On Semi-classical Limit of Spatially Homogeneous Quantum Boltzmann Equation: Weak Convergence

Abstract

It is expected in physics that the homogeneous quantum Boltzmann equation with Fermi–Dirac or Bose–Einstein statistics and with Maxwell–Boltzmann operator (neglecting effect of the statistics) for the weak coupled gases will converge to the homogeneous Fokker–Planck–Landau equation as the Planck constant $\hbar$ tends to zero. In this paper and the upcoming work (He et al. in On semi-classical limit of spatially homogeneous quantum Boltzmann equation: asymptotic expansion, preprint), we will provide a mathematical justification on this semi-classical limit. Key ingredients into the proofs are the new framework to catch the weak projection gradient, which is motivated by Villani (Arch Rational Mech Anal 143(3):273–307, 1998) to identify the H-solutions for Fokker–Planck–Landau equation, and the symmetric structure inside the cubic terms of the collision operators.

Appendix

In this appendix we prove some general properties that have been used in the previous sections, and establish the weak projection gradient.

1.1 Equivalence of definition for weak formula

We will prove

Proposition 6.1

Let $d, m\in {{{\mathbb {N}}}}$, $X={{{\mathbb {R}}}^{d}}$ or $X={{{\mathbb {R}}}}_{\ge 0}^d$, and let $\{\mu _t\}_{t\ge 0}$ be a family of positive measures on X satisfying $\sup \limits _{t\ge 0}\mu _t(X)<\infty$ and that for any $\psi \in C^m_c(X)$, $t\mapsto \int _{X}\psi ({\mathbf{x}}){\mathrm{d}}\mu _t({\mathbf{x}})$ is absolutely continuous on ${[}0,\infty )$ and

$$\begin{aligned} \frac{{\mathrm{d}}}{{\mathrm{d}}t}\int _{X}\psi ({\mathbf{x}}){\mathrm{d}}\mu _t({\mathbf{x}}) ={{\mathcal {Q}}}_t[\psi ],\quad \mathrm{a.e.}\,\,\, t\in {[}0,\infty ) \end{aligned}$$

(6.1)

where ${{\mathcal {Q}}}_t[\psi ]$ has the properties: there exist a function $0\le M\in L^1_{loc}({[}0,\infty ))$ and a null set $Z_0\subset {[}0,\infty )$ such that for any $t\in {[}0,\infty ){\setminus } Z_0$, $\psi \mapsto {{\mathcal {Q}}}_t[\psi ]$ is linear on $C^m_c(X)$ and

$$\begin{aligned} |{{\mathcal {Q}}}_t[\psi ]|\le \Vert \psi \Vert _{m,\infty }M(t)\qquad \forall \, \psi \in C_c^m(X),\quad \forall \,t\in {[}0,\infty ){\setminus } Z_0 \end{aligned}$$

(6.2)

where $\Vert \psi \Vert _{m,\infty }=\sup \limits _{\,{\mathbf{x}}\in X} \sum \limits _{|\alpha |\le m} |{\partial }^{\alpha } \psi ({\mathbf{x}})|, {\partial }^{\alpha }={\partial }^{\alpha }_{{\mathbf{x}}}={\partial }_{x_1}^{\alpha _1}{\partial }_{x_2}^{\alpha _2}\cdots {\partial }_{x_d}^{\alpha _d}, |\alpha |=\alpha _1+\alpha _2+\cdots +\alpha _d .$

Then for any $\psi \in C^m_c({[}0,\infty )\times X)$, $t\mapsto \int _{X}\psi (t,{\mathbf{x}}){\mathrm{d}}\mu _t({\mathbf{x}})$ is absolutely continuous ${[}0, \infty )$, $t\mapsto {{\mathcal {Q}}}_t[\psi (t)]$ belongs to $L^1_{loc}({[}0,\infty ))$, and

$$\begin{aligned} \frac{{\mathrm{d}}}{{\mathrm{d}}t}\int _{X}\psi (t,{\mathbf{x}}){\mathrm{d}}\mu _t({\mathbf{x}})= \int _{X}{\partial }_t\psi (t,{\mathbf{x}}){\mathrm{d}}\mu _t({\mathbf{x}}) +{\mathcal Q}_t[\psi (t)],\quad \mathrm{a.e.}\,\,\, t\in {[}0,\infty ). \end{aligned}$$

(6.3)

Proof

The proof consists of two steps.

Step 1. We prove that for any $\psi \in C_c({[}0,\infty )\times X)$, $t\mapsto \int _{X}\psi (t,{\mathbf{x}}){\mathrm{d}}\mu _t({\mathbf{x}})$ is continuous on ${[}0,\infty )$.

Suppose first that $\psi \in C_c(X)$. By extension theorem of continuous functions we can assume that $\psi \in C_c({{{\mathbb {R}}}^{d}})$. Let $0\le J\in C^{\infty }_c({{{\mathbb {R}}}^{d}})$ satisfy $\int _{{{{\mathbb {R}}}^{d}}}J({\mathbf{x}}){\mathrm{d}}{\mathbf{x}}=1$ and J is supported on the closed unite ball centered at the origin. Let $J_{\delta }({\mathbf{x}})=\delta ^{-d}J(\delta ^{-1}{\mathbf{x}}),\delta >0$, and $\psi _{\delta }=\psi *J_{\delta }$ (convolution). Then $\psi _{\delta }\in C^{\infty }_c({{{\mathbb {R}}}^{d}})$ and

$$\begin{aligned} \Vert \psi _{\delta }-\psi \Vert _{\infty }\le \Omega _{\psi }(\delta ):= \sup _{{\mathbf{x}}, {\mathbf{y}}\in {{{\mathbb {R}}}^{d}}, |{\mathbf{x}}-{\mathbf{y}}|\le \delta }|\psi ({\mathbf{x}})-\psi ({\mathbf{y}})|\rightarrow 0\quad (\delta \rightarrow 0^{+}). \end{aligned}$$

Let $C_{\mu }=\sup \limits _{t\ge 0}\mu _t(X)$. For any $t\in {[}0,\infty )$, we have

$$\begin{aligned} \bigg |\int _{X}\psi {\mathrm{d}}\mu _s- \int _{X}\psi {\mathrm{d}}\mu _{t} \bigg |\le & {} 2C_{\mu }\Omega _{\psi }(\delta )+\bigg |\int _{X}\psi _{\delta }{\mathrm{d}}\mu _s -\int _{X}\psi _{\delta }{\mathrm{d}}\mu _{t}\bigg |\\\le & {} 2C_{\mu }\Omega _{\psi }(\delta )+\Vert \psi _{\delta }\Vert _{m,\infty } \int _{t\wedge s}^{t\vee s}M(\tau ){\mathrm{d}}\tau . \end{aligned}$$

By first letting $s\rightarrow t$ and then letting $\delta \rightarrow 0^{+}$ we conclude that $t\mapsto \int _{X}\psi ({\mathbf{x}}){\mathrm{d}}\mu _t({\mathbf{x}})$ is continuous at $t\in {[}0,\infty )$.

For general case, let $\psi \in C_c({[}0,\infty )\times X)$ and take any $t\in {[}0,\infty )$. By the uniform continuity of $\psi$ on ${[}0,\infty )\times X$ and the above result we see that as $s\rightarrow t$

$$\begin{aligned}&\bigg |\int _{X}\psi (s,{\mathbf{x}}){\mathrm{d}}\mu _s({\mathbf{x}}) -\int _{X}\psi (t,{\mathbf{x}}){\mathrm{d}}\mu _{t}({\mathbf{x}})\bigg | \\&\quad \le C_{\mu }\Vert \psi (s)-\psi (t)\Vert _{\infty }+\bigg |\int _{X}\psi (t,{\mathbf{x}}){\mathrm{d}}\mu _{s}({\mathbf{x}}) -\int _{X}\psi (t,{\mathbf{x}}){\mathrm{d}}\mu _{t}({\mathbf{x}})\bigg |\rightarrow 0. \end{aligned}$$

So $t\mapsto \int _{X}\psi (t,{\mathbf{x}}){\mathrm{d}}\mu _t({\mathbf{x}})$ is continuous at $t\in {[}0,\infty )$.

Step 2. From (6.1) we see that for any $\psi \in C^m_c(X)$, $t\mapsto {{\mathcal {Q}}}_t[\psi ]$ is Lebesgue measurable on ${[}0,\infty )$. From this and using the following estimate (6.4) it is not difficult to prove that for any $\psi \in C^m_c({[}0,\infty )\times X)$, $t\mapsto {{\mathcal {Q}}}_t[\psi (t)]$ is also Lebesgue measurable on ${[}0,\infty )$. From the assumption (6.2) we have

$$\begin{aligned} \big |{{\mathcal {Q}}}_{t}[\psi (t)]-{{\mathcal {Q}}}_t[\psi (s)]\big |\le \Vert \psi (t)-\psi (s)\Vert _{m,\infty }M(t) \le C_{\psi }|t-s|M(t) \end{aligned}$$

(6.4)

for all $s\in {[}0,\infty )$ and all $t\in {[}0, \infty ){\setminus } Z_0$. Here and below $C_{\psi }=\sup \limits _{t\ge 0}\big ( \Vert \psi (t)\Vert _{m,\infty } +\Vert D_1\psi (t)\Vert _{m,\infty }\big ).$ Since

$$\begin{aligned} |{{\mathcal {Q}}}_t[\psi (t)]|\le C_{\psi }M(t)\qquad \forall \, t\in {[}0, \infty ){\setminus } Z_0 \end{aligned}$$

this implies that $t\mapsto {{\mathcal {Q}}}_t[\psi (t)]$ belongs to $L^1_{loc}({[}0,\infty ))$. Also we have for any $0\le t_1<t_2<\infty$

$$\begin{aligned} \bigg |\int _{X}\psi (t_2,{\mathbf{x}}){\mathrm{d}}\mu _{t_2}({\mathbf{x}}) -\int _{X}\psi (t_1,{\mathbf{x}}){\mathrm{d}}\mu _{t_1}({\mathbf{x}})\bigg | \le C_{\mu } C_{\psi }|t_2-t_1|+C_{\psi }\int _{t_1}^{t_2}M(s){\mathrm{d}}s. \end{aligned}$$

This implies that $t\mapsto \int _{X}\psi (t,{\mathbf{x}}){\mathrm{d}}\mu _t({\mathbf{x}})$ is absolutely continuous on ${[}0,\infty )$.

