Inward mollification and smooth finite-rank approximation on Lipschitz domains

If $D=\mathrm{\varnothing }$, $X$ is the zero space; take $F=0$, the empty cutoff family, $k_t=0$, $S_t=0$ and $t_0=1$. If $K=\mathrm{\varnothing }$ the compact approximation clause is vacuous and $F=0$ suffices; the smoothing construction below depends only on $D$ and $X$, and can be obtained by carrying it out with $K=\{0\}$. Henceforth both sets are nonempty. In particular $0<\mathrm{\mid }D\mathrm{\mid }<\mathrm{\infty }$, because $D$ contains a ball and is bounded. Fix one boundary chart $Q$, with Lipschitz graph function $\phi$ of constant $L$, and an ambient neighborhood $U$ of its boundary point with $\bar{U}\subset Q$ compact. In chart coordinates set $v=(L+2)e_d$. For $x\in \bar{D}\cap \bar{U}$, $\mathrm{\mid }z\mathrm{\mid }\le 1$, and sufficiently small $t>0$, uniformly in $x,z$, the point $y=x+tv-tz$ stays in $Q$ because $t(\mathrm{\mid }v\mathrm{\mid }+1)<\mathrm{dist}(\bar{U},Q^c)$. Moreover $y_d-\phi (y^{\mathrm{\prime }})\ge x_d-\phi (x^{\mathrm{\prime }})+t(L+2)-t\mathrm{\mid }{z}_d\mathrm{\mid }-Lt\mathrm{\mid }{z}^{\mathrm{\prime }}\mathrm{\mid }\ge t>0$. Thus $\bar{B}(x+tv,t)⋐D$. Rotate $v$ back to the original coordinates.

The compact boundary admits finitely many such $U_j$, $1\le j\le J$, covering it. The compact remainder $\bar{D}\setminus {\bigcup }_{j=1}^J{U}_j$ lies in $D$; cover it by finitely many balls with closures in $D$, and let $U_0$ be their union, omitting it if the remainder is empty. Then ${\bar{U}}_0⋐D$, so $v_0=0$ has the same ball property for sufficiently small $t$. Apply [F2] to this finite open cover of $\bar{D}$, obtaining ${\psi }_j\in C_c^{\mathrm{\infty }}(U_j)$ with ${\sum }_j{\psi }_j=1$ near $\bar{D}$. The minimum of the finitely many positive bounds on $t$ gives $t_0>0$ such that $\bar{B}(x+t{v}_j,t)⋐D$ for every $x\in \bar{D}\cap \mathrm{supp}{\psi }_j$ and $0<t<t_0$. Taking the family of all admissible chart neighborhoods before extracting a finite subcover avoids selecting a chart independently at every boundary point.

By [F9] choose a nonnegative smooth bump compactly supported in $B_1(0)$ and positive on ${\bar{B}}_{1\mathrm{/}2}(0)$, and divide it by its positive integral to obtain $\rho \in C_c^{\mathrm{\infty }}(B_1(0))$ with $\int \rho =1$. Put ${\rho }_t(w)=t^{-d}\rho (w\mathrm{/}t)$. Every $u\in X$ belongs continuously to $L^1(D)$: use [F8] for Sobolev spaces, and $\mathrm{\parallel }u{\mathrm{\parallel }}_1\le \mathrm{\mid }D\mathrm{\mid }\mathrm{\parallel }u{\mathrm{\parallel }}_{C^m}$ for $C^m$. For its zero extension $E_{0}u$ define $H_{j,t}u(x)=({\rho }_t\ast E_{0}u)(x+t{v}_j)$ and $S_{t}u={\sum }_j{\psi }_j{H}_{j,t}u$. By [F4] these are well-defined linear constructions independent of representatives and $S_{t}u\in C_c^{\mathrm{\infty }}({\mathbb{R}}^d)$.

First take $X=W^{m,p}(D)$. For every $\mathrm{\mid }\gamma \mathrm{\mid }\le m$ and $x\in \bar{D}\cap \mathrm{supp}{\psi }_j$, [F3] and step 2.1 give $D^{\gamma }{H}_{j,t}u(x)=\int \rho (z)(E_0{D}^{\gamma }u)(x+t(v_j-z))dz\mathrm{.}$ The translated center has distance greater than $t$ from $D^c$: the entire closed ball of radius $t$ is a compact subset of $D$. The identity consequently holds on a neighborhood of each indicated $x$. Equivalently the defining weak-derivative identity can be tested against $y\mapsto {\rho }_t(x+t{v}_j-y)$, whose support is compact in $D$; its two derivative signs cancel. Only interior values are used, with no assertion that derivatives commute with zero extension across $\mathrm{\partial }D$.

