= Norm bound for a Fourier multiplier on L2
{title2=$\|T_mf\|_2\le\|m\|_\infty\|f\|_2$}
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 $|m|$:
$$
\|T_m\|_{L^2\to L^2}=\|m\|_\infty,\qquad
\|T_mf\|_2\leq\|m\|_\infty\|f\|_2.
$$
The unitary normalization matters when writing the intermediate integrals. For example, $\widehat f(\xi)=\int f(x)e^{-2\pi i x\cdot\xi}\,dx$ has this normalization. The <Plancherel theorem> then turns the bound into multiplication of functions:
$$
\|T_mf\|_2^2=\int_{\mathbb R^d}|m(\xi)|^2|\widehat f(\xi)|^2\,d\xi
\leq\|m\|_\infty^2\int_{\mathbb R^d}|\widehat f(\xi)|^2\,d\xi
=\|m\|_\infty^2\|f\|_2^2.
$$
The required extension of the <Fourier transform> to all of <L2 space> is the unitary extension in the <Plancherel theorem>.<245C, Notes 2: The Fourier transform by Terence Tao>[Section 3, equation (8) and Theorem 43]{ref}
To prove sharpness, write $M=\|m\|_\infty>0$. For every $0<\varepsilon<M$, the measurable set where $|m|>M-\varepsilon$ has positive <Lebesgue measure>. Choose a subset $E$ of finite positive <Lebesgue measure> and prescribe $\widehat f=\mathbf1_E/\sqrt{|E|}$, using an <indicator function>. The inverse <Fourier transform> gives $\|f\|_2=1$ and $\|T_mf\|_2\geq M-\varepsilon$. Letting $\varepsilon$ tend to zero proves the equality of <operator norms>. The case $M=0$ is immediate.
For example, $m=\mathbf1_{[-1,1]}$ 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>.
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