Since $t\mapsto {{\mathcal {Q}}}_t[\psi (t)]$ also belongs to $L^1_{loc}({[}0,\infty ))$, there is a null set $Z_1\subset {[}0,\infty )$ such that

$$\begin{aligned} \lim _{t+h\ge 0, h\rightarrow 0}\frac{1}{h}\int _{t}^{t+h}{\mathcal Q}_{\tau }[\psi (\tau )] {\mathrm{d}}\tau ={{\mathcal {Q}}}_{t}[\psi (t)] \quad \forall \, t\in {[}0,\infty ){\setminus } Z_1. \end{aligned}$$

(6.5)

Now for any $t\in {[}0,\infty ){\setminus } Z_1$ and $0\ne h\in [-t, \infty )$, look at the difference quotient

$$\begin{aligned}&\frac{1}{h}\bigg ( \int _{X}\psi (t+h,{\mathbf{x}}){\mathrm{d}}\mu _{t+h}({\mathbf{x}})- \int _{X}\psi (t,{\mathbf{x}}){\mathrm{d}}\mu _{t}({\mathbf{x}}) \bigg )\\&\quad =\int _{X}\frac{\psi (t+h,{\mathbf{x}})-\psi (t,{\mathbf{x}})}{h}{\mathrm{d}}\mu _{t+h}({\mathbf{x}}) +\frac{1}{h}\int _{t}^{t+h}{\mathcal Q}_{\tau }[\psi (t)] {\mathrm{d}}\tau . \end{aligned}$$

Denoting $D_1\psi (t, {\mathbf{x}})={\partial }_t\psi (t,{\mathbf{x}})$ we have

$$\begin{aligned}&\bigg |\int _{X}\frac{\psi (t+h,{\mathbf{x}})-\psi (t,{\mathbf{x}})}{h}{\mathrm{d}}\mu _{t+h}({\mathbf{x}}) -\int _{X}D_1\psi (t,{\mathbf{x}}){\mathrm{d}}\mu _{t}({\mathbf{x}}) \bigg | \\&\quad \le C_{\mu }\Lambda _{D_1\psi }(|h|)+\bigg |\int _{X}D_1\psi (t+h,{\mathbf{x}}){\mathrm{d}}\mu _{t+h}({\mathbf{x}}) -\int _{X}D_1\psi (t,{\mathbf{x}}){\mathrm{d}}\mu _{t}({\mathbf{x}}) \bigg |\rightarrow 0 \end{aligned}$$

as $h\rightarrow 0$, where $\Lambda _{D_1\psi }(\delta )=\sup \limits _{t_1,t_2\ge 0, |t_1-t_2|\le \delta }\Vert D_1\psi (t_1)- D_1\psi (t_2)\Vert _{\infty }\rightarrow 0$ as $\delta \rightarrow 0^+$. Also from (6.4) we have

$$\begin{aligned}&\bigg |\frac{1}{h}\int _{t}^{t+h}{{\mathcal {Q}}}_{\tau }[\psi (t)] {\mathrm{d}}\tau -\frac{1}{h}\int _{t}^{t+h}{{\mathcal {Q}}}_{\tau }[\psi (\tau )] {\mathrm{d}}\tau \bigg |\le C_{\psi }\bigg |\int _{t}^{t+h}M(\tau ){\mathrm{d}}\tau \bigg |\rightarrow 0 \end{aligned}$$

as $h\rightarrow 0$. From these and (6.5) we conclude that the function $t\mapsto \int _{X}\psi (t,{\mathbf{x}}){\mathrm{d}}\mu _t({\mathbf{x}})$ is differentiable at $t\in {[}0,\infty ) {\setminus } Z_1$ and satisfies the equality (6.3). $\quad \square$

1.2 Construction of the potential function $\phi$

It is given in the following

Proposition 6.2

Let ${\widehat{\phi }}(r)$ satisfy (1.8), (1.10) with a constant $0\le r_0<\infty$. Then the limiting function $\phi (\rho )$ given by (1.11) is well-defined, and $\xi \mapsto {\widehat{\phi }}(|\xi |)$ is the Fourier transform of the function ${\mathbf{x}}\mapsto \phi (|{\mathbf{x}}|)$ in ${\mathcal {S}}'$ where ${\mathcal S}={{\mathcal {S}}}({{{\mathbb {R}}}^{3}})$ is the class of Schwartz functions on ${{{\mathbb {R}}}^{3}}$. Furthermore if $r_0=0$ and ${\widehat{\phi }}(r)\ge 0$ in $(0,\infty )$, then $\phi (\rho )\ge 0$ in $(0,\infty ).$

Proof

To prove the existence of the limit in (1.11), we consider

$$\begin{aligned} \phi _R(\rho )=\frac{1}{2\pi ^2\rho }\int _{0}^{R}r{\widehat{\phi }}(r)\sin (\rho r){\mathrm{d}}r,\quad \rho>0, \, R>0. \end{aligned}$$

Recalling assumptions (1.10) and (1.8) we know that $r\mapsto r{\widehat{\phi }}(r)$ is monotone in $(r_0,\infty )$ and $r{\widehat{\phi }}(r)\rightarrow 0$ as $r\rightarrow \infty$. Therefore using the second mean-value formula of integrals we have for all $R_2> R_1>r_0$ that

$$\begin{aligned} \bigg |\int _{R_1}^{R_2}r{\widehat{\phi }}(r)\sin (\rho r){\mathrm{d}}r\bigg | \le \frac{2}{\rho }\Big (|R_1{\widehat{\phi }}(R_1)|+|R_2{\widehat{\phi }}(R_2)|\Big ) \end{aligned}$$

(6.6)

which implies that for every $\rho >0$ the limit $\phi (\rho ):=\lim \limits _{R\rightarrow \infty }\phi _R(\rho )$ exists. Also from (1.8) and (6.6) we deduce that there is a constants $0<C<\infty$ such that

$$\begin{aligned} \sup _{R>0}|\phi _R(\rho )|\le C \big (1+\frac{1}{\rho ^2}\big )\quad \mathrm{hence}\quad |\phi (\rho )|\le C \big (1+\frac{1}{\rho ^2}\big )\qquad \forall \, \rho >0. \end{aligned}$$

Thus for any $\varphi \in {{\mathcal {S}}}({{{\mathbb {R}}}^{3}})$, the dominated convergence theorem can be used and we conclude that $\phi (|\cdot |){\widehat{\varphi }}\in L^1({{{\mathbb {R}}}^{3}})$ and

$$\begin{aligned} \int _{{{{\mathbb {R}}}}^3}\phi (|{\mathbf{x}}|){\widehat{\varphi }}({\mathbf{x}}){\mathrm{d}}{\mathbf{x}} =\int _{0}^{\infty }\int _{{{{\mathbb {S}}}^{2}}}\rho ^2 \phi (\rho ){\widehat{\varphi }}(\rho \sigma ){\mathrm{d}}\sigma {\mathrm{d}}\rho =\lim \limits _{R\rightarrow \infty }\int _{0}^{\infty }\int _{{{{\mathbb {S}}}^{2}}}\rho ^2 \phi _R(\rho ){\widehat{\varphi }}(\rho \sigma ){\mathrm{d}}\sigma {\mathrm{d}}\rho . \end{aligned}$$

(6.7)

On the other hand for any $0<R<\infty$ we compute

$$\begin{aligned}&\int _{|\xi |\le R}{\widehat{\phi }} (|\xi |)\varphi (\xi ){\mathrm{d}}\xi =\int _{0}^{R}r^2{\widehat{\phi }}(r)\int _{{{{\mathbb {S}}}^{2}}}(2\pi )^{-3}\int _{{{{\mathbb {R}}}^{3}}} {\widehat{\varphi }}({\mathbf{x}})e^{\mathrm{i}{\mathbf{x}}\cdot r\omega }{\mathrm{d}}{\mathbf{x}} {\mathrm{d}}\omega {\mathrm{d}}r \\&\quad =\int _{0}^{R}r^2{\widehat{\phi }}(r)(2\pi )^{-3}\int _{{{{\mathbb {R}}}^{3}}} {\widehat{\varphi }}({\mathbf{x}})\frac{4\pi }{r|{\mathbf{x}}|}\sin (r|{\mathbf{x}}|){\mathrm{d}}{\mathbf{x}} {\mathrm{d}}r=\int _{0}^{\infty }\int _{{{{\mathbb {S}}}^{2}}}\rho ^2 \phi _R(\rho ){\widehat{\varphi }}(\rho \sigma ){\mathrm{d}}\sigma {\mathrm{d}}\rho . \end{aligned}$$

Since ${\widehat{\phi }} (|\cdot |)\varphi \in L^1({{{\mathbb {R}}}^{3}})$, letting $R\rightarrow \infty$ we deduce from this and (6.7) that

$$\begin{aligned} \int _{{{{\mathbb {R}}}}^3}{\widehat{\phi }} (|\xi |)\varphi (\xi ){\mathrm{d}}\xi =\int _{{{{\mathbb {R}}}}^3}\phi (|{\mathbf{x}}|){\widehat{\varphi }}({\mathbf{x}}){\mathrm{d}}{\mathbf{x}}\qquad \forall \, \varphi \in {{\mathcal {S}}}({{{\mathbb {R}}}}^3). \end{aligned}$$

According to the Fourier transform of generalized functions, ${\widehat{\phi }}(|\xi |)$ is the Fourier transform of $\phi (|{\mathbf{x}}|)$ in ${\mathcal {S}}'$.

Now we turn to the second part of the proposition. Assume that $r_0=0$ and ${\widehat{\phi }}(r)\ge 0$ for all $r\in (0,\infty )$. Then $r\mapsto r{\widehat{\phi }}(r)$ is monotone in $(0,\infty )$. Since $\lim \limits _{r\rightarrow \infty }r{\widehat{\phi }}(r)=0$ and ${\widehat{\phi }}(r)\not \equiv 0$, this implies that $r\mapsto r{\widehat{\phi }}(r)$ is monotone non-increasing in $(0,\infty )$. Therefore using the second mean-value formula of integrals we have for any $R>\delta >0$ that

$$\begin{aligned} \int _{\delta }^{R}r{\widehat{\phi }}(r)\sin (\rho r){\mathrm{d}}r\ge \frac{\delta {\widehat{\phi }}(\delta )}{\rho } (\cos (\rho \delta )-1)\ge -\,\frac{\rho }{2}\delta ^3{\widehat{\phi }}(\delta ). \end{aligned}$$

(6.8)

Since the assumption (1.8) implies that $r\mapsto r^{3/2}{\widehat{\phi }}(r)$ belongs to $L^1((0,1))$, it follows that there is a sequence $0<\delta _n\rightarrow 0\,(n\rightarrow \infty )$ such that $\delta _n^3{\widehat{\phi }}(\delta _n)\rightarrow 0\,(n\rightarrow \infty ).$ From this and (6.8) we obtain

$$\begin{aligned} \int _{0}^{R}r{\widehat{\phi }}(r)\sin (\rho r){\mathrm{d}}r= \lim _{n\rightarrow \infty } \int _{\delta _n}^{R}r{\widehat{\phi }}(r)\sin (\rho r){\mathrm{d}}r \ge 0\qquad \forall \,\rho>0,\, R>0. \end{aligned}$$

Thus we conclude $\phi (\rho )=\lim \limits _{R\rightarrow \infty }\phi _R(\rho )\ge 0$ for all $\rho \in (0,\infty ).$ $\quad \square$

1.3 Weak projection gradient: definition and properties

Let us first list some basic notations which are widely used throughout this section.

