Fourier Multipliers and Sobolev Characterisations

Assume Countable Choice and let $n\ge 1$. The Fourier transform is the negative-sign $2\pi$-normalized transform of Fourier transform of a tempered distribution: on a Schwartz function it is the integral transform $\widehat{f}(\xi )={\int }_{{\mathbb{R}}^n}f(x)e^{-2\pi ix\cdot \xi }dx$, and on ${\mathcal{S}}^{\mathrm{\prime }}({\mathbb{R}}^n)$ it is its bilinear transpose, an automorphism with inverse ${\mathcal{F}}^{-1}$ (Fourier transform is a topological automorphism of tempered distributions).

Fix a measurable symbol $m:{\mathbb{R}}^n\to \mathbb{C}$. Define its Schwartz domain $D_m=\{f\in \mathcal{S}({\mathbb{R}}^n):m\widehat{f}\text{ is locally integrable and its regular distribution is tempered}\}\mathrm{.}$ Here $m\widehat{f}$ is the ordinary pointwise product of the measurable symbol with the Schwartz function $\widehat{f}$. For a locally integrable function $g$, its regular distribution initially means $\phi \mapsto {\int }_{{\mathbb{R}}^n}g(\xi )\phi (\xi )d\xi$ on $C_c^{\mathrm{\infty }}({\mathbb{R}}^n)$. Saying that this distribution is tempered means that this functional extends continuously to $\mathcal{S}({\mathbb{R}}^n)$. The extension is unique because $C_c^{\mathrm{\infty }}$ is dense in $\mathcal{S}$ (Smooth compact supports are dense in Schwartz space); we denote it by $u_g$. Local integrability alone does not guarantee an absolutely convergent integral against every Schwartz test. For $f\in D_m$ the extension exists by hypothesis, and we set $T_{m}f:={\mathcal{F}}^{-1}\left(u_{m\widehat{f}}\right)\in {\mathcal{S}}^{\mathrm{\prime }}({\mathbb{R}}^n),T_m:\mathcal{S}\supseteq D_m\to {\mathcal{S}}^{\mathrm{\prime }}\mathrm{.}$ The domain qualification is part of the definition: no boundedness of $T_m$, no density of $D_m$ in $\mathcal{S}$, no continuity of $m\mapsto T_m$ and no action of $T_m$ on $L^p$ is asserted. In particular this definition does not assert that $D_m=\mathcal{S}$; the zero function always belongs to $D_m$.

Translation. For $u\in {\mathcal{S}}^{\mathrm{\prime }}({\mathbb{R}}^n)$ and $a\in {\mathbb{R}}^n$ define the translate ${\tau }_{a}u$ by transposition, $⟨{\tau }_{a}u,\phi ⟩:=⟨u,\phi (\cdot +a)⟩,\phi \in \mathcal{S}({\mathbb{R}}^n)\mathrm{.}$ This is a tempered distribution: translation is a continuous complex-linear endomorphism of $\mathcal{S}({\mathbb{R}}^n)$ (Basic operations are continuous on Schwartz space), so the composition is a continuous complex-linear functional (Tempered distribution). On functions, ${\tau }_{a}f(x)=f(x-a)$ for $a\in {\mathbb{R}}^n$.

