Fourier multiplier ($m(\xi )$)

A Fourier multiplier is a function $m$ on frequency space used to define an operator by multiplication after taking a Fourier transform.

For a Fourier multiplier $m\in L^{\infty }({\mathbb{R}}^d)$, define $T_{m}f={\mathcal{F}}^{-1}(m\mathcal{F}f)$ using a unitary Fourier transform on L2 space. Its operator norm is exactly the essential supremum of $|m|$:The unitary normalization matters when writing the intermediate integrals. For example, $\hat{f}(\xi )=\int f(x)e^{-2\pi ix\cdot \xi }dx$ has this normalization. The Plancherel theorem then turns the bound into multiplication of functions:The required extension of the Fourier transform to all of L2 space is the unitary extension in the Plancherel theorem.^[ref]

To prove sharpness, write $M=\Vert m{\Vert }_{\infty }>0$. For every $0<\epsilon <M$, the measurable set where $|m|>M-\epsilon$ has positive Lebesgue measure. Choose a subset $E$ of finite positive Lebesgue measure and prescribe $\hat{f}=1_E/\sqrt{|E|}$, using an indicator function. The inverse Fourier transform gives $\Vert f{\Vert }_2=1$ and $\Vert T_{m}f{\Vert }_2\ge M-\epsilon$. Letting $\epsilon$ tend to zero proves the equality of operator norms. The case $M=0$ is immediate.

For example, $m=1_{[-1,1]}$ is an ideal low-pass filter: it is an orthogonal projection onto the signals with Fourier transform supported in that interval, and has operator norm one. In contrast, $m(\xi )=e^{-{\xi }^2}$ also has operator norm one on $L^2(\mathbb{R})$, but no nonzero signal attains equality: $|m|<1$ except on a set of Lebesgue measure zero. Changing a Fourier multiplier at one frequency cannot change this operator norm.

Suppose every continuous periodic function $f$ has a unique continuous periodic $Tf$ with Fourier coefficients $\widehat{Tf}(n)={\alpha }_n\hat{f}(n)$. Uniqueness first makes $T$ linear. If $f_j\to f$ and $T{f}_j\to g$ uniformly, continuity of each coefficient functional gives $\hat{g}(n)={\alpha }_n\hat{f}(n)$ for every fixed $n$, hence $g=Tf$. Thus $T$ has a closed graph and the closed graph theorem makes it bounded. Applying it to normalized exponentials gives $|{\alpha }_n|\le \Vert T\Vert$.

For $e_n(t)=(2\pi {)}^{-1/2}{e}^{int}$, the bounded functional $\varphi (f)=\sqrt{2\pi }(Tf)(0)$ satisfies $\varphi (e_n)={\alpha }_n$. The normalization factor follows from $e_n(0)=(2\pi {)}^{-1/2}$.

The half-plane Fourier projection is $(I-i{H}_v)/2$, where $H_v$ is the directional Hilbert transform. A disk of radius $R$, modulated to put a boundary point at the origin, has symbol $1_{\{|\xi +Rv|\le R\}}$, which converges almost everywhere to this half-plane symbol. The Plancherel theorem makes this an $L^2$ multiplier limit.

Using the cycles-per-unit inverse convention ${\mathcal{F}}^{-1}g(x)=\int g(\xi )e^{2\pi ix\cdot \xi }d\xi$, the inverse Fourier transform of $(1-|\xi |{)}_{+}$ in three dimensions, with $a=2\pi |x|$, is $-4\pi \sin a/a^3+8\pi (1-\cos a)/a^4$. Its leading oscillatory $|x{|}^{-3}$ tail is not absolutely integrable against the radial volume element $4\pi |x{|}^{2}d|x|$. Multiplying by a Gaussian Fourier transform retains a nonzero boundary-derivative tail, so the radial cutoff need not map even Schwartz functions into $L^1$. A tensor product of one-dimensional triangular cutoffs is a valid approximate identity, and must not be confused with this Euclidean radial cutoff.

A Fourier multiplier operator $T_m$ is defined by $\widehat{T_{m}f}(\xi )=m(\xi )\hat{f}(\xi )$. On $L^2$, a bounded symbol gives a bounded operator by the Plancherel theorem, while an unbounded symbol defines a closed linear operator on the functions for which $m\hat{f}$ remains square-integrable.