A bounded right inverse of the half-space trace by normal mollification

The identities on the smooth class. First $\int \psi =0$, because $\int {\mathrm{\partial }}_i(y_i\phi )=0$ for the compactly supported function $y_i\phi$. Next ${\psi }_t(y)=t^{-d}\psi (y\mathrm{/}t)$ satisfies ${\mathrm{\partial }}_t{\phi }_t(y)=-t^{-1}(d{\phi }_t(y)+y\cdot \mathrm{\nabla }{\phi }_t(y))=t^{-1}{\psi }_t(y)$: differentiating ${\phi }_t(y)=t^{-d}\phi (y\mathrm{/}t)$ in $t$, the two contributions combine into $-t^{-1}[d\phi (z)+\mathrm{\nabla }\phi (z)\cdot z]$ times $t^{-d}$, with $z=y\mathrm{/}t$, which is exactly $t^{-1}{\psi }_t(y)$ by the definition of $\psi$. Let $g\in C_c^{\mathrm{\infty }}({\mathbb{R}}^d)$ and extend $R_{+}g$ to $t=0$ by $\eta (0)\cdot g=g$. The function is smooth on $(0,\mathrm{\infty })$, and differentiating the convolution gives ${\mathrm{\partial }}_{x_i}(g\ast {\phi }_t)=g\ast {\mathrm{\partial }}_i{\phi }_t$ and ${\mathrm{\partial }}_t(g\ast {\phi }_t)=g\ast {\mathrm{\partial }}_t{\phi }_t=t^{-1}(g\ast {\psi }_t)$; hence ${\mathrm{\partial }}_{x_i}(R_{+}g)=\eta (g\ast {\mathrm{\partial }}_i{\phi }_t)$ and ${\mathrm{\partial }}_t(R_{+}g)={\eta }^{\mathrm{\prime }}(g\ast {\phi }_t)+\eta t^{-1}(g\ast {\psi }_t)$ as classical derivatives. Finally $g\ast {\phi }_t\to g$ uniformly as $t\downarrow 0$ for $g\in C_c^{\mathrm{\infty }}$, so the extension is continuous up to $t=0$ with boundary value $g$.

The norm estimates. For smooth compactly supported $g$, Young's inequality [F4] gives ${\int }_0^{\mathrm{\infty }}\mathrm{\parallel }{R}_{+}g(\cdot ,t){\mathrm{\parallel }}_p^{p}dt\le 2\mathrm{\parallel }\eta {\mathrm{\parallel }}_{\mathrm{\infty }}^p\mathrm{\parallel }\phi {\mathrm{\parallel }}_1^p\mathrm{\parallel }g{\mathrm{\parallel }}_p^p$, and the term ${\eta }^{\mathrm{\prime }}(g\ast {\phi }_t)$ is bounded by $\mathrm{\parallel }{\eta }^{\mathrm{\prime }}{\mathrm{\parallel }}_{\mathrm{\infty }}^p\mathrm{\parallel }\phi {\mathrm{\parallel }}_1^p\mathrm{\parallel }g{\mathrm{\parallel }}_p^p$, since ${\eta }^{\mathrm{\prime }}$ is supported in $[1,2]$. For $K_i={\mathrm{\partial }}_i\phi$, compact support gives $\int K_i=0$, and ${\mathrm{\partial }}_i{\phi }_t=t^{-1}(K_i{)}_t$. Thus [F1] bounds ${\int }_0^2\mathrm{\parallel }\eta (t)(g\ast {\mathrm{\partial }}_i{\phi }_t){\mathrm{\parallel }}_p^{p}dt$ by $C_i\mathrm{\parallel }\eta {\mathrm{\parallel }}_{\mathrm{\infty }}^p[g{]}_{\theta ,p}^p$. The same estimate with $K=\psi$ controls the normal term $\eta t^{-1}(g\ast {\psi }_t)$. Combining with $\mathrm{\mid }a+b{\mathrm{\mid }}^p\le 2^{p-1}(\mathrm{\mid }a{\mathrm{\mid }}^p+\mathrm{\mid }b{\mathrm{\mid }}^p)$ gives $\mathrm{\parallel }{R}_{+}g{\mathrm{\parallel }}_{W^{1,p}}^p\le C(\mathrm{\parallel }g{\mathrm{\parallel }}_p^p+[g{]}_{\theta ,p}^p)\le C\mathrm{\parallel }g{\mathrm{\parallel }}_{W^{\theta ,p}}^p$. These smooth interior derivatives are weak derivatives by integration against compactly supported tests.

Extension to $W^{\theta ,p}$ and the right-inverse identity. Let now $g\in W^{\theta ,p}({\mathbb{R}}^d)$ and choose $g_m\in C_c^{\mathrm{\infty }}({\mathbb{R}}^d)$ with $g_m\to g$ in $W^{\theta ,p}$ by [F3]. By step 2.1 the sequence $(R_{+}{g}_m)$ is Cauchy in the complete space $W^{1,p}(H)$ [F6]; define $R_{+}g$ as its limit. The value is independent of the approximating sequence and the resulting operator is linear and bounded with the constant of step 2.1, because any two approximating sequences can be interleaved. The weak derivatives of the limit are the limits of the weak derivatives, which by step 1.1 converge to the displayed convolution expressions in $L^p(H)$ by [F1] for the mean-zero tangential and normal terms, and by [F4] for the cutoff term; hence the limit satisfies the same two derivative identities. The values themselves converge to $\eta (t)(g\ast {\phi }_t)$ in $L^p(H)$ by [F4], so this limit is the formula specified in the Statement. For the trace, step 1.1 and [F2] give $T_{+}(R_{+}{g}_m)=g_m$ for each smooth $g_m$; since $T_{+}$ and $R_{+}$ are bounded, $T_{+}(R_{+}g)={\lim }_m{T}_{+}(R_{+}{g}_m)={\lim }_m{g}_m=g$ in $L^p({\mathbb{R}}^d)$. Thus $T_{+}\circ R_{+}=\mathrm{i}\mathrm{d}$ on $W^{\theta ,p}({\mathbb{R}}^d)$ and $R_{+}$ is a bounded linear right inverse.