Translation invariant operators

In this notes we show that bounded translation invariant operators from {L^{p}(\mathbb{R}^{d})} to {L^{q}(\mathbb{R}^{d})} are essentially convolution singular integral operators for certain p,q. More precisely, let {A} be a bounded linear operator from {L^{p}(\mathbb{R}^{d})} to {L^{q}(\mathbb{R}^{d})}, we say that {A} is a translation invariant operator, if

\displaystyle \tau_{y}A=A\tau_{y} \ \ \ \ \ (1)

where {(\tau_{y}\varphi)(x)=\varphi(x-y)} denotes the translation operator on , \mathbb{R}^{d} and \varphi is a  Schwartz function. The object of this notes is only find a tempered distribution {T\in \mathcal{S}'(\mathbb{R}^{d})} such that

\displaystyle (A\varphi)(x) =(\varphi\ast T)(x), \ \ \ \ \ (2)

if {A} is translation invariant operator and {A} is bounded on certain Lebesgue {L^{q}(\mathbb{R}^{d})} spaces. Firstly, let us recall some well know results.

  • Let {\alpha=(\alpha_1,\cdots,\alpha_d)\in (\mathbb{N}\cup\{0\}})^{d}, {x^{\alpha}=x_1^{\alpha_1}x_2^{\alpha_2}\cdots x_d^{\alpha_d}}. Then there exist a positive constant {C_{d,\alpha}} such that

    \displaystyle \vert x^{\alpha}\vert \leq C_{d,\alpha}\vert x\vert^{\vert \alpha\vert}, \;\; \vert \alpha\vert =\alpha_1+\cdots+\alpha_{d}. \ \ \ \ \ (3)

    In fact, let {\mathbb{S}^{d-1}} be an unity sphere and let {f:\mathbb{S}^{d-1}\rightarrow \mathbb{R}} which is given by  {f(x_1,\cdots,x_d)=\vert x_{1}^{\alpha_1}\cdots x_{d}^{\alpha_d}\vert}. Then the function {f} is a continuous function in {\mathbb{R}^{d}} and for

    \displaystyle C_{d,\alpha}=\sup_{x\in \mathbb{S}^{d-1}}\vert f(x)\vert,\nonumber \ \ \ \ \ (4)

    we get {\vert y^{\alpha}\vert \leq C_{d,\alpha}}, for all {y=\frac{x}{\vert x\vert}\in \mathbb{S}^{d-1}} and {x\in\mathbb{R}^{d}}. Since

    \displaystyle y^{\alpha}=\left(\frac{x_{1}}{\vert x\vert}\right)^{\alpha_1}\cdots \left(\frac{x_{d}}{\vert x\vert}\right)^{\alpha_d}=\frac{x_1^{\alpha_1}x_2^{\alpha_2}\cdots x_d^{\alpha_d}}{\vert x\vert^{\alpha_1+\cdots+\alpha_{d}}},\nonumber \ \ \ \ \ (5)

    we obtain (3).

  • Let {k\in \mathbb{Z}^{+}}, we have

    \displaystyle \vert x\vert^{k} \leq C_{d,k}\sum_{\vert \beta\vert=k}\vert x^{\beta}\vert, \ \ \ \ \ (6)

    for all {x\in \mathbb{R}^{d}}.

  • \displaystyle \frac{C_{d, d+1}}{(2\pi) ^{d+1}}\sum_{\vert \alpha\vert \leq d+1}\vert (-2\pi i x)^{\alpha}\vert \geq (1+\vert x\vert)^{-d-1}. \ \ \ \ \ (8)

    Indeed, by inequality (6) we have

    \displaystyle \begin{array}{rcl} \frac{C_{d, d+1}}{(2\pi)^{d+1}}\sum_{\vert \alpha\vert \leq d+1}\vert (-2\pi x)^{\alpha}\vert \geq \vert x\vert ^{d+1}\geq \left(\frac{1}{2}\right)^{d+1}\geq (1+\vert x\vert)^{-d-1}. \end{array}

Lemma 1 Let {u} be a tempered distribution. Suppose that {D^{\alpha} u\in L^{p}_{loc}(\mathbb{R}^{d})} for all {\vert \alpha\vert \leq d+1}. Then {u} is a function and there exists a positive constant {C} such that

\displaystyle \vert u(x)\vert \leq C\sum_{\vert \alpha\vert \leq d+1}\left(\int_{\mathbb{R}^{d}}\vert D^{\alpha}u(y)\vert^{p}dy\right)^{\frac{1}{p}}, \;\;\vert x\vert\geq 2. \ \ \ \ \ (9)