For $f\in L^p({\mathbb{R}}^d)$ define $T_{j,t}f(x)=\int \rho (z)f(x+t(v_j-z))dz$ and ${\omega }_f(r)={\sup }_{\mathrm{\mid }h\mathrm{\mid }\le r}\mathrm{\parallel }f(\cdot +h)-f{\mathrm{\parallel }}_p$. Translation invariance and [F5] give $\mathrm{\parallel }{T}_{j,t}f{\mathrm{\parallel }}_p\le \mathrm{\parallel }f{\mathrm{\parallel }}_p$ and $\mathrm{\parallel }{T}_{j,t}f-f{\mathrm{\parallel }}_p\le {\omega }_f((\mathrm{\mid }{v}_j\mathrm{\mid }+1)t)\mathrm{.}$ By [F4], ${\omega }_f(r)\to 0$ as $r\downarrow 0$. The integral estimates also apply when $p=1$, where [F5] follows directly from Tonelli.

Fix $0<t<t_0$. Write $k_t(x,y)={\sum }_j{\psi }_j(x){\rho }_t(x+t{v}_j-y)$, so $S_{t}u(x)={\int }_D{k}_t(x,y)u(y)dy$. Fix a closed box $Q$ containing $\bar{D}$. The finitely many $D_x^{\alpha }{k}_t$, $\mathrm{\mid }\alpha \mathrm{\mid }\le m$, are smooth and uniformly continuous on the compact set $\bar{D}\times Q$ by [F7]. For any $\eta >0$, partition $Q$ into finitely many measurable cells $Q_{\mathrm{\ell }}$ of small enough diameter, assigning shared and outer faces consistently, and choose $y_{\mathrm{\ell }}\in Q_{\mathrm{\ell }}$. Their diameters can be chosen so that $\mathrm{\mid }{D}_x^{\alpha }{k}_t(x,y)-D_x^{\alpha }{k}_t(x,y_{\mathrm{\ell }})\mathrm{\mid }<\eta$ for $x\in \bar{D}$, $y\in Q_{\mathrm{\ell }}$ and $\mathrm{\mid }\alpha \mathrm{\mid }\le m$. The point $y_{\mathrm{\ell }}$ need not lie in $D$.

Set $f_{\gamma }=E_0{D}^{\gamma }u$. The finite Leibniz rule [F6], step 4.1 and ${\sum }_j{\psi }_j=1$ near $\bar{D}$ give, for $\mathrm{\mid }\alpha \mathrm{\mid }\le m$, the a.e. identity on $D$ $D^{\alpha }(S_{t}u-u)={\sum }_j{\sum }_{\beta \le \alpha }\left(\genfrac{}{}{0px}{}{\alpha }{\beta }\right)(D^{\beta }{\psi }_j)(T_{j,t}{f}_{\alpha -\beta }-D^{\alpha -\beta }u)\mathrm{.}$ Every derivative $D^{\beta }{\psi }_j$ vanishes off $\mathrm{supp}{\psi }_j$, so step 4.1 supplies exactly the identities used in this formula. All these fixed multiplier derivatives are bounded. Step 4.2 and the finite norm sum [F1] therefore imply $\mathrm{\parallel }{S}_{t}u-u{\mathrm{\parallel }}_{W^{m,p}}\to 0$. The same Leibniz formula without subtracting $u$, and $\mathrm{\parallel }{T}_{j,t}{f}_{\gamma }{\mathrm{\parallel }}_p\le \mathrm{\parallel }{f}_{\gamma }{\mathrm{\parallel }}_p\le \mathrm{\parallel }u{\mathrm{\parallel }}_{W^{m,p}}$, give ${\sup }_{0<t<t_0}\mathrm{\parallel }{S}_t{\mathrm{\parallel }}_{X\to X}\le M_X<\mathrm{\infty }$. For example take the ${\mathrm{\ell }}^p$ norm over $\alpha$ of ${\sum }_{j,\beta \le \alpha }\left(\genfrac{}{}{0px}{}{\alpha }{\beta }\right)\mathrm{\parallel }{D}^{\beta }{\psi }_j{\mathrm{\parallel }}_{\mathrm{\infty }}$. This argument includes $m=0$ and $p=1$.

