Norm bound for a Fourier multiplier on L2 ($\Vert T_{m}f{\Vert }_2\le \Vert m{\Vert }_{\infty }\Vert f{\Vert }_2$)

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.