Mihlin smoothness convention above half the dimension

Assume Countable Choice and let $n\ge 1$. Put $q:=\lfloor n\mathrm{/}2\rfloor +1.$ A measurable $m:{\mathbb{R}}^n\to \mathbb{C}$ is a Mihlin symbol in this convention when there is a function $m_0\in C^q({\mathbb{R}}^n\setminus \{0\})$ such that $m=m_0$ Lebesgue almost everywhere and there are constants $C_{\alpha }\ge 0$, indexed by the multi-indices $\alpha$ of $C^k$ maps and multi-index derivative notation in Euclidean space with $\mathrm{\mid }\alpha \mathrm{\mid }\le q$, for which $\mathrm{\mid }{\mathrm{\partial }}^{\alpha }{m}_0(\xi )\mathrm{\mid }\le C_{\alpha }\mathrm{\mid }\xi {\mathrm{\mid }}^{-\mathrm{\mid }\alpha \mathrm{\mid }}(\xi \mathrm{\ne }0,\mathrm{\mid }\alpha \mathrm{\mid }\le q)\mathrm{.}$ The derivatives are taken in the punctured open set ${\mathbb{R}}^n\setminus \{0\}$; the value $m(0)$ is not constrained.

Consequences recorded here. The case $\alpha =0$ gives $\mathrm{\mid }{m}_0(\xi )\mathrm{\mid }\le C_0$ for every $\xi \mathrm{\ne }0$, so $\mathrm{\mid }m\mathrm{\mid }\le C_0$ Lebesgue almost everywhere, because the singleton $\{0\}$ is Lebesgue null (Every at most countable subset of ${\mathbb{R}}^n$ is Lebesgue null; in particular ${\lambda }_1(\mathbb{Q})=0$); thus a Mihlin symbol is essentially bounded and $\mathrm{\parallel }m{\mathrm{\parallel }}_{\mathrm{\infty }}=\mathrm{ess\hspace{0.17em}sup}\mathrm{\mid }m\mathrm{\mid }\le C_0$ (The essential supremum of a measurable function with respect to a measure). Consequently the exact $L^2$ multiplier lemma applies: every Schwartz function lies in the Schwartz domain of $m$, and the operator $T_m$ has a unique bounded extension to $L^2({\mathbb{R}}^n;\mathbb{C})$ of norm $\mathrm{\parallel }m{\mathrm{\parallel }}_{\mathrm{\infty }}\le C_0$ (Exact L2 Fourier multiplier norm). This definition is a sufficient symbol condition: $L^p({\mathbb{R}}^n)$ boundedness for $1<p<\mathrm{\infty }$ is a separate theorem, assigned in this library to the later Mihlin multiplier theorem built on singular-integral estimates, and no such boundedness is asserted here.

Derivative count. The count $q=\lfloor n\mathrm{/}2\rfloor +1$ is the classical "more than half the dimension" requirement used by multiplier theory. Grafakos assumes $m_0\in C^{[n\mathrm{/}2]+1}$ away from the origin with the pointwise inequalities above and derives the annular estimates used in his proof; Exact L2 Fourier multiplier norm supplies only the $L^2$ part of that theorem. Williams states the same conclusion under the stronger count $\mathrm{\mid }\alpha \mathrm{\mid }\le d+2$, so his theorem does not reduce the derivative count adopted here. The bounds are required for $\xi \mathrm{\ne }0$ only: homogeneity of order zero near the origin, such as the signum symbol in one dimension, is compatible with the condition, while a jump at a nonzero frequency is not, since $C^q$ functions on the punctured space are continuous there.