Measurability and lower semicontinuity of the smooth maximal functions

Continuity of the parameter family in Schwartz space. Fix $(y_0,t_0)\in {\mathbb{R}}^n\times (0,\mathrm{\infty })$ and a compact interval $[a,b]\subseteq (0,\mathrm{\infty })$ containing $t_0$ in its interior. For $h_{y,t}(z)={\phi }_t(y-z)$ and multi-indices $\alpha ,\beta$, a first-order Taylor expansion of ${\mathrm{\partial }}^{\beta }\phi$ along the segment from $(y_0-z)\mathrm{/}t$ to $(y-z)\mathrm{/}t$ gives, for $t\in [a,b]$ and $\epsilon :=\mathrm{\mid }y-y_0\mathrm{\mid }\mathrm{/}a\le 1$, $p_{\alpha \beta }(h_{y,t}-h_{y_0,t})\le C_{a,b}(1+\mathrm{\mid }{y}_0\mathrm{\mid }{)}^{\mathrm{\mid }\alpha \mathrm{\mid }}\frac{\mathrm{\mid }y-y_0\mathrm{\mid }}{a}{\sum }_{\mathrm{\mid }\gamma \mathrm{\mid }\le \mathrm{\mid }\alpha \mathrm{\mid }}{\sum }_{\mathrm{\mid }e\mathrm{\mid }=1}{p}_{\gamma ,\beta +e}(\phi )\mathrm{.}$ For the scale variation put $u=(y_0-z)\mathrm{/}t$. Differentiating ${\mathrm{\partial }}_z^{\beta }{h}_{y_0,t}(z)=(-1{)}^{\mathrm{\mid }\beta \mathrm{\mid }}{t}^{-n-\mathrm{\mid }\beta \mathrm{\mid }}{\mathrm{\partial }}^{\beta }\phi (u)$ gives ${\mathrm{\partial }}_t{\mathrm{\partial }}_z^{\beta }{h}_{y_0,t}(z)=(-1{)}^{\mathrm{\mid }\beta \mathrm{\mid }+1}{t}^{-n-\mathrm{\mid }\beta \mathrm{\mid }-1}((n+\mathrm{\mid }\beta \mathrm{\mid }){\mathrm{\partial }}^{\beta }\phi (u)+u\cdot \mathrm{\nabla }{\mathrm{\partial }}^{\beta }\phi (u))\mathrm{.}$ Since $z=y_0-tu$ and $t\in [a,b]$, $\mathrm{\mid }z{\mathrm{\mid }}^{\mathrm{\mid }\alpha \mathrm{\mid }}\le C_{a,b,\alpha }(1+\mathrm{\mid }{y}_0\mathrm{\mid }{)}^{\mathrm{\mid }\alpha \mathrm{\mid }}(1+\mathrm{\mid }u\mathrm{\mid }{)}^{\mathrm{\mid }\alpha \mathrm{\mid }}$. The factor $u$ in the scale derivative therefore requires one additional polynomial weight, and the mean-value estimate gives $p_{\alpha \beta }(h_{y_0,t}-h_{y_0,t_0})\le C_{a,b,n,\alpha ,\beta }(1+\mathrm{\mid }{y}_0\mathrm{\mid }{)}^{\mathrm{\mid }\alpha \mathrm{\mid }}\mathrm{\mid }t-t_0\mathrm{\mid }({\sum }_{\mathrm{\mid }\gamma \mathrm{\mid }\le \mathrm{\mid }\alpha \mathrm{\mid }}{p}_{\gamma ,\beta }(\phi )+{\sum }_{\mathrm{\mid }\gamma \mathrm{\mid }\le \mathrm{\mid }\alpha \mathrm{\mid }+1}{\sum }_{\mathrm{\mid }e\mathrm{\mid }=1}{p}_{\gamma ,\beta +e}(\phi ))\mathrm{.}$ The two estimates tend to $0$ as $(y,t)\to (y_0,t_0)$; hence $(y,t)\mapsto h_{y,t}$ is continuous from ${\mathbb{R}}^n\times (0,\mathrm{\infty })$ into $\mathcal{S}$.

Joint continuity of the convolution. Since $(f\ast {\phi }_t)(y)=⟨f,h_{y,t}⟩$ by [F1] and $h_{y,t}\to h_{y_0,t_0}$ in $\mathcal{S}$ by step 1.1, the continuity of $f$ on $\mathcal{S}$ [F3] gives $(f\ast {\phi }_t)(y)\to (f\ast {\phi }_{t_0})(y_0)$ as $(y,t)\to (y_0,t_0)$. This proves the first clause, and in particular each function $y\mapsto (f\ast {\phi }_t)(y)$ is continuous on ${\mathbb{R}}^n$ for every fixed $t$.

