Vert T_{\sigma}\mathcal{F}^{-1}\varphi_r\Vert_{L^2}\\ \\ &\Rightarrow& \sigma\in L^{2}(D_r(x))\\ \\ &\Rightarrow& f\sigma\in L^{2}(\mathbb{R}^{d}),\,f(x)=\vert D_r(x)\vert^{-1}\chi_{D_r(x)}\\ \\ &\Rightarrow& \int_{\mathbb{R}^{d}}\vert f\sigma\vert^2d\xi=\int_{\mathbb{R}^{d}}\vert T_{\sigma}f\vert^2d\xi\leq \Vert T_{\sigma}\Vert_{L^2\rightarrow L^2}^{2} \int_{\mathbb{R}^{d}}\vert f\vert^2d\xi. \end{array}

The Lebesgue differentiation’s theorem yields

\displaystyle \begin{array}{rcl} \Vert T_{\sigma}\Vert_{L^2\rightarrow L^2}^{2}-\vert\sigma(x)\vert^{2} &=&\lim_{r\rightarrow0}\frac{1}{\vert D_{r}(x)\vert}\int_{D_r(x)}(\Vert T_{\sigma}\Vert_{L^2\rightarrow L^2}^{2}-\vert\sigma(\xi)\vert^{2})d\xi\\ \\ &=&\lim_{r\rightarrow0}\int_{\mathbb{R}^{d}}(\Vert T_{\sigma}\Vert_{L^2\rightarrow L^2}^{2}-\vert\sigma(\xi)\vert^{2})\vert f(\xi)\vert^{2}d\xi\geq 0, \end{array}

in other words, {\Vert\sigma\Vert_{L^{\infty}(\mathbb{R}^{d})}\leq \Vert T_{\sigma}\Vert_{L^2(\mathbb{R}^{d})\rightarrow L^2(\mathbb{R}^{d})}}.

Let {\Sigma_{0}^{\beta}(\mathbb{R}^{d})} be the set of all tempered distributions {\sigma} away from origin of {\mathbb{R}^{d}} satisfying the estimate

\displaystyle \sup_{\xi\neq0}|\xi|^{|\alpha|+\beta}|\partial_{\xi}^{\alpha}\sigma(\xi)|\leq L, \;\beta\geq 0 \ \ \ \ \ (4)

for all multi-index {\alpha} with {\vert \alpha\vert\leq k}. In the next sections we give a much weaker hypotheses on symbol {\sigma}, via Littlewood-Paley theory, which imply such as so-called symbol-H\”{o}rmader condition and {\sigma\in\Sigma_{0}^{\beta}(\mathbb{R}^{d})}. In all that follows, the class {OP\Sigma_{0}^{\beta}(\mathbb{R}^{d})} denotes the set of multipliers operators with symbol {\sigma(T_{\sigma})=\sigma(\xi)} belonging to {\Sigma_{0}^{\beta}(\mathbb{R}^{d})}.

— 1.1. Littlewood-Paley theory —

Let {\varphi} be radial Schwartz function supported in {\{\xi\in\mathbb{R}^{d}\,:\, 0<\vert\xi\vert\leq 2\}} which is identically {1} on {\{\xi\in\mathbb{R}^{d}\,:\, 0\leq\vert\xi\vert\leq 1\}}. To get that, recall of the topological Urysohn’s lemma. For all {\xi\in\mathbb{R}^{d}}, set

\displaystyle \begin{array}{rcl} \widehat{\psi}(\xi)=\varphi(\xi)-\varphi(2\xi). \end{array}

Notice that {supp(\widehat{\psi})\subset \{\xi\in\mathbb{R}^{d}\,:\, 1/2\leq \vert\xi\vert\leq 2\}}, because {\varphi(\xi)=\varphi(2\xi)=0} and {\varphi(\xi)=\varphi(2\xi)=1} for, respectively, {\vert \xi\vert >2} and {\vert \xi\vert <1/2}. Moreover, the family of functions {\{\widehat{\psi}(\xi/2^{j})\}_{j\in\mathbb{Z}}} forms a partition of unity

\displaystyle \begin{array}{rcl} \sum_{j\in\mathbb{Z}} \widehat{\psi}(\xi/2^{j}) =1, \; \xi\in\mathbb{R}^{d}\backslash\{0\}. \end{array}

Next, for {j\in\mathbb{Z}} we define the cut-off operator or Littlewood-Palay operator as

\displaystyle \begin{array}{rcl} \widehat{\Delta_j(f)}(\xi)=\widehat{\psi}_j(\xi)\widehat{f}(\xi):=\widehat{\psi}(\xi/2^{j})\widehat{f}(\xi), \end{array}

where {f\in\mathcal{S}'(\mathbb{R}^{d})}. Notice that {\Delta_j} is supported in {\{x\in\mathbb{R}^{d}\,:\, 2^{j-1}\leq \vert x\vert\leq 2^{j+1}\}}. Indeed, by Fourier property

\displaystyle \begin{array}{rcl} \widehat{\textnormal{Dil}_{2^{-j}}^{1}\psi}(\xi)=\int_{\mathbb{R}^{d}}e^{-ix\cdot\xi}2^{jd}\psi(2^{j}x)dx=\widehat{\psi}(2^{-j}\xi) \end{array}

it follows that {\psi(2^{-j}x)=\textnormal{Dil}_{2^{-j}}^{1}\psi(x)=2^{jd}\psi(2^{j}x)}. Therefore, if {\widehat{\psi}} is supported in the annulus {\{\xi\in\mathbb{R}^{d}\,:\, 1/2\leq \vert\xi\vert\leq 2\}} then the inverse Fourier transform is supported in the annulus

\displaystyle \vert x\vert\simeq2^{j}:=\{x\in\mathbb{R}^{d}\,:\, 2^{j-1}\leq \vert x\vert\leq 2^{j+1}\}.