Proof: Let {C_{0}^{\infty}(\mathbb{R}^{d})} be the space of all bump functions {\varphi_{r}:\mathbb{R}^{d}\rightarrow\mathbb{R}} with support on  euclidean sphere {\mathbb{S}_r} of radius {r}. Since {u\in L^{p}(\mathbb{R}^{d})}, it follows from Hölder inequality that {w:=\varphi_{r}u\in L^{1}(\mathbb{R}^{d})}. Moreover, we have that  {\hat{w}} lives in {L^{1}(\mathbb{R}^{d})} too. This follows by means of the inequality (8) and some properties of Fourier transform on euclidean spaces. Indeed,

\displaystyle \begin{array}{rcl} \vert\widehat{w}(x)\vert &\leq & C_{d}(1+\vert x\vert)^{-(d+1)}\sum_{\vert\alpha\vert\leq d+1}\vert (-2\pi i x)^{\alpha}\widehat{w}(x)\vert\\    \\    &\leq &C_{d}(1+\vert x\vert)^{-(d+1)}\sum_{\vert\alpha\vert\leq d+1}\Vert\widehat{D^{\alpha}w}\Vert_{\infty}\\    \\    &\leq &C_{d}(1+\vert x\vert)^{-(d+1)}\sum_{\vert\alpha\vert\leq d+1}\Vert D^{\alpha}w\Vert_{L^{1}(\mathbb{R}^{d})}\\    \\    &\leq &C_{d}(1+\vert x\vert)^{-(d+1)}\sum_{\vert\alpha\vert\leq d+1}\Vert \sum_{\gamma\leq \alpha}\binom{\alpha}{\gamma}(D^{\gamma}u)(D^{\alpha-\gamma}\varphi_{r})\Vert_{L^{1}}\\    \\    &\leq &C_{d,\alpha, p'}(1+\vert x\vert)^{-(d+1)}\sum_{\vert\alpha\vert\leq d+1}\Vert D^{\alpha}u\Vert_{L^{p}}. \end{array}

Integrating the last inequality, we obtain

\displaystyle \Vert \widehat{w}\Vert_{L^{1}}\leq C\sum_{\vert\alpha\vert\leq d+1}\Vert D^{\alpha}u\Vert_{L^{p}(\mathbb{R}^{d})}.\nonumber \ \ \ \ \ (10)

It follows that Fourier inverse of {\widehat{w}} agrees, almost everywhere, with an uniformly continuous function (see Folland p. 239). Since {\varphi_{r}(x)=1} on {\mathbb{S}_r}, so the tempered distribution {u} can be re-defined as an uniformly continuous function, namely {\psi}, which is defined on {B_{r}(y)} for all {r>0}. Since the choice of {r} is not relevant, we get

\displaystyle \begin{array}{rcl} \Vert u\Vert_{\infty}=\Vert \psi \Vert_{L^{\infty}}\leq \Vert \widehat{\psi}\Vert_{L^{1}}= \Vert \widehat{w}\Vert_{L^{1}}\leq C\sum_{\vert\alpha\vert\leq d+1}\Vert D^{\alpha}u\Vert_{L^{p}(\mathbb{R}^{d})}. \end{array}


Lemma 2 Let {u\in\mathcal{S}'(\mathbb{R}^{d})} and let  {D^{\alpha}u\in L^{p}(\mathbb{R}^{d})}. If {A:L^{p}(\mathbb{R}^{d})\rightarrow L^q(\mathbb{R}^{d})} is a bounded linear operator which is translation invariant, then {A} commutes with derivatives, that is,

\displaystyle A(D^{\alpha}u)=D^{\alpha}(Au), \ \ \ \ \ (11)

for all multi-index {\alpha}.

Proof: For readers convenience, let {\alpha=e_j:=(0,\cdots,0, 1,0,\cdots,0)}. The general case, follows by interactions. Firstly, let {h\in\mathbb{R}} and set

\displaystyle u_{h}(x):=\tau_{-he_j}(x)=u(x_1,\cdots, x_{j}+h,\cdots,x_{d}) .\nonumber \ \ \ \ \ (12)

Let v=Au and note that Au_{h}=A\tau_{-he_j}=\tau_{-he_j}A=v_{h}. It follows that,

\displaystyle A(\frac{u_h-u}{h})=\frac{v_h-v}{h}\nonumber \ \ \ \ \ (14)

which converges to {D_{x_j}v} in {L^q(\mathbb{R}^{d})} as {h\rightarrow 0}. The continuity of {A} in join with {D_{x_j} u\in L^{p}(\mathbb{R}^{d})} give us