Lower semicontinuity. Let $\lambda >0$. For the radial and nontangential functions assume $\int \phi \mathrm{\ne }0$, and suppose $M_{\phi }^{\ast ,a}f(x_0)>\lambda$. Since by step 2.1 the function $(y,t)\mapsto \mathrm{\mid }(f\ast {\phi }_t)(y)\mathrm{\mid }$ is continuous and the closed cone $\{\mathrm{\mid }y-x_0\mathrm{\mid }\le at\}$ is the closure of the open cone $\{\mathrm{\mid }y-x_0\mathrm{\mid }<at\}$, the supremum over the open cone equals the supremum over the closed one: a witness in the closed cone with value $>\lambda$ can be moved slightly along the segment towards $x_0$ to a witness with $\mathrm{\mid }y-x_0\mathrm{\mid }<at$ and value still $>\lambda$. Fix such $t>0$, $y$ with $\mathrm{\mid }y-x_0\mathrm{\mid }<at$ and $\mathrm{\mid }(f\ast {\phi }_t)(y)\mathrm{\mid }>\lambda$; by step 2.1 there is a neighbourhood $U$ of $y$ on which $\mathrm{\mid }(f\ast {\phi }_t)\mathrm{\mid }>\lambda$. The set of $x$ with $\mathrm{\mid }y-x\mathrm{\mid }<at$ is open and contains $x_0$, so $U^{\mathrm{\prime }}=\{x:\mathrm{\mid }y-x\mathrm{\mid }<at\}$ is a neighbourhood of $x_0$ on which $M_{\phi }^{\ast ,a}f(x)\ge \mathrm{\mid }(f\ast {\phi }_t)(y)\mathrm{\mid }>\lambda$ for every $y\in U\cap$ the ball of radius $at$ centred at $x$: more precisely, for $x\in U^{\mathrm{\prime }}$ choose $y^{\mathrm{\prime }}\in U\subset B(x,at)$ (possible because $U$ is a neighbourhood of $y$ and $\mathrm{\mid }y-x\mathrm{\mid }<at$, so $U\cap B(x,at)\mathrm{\ne }\mathrm{\varnothing }$), and then $M_{\phi }^{\ast ,a}f(x)>\lambda$. Hence $\{M_{\phi }^{\ast ,a}f>\lambda \}$ is open and $M_{\phi }^{\ast ,a}f$ is lower semicontinuous. The radial case is identical with $y=x$ and the value $\mathrm{\mid }(f\ast {\phi }_t)(x_0)\mathrm{\mid }>\lambda$: step 2.1 gives a neighbourhood of $x_0$ on which $\mathrm{\mid }(f\ast {\phi }_t)\mathrm{\mid }>\lambda$, so $M_{\phi }^{0}f>\lambda$ there. For the grand maximal function, with no integral restriction on $\psi \in {\mathcal{F}}_N$, if $M_{N}f(x_0)>\lambda$, the direct definition gives $\psi \in {\mathcal{F}}_N$, $t>0$ and $y$ with $\mathrm{\mid }y-x_0\mathrm{\mid }\le t$ and $\mathrm{\mid }(f\ast {\psi }_t)(y)\mathrm{\mid }>\lambda$. If $\mathrm{\mid }y-x_0\mathrm{\mid }=t$, continuity from step 2.1 lets us move $y$ slightly toward $x_0$ while keeping the value above $\lambda$, so we may assume $\mathrm{\mid }y-x_0\mathrm{\mid }<t$. Then $U=\{x:\mathrm{\mid }y-x\mathrm{\mid }<t\}$ is an open neighbourhood of $x_0$, and for every $x\in U$ the same $\psi ,t,y$ is admissible in the defining supremum, giving $M_{N}f(x)>\lambda$. Thus $\{M_{N}f>\lambda \}$ is open and $M_{N}f$ is lower semicontinuous.

Measurability. An extended-real lower semicontinuous function is Borel: for each real $\lambda$ the set $\{g>\lambda \}$ is open, hence Borel, and the Borel structure of $[0,\mathrm{\infty }]$ is generated by the open (or by the intervals $(\lambda ,\mathrm{\infty }]$ and $[0,\lambda )$) sets. Applying this to $M_{\phi }^{0}f$, $M_{\phi }^{\ast ,a}f$ and $M_{N}f$ by step 3.1 gives the stated Borel measurability, with values in $[0,\mathrm{\infty }]$; the value $+\mathrm{\infty }$ is not excluded, and if it occurs it occurs on a measurable set. This proves the lemma.