Exact L2 Fourier multiplier norm

Put $M=\mathrm{\parallel }m{\mathrm{\parallel }}_{\mathrm{\infty }}$, so $\mathrm{\mid }m\mathrm{\mid }\le M$ almost everywhere and no smaller constant has this property [F1]. For an $L^2$ class $g$ the product $mg$ is measurable, and $\mathrm{\mid }mg\mathrm{\mid }\le M\mathrm{\mid }g\mathrm{\mid }$ almost everywhere, so $mg\in L^2$ with $\mathrm{\parallel }mg{\mathrm{\parallel }}_2\le M\mathrm{\parallel }g{\mathrm{\parallel }}_2$; the assignment $M_{m}g=mg$ is complex-linear and depends only on the classes of $m$ and $g$, since changing either on a null set changes $mg$ only on a null set.

Assume $M>0$ and fix $0<\epsilon <M$. Were $\mathrm{\mid }m\mathrm{\mid }\le M-\epsilon$ almost everywhere, [F1] would give $M\le M-\epsilon$, a contradiction; hence $E_{\epsilon }=\{x:\mathrm{\mid }m(x)\mathrm{\mid }>M-\epsilon \}$ has positive Lebesgue measure.

The operator $S_m:={\mathcal{F}}_2^{-1}{M}_m{\mathcal{F}}_2$ is a bounded complex-linear operator on $L^2({\mathbb{R}}^n;\mathbb{C})$ with $\mathrm{\parallel }{S}_m\mathrm{\parallel }\le \mathrm{\parallel }{M}_m\mathrm{\parallel }\le M$, by [F2] and step 1.1.

Let $f\in \mathcal{S}({\mathbb{R}}^n)$ and let $\widehat{f}$ be its integral transform, a Schwartz class; then $m\widehat{f}\in L^2$ with $\mathrm{\parallel }m\widehat{f}{\mathrm{\parallel }}_2\le M\mathrm{\parallel }\widehat{f}{\mathrm{\parallel }}_2$, so $m\widehat{f}$ is locally integrable and its regular distribution is tempered by [F5]; hence $f\in D_m$ and $T_{m}f={\mathcal{F}}^{-1}(u_{m\widehat{f}})$ by [F7].

The sets $E_{\epsilon }\cap B(0,k)$ increase to $E_{\epsilon }$, so [F8] gives $0<\mathrm{\mid }{E}_{\epsilon }\mathrm{\mid }={\sup }_k\mathrm{\mid }{E}_{\epsilon }\cap B(0,k)\mathrm{\mid }$; choose $k$ with $0<\mathrm{\mid }{E}_{\epsilon }\cap B(0,k)\mathrm{\mid }<\mathrm{\infty }$, possible because balls have finite measure.

Applying [F6] with $h={\mathcal{F}}_2^{-1}(m{\mathcal{F}}_{2}f)\in L^2$ identifies ${\mathcal{F}}^{-1}(u_{m{\mathcal{F}}_{2}f})=u_h$; since $\widehat{f}={\mathcal{F}}_{2}f$ as $L^2$ classes by [F2] and [F6], step 2.2 gives $T_{m}f=u_h$ and therefore the $L^2$ class identity $T_{m}f={\mathcal{F}}_2^{-1}(m{\mathcal{F}}_{2}f)=S_{m}f$, with $\mathrm{\parallel }{T}_{m}f{\mathrm{\parallel }}_2=\mathrm{\parallel }m{\mathcal{F}}_{2}f{\mathrm{\parallel }}_2\le M\mathrm{\parallel }{\mathcal{F}}_{2}f{\mathrm{\parallel }}_2=M\mathrm{\parallel }f{\mathrm{\parallel }}_2$.

Claim $\mathrm{\parallel }{S}_m\mathrm{\parallel }=M$. In the degenerate case $M=0$, [F1] gives $m=0$ almost everywhere, so $M_m=0$ and $\mathrm{\parallel }{S}_m\mathrm{\parallel }=0=M$.

Put $g=\mathrm{\mid }{E}_{\epsilon }\cap B(0,k){\mathrm{\mid }}^{-1\mathrm{/}2}{\mathbf{1}}_{E_{\epsilon }\cap B(0,k)}\in L^2$, a unit vector. On $E_{\epsilon }\cap B(0,k)$ one has $\mathrm{\mid }m\mathrm{\mid }>M-\epsilon$ and $g\mathrm{\ne }0$, so $\mathrm{\parallel }{M}_{m}g{\mathrm{\parallel }}_2^2={\int }_{E_{\epsilon }\cap B(0,k)}\mathrm{\mid }m{\mathrm{\mid }}^2\mathrm{\mid }g{\mathrm{\mid }}^2>(M-\epsilon {)}^2\int g^2=(M-\epsilon {)}^2$; therefore, putting $f={\mathcal{F}}_2^{-1}g$, [F2] gives $\mathrm{\parallel }f{\mathrm{\parallel }}_2=1$ and $\mathrm{\parallel }{S}_{m}f{\mathrm{\parallel }}_2=\mathrm{\parallel }{M}_{m}g{\mathrm{\parallel }}_2>M-\epsilon$; and $\mathrm{\parallel }{S}_m\mathrm{\parallel }\ge M-\epsilon$ for every such $\epsilon$.

By step 3.1 the operator $S_m$ agrees on the dense subspace $\mathcal{S}({\mathbb{R}}^n)$ with the Schwartz-core action $f\mapsto T_{m}f$ of [F7]; since $\mathcal{S}$ is dense in $L^2$ by [F3] and $S_m$ is bounded linear by step 2.1, [F4] makes $S_m$ the unique bounded linear extension of the core action, with the same operator norm.

Step 3.2 gives $\mathrm{\parallel }{S}_m\mathrm{\parallel }=0=M$ when $M=0$, and when $M>0$ step 3.3 gives $\mathrm{\parallel }{S}_m\mathrm{\parallel }\ge M-\epsilon$ for every $0<\epsilon <M$, hence $\mathrm{\parallel }{S}_m\mathrm{\parallel }\ge M$ after letting $\epsilon \downarrow 0$; step 2.1 gives $\mathrm{\parallel }{S}_m\mathrm{\parallel }\le M$; hence $\mathrm{\parallel }{S}_m\mathrm{\parallel }=M$, which together with step 4.1 proves the exact operator norm of statement 2.

Finally let $m=m^{\mathrm{\prime }}$ almost everywhere. Then for every $L^2$ class $g$ the products $mg$ and $m^{\mathrm{\prime }}g$ agree almost everywhere, so $M_m=M_{m^{\mathrm{\prime }}}$ and $S_m=S_{m^{\mathrm{\prime }}}$; the operators, and hence their norm $M$, depend only on the almost-everywhere class of $m$. Countable Choice supplies the hypotheses of [F2]-[F4], [F6]-[F7] and the Lebesgue-measure part of [F8]. The choice of a single integer $k$ in step 2.3 for a fixed $\epsilon$ requires no additional choice principle.