Now take $X=C^m(\bar{D})$. By [F6] each bounded classical derivative is its weak derivative, so $u\in W^{m,1}(D)$ and step 4.1 applies. Each $D^{\gamma }u$ is uniformly continuous on $\bar{D}$ by [F7]. For $x\in \bar{D}\cap \mathrm{supp}{\psi }_j$, every sampled $y=x+t(v_j-z)$ lies in $D$ and $\mathrm{\mid }y-x\mathrm{\mid }\le (\mathrm{\mid }{v}_j\mathrm{\mid }+1)t$. Positivity and mass one of $\rho$ thus give ${\sup }_{x\in \bar{D}\cap \mathrm{supp}{\psi }_j}\mathrm{\mid }{D}^{\gamma }{H}_{j,t}u(x)-D^{\gamma }u(x)\mathrm{\mid }\le {\sup }_{\begin{array}{c}x,y\in \bar{D}\\ \mathrm{\mid }x-y\mathrm{\mid }\le (\mathrm{\mid }{v}_j\mathrm{\mid }+1)t\end{array}}\mathrm{\mid }{D}^{\gamma }u(y)-D^{\gamma }u(x)\mathrm{\mid }⟶0.$ The formula of step 5.1, first on $D$ and then by continuity on $\bar{D}$, gives $\mathrm{\parallel }{S}_{t}u-u{\mathrm{\parallel }}_{C^m}\to 0$. Bounding each averaged derivative by $\mathrm{\parallel }{D}^{\gamma }u{\mathrm{\parallel }}_{\mathrm{\infty }}$ and summing the same fixed multiplier constants gives ${\sup }_{0<t<t_0}\mathrm{\parallel }{S}_t{\mathrm{\parallel }}_{X\to X}\le M_X<\mathrm{\infty }$ also in this case.

Define $F_{t,\eta }u(x)={\sum }_{\mathrm{\ell }}{k}_t(x,y_{\mathrm{\ell }}){\int }_{D\cap Q_{\mathrm{\ell }}}u(y)dy\mathrm{.}$ Each coefficient is a bounded linear functional by step 3.1, and there are finitely many coefficients. Each range function $k_t(\cdot ,y_{\mathrm{\ell }})$ belongs to $C_c^{\mathrm{\infty }}({\mathbb{R}}^d)$, so the operator has finite rank with the required smooth range. Comparison with the integral for $S_t$ gives ${\max }_{\mathrm{\mid }\alpha \mathrm{\mid }\le m}{\sup }_{\bar{D}}\mathrm{\mid }{D}^{\alpha }(F_{t,\eta }-S_t)u\mathrm{\mid }\le \eta \mathrm{\parallel }u{\mathrm{\parallel }}_{L^1(D)}\mathrm{.}$ For $X=C^m(\bar{D})$, this yields $\mathrm{\parallel }{F}_{t,\eta }-S_t{\mathrm{\parallel }}_{X\to X}\le \mathrm{\mid }D\mathrm{\mid }\eta$. For $X=W^{m,p}(D)$ let $A_m$ be the finite number of multi-indices of order at most $m$. Taking the derivative $L^p$ norms and then applying [F8] gives $\mathrm{\parallel }{F}_{t,\eta }-S_t{\mathrm{\parallel }}_{X\to X}\le A_m^{1\mathrm{/}p}\mathrm{\mid }D{\mathrm{\mid }}^{1\mathrm{/}p}\eta \mathrm{\mid }D{\mathrm{\mid }}^{1-1\mathrm{/}p}=A_m^{1\mathrm{/}p}\mathrm{\mid }D\mathrm{\mid }\eta$. This includes $p=1$, and it shows each $F_{t,\eta }$ is bounded because $S_t$ is bounded.

By steps 5.1 and 6.1 the operators $S_t:X\to X$ converge pointwise to the identity and satisfy $\mathrm{\parallel }{S}_t\mathrm{\parallel }\le M_X$. Choose $\delta <\epsilon \mathrm{/}[4(M_X+1)]$. Compactness gives a finite $\delta$-net $u_1,\dots ,u_N$ in $K$. Pointwise convergence permits one $t\in (0,t_0)$ such that ${\max }_i\mathrm{\parallel }{S}_t{u}_i-u_i\mathrm{\parallel }<\epsilon \mathrm{/}4$. For any $u\in K$, choose one net point within $\delta$; then $\mathrm{\parallel }{S}_{t}u-u\mathrm{\parallel }\le (M_X+1)\delta +{\max }_i\mathrm{\parallel }{S}_t{u}_i-u_i\mathrm{\parallel }<\epsilon \mathrm{/}2$. Compactness also makes $R={\sup }_{u\in K}\mathrm{\parallel }u\mathrm{\parallel }$ finite: a finite cover by unit balls bounds all these norms by one plus the largest center norm.

Choose $\eta >0$ small enough that the relevant bound in step 7.1 is below $\epsilon \mathrm{/}[2(R+1)]$. For $F=F_{t,\eta }$, ${\sup }_{u\in K}\mathrm{\parallel }Fu-u\mathrm{\parallel }\le \mathrm{\parallel }F-S_t\mathrm{\parallel }R+{\sup }_{u\in K}\mathrm{\parallel }{S}_{t}u-u\mathrm{\parallel }<\epsilon \mathrm{.}$ Its range is a finite span of ambient compactly supported smooth functions by step 7.1. Applying this conclusion to the singleton compact set $K=\{u\}$ proves smooth-restriction density for every $u\in X$. All net, partition and representative-point choices here are finite; Countable Choice enters only the analytical interfaces stated above.