$\bullet$ We recall that $\Pi ({\mathbf{z}})=\Pi ({\mathbf{n}}):=\mathrm{I}-{\mathbf{n}}\otimes {\mathbf{n}}\in {{{\mathbb {R}}}}^{3\times 3}$ where ${\mathbf{n}}={\mathbf{z}}/|{\mathbf{z}}|$.

$\bullet$ The vertical plane ${{{\mathbb {R}}}}^2({\mathbf{z}})$ is defined by

$$\begin{aligned} {{{\mathbb {R}}}}^2({\mathbf{z}})=\{ \mathbf{h}\in {{{\mathbb {R}}}^{3}}\,\,|\,\,{\mathbf{z}}\cdot \mathbf{h}=0\},\quad {\mathbf{z}}\ne \mathbf{0}. \end{aligned}$$

$\bullet$ Y is a Borel set in ${{{\mathbb {R}}}}^N$. In our research problems shown above, Y has been taken ${[}0,\infty )\times {{{\mathbb {R}}}^{3}}$ and ${[}0,\infty )$.

Test function spaces for scalar and vector-valued cases are defined by

$$\begin{aligned}&{{\mathcal {T}}}_{c}(Y\times {{{\mathbb {R}}}^{3}})=\big \{ \psi \in C_c(Y\times {{{\mathbb {R}}}^{3}})\,\,\big |\nonumber \\&\quad \,\, \mathrm{for\,\,any\,\, }\,{\mathbf{y}}\in Y,\,\,\psi ({\mathbf{y}}, \mathbf{0})=0,\,\, {\mathbf{z}}\mapsto \psi ({\mathbf{y}},{\mathbf{z}})\in C^1({{{\mathbb {R}}}^{3}}) , \nonumber \\&\mathrm{and}\,\, \sup _{{\mathbf{y}}\in Y,{\mathbf{z}}\in {{{\mathbb {R}}}^{3}}}|\nabla _{{\mathbf{z}}}\psi ({\mathbf{y}},{\mathbf{z}})|<\infty \big \}, \end{aligned}$$

(6.9)

$$\begin{aligned}&{{\mathcal {T}}}_{c}(Y\times {{{\mathbb {R}}}^{3}}, {{{\mathbb {R}}}^{3}})=\big \{ \Psi =(\Psi _1, \Psi _2, \Psi _3)^{\tau }\in C_c(Y\times {{{\mathbb {R}}}^{3}}, {{{\mathbb {R}}}^{3}})\,\,\big |\nonumber \\&\quad \,\, \mathrm{for\,\,any\,\, }\,{\mathbf{y}}\in Y,\,\,\Psi ({\mathbf{y}}, \mathbf{0})=\mathbf{0},\nonumber \\&{\mathbf{z}}\mapsto \Psi ({\mathbf{y}},{\mathbf{z}})\in C^1({{{\mathbb {R}}}^{3}}, {{{\mathbb {R}}}^{3}}),\,\mathrm{and}\,\, \sup _{{\mathbf{y}}\in Y,{\mathbf{z}}\in {{{\mathbb {R}}}^{3}}}|\Psi '_{{\mathbf{z}}}({\mathbf{y}},{\mathbf{z}})|<\infty \big \} \end{aligned}$$

(6.10)

where $\Psi '_{{\mathbf{z}}}({\mathbf{y}},{\mathbf{z}})=\big (\frac{\partial \Psi _i({\mathbf{y}},{\mathbf{z}})}{\partial z_j}\big )_{3\times 3}$, $|\Psi '_{{\mathbf{z}}}({\mathbf{y}},{\mathbf{z}})|=\Big (\sum \limits _{i=1}^3\sum \limits _{j=1}^3|\frac{\partial \Psi _i({\mathbf{y}},{\mathbf{z}})}{\partial z_j}|^2\Big )^{1/2}.$

Remark 6.3

(1) Denote by $\nabla _{\mathbf{z}}=(\partial _{z_1}, \partial _{z_2},\partial _{z_3})^{\tau }$ the gradient operator of functions on ${{{\mathbb {R}}}^{3}}$. Using the projection $\Pi ({\mathbf{z}})$ we have an orthogonal decomposition $\nabla _{\mathbf{z}} =\Pi ({\mathbf{z}})\nabla _{\mathbf{z}}+({\mathbf{n}}\cdot \nabla _{\mathbf{z}} ){\mathbf{n}}$, i.e.

$$\begin{aligned} \nabla _{\mathbf{z}} F({\mathbf{z}}) =\Pi ({\mathbf{z}})\nabla _{\mathbf{z}} F({\mathbf{z}})+({\mathbf{n}}\cdot \nabla _{\mathbf{z}} F({\mathbf{z}}) ){\mathbf{n}},\quad \Pi ({\mathbf{z}})\nabla _{\mathbf{z}} F({\mathbf{z}})\perp {\mathbf{n}} \end{aligned}$$

(6.11)

for all $F\in C^1({{{\mathbb {R}}}^{3}})$. This gives the differentiation of F at ${\mathbf{z}}$ along the vertical plane ${{{\mathbb {R}}}}^2({\mathbf{z}})$:

$$\begin{aligned} F({\mathbf{z}}+\mathbf{h})-F({\mathbf{z}})=\Pi ({\mathbf{z}})\nabla _{\mathbf{z}} F({\mathbf{z}})\cdot \mathbf{h} + o(\mathbf{h}),\quad \mathbf{h}\in {{{\mathbb {R}}}}^2({\mathbf{z}}) \end{aligned}$$

(6.12)

In view of (6.12), we call $\Pi ({\mathbf{z}})\nabla _{\mathbf{z}} F({\mathbf{z}})$ the gradient of F at ${\mathbf{z}}$ along the vertical plan ${{{\mathbb {R}}}}^2({\mathbf{z}})$, or simply, the projection gradient of F at ${\mathbf{z}}$. Let $\Psi \in C^1_c({{{\mathbb {R}}}^{3}},{{{\mathbb {R}}}^{3}})$. Using the identity

$$\begin{aligned} \mathbf{a}\cdot \Pi ({\mathbf{z}})\mathbf{b}= \Pi ({\mathbf{z}})\mathbf{a}\cdot \mathbf{b}=\mathbf{b}\cdot \Pi ({\mathbf{z}})\mathbf{a},\quad \mathbf{a},\mathbf{b}\in {{{\mathbb {R}}}^{3}}\end{aligned}$$

(6.13)

and integration by parts we have

$$\begin{aligned}\int _{{{{\mathbb {R}}}^{3}}}\Psi ({\mathbf{z}})\cdot \Pi ({\mathbf{z}})\nabla _{\mathbf{z}}F({\mathbf{z}}) {\mathrm{d}}{\mathbf{z}} =-\int _{{{{\mathbb {R}}}^{3}}}F({\mathbf{z}})\nabla _{\mathbf{z}}\cdot \Pi ({\mathbf{z}})\Psi ({\mathbf{z}}) {\mathrm{d}}{\mathbf{z}}.\end{aligned}$$

(2) For all $\Psi \in {{\mathcal {T}}}_{c}(Y\times {{{\mathbb {R}}}^{3}}, {{{\mathbb {R}}}^{3}})$, we have

$$\begin{aligned} \nabla _{{\mathbf{z}}} \cdot \Pi ({\mathbf{z}})\Psi ({\mathbf{y}}, {\mathbf{z}})=\nabla _{{\mathbf{z}}} \cdot \Psi ({\mathbf{y}},{\mathbf{z}})-{\mathbf{n}}^{\tau }\Psi '_{{\mathbf{z}}}({\mathbf{y}},{\mathbf{z}}){\mathbf{n}}- 2\frac{\Psi ({\mathbf{y}},{\mathbf{z}})}{|{\mathbf{z}}|}\cdot {\mathbf{n}}, \end{aligned}$$

(6.14)

where ${\mathbf{n}}={\mathbf{z}}/|{\mathbf{z}}|,\,{\mathbf{z}}\in {{{\mathbb {R}}}^{3}}{\setminus }\{\mathbf{0}\}$. Since $\Psi ({\mathbf{y}}, \mathbf{0})=\mathbf{0}$, the quotient $\frac{\Psi ({\mathbf{y}},{\mathbf{z}})}{|{\mathbf{z}}|}$ makes sense and is bounded:

$$\begin{aligned} \sup _{{\mathbf{y}}\in Y,{\mathbf{z}}\in {{{\mathbb {R}}}^{3}}{\setminus }\{\mathbf{0}\}} \frac{|\Psi ({\mathbf{y}},{\mathbf{z}})|}{|{\mathbf{z}}|}\le \sup _{{\mathbf{y}}\in Y,{\mathbf{z}}\in {{{\mathbb {R}}}^{3}}}|\Psi _{{\mathbf{z}}}'({\mathbf{y}},{\mathbf{z}})|<\infty . \end{aligned}$$

Thus $\nabla _{{\mathbf{z}}} \cdot \Pi ({\mathbf{z}})\Psi ({\mathbf{y}}, {\mathbf{z}})$ is bounded and has compact support in $Y\times {{{\mathbb {R}}}}^3$.

1.3.1 Definition of the weak projection gradients

Definition 6.4

Let $F\in L^1_{loc}(Y\times {{{\mathbb {R}}}^{3}})$. We say that $F({\mathbf{y,z}})$ has a weak projection gradient $\mathbf{D}({\mathbf{y}},{\mathbf{z}})\in {{{\mathbb {R}}}}^2({\mathbf{z}})$ in ${\mathbf{z}}\in {{{\mathbb {R}}}^{3}}{\setminus }\{\mathbf{0}\}$, if $\mathbf{D} \in L^1_{loc} (Y\times {{{\mathbb {R}}}^{3}}, {{{\mathbb {R}}}^{3}})$ and

$$\begin{aligned} \int _{Y\times {{{\mathbb {R}}}^{3}}} \Psi ({\mathbf{y}}, {\mathbf{z}})\cdot \mathbf{D}({\mathbf{y}}, {\mathbf{z}}) {\mathrm{d}}{\mathbf{z}} {\mathrm{d}}{\mathbf{y}} =-\int _{Y\times {{{\mathbb {R}}}^{3}}}F({\mathbf{y}}, {\mathbf{z}})\nabla _{\mathbf{z}} \cdot \Pi ({\mathbf{z}})\Psi ({\mathbf{y}}, {\mathbf{z}}) {\mathrm{d}}{\mathbf{z}}{\mathrm{d}}{\mathbf{y}} \end{aligned}$$

(6.15)

for all $\Psi \in {{\mathcal {T}}}_{c}(Y\times {{{\mathbb {R}}}^{3}}, {{{\mathbb {R}}}^{3}}).$ In this case, we denote

$$\begin{aligned}{} \mathbf{D}({\mathbf{y}}, {\mathbf{z}})= \Pi ({\mathbf{z}})\nabla _{\mathbf{z}} F({\mathbf{y}}, {\mathbf{z}}).\end{aligned}$$

Several remarks are in order.