\displaystyle \begin{array}{rcl} \Vert \frac{v_h-v}{h} -A(D_{x_j} u)\Vert_{L^q(\mathbb{R}^{d})}&=&\Vert A(\frac{u_h-u}{h} -D_{x_j} u)\Vert_{L^q(\mathbb{R}^{d})}\\\\&\leq & \Vert A\Vert\, \Vert \frac{u_h-u}{h} -D_{x_j} u\Vert_{L^p(\mathbb{R}^{d})}. \end{array}

Therefore, by uniqueness of limit in {L^q(\mathbb{R}^{d})} one has

\displaystyle D_{x_j}Au =A(D_{x_j} u),\nonumber \ \ \ \ \ (15)

as we desired. \Box

Let us to introduce the reflection operator in {\mathcal{S}(\mathbb{R}^{d})} which is given by,

\displaystyle (R\varphi)(x)=\varphi(-x), \ \ \ \ \ (16)

for all {\varphi\in\mathcal{S}(\mathbb{R}^{d})}. Notice that, for {\varphi\in L^{1}_{loc}(\mathbb{R}^{d})} the integral

\displaystyle \begin{array}{rcl} (\varphi\ast u)(x)&=&\int u(y) \varphi(-(y-x)) dy\\\\ &=&\int u(y) (R\varphi)(y-x) dy\\\\ &=&\int u(y) (\tau_{x}R\varphi)(y) dy \end{array}

for all {u\in C_{0}^{\infty}(\mathbb{R}^{d})}. This motivates to  define convolution operator between a Schwartz function {\varphi\in\mathcal{S}(\mathbb{R}^{d})} and a tempered distribution {u\in\mathcal{S}'(\mathbb{R}^{d})} as follows

\displaystyle (\varphi\ast u )(x)= u(\tau_{x}R\varphi). \ \ \ \ \ (17)


Theorem 3 Let {A: L^p(\mathbb{R}^{d})\rightarrow L^q(\mathbb{R}^{d})} be a bounded linear operator. If {A} is translation invariant, there exists {T\in \mathcal{S}'(\mathbb{R}^{d})} such that (2) holds.

Proof: Let {\varphi\in\mathcal{S}(\mathbb{R}^{d})} be a Schwartz function such that {D^{\alpha}\varphi\in L^{p}(\mathbb{R}^{d})}. In view of Lemma 1 and Lemma 2, {A\varphi} is a continuous function which satisfy the estimate

\displaystyle \begin{array}{rcl} \vert (A\varphi)(0)\vert &\leq& C \sum_{\vert \alpha\vert\leq d+1} \Vert D^{\alpha} (A\varphi)\Vert_{L^{q}(\mathbb{R}^{d})}\\ \\&\leq & C\,\Vert A\Vert \sum_{\vert \alpha\vert\leq d+1} \Vert D^{\alpha}\varphi\Vert_{L^{p}(\mathbb{R}^{d})}\\ \\&\leq & C\,\Vert A\Vert \left(\int (1+\vert x\vert^{2})^{-Np}\right)^{1/p}\sum_{\vert \alpha\vert\leq d+1} \Vert \varphi\Vert_{(N,\alpha)}\\ \\&\leq & C\,\Vert A\Vert \sum_{\vert \alpha\vert\leq d+1} \Vert \varphi\Vert_{(N,\alpha)} \end{array}

for {N>d/p}. Here {\Vert \cdot\Vert_{(N,\alpha)}} stands for Schwartz norm

\displaystyle \Vert \varphi\Vert_{(N,\alpha)}=\sup_{x\in \mathbb{R}^{d}}\,(1+\vert x\vert^{2})^{N}\vert D^{\alpha}\varphi(x)\vert. \ \ \ \ \ (18)

In other words, there exists a tempered distribution {T} such that \displaystyle (A\varphi)(0)=T(R\varphi). As  {A} is translation invariance, in view of (17) we have

\displaystyle \begin{array}{rcl} (A\varphi)(x)&=&\tau_{-x}(A\varphi)(0)=(A\tau_{-x}\varphi)(0)\\ \\&=&T(R\tau_{-x}\varphi)=T(\tau_{x}R\varphi)=(\varphi\ast T)(x) \end{array}

because of {(R\tau_{x}\varphi)(z)=(\tau_{-x}R\varphi)(z)}. \\\\\\\Box


2 thoughts on “Translation invariant operators

  1. In the very last part of the post there is a typo: you say “Hence $(A\phi)(0)$ is a bounded linear functional … and so $(A\phi)(0)=T(R\phi)$ for **any** tempered distribution $T$”. The correct wording is something like “and so *there exists* a tempered distribution $T$ such that …”.

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