In other words, the operator {\Delta_j} isolate the part of a function being each one concetrated near annulus {\vert x\vert\simeq2^{j}}.

Proposition 2 The cut-off operator is a self-adjoint operator,

\displaystyle \begin{array}{rcl} \Delta_j\phi(\varphi)=\phi(\Delta_j\varphi) \end{array}

for any {\varphi\in\mathcal{S}(\mathbb{R}^{d})} and {\phi\in\mathcal{S}'(\mathbb{R}^{d})}.

Proof: The prove is a consequence of the radial property of {\psi}. For that, firstly, shall be observed that

\displaystyle \begin{array}{rcl} \Delta_j\phi(\varphi) &=&\mathcal{F}(\Delta_j\phi)(\mathcal{F}^{-1}{\varphi})\notag\\ \\ &=&\widehat{\psi}_j\widehat{\phi}(\mathcal{F}^{-1}{\varphi})\notag\\ \\ &=&\phi(\mathcal{F}(\psi_j\mathcal{F}^{-1}{\varphi})), \end{array}

when we have in mind (1). Now, radial property of {\psi} and (4) implies that

\displaystyle \begin{array}{rcl} \Delta_j\phi(\varphi) &=&\phi\left(\int e^{-i\xi\cdot x} \psi_{j}(x)\widehat{\varphi}(-x)dx\right)\\ \\ &=&\phi\left(\int e^{i\xi\cdot x} \psi_{j}(-x)\widehat{\varphi}(x)dx\right)\\ \\ &=&\phi(\mathcal{F}^{-1}(\psi_{j}(\cdot)\widehat{\varphi})(\xi))\\ \\ &=&\phi(\Delta_j\varphi). \end{array}

\Box

The decomposition of a tempered distribution {f} into a sum of “long pieces” {\Delta_j(f)} supported in the annulus {\vert x\vert \simeq 2^{j}} is known as Littlewood-Paley decomposition, because of the famous works de Littlewood and Paley related to Fourier and Power series, see xx. More precisely, Littlewood-Paley decomposition of a tempered distribution {f\in\mathcal{S}'(\mathbb{R}^{d})} is defined as follows

\displaystyle f=\sum_{j\in\mathbb{Z}} \Delta_j(f) \ \ \ \ \ (5)

However, there is tempered distributions for which Littlewood-Paley decomposition fail on {\mathcal{S}'(\mathbb{R}^{d})}. For instance take the distribution {f(x)=1} and note by from Proposition 2 and {\widehat{1}(\varphi)=\varphi(0)} that

\displaystyle \begin{array}{rcl} \Delta_jf(\varphi)=f(\Delta_j\varphi)=f(\mathcal{F}^{-1}(\widehat{\Delta_j\varphi}))=\widehat{f}(\widehat{\Delta_j\varphi})=\widehat{\Delta_j\varphi}(0)=0 \end{array}

that is, {\Delta_jf=0}, {j\in\mathbb{Z}}. To overcome this discomfort, we introduce in {\mathcal{S}'(\mathbb{R}^{d})} the equivalence class {\sim} given by

\displaystyle \begin{array}{rcl} u\sim v\Longleftrightarrow u-v\in\mathcal{P} \end{array}

where {\mathcal{P}} denotes the set of polynomials {p},

\displaystyle \begin{array}{rcl} p(x)=\sum_{\vert\alpha\vert\leq m}c_{\beta}(x)x^{\alpha} \end{array}

with complex coefficients {c_{\alpha}} and {m} a non-negative integer. The space {\mathcal{S}'(\mathbb{R}^{d})} endowed with {\sim} will be denoted by {\mathcal{S'}/\mathcal{P}}.

Proposition 3 Let {\mathcal{S}_{\infty}(\mathbb{R}^{d})} be the space of Schwartz functions such that

\displaystyle \begin{array}{rcl} \int_{\mathbb{R}^{d}} x^{\alpha}\varphi(x)dx=0. \end{array}

Then {\mathcal{S}_{\infty}(\mathbb{R}^{d})} is a subspace of {\mathcal{S}(\mathbb{R}^{d})} which has

\displaystyle \mathcal{S}_{\infty}'(\mathbb{R}^{d})=\mathcal{S'}/\mathcal{P} \ \ \ \ \ (6)

in sense of isomophism.

Proof: The prove is standard, consider the identification map {\mathcal{J}} which goes an element {\mathcal{S}'(\mathbb{R}^{d})} into the equivalence class {\mathcal{S'}/\mathcal{P}} that contains it. The kernel of the map {\mathcal{J}} is {\mathcal{P}}. \Box

The aim is that, for any {u\in\mathcal{S}'_{\infty}(\mathbb{R}^{d})} we have

\displaystyle \begin{array}{rcl} (\psi_{j}\widehat{u})^{\vee}\rightarrow 0 \text{ in } \mathcal{S}'_{\infty} \text{ as } j\rightarrow+\infty. \end{array}

This follows by observing that

\displaystyle \varphi(2^{-j}\cdot)\widehat{u}\rightarrow\widehat{u} \text{ in } \mathcal{S}'_{\infty}(\mathbb{R}^{d}) \text{ as } j\rightarrow+\infty \ \ \ \ \ (7)

where {\varphi} is that radial Schwartz function suported into a compact set and equal to {1} in {D_{1}(0)}. To show (7), recall that a distribution {u} is a tempered distribution if