Remark 6.5

 

• (1) The perpendicular property $\mathbf{D}({\mathbf{y}},{\mathbf{z}})\in {{{\mathbb {R}}}}^2({\mathbf{z}})$ is necessary for $\mathbf{D} \in L^1_{loc}(Y\times {{{\mathbb {R}}}^{3}}, {{{\mathbb {R}}}^{3}})$ satisfying (6.15). In fact, by inserting $\Psi _{\varepsilon }({\mathbf{y,z}}):=\big (\mathrm{I}-\frac{{\mathbf{z}}{\mathbf{z}}^{\tau }}{\varepsilon ^2+|{\mathbf{z}}|^2} \big )\Psi ({\mathbf{y,z}})$ with $\Psi \in {{\mathcal {T}}}_{c}(Y\times {{{\mathbb {R}}}^{3}}, {{{\mathbb {R}}}^{3}})$ into (6.15) and letting $\varepsilon \rightarrow 0^{+}$ we conclude $\mathbf{D}({\mathbf{y}},{\mathbf{z}})\cdot {\mathbf{z}}=0$ for a.e. $({\mathbf{y}},{\mathbf{z}})\in Y\times {{{\mathbb {R}}}^{3}}$.

• (2) It is obvious that, up to a set of measure zero, $\mathbf{D}({\mathbf{y}}, {\mathbf{z}})=\Pi ({\mathbf{z}})\nabla _{\mathbf{z}} F({\mathbf{y}}, {\mathbf{z}})$ is uniquely determined by the identity (6.15).

• (3) From (6.14) one sees that the condition $\Psi ({\mathbf{y}},\mathbf{0})=\mathbf{0}$ for the test functions is only used to bound $\nabla _{{\mathbf{z}}} \cdot \Pi ({\mathbf{z}})\Psi ({\mathbf{y}}, {\mathbf{z}})$; it can be removed if F satisfies that $F({\mathbf{y}},{\mathbf{z}})/|{\mathbf{z}}|\in L^1_{loc}(Y\times {{{\mathbb {R}}}^{3}})$.

To adapt Definition 6.4 to our problem, we observe that for a smooth function $F({\mathbf{v}}, {\mathbf{v}}_*)$, let ${\bar{F}}({{\mathbf{w}}},{\mathbf{z}})=F\big (\frac{\mathbf{w}+\mathbf{z}}{2}, \frac{\mathbf{w}-\mathbf{z}}{2}\big ),$ then it holds the following relation:

$$\begin{aligned} \Pi ({{\mathbf{v}}-{\mathbf{v}}}_*)\big (\nabla _{{\mathbf{v}}}-\nabla _{{\mathbf{v}}_*}\big ) F({\mathbf{v}}, {\mathbf{v}}_*)=2\Pi ({\mathbf{z}})\nabla _{{\mathbf{z}}}{\bar{F}}({{\mathbf{w}}},{\mathbf{z}})\big |_{ {\mathbf{w}}={\mathbf{v}}+{\mathbf{v}}_*,\,{\mathbf{z}}={{\mathbf{v}}-{\mathbf{v}}}_*}. \end{aligned}$$

Definition 6.6

Let $F\in L^1_{loc} (Y\times {{{{\mathbb {R}}}^{3}}\times {{{\mathbb {R}}}^{3}}})$. We say that $F({\mathbf{y}},{\mathbf{v}}, {\mathbf{v}}_*)$ has the weak projection gradient $\Pi ({{\mathbf{v}}-{\mathbf{v}}}_*)\nabla _{{{\mathbf{v}}-{\mathbf{v}}}_*}F({\mathbf{y}},{\mathbf{v}}, {\mathbf{v}}_*)$ in $\mathbf{v -v}_*\ne \mathbf{0}$, if the function ${\bar{F}}({\mathbf{y}}, {\mathbf{w}},{\mathbf{z}}):=F\big ({\mathbf{y}},\frac{{{\mathbf{w}}}+{\mathbf{z}}}{2},\frac{{{\mathbf{w}}}-{\mathbf{z}}}{2}\big )$ has the weak projection gradient $\Pi ({\mathbf{z}})\nabla _{\mathbf{z}} {\bar{F}}({\mathbf{y}},{\mathbf{w}}, {\mathbf{z}})$ in ${\mathbf{z}}\in {{{\mathbb {R}}}^{3}}{\setminus }\{\mathbf{0}\}$. In this case we define

$$\begin{aligned} \Pi ({{\mathbf{v}}-{\mathbf{v}}}_*)\nabla _{{{\mathbf{v}}-{\mathbf{v}}}_*}F({\mathbf{y}},{\mathbf{v}}, {\mathbf{v}}_*) :=2\Pi ({\mathbf{z}})\nabla _{\mathbf{z}} {\bar{F}}({\mathbf{y}},{\mathbf{w}}, {\mathbf{z}})\big |_{{\mathbf{w}}={\mathbf{v}}+{\mathbf{v}}_*,\,{\mathbf{z}}={{\mathbf{v}}-{\mathbf{v}}}_*}. \end{aligned}$$

(6.16)

From the relation between the two types of weak projection gradients one sees that in order to study the weak projection gradients defined in the second definition, it needs only to study the weak projection gradients defined in the first definition.

Let us first show that the vector-valued test function space in Definition 6.4 can be reduced to the scalar test function space.

Lemma 6.7

A function $F\in L^1_{loc} (Y\times {{{\mathbb {R}}}^{3}})$ has the weak projection gradient $\Pi ({\mathbf{z}})\nabla _{{\mathbf{z}}}F({\mathbf{y}}, {\mathbf{z}})$ in ${\mathbf{z}}\in {{{\mathbb {R}}}^{3}}{\setminus }\{\mathbf{0}\}$, if and only if there exists a vector-valued function $\mathbf{D}({\mathbf{y}},{\mathbf{z}})\in {{{\mathbb {R}}}}^2({\mathbf{z}})$ such that $\mathbf{D}\in L^1_{loc}(Y\times {{{\mathbb {R}}}^{3}}, {{{\mathbb {R}}}^{3}})$ and the equality

$$\begin{aligned} \int _{Y\times {{{\mathbb {R}}}^{3}}} \psi ({\mathbf{y}}, {\mathbf{z}})\mathbf{D}({\mathbf{y}}, {\mathbf{z}}) {\mathrm{d}}{\mathbf{z}} {\mathrm{d}}{\mathbf{y}} =-\int _{Y\times {{{\mathbb {R}}}^{3}}}F({\mathbf{y}},{\mathbf{z}})\Big (\Pi ({\mathbf{z}})\nabla _{\mathbf{z}} \psi ({\mathbf{y}},{\mathbf{z}})- 2\frac{\psi ({\mathbf{y}},{\mathbf{z}})}{|{\mathbf{z}}|}{\mathbf{n}} \Big ) {\mathrm{d}}{\mathbf{z}}{\mathrm{d}}{\mathbf{y}} \end{aligned}$$

(6.17)

holds for all $\psi \in {{\mathcal {T}}}_{c}(Y\times {{{\mathbb {R}}}^{3}})$. Moreover, if (6.17) holds for all $\psi \in {{\mathcal {T}}}_{c}(Y\times {{{\mathbb {R}}}^{3}})$, then $\mathbf{D}({\mathbf{y}}, {\mathbf{z}})= \Pi ({\mathbf{z}})\nabla _{\mathbf{z}} F({\mathbf{y}}, {\mathbf{z}})$ a.e. on $Y\times {{{\mathbb {R}}}^{3}}$.

The proof follows easily from Definition 6.4 and the implication relation that if $\psi \in {{\mathcal {T}}}_{c}(Y\times {{{\mathbb {R}}}^{3}})$ then $\psi ({\mathbf{y}},{\mathbf{z}})\mathbf{a}$ belongs to ${\mathcal T}_{c}(Y\times {{{\mathbb {R}}}^{3}}, {{{\mathbb {R}}}^{3}})$ for all constant vector $\mathbf{a}\in {{{\mathbb {R}}}^{3}}$. We omit the details here.

By definition of the projection $\Pi ({\mathbf{z}})$ we have $\Pi ({\mathbf{z}}){\mathbf{z}}=\mathbf{0}$ for all ${\mathbf{z}}\in {{{\mathbb {R}}}^{3}}{\setminus }\{\mathbf{0}\}$. From this it is easily deduced the following equalities that are often used: For any $a\in C^1((0,\infty ))$ and any ${\mathbf{z}}\in {{{\mathbb {R}}}^{3}}{\setminus }\{\mathbf{0}\}$,

$$\begin{aligned}&\Pi ({\mathbf{z}})\nabla _{{\mathbf{z}}}\big (a(|{\mathbf{z}}|)\psi ({\mathbf{z}})\big )=a(|{\mathbf{z}}|) \Pi ({\mathbf{z}})\nabla _{{\mathbf{z}}} \psi ({\mathbf{z}})\qquad \forall \, \psi \in C^1({{{\mathbb {R}}}^{3}}) \end{aligned}$$

(6.18)

$$\begin{aligned}&\nabla _{{\mathbf{z}}}\cdot \Pi ({\mathbf{z}})a(|{\mathbf{z}}|)\psi ({\mathbf{z}})=a(|{\mathbf{z}}|) \nabla _{{\mathbf{z}}}\cdot \Pi ({\mathbf{z}})\psi ({\mathbf{z}})\qquad \forall \, \psi \in C^1({{{\mathbb {R}}}^{3}},{{{\mathbb {R}}}^{3}}). \end{aligned}$$

(6.19)

1.3.2 Some properties of weak projection gradients

The first part of the following lemma shows that if in addition that $F({\mathbf{y}}, {\mathbf{z}})/|{\mathbf{z}}|\in L^1_{loc}(Y\times {{{\mathbb {R}}}^{3}})$, then the condition $\psi ({\mathbf{y}}, \mathbf{0})=0$ in test function space ${{\mathcal {T}}}_{c}(Y\times {{{\mathbb {R}}}^{3}})$ can be removed.

