---
tag: 41D1KJWM78
kind: definition
name: Lp multiplier
msc: [42B15, 42A45]
defines: RAGR38Q40B
uses: [ZQEFCY61T7, DPVM34ADZV, SASP14MA1Z, CG9JEATW3Q]
sources:
  - url: https://en.wikipedia.org/wiki/Multiplier_(Fourier_analysis)
---

## Definition

Let $n \ge 1$ and $1 \le p \le \infty$.
Let $\mathcal{S}(\R^n)$ be the [[Schwartz space|CG9JEATW3Q]], let $\hat f$ be the [[Fourier transform|DPVM34ADZV]] of $f$, and let $g^\vee$ be the inverse Fourier transform of $g$.
Let $m \colon \R^n \to \C$ be a bounded measurable function, and let $T_m f = (m \hat f)^\vee$ for $f \in \mathcal{S}(\R^n)$.
The function $m$ is an *$L^p$ multiplier* if there is a constant $C$ such that
$$\norm{T_m f}_{L^p(\R^n)} \le C \norm{f}_{L^p(\R^n)} \quad \text{for all } f \in \mathcal{S}(\R^n).$$
The smallest such $C$ is the multiplier norm of $m$, and $\mathcal{M}_p$ denotes the space of $L^p$ multipliers.

## Remarks

Example: for $h \in \R^n$, let $m(\xi) = e^{2\pi i \xi \cdot h}$, with the convention $\hat f(\xi) = \int f(x) e^{-2\pi i x \cdot \xi} \, dx$.
Then $T_m f(x) = f(x + h)$, so $\norm{T_m f}_{L^p} = \norm{f}_{L^p}$ and $m \in \mathcal{M}_p$ with norm $1$ for every $p$.

Non-example: in $\R^1$, $m(\xi) = \xi^2$ is not bounded, so it is not an $L^p$ multiplier.
Indeed, for $p = 2$ the Plancherel theorem gives $\norm{T_m f}_{L^2} = \norm{\xi^2 \hat f}_{L^2}$, and its ratio to $\norm{f}_{L^2} = \norm{\hat f}_{L^2}$ is unbounded when $\hat f$ is a fixed bump moved far out.

By the Plancherel theorem, $\mathcal{M}_2 = L^\infty(\R^n)$ with equal norms.
By duality, $\mathcal{M}_p = \mathcal{M}_{p'}$ when $1/p + 1/p' = 1$.
The space $\mathcal{M}_1 = \mathcal{M}_\infty$ consists of the Fourier transforms of finite Borel measures.
By the [[Riesz–Thorin theorem|AGRKRG1TMJ]], $\mathcal{M}_p \subseteq \mathcal{M}_q$ when $q$ lies between $p$ and $2$.