Rough composition on a flat closed set

Put $\theta =r-s\ge 1$. Work first on a compact $K\subseteq A^{\ast }$ contained in a relatively compact ball of $W$. For $x\in K$ let $F_x$ be the order-$r$ Taylor polynomial of $f$ at $g(x)$ and $G_x$ the order-$\theta$ Taylor polynomial of $g$ at $x$. Define $P_x$ to be the degree-at-most-$r$ Taylor polynomial at $x$ of the polynomial composition $F_x\circ G_x$. Since $G_x(x)=g(x)$ and the derivatives of $f$ of orders $1,\dots ,s$ vanish at $g(x)\in A$, $P_x(x)=f(g(x)),D^j{P}_x(x)=0(1\le j\le s)\mathrm{.}$ Although $g$ has only $\theta$ derivatives, $P_x$ has a well-defined order-$r$ jet: every nonconstant term uses at least $s+1$ factors from $G_x-g(x)$, so a coefficient of degree at most $r$ cannot use a derivative of $g$ above order $r-s=\theta$.

We verify the exact Whitney condition in [F1] using only function-value Taylor estimates. Uniformly for $x$ in a compact part of $A^{\ast }$ and small $h$, [F2] gives $g(x+h)-G_x(x+h)=o(\mathrm{\mid }h{\mathrm{\mid }}^{\theta })$ and $G_x(x+h)-g(x)=O(\mathrm{\mid }h\mathrm{\mid })$. Since $Df$ and its first $s-1$ derivatives vanish at $g(x)$, [F2] gives $\mathrm{\mid }Df(v)\mathrm{\mid }=O(\mathrm{\mid }h{\mathrm{\mid }}^s)$ on the segment joining $g(x+h)$ and $G_x(x+h)$. The mean-value integral therefore gives $f(g(x+h))-f(G_x(x+h))=o(\mathrm{\mid }h{\mathrm{\mid }}^{s+\theta })=o(\mathrm{\mid }h{\mathrm{\mid }}^r)\mathrm{.}$ Taylor expansion of $f$ at $g(x)$ gives $f(G_x(x+h))-F_x(G_x(x+h))=o(\mathrm{\mid }h{\mathrm{\mid }}^r)$, and truncating the polynomial composition to degree $r$ costs $O(\mathrm{\mid }h{\mathrm{\mid }}^{r+1})$. Consequently $f(g(x+h))-P_x(x+h)=o(\mathrm{\mid }h{\mathrm{\mid }}^r)$ uniformly on compact parts of $A^{\ast }$. The same estimate holds when $s=0$: then $Df$ is merely bounded and $\theta =r$.

Take $x,y\in K$, $d=\mathrm{\mid }x-y\mathrm{\mid }$, and let $z$ range over the ball $B(y,d)$. Then $\mathrm{\mid }z-x\mathrm{\mid }\le 2d$ and $\mathrm{\mid }z-y\mathrm{\mid }\le d$, so step 2.1 gives $\mathrm{\mid }{P}_x(z)-P_y(z)\mathrm{\mid }=o(d^r)$ uniformly throughout that ball. The difference is a polynomial of degree at most $r$. Rescale $z=y+dw$: on the finite-dimensional space of degree-$r$ polynomials, each coefficient functional is bounded by a constant times the supremum on the unit ball. One can see this directly by evaluating on a fixed finite tensor-product grid of distinct points and inverting its fixed Vandermonde matrix. Thus for every multi-index $\mathrm{\mid }\alpha \mathrm{\mid }\le r$, $D^{\alpha }{P}_x(y)-D^{\alpha }{P}_y(y)=o(d^{r-\mathrm{\mid }\alpha \mathrm{\mid }})\mathrm{.}$ This is the compatibility hypothesis of [F1], locally uniformly.

To pass from compact $K$ to relatively closed $A^{\ast }$, choose a compact exhaustion $K_j\subset \mathrm{int}{K}_{j+1}$ of $W$, namely $K_j=\{x:\mathrm{\mid }x\mathrm{\mid }\le j,\mathrm{dist}(x,{\mathbb{R}}^d\setminus W)\ge 1\mathrm{/}j\}$ (with distance to the empty set interpreted as infinite). Choose smooth bumps $b_j$ supported in $\mathrm{int}{K}_{j+1}$, equal to one on $K_j$, by convolving distance cutoffs in the positive gap between $K_j$ and ${\mathbb{R}}^d\setminus \mathrm{int}{K}_{j+1}$. Set ${\chi }_j=1-{\prod }_{i\le j}(1-b_i)$, so ${\chi }_j$ is supported in $\mathrm{int}{K}_{j+1}$, equals one on $K_j$, and $0\le {\chi }_j\le {\chi }_{j+1}\le 1$. Put ${\rho }_1={\chi }_1$, ${\rho }_j={\chi }_j-{\chi }_{j-1}$ for $j>1$. The nonnegative ${\rho }_j$ form a locally finite smooth partition of unity on $W$, and $\mathrm{supp}{\rho }_j\subseteq K_{j+1}$. Apply [F1] on the closed compact set $A^{\ast }\cap K_{j+1}$ to obtain an extension $H_j$ of its jets. The locally finite sum $H={\sum }_j{\rho }_j{H}_j$ is $C^r$ on $W$. At any $x\in A^{\ast }$ with ${\rho }_j(x)\mathrm{\ne }0$, the full order-$r$ jet of $H_j$ is $P_x$. Leibniz's rule and ${\sum }_j{\rho }_j=1$ show that the full jet of $H$ at $x$ is also $P_x$. Step 1.1 therefore gives $H=f\circ g$ and $D^{j}H=0$ for $1\le j\le s$ on $A^{\ast }$.