Translation invariance of the domain and of the operator. If $f\in D_m$ and $a\in {\mathbb{R}}^n$, then ${\tau }_{a}f\in D_m$ and $T_m{\tau }_{a}f={\tau }_a{T}_{m}f\mathrm{.}$ Justification. Write $e_{-a}(\xi )=e^{-2\pi ia\cdot \xi }$. First, $\mathcal{F}({\tau }_{a}f)=e_{-a}\widehat{f}$: the integral formula is the translation law for the $L^1$ transform, applicable because Schwartz functions are integrable (Schwartz derivatives are integrable, Translation, modulation, linear dilation and reflection laws), and the distributional transform of the integrable function ${\tau }_{a}f$ is the regular distribution of its integral transform (Fourier transform agrees with l one and plancherel transforms). Second, with $g=m\widehat{f}$, the product $m\mathcal{F}({\tau }_{a}f)=e_{-a}g$ is locally integrable, and its regular distribution satisfies $⟨u_{e_{-a}g},\phi ⟩=⟨u_g,e_{-a}\phi ⟩$ for every compactly supported smooth test $\phi$; since $e_{-a}$ is smooth with polynomially bounded derivatives, Smooth polynomially bounded multipliers on schwartz space makes $e_{-a}{u}_g$ a tempered extension of the regular distribution of $e_{-a}g$. Uniqueness of extension gives $u_{e_{-a}g}=e_{-a}{u}_g$ on all Schwartz tests. Hence ${\tau }_{a}f\in D_m$. Third, for $u\in {\mathcal{S}}^{\mathrm{\prime }}$ one has $\mathcal{F}({\tau }_{a}u)=e_{-a}\mathcal{F}u$: pairing both sides with a Schwartz test $\phi$ and using the definition of $\mathcal{F}$ on ${\mathcal{S}}^{\mathrm{\prime }}$ gives $⟨\mathcal{F}({\tau }_{a}u),\phi ⟩=⟨{\tau }_{a}u,\mathcal{F}\phi ⟩=⟨u,{\tau }_{-a}\mathcal{F}\phi ⟩=⟨u,\mathcal{F}(e_{-a}\phi )⟩=⟨e_{-a}\mathcal{F}u,\phi ⟩,$ where the middle identity is the modulation law $\mathcal{F}(e^{-2\pi ia\cdot x}\phi )(\xi )=\widehat{\phi }(\xi +a)$ of Translation, modulation, linear dilation and reflection laws and the last identity is the definition of multiplication of a distribution by the smooth symbol $e_{-a}$ (Smooth polynomially bounded multipliers on schwartz space). Applying this to $u=T_{m}f$ and combining the three computations, $\mathcal{F}(T_m{\tau }_{a}f)=u_{e_{-a}g}=e_{-a}{u}_g=e_{-a}\mathcal{F}(T_{m}f)=\mathcal{F}({\tau }_a{T}_{m}f)\mathrm{.}$ Injectivity of $\mathcal{F}$ on ${\mathcal{S}}^{\mathrm{\prime }}$ now gives $T_m{\tau }_{a}f={\tau }_a{T}_{m}f$.

The Countable Choice hypothesis is inherited only from the cited Fourier interfaces; the transposition defining ${\tau }_a$ uses none.

Assume Countable Choice and let $n\ge 1$. Let $m:{\mathbb{R}}^n\to \mathbb{C}$ be measurable with finite essential supremum $M:=\mathrm{\parallel }m{\mathrm{\parallel }}_{\mathrm{\infty }}=\mathrm{ess\hspace{0.17em}sup}\mathrm{\mid }m\mathrm{\mid }<\mathrm{\infty }$. Then:

1. $\mathcal{S}({\mathbb{R}}^n)\subseteq D_m$, and for every Schwartz class $f$ the tempered distribution $T_{m}f$ of Translation-invariant Fourier multiplier on the Schwartz core is the regular distribution of the $L^2$ class ${\mathcal{F}}_2^{-1}(m\cdot {\mathcal{F}}_{2}f)$, so as $L^2$ classes $T_{m}f={\mathcal{F}}_2^{-1}(m{\mathcal{F}}_{2}f)$ and $\mathrm{\parallel }{T}_{m}f{\mathrm{\parallel }}_2\le M\mathrm{\parallel }f{\mathrm{\parallel }}_2$.
2. The Schwartz-core action extends uniquely to a bounded operator $T_m:L^2({\mathbb{R}}^n;\mathbb{C})\to L^2({\mathbb{R}}^n;\mathbb{C})$, namely $T_m={\mathcal{F}}_2^{-1}{M}_m{\mathcal{F}}_2$ with $M_{m}g=mg$ the multiplication operator, and its operator norm is exactly $\mathrm{\parallel }{T}_m{\mathrm{\parallel }}_{L^2\to L^2}=M={\mathrm{ess\hspace{0.17em}sup}}_{{\mathbb{R}}^n}\mathrm{\mid }m\mathrm{\mid }\mathrm{.}$
3. The operator depends only on the almost-everywhere class of $m$: if $m=m^{\mathrm{\prime }}$ almost everywhere then the two operators on $L^2$ agree. In particular the values of $m$ on Lebesgue-null sets, including the single point $\{0\}$, do not affect the operator or its norm.

This is an $L^2$ statement only: no $L^p$ boundedness for $p\mathrm{\ne }2$ is asserted, and $m$ is not assumed continuous, smooth, or polynomially bounded.