\displaystyle \begin{array}{rcl} \lim_{k\rightarrow\infty}u(\lambda_k) = 0 \text{ whenever } \lambda_k\rightarrow0 \text{ in } \mathcal{S} \text{ as } k\rightarrow\infty. \end{array}

Easily, via identification (6), this indue in nature’s way the convergence in {\mathcal{S}'_{\infty}(\mathbb{R}^{d})}. Thus, it is quite easy to show {\varphi_{j}\widehat{u}\in \mathcal{S}'_{\infty}(\mathbb{R}^{d})}. It remains to show

\displaystyle \begin{array}{rcl} \lim_{j\rightarrow+\infty}\varphi(2^{-j}\cdot)\widehat{u}(\lambda)=\widehat{u}(\lambda). \end{array}

Let us fixe {i\in\mathbb{N}} such that {2^{i-1}\leq \vert \xi\vert\leq 2^{i}} and write

\displaystyle \begin{array}{rcl} \lim_{j\rightarrow+\infty}\varphi(2^{-j}\xi)\widehat{u}(\lambda)=\lim_{j\geq i,i\rightarrow+\infty}\varphi(2^{-j}\xi)\widehat{u}(\lambda)=\widehat{u}(\lambda) \end{array}

because for {j\geq i} we have {1/2\leq \vert\xi\vert \leq 1} and {\varphi \equiv1} on ball {D_{1}(0)}. Thus, given {p\in\mathcal{P}(\mathbb{R}^{d})} one has {p=0} and the its Littlewood-Paley decomposition is null in {\mathcal{S}'_{\infty}(\mathbb{R}^{d})}, because of

\displaystyle \begin{array}{rcl} \lim_{j\rightarrow+\infty}\psi_{j}\widehat{p}(\lambda)=0 \end{array}

and

\displaystyle \begin{array}{rcl} \lim_{j\rightarrow-\infty}\psi_{j}\widehat{p}(\lambda)=0. \end{array}

The last equality is obtained via

\displaystyle \begin{array}{rcl} \lim_{j\rightarrow-\infty}\varphi(2^{-j}\cdot)\widehat{p}(\lambda) &=&\sum_{\vert\alpha\vert\leq m}c_{\alpha}\lim_{j\rightarrow-\infty}\varphi(2^{-j}\cdot)\partial^{\alpha}\delta_{0}(\lambda)\\ \\ &=&\sum_{\vert\alpha\vert\leq m}c_{\alpha}\lim_{j\rightarrow-\infty}\varphi(2^{-j}\cdot)\delta_{0}(\partial^{\alpha}\lambda)=0, \end{array}

where {\delta_0} is the Dirac mass and don’t forget that {\mathcal{F}^{-1}(\partial^{\beta}\delta_{0})(x)=(-2\pi i x)^{\beta}}, which is consequence of properties

  • (i){(\partial_{x}^{\alpha}\varphi)^{\wedge}(\xi)=\xi^{\alpha}\widehat{\varphi}(\xi)} for every multi-index {\alpha\in (\mathbb{N}\cup\{0\})^{d}}
  • (ii){(\partial^{\alpha}_{\xi}\widehat{\varphi})(\xi)=((-x)^{\alpha}\varphi(x))^{\wedge}(\xi)} for every multi-index {\alpha\in (\mathbb{N}\cup\{0\})^{d}}

for all {\varphi\in\mathcal{S}(\mathbb{R}^{d})}.

Proposition 4 If {\sigma\in \Sigma_{0}^{1}(\mathbb{R}^{d})}, then {k_{\sigma}} agrees with {C^{k}(\mathbb{R}^{d})} function away from the origin and satisfies the following pointwise estimate

\displaystyle \vert \partial_{z}^{\beta}k_{\sigma}(z)\vert\leq C L |z|^{-d-\vert\beta\vert},\; z\neq 0 \ \ \ \ \ (8)

for all multi-index {\beta\in(\mathbb{N}\cup\{0\})^{d}}, where {k_{\sigma}(z)=(2\pi)^{-d}\int_{\mathbb{R}^{d}}e^{iz\cdot\xi}\sigma(\xi)d\xi}.

Proof: Let {\sigma_j(\xi)=\psi_j(\xi)\sigma(\xi)}, then for all {\xi\neq 0} this give us

\displaystyle \begin{array}{rcl} \sum_{j\in\mathbb{Z}}\sigma_j \,=\,\sigma \text{ in } \mathcal{S}'/\mathcal{P} \end{array}

where each peace {\sigma_j} is supported on annulus {\vert\xi\vert \simeq 2^{j}:=\{\xi\in\mathbb{R}^{d}\,:\, 2^{j-1}\leq\vert\xi\vert\leq 2^{j+1}\}}. So makes sense to define,

\displaystyle \begin{array}{rcl} k_j(x)=(2\pi)^{-d}\int_{\mathbb{R}^{d}}e^{ix\cdot\xi}\sigma_j(\xi)d\xi. \end{array}

If {\sigma\in\Sigma_{0}^{1}(\mathbb{R}^{d})}, for {\alpha\in (\mathbb{N}\cup\{0\})^{d}} one has

\displaystyle \vert \partial_{\xi}^{\alpha}\sigma_j(\xi)\vert \leq\vert \sum_{j\in\mathbb{Z}}\partial_{\xi}^{\alpha}\sigma_j(\xi)\vert\leq \sup_{\vert\xi\vert \simeq 2^{j}}\vert \partial_{\xi}^{\alpha}\sigma(\xi)\vert\leq L\, 2^{-j\vert\alpha\vert}. \ \ \ \ \ (9)