Lemma 6.8

Suppose $F\in L^1_{loc}(Y\times {{{\mathbb {R}}}^{3}})$ has the weak projection gradient $\Pi ({\mathbf{z}})\nabla _{\mathbf{z}} F({\mathbf{y}}, {\mathbf{z}})$ in ${\mathbf{z}}\in {{{\mathbb {R}}}^{3}}{\setminus }\{\mathbf{0}\}$.

• (a) If $F({\mathbf{y}}, {\mathbf{z}})/|{\mathbf{z}}|\in L^1_{loc}(Y\times {{{\mathbb {R}}}^{3}})$, then for any $\psi \in C_c^1({{{\mathbb {R}}}^{3}})$, there is a null set $Z_{\psi }\subset Y$ such that for all ${\mathbf{y}}\in Y{\setminus } Z_{\psi }$

  $$\begin{aligned} \int _{{{{\mathbb {R}}}^{3}}}\psi ({\mathbf{z}})\Pi ({\mathbf{z}})\nabla _{\mathbf{z}} F({\mathbf{y}}, {\mathbf{z}}) {\mathrm{d}}{\mathbf{z}} =-\int _{{{{\mathbb {R}}}^{3}}}F({\mathbf{y}}, {\mathbf{z}})\Big (\Pi ({\mathbf{z}})\nabla \psi ({\mathbf{z}})- 2\frac{\psi ({\mathbf{z}})}{|{\mathbf{z}}|}{\mathbf{n}}\Big ){\mathrm{d}}{\mathbf{z}}. \end{aligned}$$

  (6.20)

• (b) Let $a\in C^1((0,\infty ))$. If $a(|{\mathbf{z}}|)F({\mathbf{y}},{\mathbf{z}})$ and $a(|{\mathbf{z}}|)\Pi ({\mathbf{z}})\nabla _{{\mathbf{z}}} F({\mathbf{y}},{\mathbf{z}})$ both belong to $L^1_{loc}(Y\times {{{\mathbb {R}}}^{3}})$, then $a(|{\mathbf{z}}|)F({\mathbf{y}},{\mathbf{z}})$ also has the weak projection gradient $\Pi ({\mathbf{z}})\nabla _{{\mathbf{z}}}\big (a(|{\mathbf{z}}|) F({\mathbf{y}},{\mathbf{z}})\big )$ in ${\mathbf{z}}\in {{{\mathbb {R}}}^{3}}{\setminus }\{\mathbf{0}\}$ and it holds

  $$\begin{aligned} \Pi ({\mathbf{z}})\nabla _{{\mathbf{z}}}\big (a(|{\mathbf{z}}|) F({\mathbf{y}},{\mathbf{z}})\big )=a(|{\mathbf{z}}|)\Pi ({\mathbf{z}})\nabla _{{\mathbf{z}}}F({\mathbf{y}},{\mathbf{z}}). \end{aligned}$$

  (6.21)

Proof

(a): First of all from the assumptions we have for any $R_1,R_2\in {{{\mathbb {N}}}}$ that

$$\begin{aligned}&\int _{|{\mathbf{y}}|<R_1}\bigg (\int _{|{\mathbf{z}}|<R_2}\big |\Pi ({\mathbf{z}})\nabla _{\mathbf{z}} F({\mathbf{y}}, {\mathbf{z}}) \big |{\mathrm{d}}{\mathbf{z}}\bigg ){\mathrm{d}}{\mathbf{y}}= \int _{|{\mathbf{y}}|<R_1, |{\mathbf{z}}|<R_2}\big |\Pi ({\mathbf{z}})\nabla _{\mathbf{z}} F({\mathbf{y}}, {\mathbf{z}}) \big |{\mathrm{d}}{\mathbf{z}}{\mathrm{d}}{\mathbf{y}}<\infty ,\\&\int _{|{\mathbf{y}}|<R_1}\bigg (\int _{|{\mathbf{z}}|<R_2}|F({\mathbf{y}}, {\mathbf{z}})|\big (1+\frac{1}{|{\mathbf{z}}|}\big ) {\mathrm{d}}{\mathbf{z}}\bigg ){\mathrm{d}}{\mathbf{y}} = \int _{|{\mathbf{y}}|<R_1, |{\mathbf{z}}|<R_2}|F({\mathbf{y}}, {\mathbf{z}})|\big (1+\frac{1}{|{\mathbf{z}}|}\big ) {\mathrm{d}}{\mathbf{z}}{\mathrm{d}}{\mathbf{y}}<\infty . \end{aligned}$$

Therefore all integrals in the rest of the proof are absolutely convergent and

$$\begin{aligned} \int _{|{\mathbf{z}}|<R_2}\big |\Pi ({\mathbf{z}})\nabla _{\mathbf{z}} F({\mathbf{y}}, {\mathbf{z}}) \big |{\mathrm{d}}{\mathbf{z}}<\infty ,\quad \int _{|{\mathbf{z}}|<R_2}|F({\mathbf{y}}, {\mathbf{z}})|\big (1+\frac{1}{|{\mathbf{z}}|}\big ) {\mathrm{d}}{\mathbf{z}}<\infty \end{aligned}$$

for all $R_2\in {{{\mathbb {N}}}}$ and all ${\mathbf{y}}\in Y{\setminus } Z_0$, where $Z_0$ is a common null set in Y.

Let $\zeta \in C^{\infty }({{{\mathbb {R}}}}_{\ge 0})$ satisfy $0\le \zeta (\cdot )\le 1$ on ${{{\mathbb {R}}}}_{\ge 0}$, and $\zeta (r)=0$ for all $r\in {[}0,1]$, $\zeta (r)=1$ for all $r\ge 2$. Let $\zeta _{\delta }(|{\mathbf{z}}|)=\zeta (|{\mathbf{z}}|/\delta ),\delta >0$. Take any $\varphi \in C_c(Y)$ and $\psi \in C_c^1({{{\mathbb {R}}}^{3}})$. Let $\psi _{\delta }({\mathbf{y}},{\mathbf{z}})=\varphi ({\mathbf{y}})\zeta _{\delta }(|{\mathbf{z}}|)\psi ({\mathbf{z}}).$ Then $\psi _{\delta }\in {{\mathcal {T}}}_{c}(Y\times {{{\mathbb {R}}}^{3}})$ and thanks to the identity (6.18) we have $\Pi ({\mathbf{z}})\nabla _{\mathbf{z}} \psi _{\delta }({\mathbf{y}},{\mathbf{z}}) =\varphi ({\mathbf{y}})\zeta _{\delta }(|{\mathbf{z}}|)\Pi ({\mathbf{z}}) \nabla \psi ({\mathbf{z}})$ and so, by Lemma 6.7,

$$\begin{aligned}&\int _{Y\times {{{\mathbb {R}}}^{3}}}\varphi ({\mathbf{y}})\zeta _{\delta }(|{\mathbf{z}}|)\psi ({\mathbf{z}}) \Pi ({\mathbf{z}})\nabla _{\mathbf{z}} F({\mathbf{y}}, {\mathbf{z}}) {\mathrm{d}}{\mathbf{z}} {\mathrm{d}}{\mathbf{y}}\\&\quad =-\int _{Y\times {{{\mathbb {R}}}^{3}}}\varphi ({\mathbf{y}})\zeta _{\delta }(|{\mathbf{z}}|)F({\mathbf{y}},{\mathbf{z}})\Big (\Pi ({\mathbf{z}})\nabla _{\mathbf{z}} \psi ({\mathbf{z}})- 2\frac{\psi ({\mathbf{z}})}{|{\mathbf{z}}|}{\mathbf{n}} \Big ) {\mathrm{d}}{\mathbf{z}}{\mathrm{d}}{\mathbf{y}}. \end{aligned}$$

By definition of the test functions and the fact that $\lim \limits _{\delta \rightarrow 0^{+}}\zeta _{\delta }(|{\mathbf{z}}|)=1$ for all ${\mathbf{z}}\in {{{\mathbb {R}}}^{3}}{\setminus }\{\mathbf{0}\}$, dominated convergence theorem yields that

$$\begin{aligned}\int _{Y\times {{{\mathbb {R}}}^{3}}} \varphi ({\mathbf{y}})\psi ({\mathbf{z}}) \Pi ({\mathbf{z}})\nabla _{\mathbf{z}} F({\mathbf{y}}, {\mathbf{z}}) {\mathrm{d}}{\mathbf{z}} {\mathrm{d}}{\mathbf{y}} =-\int _{Y\times {{{\mathbb {R}}}^{3}}}F({\mathbf{y}},{\mathbf{z}})\varphi ({\mathbf{y}})\Big ( \Pi ({\mathbf{z}})\nabla \psi ({\mathbf{z}})- 2\frac{\psi ({\mathbf{z}})}{|{\mathbf{z}}|}{\mathbf{n}} \Big ) {\mathrm{d}}{\mathbf{z}}{\mathrm{d}}{\mathbf{y}}.\end{aligned}$$

is a null set $Z_{\psi }\subset Y$ such that (6.20) holds for all ${\mathbf{y}}\in Y{\setminus } Z_{\psi }$.

(b): For any $\Psi \in {{\mathcal {T}}}_{c}(Y\times {{{\mathbb {R}}}^{3}},{{{\mathbb {R}}}^{3}})$, the function $\zeta _{\delta }(|{\mathbf{z}}|)a(|{\mathbf{z}}|)\Psi ({\mathbf{y}}, {\mathbf{z}})$ ($\delta >0$) belongs to ${{\mathcal {T}}}_{c}(Y\times {{{\mathbb {R}}}^{3}},{{{\mathbb {R}}}^{3}})$ and using (6.18) we have $\nabla _{\mathbf{z}} \cdot \Pi ({\mathbf{z}})\zeta _{\delta }(|{\mathbf{z}}|)a(|{\mathbf{z}}|)\Psi ({\mathbf{y}}, {\mathbf{z}}) =\zeta _{\delta }(|{\mathbf{z}}|)a(|{\mathbf{z}}|)\nabla _{\mathbf{z}} \cdot \Pi ({\mathbf{z}})\Psi ({\mathbf{y}}, {\mathbf{z}})$ for all ${\mathbf{z}}\in {{{\mathbb {R}}}^{3}}{\setminus }\{\mathbf{0}\}$ and so

$$\begin{aligned}\int _{Y\times {{{\mathbb {R}}}^{3}}} \Psi \zeta _{\delta }(|{\mathbf{z}}|)a(|{\mathbf{z}}|)\cdot \Pi ({\mathbf{z}})\nabla _{{\mathbf{z}}} F{\mathrm{d}}{\mathbf{z}} {\mathrm{d}}{\mathbf{y}}=-\int _{Y\times {{{\mathbb {R}}}^{3}}}F\zeta _{\delta }(|{\mathbf{z}}|)a(|{\mathbf{z}}|)\nabla _{\mathbf{z}} \cdot \Pi ({\mathbf{z}})\Psi {\mathrm{d}}{\mathbf{z}}{\mathrm{d}}{\mathbf{y}}.\end{aligned}$$