Assume Countable Choice and let $n\ge 1$. Let $m:{\mathbb{R}}^n\to \mathbb{C}$ be measurable, with Schwartz domain $D_m$ and operator $T_m$ as in Translation-invariant Fourier multiplier on the Schwartz core, and fix $1\le p<\mathrm{\infty }$.

Definition of an $L^p$ Fourier multiplier. The symbol $m$ is an $L^p$ Fourier multiplier when:

1. $\mathcal{S}({\mathbb{R}}^n)\subseteq D_m$, so $T_m$ is defined on all of Schwartz space;
2. for every $f\in \mathcal{S}({\mathbb{R}}^n)$ the tempered distribution $T_{m}f$ is the regular distribution of some class in $L^p({\mathbb{R}}^n;\mathbb{C})$, in the conventions of Complex Lp classes and Euclidean test-function conventions;
3. there is a finite constant $C$ with $\mathrm{\parallel }{T}_{m}f{\mathrm{\parallel }}_{L^p}\le C\mathrm{\parallel }f{\mathrm{\parallel }}_{L^p}$ for every $f\in \mathcal{S}({\mathbb{R}}^n)$, where the norm on the left is that of the uniquely determined $L^p$ class representing $T_{m}f$.

Condition 2 is meaningful because the regular-distribution map is injective on locally integrable classes (Locally integrable functions embed in distributions), so the $L^p$ class representing $T_{m}f$ is unique. The set of admissible constants $C$ in condition 3 is nonempty by hypothesis and bounded below by $0$.

The multiplier norm. Assume $m$ is an $L^p$ multiplier. Then the assignment $f\mapsto T_{m}f$ is a complex-linear map from the dense subspace $\mathcal{S}({\mathbb{R}}^n)$ of $L^p({\mathbb{R}}^n;\mathbb{C})$ into the Banach space $L^p({\mathbb{R}}^n;\mathbb{C})$: the compactly supported smooth functions are contained in Schwartz space and are dense in $L^p$ for finite $p$ (Complex finite-simple and smooth compact-support density for finite p), and $L^p$ is complete (Complex completeness, density, and inner product: the consumer interface). By A bounded linear map from a dense normed subspace into a Banach space extends uniquely with the same norm there is a unique bounded linear $\widetilde{T_m}:L^p({\mathbb{R}}^n;\mathbb{C})\to L^p({\mathbb{R}}^n;\mathbb{C})$ extending it, with $\mathrm{\parallel }\widetilde{T_m}\mathrm{\parallel }=\inf \{C:\mathrm{\parallel }{T}_{m}f{\mathrm{\parallel }}_p\le C\mathrm{\parallel }f{\mathrm{\parallel }}_p\text{ for all }f\in \mathcal{S}({\mathbb{R}}^n)\}\mathrm{.}$ We write $T_m$ for this extension as well and define the multiplier norm $\mathrm{\parallel }m{\mathrm{\parallel }}_{M_p}:=\mathrm{\parallel }{T}_m{\mathrm{\parallel }}_{L^p\to L^p}=\mathrm{\parallel }\widetilde{T_m}\mathrm{\parallel }\mathrm{.}$ The infimum is a minimum, attained by the operator norm of the extension. This definition asserts nothing about which symbols are multipliers: no Mihlin type condition, no endpoint $p=1$ or $p=\mathrm{\infty }$ boundedness claim, and no algebraic property of the set $M_p=\{m:\mathrm{\parallel }m{\mathrm{\parallel }}_{M_p}<\mathrm{\infty }\}$ is stated here.

The case $p=\mathrm{\infty }$. At $p=\mathrm{\infty }$ the Schwartz core is not dense. A Schwartz function satisfies $\mathrm{\mid }\phi (x)\mathrm{\mid }\le C_N(1+\mathrm{\mid }x\mathrm{\mid }{)}^{-N}$ for every $N$, because $(1+\mathrm{\mid }x{\mathrm{\mid }}^2{)}^N$ is a finite combination of monomials and each $\mathrm{\mid }{x}^{\alpha }\phi (x)\mathrm{\mid }$ is a finite Schwartz seminorm (Schwartz space and its seminorms), so every Schwartz class has a $C_0$ representative; the $L^{\mathrm{\infty }}$-closure of $C_c({\mathbb{R}}^n;\mathbb{C})$ is exactly the set of classes with a $C_0$ representative, and it does not contain the class of the constant function $1$ (Complex finite-simple and smooth compact-support density for finite p). Hence uniqueness of a bounded extension of the Schwartz-core action cannot be inferred, and this definition attaches no intrinsic norm $\mathrm{\parallel }m{\mathrm{\parallel }}_{M_{\mathrm{\infty }}}$ to the core action alone. Whether a chosen bounded extension exists on $L^{\mathrm{\infty }}$, and which one is intended, are separate specification decisions not made here.