It follows that

\displaystyle \begin{array}{rcl} \vert (-2\pi iz)^{\alpha}\partial_{z}^{\beta}k_{j}(z)\vert &=&\vert\int_{\mathbb{R}^{d}}e^{2\pi i\xi\cdot z} \partial_{\xi}^{\alpha}(\xi^{\beta}\sigma_{j}(\xi))d\xi\vert\nonumber\\ \notag\\ &=&\vert\int_{\vert\xi\vert \simeq 2^{j}}e^{2\pi i\xi\cdot z} \partial_{\xi}^{\alpha}(\xi^{\beta}\sigma_{j}(\xi))d\xi\vert\nonumber\\ \notag\\ &=&\vert\sum_{\gamma\leq\alpha}\binom{\alpha}{\gamma}\int_{\vert\xi\vert \simeq 2^{j}}e^{ 2\pi i\xi\cdot z}(\partial_{\xi}^{\gamma}\xi^{\beta})(\partial_{\xi}^{\alpha-\gamma}\sigma_{j}(\xi))d\xi\vert\\ \notag\\ &=&C_{\alpha,\beta,\gamma}\int_{\vert\xi\vert \simeq 2^{j}}e^{2\pi i\xi\cdot z} (\xi^{\beta-\gamma})(\partial_{\xi}^{\alpha-\gamma}\sigma_{j}(\xi))d\xi\vert\\ \notag\\ &\leq& C \,2^{jd}2^{j(\vert\beta\vert-\vert\gamma\vert)}L\,2^{j(-\vert\alpha\vert+\vert\gamma\vert)}\\ \notag\\ &=&C L\,2^{j(d+\vert\beta\vert -\vert\alpha\vert)}. \end{array}

If we put {\vert \alpha\vert = m} into the last equality above, for all multi-indices {\beta} and non-negative {m} we have

\displaystyle \vert \partial_{z}^{\beta}k_{j}(z)\vert\leq CL\,\min\{2^{j(d+\vert\beta\vert)}, \vert z\vert^{-m}2^{j(d+\vert\beta\vert -m)}\}. \ \ \ \ \ (10)

We split {\sum_{2^{j}}\vert \partial_{z}^{\beta}k_{j}(z)\vert} into two sums as

\displaystyle \begin{array}{rcl} \sum_{2^{j}}\vert \partial_{z}^{\beta}k_{j}(z)\vert=\sum_{2^j\leq \vert z\vert^{-1}}\vert \partial_{z}^{\beta}k_{j}(z)\vert+\sum_{2^j\geq \vert z\vert^{-1}}\vert \partial_{z}^{\beta}k_{j}(z)\vert. \end{array}

For {m=0} and {m>d+\vert\beta\vert} into (10), respectively, we get

\displaystyle \begin{array}{rcl} \sum_{2^j\leq \vert z\vert^{-1}}\vert \partial_{z}^{\beta}k_{j}(z)\vert&\leq& CL\, \sum_{2^j\leq \vert z\vert^{-1}}2^{j(d+\vert\beta\vert)}\leq CL\, \vert z\vert^{-(d+\vert\beta\vert)} \end{array}

and

\displaystyle \begin{array}{rcl} \sum_{2^j\geq \vert z\vert^{-1}}\vert \partial_{z}^{\beta}k_{j}(z)\vert&\leq& CL\, \vert z\vert^{-m}\sum_{2^j\geq \vert z\vert^{-1}}2^{j(d+\vert\beta\vert -m)}\notag\\ \\ &\leq&CL\, \vert z\vert^{-m}\vert z\vert^{-(d+\vert\beta\vert -m)}=CL\,\vert z\vert^{-(d+\vert\beta\vert)}. \end{array}

This inequalities says us that {\sum_{j}\partial_{z}^{\beta}k_{j}(z)} converges absolutely and uniforms in {z\neq 0}. It follows that {\sum_{j}k_{j}(z)} converges in {C^{k}(\mathbb{R}^{d}\backslash\{0\})} to a function {\tilde{k}} which also satisfies the estimate

\displaystyle \vert \partial_{z}^{\beta}\tilde{k}(z)\vert\leq CL\, \vert z\vert^{-(d+\vert\beta\vert)}. \ \ \ \ \ (11)

As {\sum_{j}\sigma_j=\sum_{j}\widehat{k}_{\sigma}} converges to {\sigma=\widehat{k}_{\sigma}}, then {k_{\sigma}=\tilde{k}} and we obtain the desired estimate. \Box

In what follows, {L^{p}(\mathbb{R}^{d}\rightarrow X)} denotes the space of {p-}Bochner integrable functions {g} from {\mathbb{R}^{d}} to a Banach space {X} and we write

\displaystyle \Vert g\Vert_{L^{p}(\mathbb{R}^{d}\rightarrow X)}=\left(\int_{\mathbb{R}^{d}}\Vert g(x)\Vert_{X}^{p}dx\right)^{\frac{1}{p}}. \ \ \ \ \ (12)

Theorem 5 (Littlewood-Paley Theorem) Let {\Delta_{j}} the Littlewood-Paley operator. If {1<p<\infty}, there exists a positive constant {C} such that

\displaystyle \Vert \Delta_{j}f\Vert_{L^p(\mathbb{R}^{d}\rightarrow l^{2}(\mathbb{Z}))}\leq C \Vert f\Vert_{L^p(\mathbb{R}^{d})}. \ \ \ \ \ (13)

Moreover,

\displaystyle \Vert f\Vert_{L^p(\mathbb{R}^{d})}\leq C \Vert \Delta_{j}f\Vert_{L^p(\mathbb{R}^{d}\rightarrow l^{2}(\mathbb{Z}))}. \ \ \ \ \ (14)

