Lp Fourier multiplier and its norm

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.