Assume Countable Choice and let $\mathbb{T}=\mathbb{R}\mathrm{/}\mathbb{Z}$ carry its normalized Haar measure $m_{\mathbb{T}}$, so that $m_{\mathbb{T}}(\mathbb{T})=1$. Let $1\le p\le 2$ and let $p^{\mathrm{\prime }}$ be the conjugate exponent, $1\mathrm{/}p+1\mathrm{/}{p}^{\mathrm{\prime }}=1$. Every complex class $f\in L^p(\mathbb{T};\mathbb{C})$ has a Fourier coefficient sequence $(\widehat{f}(k){)}_{k\in \mathbb{Z}}$ in ${\mathrm{\ell }}^{p^{\mathrm{\prime }}}(\mathbb{Z};\mathbb{C})$ and $({\sum }_{k\in \mathbb{Z}}\mathrm{\mid }\widehat{f}(k){\mathrm{\mid }}^{p^{\mathrm{\prime }}}{)}^{1\mathrm{/}{p}^{\mathrm{\prime }}}\le \mathrm{\parallel }f{\mathrm{\parallel }}_{L^p(\mathbb{T})},1\le p<2,$ with the supremum reading at $p=1$, ${\sup }_{k\in \mathbb{Z}}\mathrm{\mid }\widehat{f}(k)\mathrm{\mid }\le \mathrm{\parallel }f{\mathrm{\parallel }}_{L^1(\mathbb{T})},$ and at $p=2$ the Parseval equality $({\sum }_{k\in \mathbb{Z}}\mathrm{\mid }\widehat{f}(k){\mathrm{\mid }}^2{)}^{1\mathrm{/}2}=\mathrm{\parallel }f{\mathrm{\parallel }}_{L^2(\mathbb{T})}\mathrm{.}$ The Fourier coefficients are those of Fourier coefficients and trigonometric polynomials on the torus. No reverse inequality for $p>2$ and no endpoint statement beyond $p=1,2$ is claimed.

Assume Countable Choice, let $n\ge 1$, and let $\widehat{f}(\xi )={\int }_{{\mathbb{R}}^n}f(x)e^{-2\pi ix\cdot \xi }dx$ be the integral Fourier transform on $L^1({\mathbb{R}}^n;\mathbb{C})$. For $1\le p\le 2$ with conjugate exponent $p^{\mathrm{\prime }}$, the transform extends compatibly to a complex-linear bounded map ${\mathcal{F}}_p:L^p({\mathbb{R}}^n;\mathbb{C})⟶L^{p^{\mathrm{\prime }}}({\mathbb{R}}^n;\mathbb{C}),\mathrm{\parallel }{\mathcal{F}}_{p}f{\mathrm{\parallel }}_{p^{\mathrm{\prime }}}\le \mathrm{\parallel }f{\mathrm{\parallel }}_p\mathrm{.}$ At $p=1$ this map is the integral transform of The L1 transform is bounded and uniformly continuous, at $p=2$ it is the unitary Plancherel transform of Plancherel theorem, and on $L^1\cap L^2$ the two interpretations agree almost everywhere (Agreement of the integral and L2 transforms). For $1<p<2$ the extension agrees almost everywhere with the integral transform on its intersection with $L^1$ and with the Plancherel transform on its intersection with $L^2$. No statement for $p>2$ and no pointwise representative identity is claimed.