Proof: The inequality (13) follows from Theorem 5.4 in [1]. In fact,

\displaystyle \begin{array}{rcl} \Vert (\Sigma_{j\in\mathbb{Z}}\vert \Delta_{j}f\vert^{2})^{\frac{1}{2}}\Vert_{L^2(\mathbb{R}^{d})}^{2} &=&\int_{\mathbb{R}^{d}}\Sigma_{j}\vert \Delta_{j}f(x)\vert^{2}dx=\sum_{j}\int_{\mathbb{R}^{d}}\vert \Delta_{j}f(x)\vert^{2}dx\nonumber\\ \\ &=&\sum_{j}\int_{\mathbb{R}^{d}}\vert \widehat{\psi}_{j}(\xi)\widehat{f}(\xi)\vert^{2} d\xi\nonumber\\ \\ &=&\int_{\mathbb{R}^{d}}\Sigma_{j}\vert \widehat{\psi}_{j}(\xi)\vert^{2}\vert \widehat{f}(\xi)\vert^{2} d\xi\nonumber\\ \\ &=&\int_{\mathbb{R}^{d}}\vert \widehat{f}(\xi)\vert^{2} d\xi=\Vert f\Vert_{L^2}^{2}, \end{array}

in other words, the map {f\mapsto \Delta_j f} is bounded from {L^{2}(\mathbb{R}^{d})} to {L^{2}(\mathbb{R}^{d}\rightarrow l^{2}(\mathbb{Z}))}. It remains to show that the kernel {\psi_{j}} of the map {f\mapsto \Delta_j f} satisfies the {l^2(\mathbb{Z})-}H\”{o}rmander condition, which can be obtained from estimate

\displaystyle \begin{array}{rcl} \Vert \nabla \psi_j(x)\Vert_{l^{2}(\mathbb{Z})}\leq C\vert x\vert^{-d-1}. \end{array}

Indeed, by mean value theorem one has

\displaystyle \begin{array}{rcl} \int_{\vert x\vert\geq 2\vert y\vert }\Vert \psi_{j}(y-x) -\psi_j(x)\Vert_{l^{2}}dx &\leq& \int_{\vert x\vert\geq 2\vert y\vert }\vert y\vert\Vert\nabla\psi_j(x-\theta y)\Vert_{l^{2}}dx,\,0<\theta <1\\ \\ &\leq& C\int_{\vert x\vert\geq 2\vert y\vert }\vert y\vert\vert x-\theta y\vert^{-d-1} dx\\ \\ &\leq& C \int_{\vert x\vert\geq 2\vert y\vert }\vert x\vert^{-d} dx,\,\vert x-\theta y\vert\geq (1-\theta)\frac{1}{2}\vert x\vert\\ \\ &\leq& C. \end{array}

The inequalities above show us that the kernel {K(x,y):=\{\psi_j(x-y)\}_{j\in\mathbb{Z}}} is a singular kernel from {\mathbb{R}} to {l^{2}(\mathbb{Z})} and the inequality (13) follows from Theorem 5.4 in [1].

From now on, will be showed the estimate (14). To this end, we remember of {l^1(\mathbb{Z})\subset l^2(\mathbb{Z})} which give us

\displaystyle (\Sigma_j\vert \nabla \psi_j(x)\vert ^{2})^{\frac{1}{2}}\leq \Sigma_j\vert \nabla \psi_j(x)\vert. \ \ \ \ \ (15)

Recall that {\psi_j(x)=2^{jd}\psi(2^{j}x)}, where {\widehat{\psi}} is a Schwartz function. Therefore, {\vert\xi\vert^{\vert\alpha\vert}\vert\partial_{\xi}^{\alpha}\widehat{\psi}(\xi)\vert \leq C} for all multi-index {\alpha}, that is, {\widehat{\psi}\in\Sigma_{0}^{1}(\mathbb{R}^{d})}. And by Proposition 4 follows that there exists a positive constant {C>0} for which

\displaystyle \vert \nabla \psi(x)\vert \leq C\vert x\vert^{-d-1}. \ \ \ \ \ (16)

Setting

\displaystyle \begin{array}{rcl} \sum_j\vert \nabla \psi_j(x)\vert= \sum_{j\geq l }\vert 2^{j(d+1)}\nabla \psi(2^{j}x)\vert+\sum_{j\leq l}\vert 2^{j(d+1)}\nabla \psi(2^{j}x)\vert \text{ for a fixed } l>0 \end{array}

and recalling that {supp(\psi_j(x))\subset \vert x\vert \simeq 2^{j}}, we have via inequalities (15) and (16), respectively, that

\displaystyle \begin{array}{rcl} (\sum_j\vert \nabla \psi_j(x)\vert ^{2})^{\frac{1}{2}} &\leq& C\sum_{j\geq l}2^{j(d+1)-2j(d+1)}+ C\sum_{j\leq l}2^{j(d+1)-j(d+1)}\vert x\vert^{-d-1}\\ \\ &=&C\sum_{j\geq l}2^{-j(d+1)}+ C\sum_{j\leq l}\vert x\vert^{-d-1}\\ \\ &\leq& C \vert x\vert ^{-d-1}. \end{array}

Let {f} be in {L^{p'}(\mathbb{R}^{d})/\mathcal{P}}, where {1=\frac{1}{p}+\frac{1}{p'}}. For each {x\in\mathbb{R}^{d}}, let {g(x)=\{g_{j}(x)\}_{j}\in l^{2}(\mathbb{Z})} and {g\in L^{p}(\mathbb{R}^{d}\rightarrow l^2(\mathbb{Z}))}. From self-adjointness of {\Delta_{j}}, follows that