By the assumptions in (b) and recalling that $\Psi \in {{\mathcal {T}}}_{c}(Y\times {{{\mathbb {R}}}^{3}},{{{\mathbb {R}}}^{3}})$, letting $\delta \rightarrow 0^+$ we obtain

$$\begin{aligned}\int _{Y\times {{{\mathbb {R}}}^{3}}} \Psi ({\mathbf{y}}, {\mathbf{z}}) a(|{\mathbf{z}}|)\cdot \Pi ({\mathbf{z}})\nabla _{{\mathbf{z}}} F({\mathbf{y}}, {\mathbf{z}}){\mathrm{d}}{\mathbf{z}} {\mathrm{d}}{\mathbf{y}}= -\int _{Y\times {{{\mathbb {R}}}^{3}}}F({\mathbf{y}}, {\mathbf{z}}) a(|{\mathbf{z}}|)\nabla _{\mathbf{z}} \cdot \Pi ({\mathbf{z}})\Psi ({\mathbf{y}}, {\mathbf{z}}) {\mathrm{d}}{\mathbf{z}}{\mathrm{d}}{\mathbf{y}}. \end{aligned}$$

This proves (6.21) by definition of weak projection gradients. $\quad \square$

Note that for the case $a(r)=r^{\alpha }$ with $\alpha >0$, the condition “ $|{\mathbf{z}}|^{\alpha }F({\mathbf{y}},{\mathbf{z}})$ and $|{\mathbf{z}}|^{\alpha }\Pi ({\mathbf{z}})\nabla _{{\mathbf{z}}} F({\mathbf{y}},{\mathbf{z}})$ both belong to $L^1_{loc}(Y\times {{{\mathbb {R}}}^{3}})$" is satisfied automatically and thus for this case the conclusion of the lemma holds true. Whereas for the case $a(r)=r^{-\alpha }$ ($\alpha >0$), the condition “ $|{\mathbf{z}}|^{-\alpha }F({\mathbf{y}},{\mathbf{z}})$ and $|{\mathbf{z}}|^{-\alpha }\Pi ({\mathbf{z}})\nabla _{{\mathbf{z}}} F({\mathbf{y}},{\mathbf{z}})$ both belong to $L^1_{loc}(Y\times {{{\mathbb {R}}}^{3}})$" should be checked out in practice.

The following two lemmas are important:

Lemma 6.9

Suppose that $F\in L^2_{loc}(Y\times {{{\mathbb {R}}}^{3}})$ has the weak projection gradient $\Pi ({\mathbf{z}})\nabla _{{\mathbf{z}}} F({\mathbf{y}},{\mathbf{z}})$ in ${\mathbf{z}}\in {{{\mathbb {R}}}^{3}}{\setminus }\{\mathbf{0}\}$ and that $\Pi ({\mathbf{z}})\nabla _{{\mathbf{z}}} F({\mathbf{y}},{\mathbf{z}})$ belongs to $L^2_{loc}(Y\times {{{\mathbb {R}}}^{3}}, {{{\mathbb {R}}}^{3}})$. Then $F^2$ also has the weak projection gradient $\Pi ({\mathbf{z}})\nabla _{{\mathbf{z}}}F^2({\mathbf{y}},{\mathbf{z}})$ in ${\mathbf{z}}\in {{{\mathbb {R}}}^{3}}{\setminus }\{\mathbf{0}\}$ and

$$\begin{aligned} \Pi ({\mathbf{z}})\nabla _{{\mathbf{z}}}F^2({\mathbf{y}},{\mathbf{z}})=2F({\mathbf{y}},{\mathbf{z}})\Pi ({\mathbf{z}})\nabla _{{\mathbf{z}}}F({\mathbf{y}},{\mathbf{z}}). \end{aligned}$$

Lemma 6.10

Suppose that $0< F\in L^1_{loc}(Y\times {{{\mathbb {R}}}^{3}})$ has the weak projection gradient $\Pi ({\mathbf{z}})\nabla _{{\mathbf{z}}} F({\mathbf{y}},{\mathbf{z}})$ in ${\mathbf{z}}\in {{{\mathbb {R}}}^{3}}{\setminus }\{\mathbf{0}\}$ and satisfies

$$\begin{aligned} \inf _{{\mathbf{y}}\in Y, |{\mathbf{y}}|\le R_1, |{\mathbf{z}}|\le R_2}F({\mathbf{y}}, {\mathbf{z}})>0 \qquad \forall \, 0<R_1, R_2<\infty , \end{aligned}$$

$$\begin{aligned} \frac{F({\mathbf{y}},{\mathbf{z}})}{|{\mathbf{z}}|}\,\,\mathrm{belongs \,\,to}\,\,\in L^1_{loc}(Y\times {{{\mathbb {R}}}^{3}}). \end{aligned}$$

(6.22)

Then $\sqrt{F({\mathbf{y}},{\mathbf{z}})}$ has also the weak projection gradient $\Pi ({\mathbf{z}})\nabla _{{\mathbf{z}}} \sqrt{F({\mathbf{y}},{\mathbf{z}})}$ in ${\mathbf{z}}\in {{{\mathbb {R}}}^{3}}{\setminus }\{\mathbf{0}\}$ and it holds

$$\begin{aligned} \Pi ({\mathbf{z}})\nabla _{{\mathbf{z}}} \sqrt{F({\mathbf{y}},{\mathbf{z}})}=\frac{\Pi ({\mathbf{z}})\nabla _{{\mathbf{z}}} F({\mathbf{y}},{\mathbf{z}})}{2\sqrt{F({\mathbf{y}},{\mathbf{z}})}}. \end{aligned}$$

To prove the two lemmas, we first make a common preparation. Note that the assumption $F\in L^2_{loc}(Y\times {{{\mathbb {R}}}^{3}})$ implies that F satisfies (6.22).

Lemma 6.11

Let $j(t)=c\exp (\frac{-1}{1-t})1_{\{t<1\}}$ and $J(\mathbf{u})=j(|\mathbf{u}|^2)$ with $c>0$ such that $\int _{{{{\mathbb {R}}}^{3}}}J(\mathbf{u}){\mathrm{d}}\mathbf{u}=1.$ Set $J_{\delta }(\mathbf{u})=\delta ^{-3}J(\delta ^{-1}{} \mathbf{u}), \delta >0.$ Suppose $F\in L^1_{loc}(Y\times {{{\mathbb {R}}}^{3}})$ satisfies (6.22) and has the weak projection gradient $\Pi ({\mathbf{z}})\nabla _{{\mathbf{z}}} F({\mathbf{y}},{\mathbf{z}})$ in ${\mathbf{z}}\in {{{\mathbb {R}}}^{3}}{\setminus }\{\mathbf{0}\}$. Let $F_{\delta }({\mathbf{y}},{\mathbf{z}})=(J_{\delta }*F({\mathbf{y}},\cdot ))({\mathbf{z}}).$ Then for any any $\delta >0$ and any ${\mathbf{z}}\in {{{\mathbb {R}}}^{3}}{\setminus }\{\mathbf{0}\}$, there is a null set $Z_{\delta ,{\mathbf{z}}}\subset Y$, such that for all ${\mathbf{y}}\in Y{\setminus } Z_{\delta ,{\mathbf{z}}}$,

$$\begin{aligned}&\Pi ({\mathbf{z}})\nabla _{\mathbf{z}} F_{\delta }({\mathbf{y}},{\mathbf{z}})-\Pi ({\mathbf{z}})\nabla _{\mathbf{z}} F({\mathbf{y}},{\mathbf{z}})\nonumber \\&\quad = \int _{|\mathbf{u}|\le 1}J(\mathbf{u})\Pi ({\mathbf{z}})\Big (\Pi ({\mathbf{z}}-\delta \mathbf{u})\nabla _{\mathbf{z}} F({\mathbf{y}},{\mathbf{z}}-\delta \mathbf{u}) -\Pi ({\mathbf{z}})\nabla _{\mathbf{z}} F({\mathbf{y}},{\mathbf{z}})\Big ) {\mathrm{d}}{} \mathbf{u}\nonumber \\&\qquad + \int _{|\mathbf{u}|\le 1}\big (F({\mathbf{y}},{\mathbf{z}}-\delta \mathbf{u})-F({\mathbf{y}},{\mathbf{z}})\big )\Big \{\big (\nabla J(\mathbf{u}) \nonumber \\&\qquad \cdot \frac{{\mathbf{z}}-\delta \mathbf{u}}{|{\mathbf{z}}-\delta \mathbf{u}|}\big )\frac{-\Pi ({\mathbf{z}})\mathbf{u}}{|{\mathbf{z}}-\delta \mathbf{u}|} -2J(\mathbf{u})\Pi ({\mathbf{z}})\frac{{\mathbf{z}}-\delta \mathbf{u}}{|{\mathbf{z}}-\delta \mathbf{u}|^2}\Big \}{\mathrm{d}}\mathbf{u}, \end{aligned}$$

(6.23)

and

$$\begin{aligned}&\Pi ({\mathbf{z}})\nabla _{\mathbf{z}} F_{\delta }({\mathbf{y}},{\mathbf{z}})-\Pi ({\mathbf{z}})\nabla _{\mathbf{z}} F({\mathbf{y}},{\mathbf{z}}) \nonumber \\&\quad = \int _{|\mathbf{u}|\le 1}J(\mathbf{u})\Pi ({\mathbf{z}})\Big (\Pi ({\mathbf{z}}-\delta \mathbf{u})\nabla _{\mathbf{z}} F({\mathbf{y}},{\mathbf{z}}-\delta \mathbf{u}) -\Pi ({\mathbf{z}})\nabla _{\mathbf{z}} F({\mathbf{y}},{\mathbf{z}})\Big ) {\mathrm{d}}{} \mathbf{u} \nonumber \\&\quad \quad + \int _{|\mathbf{u}|\le 1}\Big (\frac{F({\mathbf{y}},{\mathbf{z}}-\delta \mathbf{u})}{|{\mathbf{z}}-\delta \mathbf{u}|}- \frac{F({\mathbf{y}},{\mathbf{z}})}{|{\mathbf{z}}|}\Big )\Big \{\big (\nabla J(\mathbf{u}) \cdot \frac{{\mathbf{z}}-\delta \mathbf{u}}{|{\mathbf{z}}-\delta \mathbf{u}|}\big )\big (-\Pi ({\mathbf{z}})\mathbf{u}\big ) \nonumber \\&\qquad -2J(\mathbf{u})\Pi ({\mathbf{z}})\frac{{\mathbf{z}}-\delta \mathbf{u}}{|{\mathbf{z}}-\delta \mathbf{u}|}\Big \}{\mathrm{d}}{} \mathbf{u}. \end{aligned}$$