Assume Countable Choice and let $n\ge 1$. Put $q:=\lfloor n\mathrm{/}2\rfloor +1.$ A measurable $m:{\mathbb{R}}^n\to \mathbb{C}$ is a Mihlin symbol in this convention when there is a function $m_0\in C^q({\mathbb{R}}^n\setminus \{0\})$ such that $m=m_0$ Lebesgue almost everywhere and there are constants $C_{\alpha }\ge 0$, indexed by the multi-indices $\alpha$ of $C^k$ maps and multi-index derivative notation in Euclidean space with $\mathrm{\mid }\alpha \mathrm{\mid }\le q$, for which $\mathrm{\mid }{\mathrm{\partial }}^{\alpha }{m}_0(\xi )\mathrm{\mid }\le C_{\alpha }\mathrm{\mid }\xi {\mathrm{\mid }}^{-\mathrm{\mid }\alpha \mathrm{\mid }}(\xi \mathrm{\ne }0,\mathrm{\mid }\alpha \mathrm{\mid }\le q)\mathrm{.}$ The derivatives are taken in the punctured open set ${\mathbb{R}}^n\setminus \{0\}$; the value $m(0)$ is not constrained.

Consequences recorded here. The case $\alpha =0$ gives $\mathrm{\mid }{m}_0(\xi )\mathrm{\mid }\le C_0$ for every $\xi \mathrm{\ne }0$, so $\mathrm{\mid }m\mathrm{\mid }\le C_0$ Lebesgue almost everywhere, because the singleton $\{0\}$ is Lebesgue null (Every at most countable subset of ${\mathbb{R}}^n$ is Lebesgue null; in particular ${\lambda }_1(\mathbb{Q})=0$); thus a Mihlin symbol is essentially bounded and $\mathrm{\parallel }m{\mathrm{\parallel }}_{\mathrm{\infty }}=\mathrm{ess\hspace{0.17em}sup}\mathrm{\mid }m\mathrm{\mid }\le C_0$ (The essential supremum of a measurable function with respect to a measure). Consequently the exact $L^2$ multiplier lemma applies: every Schwartz function lies in the Schwartz domain of $m$, and the operator $T_m$ has a unique bounded extension to $L^2({\mathbb{R}}^n;\mathbb{C})$ of norm $\mathrm{\parallel }m{\mathrm{\parallel }}_{\mathrm{\infty }}\le C_0$ (Exact L2 Fourier multiplier norm). This definition is a sufficient symbol condition: $L^p({\mathbb{R}}^n)$ boundedness for $1<p<\mathrm{\infty }$ is a separate theorem, assigned in this library to the later Mihlin multiplier theorem built on singular-integral estimates, and no such boundedness is asserted here.

Derivative count. The count $q=\lfloor n\mathrm{/}2\rfloor +1$ is the classical "more than half the dimension" requirement used by multiplier theory. Grafakos assumes $m_0\in C^{[n\mathrm{/}2]+1}$ away from the origin with the pointwise inequalities above and derives the annular estimates used in his proof; Exact L2 Fourier multiplier norm supplies only the $L^2$ part of that theorem. Williams states the same conclusion under the stronger count $\mathrm{\mid }\alpha \mathrm{\mid }\le d+2$, so his theorem does not reduce the derivative count adopted here. The bounds are required for $\xi \mathrm{\ne }0$ only: homogeneity of order zero near the origin, such as the signum symbol in one dimension, is compatible with the condition, while a jump at a nonzero frequency is not, since $C^q$ functions on the punctured space are continuous there.

Assume Countable Choice and let $n\ge 1$. For every $u\in {\mathcal{S}}^{\mathrm{\prime }}({\mathbb{R}}^n)$ and every multi-index $\alpha \in {\mathbb{N}}_0^n$, $\mathcal{F}(D^{\alpha }u)=(2\pi i\xi {)}^{\alpha }\mathcal{F}u\text{in }{\mathcal{S}}^{\mathrm{\prime }}({\mathbb{R}}^n),$ where $D^{\alpha }$ is the distributional derivative of $C^k$ maps and multi-index derivative notation in Euclidean space and $\mathcal{F}$ is the negative-sign $2\pi$-normalized transform. Moreover, if $u=u_f$ and $D^{\alpha }u=u_g$ for $L^2$ classes $f,g\in L^2({\mathbb{R}}^n)$ and their regular distributions, then the unitary Plancherel transforms satisfy ${\mathcal{F}}_{2}g(\xi )=(2\pi i\xi {)}^{\alpha }{\mathcal{F}}_{2}f(\xi )\text{for almost every }\xi \in {\mathbb{R}}^n\mathrm{.}$ The second assertion compares the Plancherel classes only; it neither asserts pointwise values of arbitrary representatives nor presupposes any Sobolev-space notation.