\displaystyle \begin{array}{rcl} \int_{\mathbb{R}^{d}}\langle \Delta_{j}(f), g\rangle_{l^2(\mathbb{Z})}(x)dx=\int_{\mathbb{R}^{d}}\langle f, \Delta_{j}(g_{j})\rangle_{l^2(\mathbb{Z})}(x)dx:= \int_{\mathbb{R}^{d}}f(x)\Sigma_{j}\bar{\Delta_j}(g_{j})(x)dx. \end{array}

By duality of {L^{p}(\mathbb{R}^{d})} and Cauchy-Schwartz inequality, respectively, one has

\displaystyle \begin{array}{rcl} \Vert \Sigma_j\Delta_{j}(g_j)\Vert_{L^{p}} &=&\sup_{\Vert f\Vert_{{p'}}\leq 1}\left\vert \int_{\mathbb{R}^{d}}\langle \Delta_{j}(f), g\rangle_{l^2(\mathbb{Z})}(x)dx\right\vert\\ \\ &\leq& \sup_{\Vert f\Vert_{{p'}}\leq 1}\int_{\mathbb{R}^{d}}\Vert \Delta_{j}(f)(x)\Vert_{l^{2}} \Vert g(x)\Vert_{l^2}dx\\ \\ &\leq& \sup_{\Vert f\Vert_{{p'}}\leq 1}\Vert \Delta_{j}(f)\Vert_{L^{p'}(\mathbb{R}^{d}\rightarrow l^{2})} \Vert g(x)\Vert_{{L^{p}(\mathbb{R}^{d}\rightarrow l^{2})}}\\ \\ &\leq& C\sup_{\Vert f\Vert_{{p'}}\leq 1}\Vert f\Vert_{L^{p'}} \Vert g(x)\Vert_{{L^{p}(\mathbb{R}^{d}\rightarrow l^{2})}}\\ \\ &\leq& C\Vert g(x)\Vert_{{L^{p}(\mathbb{R}^{d}\rightarrow l^{2})}} \end{array}

where we use the Hölder inequality and first Littlewood-Paley inequality (13) above. If we replace {\Delta_j} by {\tilde{\Delta}_j} in (16) such that {\tilde{\Delta}_{j}\Delta_j=\Delta_j} and we put {g=\{\Delta_j(f)\}_{j\in\mathbb{Z}}}, it follows that

\displaystyle \Vert f\Vert_{L^{p}}=\Vert \Sigma_j\Delta_j(f)\Vert_{L^{p}}(\mathbb{R}^{d})\leq C\, \Vert \Delta_{j}(f)\Vert_{L^p(\mathbb{R}^{d}\rightarrow l^{2} )}, \ \ \ \ \ (17)

as desired. It remains to get {\tilde{\Delta}_{j}}. To do this, let

\displaystyle \widehat{\tilde{\psi}}(\xi)=\varphi(\xi/4)-\varphi(4\xi) \ \ \ \ \ (18)

and set

\displaystyle \widehat{\tilde{\Delta}}_{j}(f)(\xi)=\widehat{\tilde{\psi}}_{j}(\xi)\widehat{f}(\xi)=(\varphi(\xi/2^{j+2})-\varphi(\xi/2^{j-2}))\widehat{f}(\xi). \ \ \ \ \ (19)

Then {\widehat{\tilde{\psi}}_{j}\widehat{\psi}_{j}=\widehat{\psi}_{j}} on annulus {\{\xi\,:\, 2^{j-1}<\vert\xi\vert<2^{j+1}\}}, that is, {\tilde{\Delta}_{j}{\Delta}_{j}={\Delta}_{j}}. Indeed, {\varphi(\xi/2^{j+2})=1} since {\vert\xi\vert/2^{k-2}> 2} and {\varphi(\xi/2^{j-2})=0} since {\vert\xi\vert/2^{k+2}<1/2}. This finish our prove. \Box

Lemma 6 Let {\sigma\in L^2_s(\mathbb{R}^{d})} and {2s>d\geq1}. Set {\widehat {T_{\sigma(2^{j}\cdot)}f}(\xi)=\sigma(2^{j}\xi)\widehat{f}(\xi)}. If {u\in L^{1}_{loc}(\mathbb{R}^{d})}, then there exists a positive constant {C>0} (independent of {\Vert\sigma\Vert_{L^{2}_s}}, {f} and {u}) such that

\displaystyle \int_{\mathbb{R}^{d} } \vert T_{\sigma(2^{j}\cdot)}f(x)\vert^{2}u(x)dx\leq C \Vert \sigma \Vert_{L^{2}_{s}}^{2}\int_{\mathbb{R}^{d}}\vert f(y)\vert^{2}Mu(y)dy \ \ \ \ \ (20)

where {Mu} denotes the Hardy-Littlewood maximal function,

\displaystyle Mu(x)=\sup_{r>0}\frac{1}{\vert D_{r}(x_0)\vert}\int_{D_{r}(x_0)}\vert u(x)\vert dx. \ \ \ \ \ (21)

Proof: Let {\widehat{k}_{\sigma}(\xi)=\sigma(\xi)}, then the kernel of {T_{\sigma(2^{j}\cdot)}} satisfy {k_{\sigma}(2^{j}x)=\check{\sigma}(2^{j}x)=2^{-jd}k_{\sigma}(2^{-j}x)}. Therefore, by Hölder inequality and Plancharel theorem we obtain

\displaystyle \begin{array}{rcl} \vert T_{\sigma(2^{j}\cdot)}f(x)\vert &\leq& 2^{-jd}\int_{\mathbb{R}^{d}}\vert k_{\sigma}(2^{-j}y)f(x-y)\vert dy\\ \\ &=&2^{-jd}\int_{\mathbb{R}^{d}}\vert (1+\vert 2^{-j}y\vert^{2})^{\frac{s}{2}}k_{\sigma}(2^{-j}y)\frac{f(x-y)}{(1+\vert 2^{-j}y\vert^{2})^{\frac{s}{2}}}\vert dy \\ \\ &\leq& 2^{-jd}\Vert (1+\vert 2^{-j}\cdot \vert^{2})^{\frac{s}{2}}k_{\sigma}(2^{-j}\cdot )\Vert_{L^{2}}\left(\int_{\mathbb{R}^{d}}\frac{\vert f(x-y)\vert^{2}}{(1+\vert 2^{-j}y\vert^{2})^{s}}dy\right)^{\frac{1}{2}}\\ \\ &=&2^{-jd}2^{jd/2}\Vert (1+\vert\cdot \vert^{2})^{\frac{s}{2}}k_{\sigma}(\cdot )\Vert_{L^{2}}\left(\int_{\mathbb{R}^{d}}\frac{\vert f(x-y)\vert^{2}}{(1+\vert 2^{-j}y\vert^{2})^{s}}dy\right)^{\frac{1}{2}}\\ \\ &=&\Vert \sigma \Vert_{L^{2}_{s}(\mathbb{R}^{d})}\left(\int_{\mathbb{R}^{d}}\varphi_{2^{j}}(y)\vert f(x-y)\vert^{2}dy\right)^{\frac{1}{2}} \end{array}

where {\varphi_{t}(x)=t^{-d}\varphi(x/t)} and {\varphi(x)=(1+\vert x\vert^{2})^{s}}. It follows by Fubini theorem that

\displaystyle \begin{array}{rcl} \int_{\mathbb{R}^{d} } \vert T_{\sigma(2^{j}\cdot)}f(x)\vert^{2}u(x)dx &\leq& \Vert \sigma \Vert_{L^{2}_{s}}^{2}\int_{\mathbb{R}^{d}}\vert f(y)\vert^{2}\int_{\mathbb{R}^{d}}\varphi_{2^{j}}(x-y)u(x)dxdy\\ \\ &\leq& \Vert \sigma \Vert_{L^{2}_{s}}^{2}\int_{\mathbb{R}^{d}}\vert f(y)\vert^{2}\varphi_{2^{j}}\ast u(y)dy\\ \\ &\leq& C \Vert \sigma \Vert_{L^{2}_{s}}^{2}\int_{\mathbb{R}^{d}}\vert f(y)\vert^{2}Mu(y)dy. \end{array}

The last inequality is obtaind from splitting {\mathbb{R}^{d}} into lattices {I_{q}} of length {\sqrt{\lambda}}, see figure below.

Indeed,

\displaystyle \begin{array}{rcl} \varphi_{\sqrt{\lambda}}\ast u(y) &=&\int_{\mathbb{R}^{d}}\varphi_{\sqrt{\lambda}}(x-y) u(y)dy\\ \\ &=&\sum_{q\in\mathbb{Z}^{d}}\int_{I_{q}}\varphi_{\sqrt{\lambda}}(x-y)u(y)dy\\ \\ &=&\sum_{q\in\mathbb{Z}^{d}}\lambda^{-\frac{d}{2}}\int_{I_{q}}\varphi(\frac{x-y}{\sqrt{\lambda}})u(y)dy\\ \\ &=&\sum_{q\in\mathbb{Z}^{d}}\frac{1}{vol(I_{q})}\int_{I_{q}}\varphi(\frac{x-y}{\sqrt{\lambda}})u(y)dy \end{array}

where {q_i} is a integer such that {y_i=x_i+\sqrt{\lambda}q_i}. Setting {I_{q}=\prod_{i=1}^{d}[q_i\sqrt{\lambda},(1+q_i)\sqrt{\lambda}]} it leads us to {vol(I_q)=\lambda^{\frac{d}{2}}}. By definition of {\varphi} and (21), it follows that

\displaystyle \begin{array}{rcl} \int_{\mathbb{R}^{d}}\varphi_{\sqrt{\lambda}}(x-y) u(y)dy&\leq& \sum_{q\in\mathbb{Z}^{d}}\frac{1}{(1+\vert q\vert ^{2})^{s}}Mu(y)\\ \\ &\leq& \sum_{q\in\mathbb{Z}^{d}}\frac{1}{\vert q\vert^{2s}}Mu(y) \end{array}

the last {2s-}series converges, because of {2s>d\geq 1}. \Box

The next theorem is a fractional variant of Hörmander multiplier theorem.

Theorem 7 Let {\sigma_{j}(\xi)=\widehat{\psi}_{j}(\xi)\sigma(\xi)} be such that

\displaystyle \sup_{j\in\mathbb{Z}}\Vert \sigma_{j}\Vert_{L^2_{s}(\mathbb{R}^{d})}<\infty \ \ \ \ \ (22)

for {s>d/2} and let {1<p<\infty}. Then the operator {T_{\sigma}} is a multiplier on {L^{p}(\mathbb{R}^{d})}.

Proof: From Littlewood-Paley inequality (14),

\displaystyle \begin{array}{rcl} \Vert T_{\sigma}f\Vert_{L^{p}(\mathbb{R}^{d})} &\leq& C\Vert \Delta_{j}T_{\sigma}f\Vert_{L^p(\mathbb{R}^{d}\rightarrow l^{2}(\mathbb{Z}))}\\ \\ &=&C\Vert \tilde{\Delta}_{j}\Delta_{j}T_{\sigma}f\Vert_{L^p(\mathbb{R}^{d}\rightarrow l^{2}(\mathbb{Z}))}\\ \\ &=&C\Vert \Delta_{j}T_{\sigma}\tilde{\Delta}_{j}f\Vert_{L^p(\mathbb{R}^{d}\rightarrow l^{2}(\mathbb{Z}))}. \end{array}

Thus,

\displaystyle \Vert T_{\sigma}f\Vert_{L^{p}(\mathbb{R}^{d})}\leq C\Vert \Delta_{j}T_{\sigma}g_j\Vert_{L^p(\mathbb{R}^{d}\rightarrow l^{2}(\mathbb{Z}))} \ \ \ \ \ (23)

where {g_j=\tilde{\Delta}_{j}f} and {\tilde{\Delta}_{j}} is a Littlewood-Paley operator such that {\Delta_j\tilde{\Delta}_{j}=\tilde{\Delta}_{j}\Delta_j=\Delta_j} given in paragraphs above. Now, we write

\displaystyle \Vert \Delta_{j}T_{\sigma}g_j\Vert_{L^p(\mathbb{R}^{d}\rightarrow l^{2}(\mathbb{Z}))}^{2}=\Vert \left(\Sigma_{j}\vert \Delta_{j}T_{\sigma}g_j\vert^{2}\right)^{\frac{1}{2}}\Vert_{L^p(\mathbb{R}^{d})}^{2}=\Vert \Sigma_{j}\vert \Delta_{j}T_{\sigma}g_j\vert^{2}\Vert_{L^{\frac{p}{2}}(\mathbb{R}^{d})}. \ \ \ \ \ (24)

If {p>2}, by Riesz representation theorem there exists {r} satisfying {\frac{1}{r}+\frac{2}{p}=1} and {u\in L^{r}(\mathbb{R}^{d})} for which

\displaystyle \Vert \Sigma_{j}\vert \Delta_{j}T_{\sigma}g_j\vert^{2}\Vert_{L^{\frac{p}{2}}(\mathbb{R}^{d})}=\sup_{\Vert u\Vert_{L^r}\leq 1}\int_{\mathbb{R}^{d}} \Sigma_{j}\vert \Delta_{j}T_{\sigma}g_j(x)\vert^{2}\, u(x)dx. \ \ \ \ \ (25)

As the symbol of {\Delta_{j}T_{\sigma}} is {\sigma_{j}=\psi_{j}(\xi)\sigma(\xi)=\psi(\tilde{\xi})\sigma(2^{j}\tilde{\xi})} and {\sigma_j\in L^2_s(\mathbb{R}^{d})}, for all {j\in\mathbb{Z}}. The Lemma 6 gives

\displaystyle \Vert \Sigma_{j}\vert \Delta_{j}T_{\sigma}g_j\vert^{2}\Vert_{L^{\frac{p}{2}}(\mathbb{R}^{d})}\leq C\Vert \sigma_j\Vert_{L^2_{s}(\mathbb{R}^{d})}\sup_{\Vert u\Vert_{L^r}\leq 1}\int_{\mathbb{R}^{d}} \Sigma_{j}\vert g_j(x)\vert^{2}\, Mu(x)dx. \ \ \ \ \ (26)

As {1<r<\infty}, by Hardy-Littlewood theorem and Hölder’s inequality we get

\displaystyle \begin{array}{rcl} \Vert \Sigma_{j}\vert \Delta_{j}T_{\sigma}g_j\vert^{2}\Vert_{L^{\frac{p}{2}}(\mathbb{R}^{d})} &\leq& C\Vert \sigma_j\Vert_{L^2_{s}(\mathbb{R}^{d})}\sup_{\Vert u\Vert_{L^r}\leq 1}\Vert \Sigma_{j}\vert g_j\vert^{2}\Vert_{L^{\frac{p}{2}}(\mathbb{R}^{d})} \Vert Mu\Vert_{L^{r}(\mathbb{R}^{d})}\\ \\ &\leq& C\Vert \sigma_j\Vert_{L^2_{s}}\Vert \Sigma_{j}\vert \tilde{\Delta}_{j}f\vert^{2}\Vert_{L^{\frac{p}{2}}}\\ \\ &=&C\Vert \sigma_j\Vert_{L^2_{s}}\Vert \tilde{\Delta}_{j}f\Vert_{L^p(\mathbb{R}^{d}\rightarrow l^{2}(\mathbb{Z}))}^{2}\\ \\ &\leq& C\Vert \sigma_j\Vert_{L^2_{s}}\Vert f\Vert_{L^p(\mathbb{R}^{d})}^{2}, \end{array}

because the Littlewood-Paley inequality (13) still hold with {\tilde{\Delta}_{j}} in place of {\Delta_j}. Inserting the last inequality into (23) we have

\displaystyle \Vert T_{\sigma}f\Vert_{L^{p}(\mathbb{R}^{d})}\leq C\Vert \sigma_j\Vert_{L^2_{s}(\mathbb{R}^{d})}\Vert f\Vert_{L^p(\mathbb{R}^{d})} \ \ \ \ \ (27)

for all {p>2}. The case {p<2} follows by standard duality argument and shall be remarked only that the symbol of form adjuint operator {T^{\ast}_{\lambda}} is {\lambda(\xi)=\sigma(-\xi)} and hence is bounded by arguments above. The case {p=2} it’s a small exercise to reader. \Box

In the next post shall be showed that fractional condition (22) implies so-called symbol Hörmander condition in Theorem 2.5 and also Minhklin condition for {d>1}. Some small applications to PDEs equations will be mentioned finally.

2 thoughts on “A fractional Hörmander type multiplier theorem

  1. Pingback: Heat-wave equation in Morrey spaces | Being simple

  2. Pingback: Hormander-Minlin | Being simple

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