(6.24)

Proof

Since $({\mathbf{y}},\mathbf{u})\mapsto \frac{F({\mathbf{y}},\mathbf{u})}{|\mathbf{u}|}$ belongs to $L^1_{loc}(Y\times {{{\mathbb {R}}}^{3}})$, applying Lemma 6.8(a) to $({\mathbf{y}}, \mathbf{u})\mapsto F({\mathbf{y}}, \mathbf{u})$ and taking $\mathbf{u}\mapsto J_{\delta }({\mathbf{z}}-\mathbf{u})$ as a test function with ${\mathbf{z}}, \delta$ fixed, we have, for a null set $Z_{\delta ,{\mathbf{z}}}\subset Y$ that

$$\begin{aligned}&\int _{{{{\mathbb {R}}}^{3}}} J_{\delta }({\mathbf{z}}-\mathbf{u})\Pi (\mathbf{u})\nabla _\mathbf{u} F({\mathbf{y}},\mathbf{u}) {\mathrm{d}}{} \mathbf{u}\\&\quad =-\int _{{{{\mathbb {R}}}^{3}}}F({\mathbf{y}},\mathbf{u})\Big (\Pi (\mathbf{u})\nabla _{\mathbf{u}}( J_{\delta }({\mathbf{z}}-\mathbf{u})) -2\frac{J_{\delta }({\mathbf{z}}-\mathbf{u})}{|\mathbf{u}|^2}{} \mathbf{u}\Big ){\mathrm{d}}{} \mathbf{u}\\&\quad =\int _{{{{\mathbb {R}}}^{3}}}F({\mathbf{y}},\mathbf{u}) \nabla _{{\mathbf{z}}}J_{\delta }({\mathbf{z}}-\mathbf{u}){\mathrm{d}}{} \mathbf{u}-\int _{{{{\mathbb {R}}}^{3}}}F({\mathbf{y}},\mathbf{u}) \Big (\big (\nabla _{{\mathbf{z}}}J_{\delta }({\mathbf{z}}-\mathbf{u})\cdot \frac{\mathbf{u}}{|\mathbf{u}|}\big )\frac{\mathbf{u}}{|\mathbf{u}|}\\&\qquad \qquad -2J_{\delta }({\mathbf{z}}-\mathbf{u})\frac{\mathbf{u}}{|\mathbf{u}|^2}\Big ){\mathrm{d}}{} \mathbf{u} \end{aligned}$$

for all ${\mathbf{y}}\in Y{\setminus } Z_{\delta ,{\mathbf{z}}}$. This gives

$$\begin{aligned}&\nabla _{{\mathbf{z}}} F_{\delta }({\mathbf{y}},{\mathbf{z}})=\int _{{{{\mathbb {R}}}^{3}}}F({\mathbf{y}},\mathbf{u}) \nabla _{{\mathbf{z}}}J_{\delta }({\mathbf{z}}-\mathbf{u}){\mathrm{d}}{} \mathbf{u} \\&\quad = \int _{|\mathbf{u}|<1}J(\mathbf{u})\Pi ({\mathbf{z}}-\delta \mathbf{u})\nabla _{\mathbf{z}}F({\mathbf{y}},{\mathbf{z}}-\delta \mathbf{u}) {\mathrm{d}}{} \mathbf{u} \\&\qquad + \int _{|\mathbf{u}|<1}F({\mathbf{y}},{\mathbf{z}}-\delta \mathbf{u})\Big (\big (\delta ^{-1}\nabla J(\mathbf{u}) \cdot \frac{{\mathbf{z}}-\delta \mathbf{u}}{|{\mathbf{z}}-\delta \mathbf{u}|}\big )\frac{{\mathbf{z}}-\delta \mathbf{u}}{|{\mathbf{z}}-\delta \mathbf{u}|} -2J(\mathbf{u}) \frac{{\mathbf{z}}-\delta \mathbf{u}}{|{\mathbf{z}}-\delta \mathbf{u}|^2}\Big ){\mathrm{d}}{} \mathbf{u}. \end{aligned}$$

and so

$$\begin{aligned}&\Pi ({\mathbf{z}}) \nabla _{{\mathbf{z}}} F_{\delta }({\mathbf{y}},{\mathbf{z}})-\Pi ({\mathbf{z}}) \nabla _{{\mathbf{z}}} F({\mathbf{y}},{\mathbf{z}})\\&\quad = \int _{|\mathbf{u}|<1}J(\mathbf{u})\Pi ({\mathbf{z}})\Big (\Pi ({\mathbf{z}}-\delta \mathbf{u})\nabla _{\mathbf{z}} F({\mathbf{y}},{\mathbf{z}}-\delta \mathbf{u}) -\Pi ({\mathbf{z}})\nabla _{\mathbf{z}} F({\mathbf{y}},{\mathbf{z}})\Big ) {\mathrm{d}}{} \mathbf{u} \\&\quad \quad + \int _{|\mathbf{u}|<1}F({\mathbf{y}},{\mathbf{z}}-\delta \mathbf{u})\Big \{\big (\nabla J(\mathbf{u}) \cdot \frac{{\mathbf{z}}-\delta \mathbf{u}}{|{\mathbf{z}}-\delta \mathbf{u}|}\big )\delta ^{-1}\Pi ({\mathbf{z}})\frac{{\mathbf{z}}-\delta \mathbf{u}}{|{\mathbf{z}}-\delta \mathbf{u}|} -2J(\mathbf{u})\Pi ({\mathbf{z}}) \frac{{\mathbf{z}}-\delta \mathbf{u}}{|{\mathbf{z}}-\delta \mathbf{u}|^2}\Big \}{\mathrm{d}}{} \mathbf{u} \end{aligned}$$

for all ${\mathbf{y}}\in Y{\setminus } Z_{\delta ,{\mathbf{z}}}$. Here we have used $\Pi ({\mathbf{z}})^2=\Pi ({\mathbf{z}})$.

To prove (6.23) and (6.24), we then need only to prove the following claim: for any Borel measurable function G(r, s, t) on ${[}0,1]\times {{{\mathbb {R}}}}\times {[}0,\infty )$ satisfying $|G(r,s, t)|\le C\frac{1}{t^{\alpha }}$ for all $(r,s,t)\in {[}0,1]\times {{{\mathbb {R}}}} \times (0,\infty )$ for some constants $0<C<\infty , \alpha <3$, we have

$$\begin{aligned} \int _{|\mathbf{u}|\le 1}G\big (|\mathbf{u}|, \mathbf{u} \cdot ({\mathbf{z}}-\delta \mathbf{u}), |{\mathbf{z}}-\delta \mathbf{u}|\big )\Pi ({\mathbf{z}})({\mathbf{z}}-\delta \mathbf{u}) {\mathrm{d}}{} \mathbf{u}=\mathbf{0},\quad {\mathbf{z}}\ne \mathbf{0},\delta >0. \end{aligned}$$

(6.25)

In fact, from the assumption on G we see that the integral below is absolutely convergent, and recalling ${\mathbf{n}}={\mathbf{z}}/|{\mathbf{z}}|$, $\Pi ({\mathbf{z}}){\mathbf{z}}=\mathbf{0}$, and (2.2) we compute

$$\begin{aligned}&\int _{|\mathbf{u}|\le 1} G\big (|\mathbf{u}|, \mathbf{u} \cdot ({\mathbf{z}}-\delta \mathbf{u}), |{\mathbf{z}}-\delta \mathbf{u}|\big )\Pi ({\mathbf{z}})({\mathbf{z}}-\delta \mathbf{u}) {\mathrm{d}}{} \mathbf{u}\\&\quad =-\delta \int _{0}^{1}r^3\int _{0}^{\pi }\sin ^2(\theta )\\&\qquad \times G\big (r, r|{\mathbf{z}}|\cos (\theta )-\delta r^2, \sqrt{|{\mathbf{z}}|^2- 2\delta |{\mathbf{z}}|r \cos (\theta )+\delta ^2r^2} \big ) \bigg (\int _{{{{\mathbb {S}}}}^1({\mathbf{n}})}\sigma {\mathrm{d}}\sigma \bigg ) {\mathrm{d}}\theta {\mathrm{d}}r =\mathbf{0}, \end{aligned}$$

where in the last equality we have used $\int _{{{{\mathbb {S}}}}^1({\mathbf{n}})}\sigma {\mathrm{d}}\sigma =\mathbf{0}.$

Now since $J(\mathbf{u})=j(|\mathbf{u}|^2), \nabla J(\mathbf{u})= 2j'(|\mathbf{u}|^2)\mathbf{u}$, applying (6.25) gives

$$\begin{aligned}&\int _{|\mathbf{u}|<1}\Big \{\big (\nabla J(\mathbf{u}) \cdot \frac{{\mathbf{z}}-\delta \mathbf{u}}{|{\mathbf{z}}-\delta \mathbf{u}|}\big )\delta ^{-1}\Pi ({\mathbf{z}})\frac{{\mathbf{z}}-\delta \mathbf{u}}{|{\mathbf{z}}-\delta \mathbf{u}|} -2J(\mathbf{u})\Pi ({\mathbf{z}}) \frac{{\mathbf{z}}-\delta \mathbf{u}}{|{\mathbf{z}}-\delta \mathbf{u}|^2}\Big \}{\mathrm{d}}{} \mathbf{u} =\mathbf{0},\\&\quad \int _{|\mathbf{u}|\le 1}\Big \{\big (\nabla J(\mathbf{u}) \cdot \frac{{\mathbf{z}}-\delta \mathbf{u}}{|{\mathbf{z}}-\delta \mathbf{u}|}\big )\delta ^{-1}\Pi ({\mathbf{z}})({\mathbf{z}}-\delta \mathbf{u}) -2J(\mathbf{u})\Pi ({\mathbf{z}})\frac{{\mathbf{z}}-\delta \mathbf{u}}{|{\mathbf{z}}-\delta \mathbf{u}|}\Big \}{\mathrm{d}}{} \mathbf{u} =\mathbf{0}. \end{aligned}$$

These together with $\delta ^{-1}\Pi ({\mathbf{z}})({\mathbf{z}}-\delta \mathbf{u})= \Pi ({\mathbf{z}})(-\mathbf{u})$ yield (6.23) and (6.24). $\quad \square$

We also need the following elementary property:

Lemma 6.12

Let $1\le p<\infty , l\in {{{\mathbb {N}}}}, g\in L^p_{loc}(Y\times {{{\mathbb {R}}}^{3}}, {{{\mathbb {R}}}}^l)$. Then for all $0<R<\infty$ we have

$$\begin{aligned}&\int _{{\mathbf{y}}\in Y, |{\mathbf{y}}|\le R, |{\mathbf{z}}|\le R}|g({\mathbf{y}},{\mathbf{z}}+\mathbf{h})-g({\mathbf{y}},{\mathbf{z}})|^p{\mathrm{d}}{\mathbf{z}}{\mathrm{d}}{\mathbf{y}}\rightarrow 0\quad \mathrm{as}\quad \mathbf{h}\rightarrow \mathbf{0},\\&\sup _{|\mathbf{h}|<1} \int _{{\mathbf{y}}\in Y, |{\mathbf{y}}|\le R, |{\mathbf{z}}|\le R} |g({\mathbf{y}},{\mathbf{z}}+\mathbf{h})-g({\mathbf{y}},{\mathbf{z}})|^p{\mathrm{d}}{\mathbf{z}}{\mathrm{d}}{\mathbf{y}}\le 2^p\int _{{\mathbf{y}}\in Y, |{\mathbf{y}}|\le R, |{\mathbf{z}}|\le R+1} |g({\mathbf{y}}, {\mathbf{z}})|^p{\mathrm{d}}{\mathbf{z}}{\mathrm{d}}{\mathbf{y}}. \end{aligned}$$

Proof of Lemma 6.9

Let F satisfy the assumptions in Lemma 6.9 and let $F_{\delta }({\mathbf{y}},{\mathbf{z}})=(J_{\delta }*F({\mathbf{y}},\cdot ))({\mathbf{z}})$. Using Lemma 6.12, one has

$$\begin{aligned}&\sup _{0<\delta <1} \int _{{\mathbf{y}}\in Y, |{\mathbf{y}}|\le R, |{\mathbf{z}}|\le R}|F_{\delta }({\mathbf{y}},{\mathbf{z}})|^2{\mathrm{d}}{\mathbf{z}}{\mathrm{d}}{\mathbf{y}} \le \int _{{\mathbf{y}}\in Y, |{\mathbf{y}}|\le R,|{\mathbf{z}}|\le R+1}|F({\mathbf{y}},{\mathbf{z}})|^2{\mathrm{d}}{\mathbf{z}}{\mathrm{d}}{\mathbf{y}},\\&\int _{{\mathbf{y}}\in Y, |{\mathbf{y}}|\le R, |{\mathbf{z}}|\le R}|F_{\delta }({\mathbf{y}},{\mathbf{z}})-F({\mathbf{y}},{\mathbf{z}})|^2{\mathrm{d}}{\mathbf{z}}{\mathrm{d}}{\mathbf{y}}\rightarrow 0\quad \mathrm{as}\quad \delta \rightarrow 0^{+}. \end{aligned}$$

Since $F_{\delta }({\mathbf{y}},{\mathbf{z}})$ is smooth in ${\mathbf{z}}$, we have $\nabla _{{\mathbf{z}}}(F_{\delta }({\mathbf{y}},{\mathbf{z}})^2) =2F_{\delta }({\mathbf{y}},{\mathbf{z}})\nabla _{{\mathbf{z}}}F_{\delta }({\mathbf{y}},{\mathbf{z}})$ and so for any $\Psi \in {{\mathcal {T}}}_{c}(Y\times {{{\mathbb {R}}}^{3}}, {{{\mathbb {R}}}^{3}})$

$$\begin{aligned} \int _{Y\times {{{\mathbb {R}}}^{3}}} \Psi ({\mathbf{y}}, {\mathbf{z}})\cdot 2F_{\delta }({\mathbf{y}},{\mathbf{z}})\Pi ({\mathbf{z}})\nabla _{{\mathbf{z}}}F_{\delta }({\mathbf{y}},{\mathbf{z}}) {\mathrm{d}}{\mathbf{z}}{\mathrm{d}}{\mathbf{y}} =- \int _{Y\times {{{\mathbb {R}}}^{3}}} F^2_{\delta }({\mathbf{y}},{\mathbf{z}})\nabla _{{\mathbf{z}}}\cdot \Pi ({\mathbf{z}})\Psi ({\mathbf{y}},{\mathbf{z}}) {\mathrm{d}}{\mathbf{z}}{\mathrm{d}}{\mathbf{y}}. \end{aligned}$$

(6.26)

Convergence of l.h.s of (6.26):  By change of variables and simple estimates it is not difficult to prove that there is a constant $0<C<\infty$ such that

$$\begin{aligned} \int _{|\mathbf{u}|<1}\Big |\frac{\Pi ({\mathbf{z}})\mathbf{u}}{|{\mathbf{z}}-\delta \mathbf{u}|}\Big |^2{\mathrm{d}}{} \mathbf{u} \le \frac{C}{|{\mathbf{z}}|^2},\quad \int _{|\mathbf{u}|<1}\Big | \Pi ({\mathbf{z}})\frac{{\mathbf{z}}-\delta \mathbf{u}}{|{\mathbf{z}}-\delta \mathbf{u}|^2}\Big |^2{\mathrm{d}}{} \mathbf{u} \le \frac{C}{|{\mathbf{z}}|^2}, \quad \forall \, \delta >0,\, {\mathbf{z}}\ne \mathbf{0}. \end{aligned}$$

Then using (6.23) and $|\Pi ({\mathbf{z}})\mathbf{h}|\le |\mathbf{h}|$, we obtain that, for all ${\mathbf{z}}\in {{{\mathbb {R}}}^{3}}{\setminus }\{\mathbf{0}\}, 0<\delta <1$ and all ${\mathbf{y}}\in Y{\setminus } Z_{\delta ,{\mathbf{z}}}$,

$$\begin{aligned}&\big |\Pi ({\mathbf{z}})\nabla _{\mathbf{z}} F_{\delta }({\mathbf{y}},{\mathbf{z}})-\Pi ({\mathbf{z}})\nabla _{\mathbf{z}} F({\mathbf{y}},{\mathbf{z}})\big |^2 \\&\quad \le C\int _{|\mathbf{u}|\le 1}\big |\Pi ({\mathbf{z}}-\delta \mathbf{u})\nabla _{\mathbf{z}} F({\mathbf{y}},{\mathbf{z}}-\delta \mathbf{u}) -\Pi ({\mathbf{z}})\nabla _{\mathbf{z}} F({\mathbf{y}},{\mathbf{z}})\big |^2 {\mathrm{d}}{} \mathbf{u}\\&\quad \quad +\frac{C}{|{\mathbf{z}}|^2}\int _{|\mathbf{u}|\le 1}\big |F({\mathbf{y}},{\mathbf{z}}-\delta \mathbf{u})-F({\mathbf{y}},{\mathbf{z}})\big |^2 {\mathrm{d}}{} \mathbf{u}. \end{aligned}$$

From this and definition of $\Psi \in {{\mathcal {T}}}_{c}(Y\times {{{\mathbb {R}}}^{3}}, {{{\mathbb {R}}}^{3}})$ which implies that $|\Psi ({\mathbf{y}},{\mathbf{z}})|\le C_{\Psi }|{\mathbf{z}}|$, we have

$$\begin{aligned}&\int _{Y\times {{{\mathbb {R}}}^{3}}} |\Psi ({\mathbf{y}}, {\mathbf{z}})|^2 \big |\Pi ({\mathbf{z}})\nabla _{{\mathbf{z}}}F_{\delta }({\mathbf{y}},{\mathbf{z}}) -\Pi ({\mathbf{z}})\nabla _{{\mathbf{z}}}F({\mathbf{y}},{\mathbf{z}})\big |^2 {\mathrm{d}}{\mathbf{z}}{\mathrm{d}}{\mathbf{y}} \\&\quad \le C_{\Psi }\int _{|\mathbf{u}|\le 1} \bigg (\int _{|{\mathbf{y}}|<R, |{\mathbf{z}}|<R} \big |\Pi ({\mathbf{z}}-\delta \mathbf{u})\nabla _{\mathbf{z}} F({\mathbf{y}},{\mathbf{z}}-\delta \mathbf{u}) -\Pi ({\mathbf{z}})\nabla _{\mathbf{z}} F({\mathbf{y}},{\mathbf{z}})\big |^2{\mathrm{d}}{\mathbf{z}}{\mathrm{d}}{\mathbf{y}}\bigg ) {\mathrm{d}}{} \mathbf{u} \\&\qquad +C_{\Psi }\int _{|\mathbf{u}|\le 1} \bigg (\int _{|{\mathbf{y}}|\le R, |{\mathbf{z}}|\le R} \big |F({\mathbf{y}},{\mathbf{z}}-\delta \mathbf{u})-F({\mathbf{y}},{\mathbf{z}})\big |^2{\mathrm{d}}{\mathbf{z}}{\mathrm{d}}{\mathbf{y}}\bigg ) {\mathrm{d}}{} \mathbf{u} \rightarrow 0\quad (\delta \rightarrow 0^{+}), \end{aligned}$$

where we have used Lemma 6.12. This implies the desired convergence:

$$\begin{aligned}&\bigg | \int _{Y\times {{{\mathbb {R}}}^{3}}} \Psi \cdot 2F_{\delta }\Pi ({\mathbf{z}})\nabla _{{\mathbf{z}}}F_{\delta } {\mathrm{d}}{\mathbf{z}}{\mathrm{d}}{\mathbf{y}}- \int _{Y\times {{{\mathbb {R}}}^{3}}} \Psi \cdot 2F\Pi ({\mathbf{z}})\nabla _{{\mathbf{z}}}F {\mathrm{d}}{\mathbf{z}}{\mathrm{d}}{\mathbf{y}}\bigg | \\&\quad \le 2 \int _{|{\mathbf{y}}|\le R, |{\mathbf{z}}|\le R} |\Psi | |F_{\delta }|\big |\Pi ({\mathbf{z}})\nabla _{{\mathbf{z}}}F_{\delta } -\Pi ({\mathbf{z}})\nabla _{{\mathbf{z}}}F\big | {\mathrm{d}}{\mathbf{z}}{\mathrm{d}}{\mathbf{y}}\\&\quad \quad +2 \int _{|{\mathbf{y}}|\le R, |{\mathbf{z}}|\le R} |\Psi | \big |F_{\delta } -F\big |\big |\Pi ({\mathbf{z}})\nabla _{{\mathbf{z}}}F \big | {\mathrm{d}}{\mathbf{z}}{\mathrm{d}}{\mathbf{y}}\\&\quad \le 2\bigg ( \int _{|{\mathbf{y}}|\le R, |{\mathbf{z}}|\le R} |F_{\delta }|^2{\mathrm{d}}{\mathbf{z}}\bigg )^{1/2} \bigg ( \int _{|{\mathbf{y}}|\le R, |{\mathbf{z}}|\le R} |\Psi |^2 \big |\Pi ({\mathbf{z}})\nabla _{{\mathbf{z}}}F_{\delta } -\Pi ({\mathbf{z}})\nabla _{{\mathbf{z}}}F\big |^2 {\mathrm{d}}{\mathbf{z}}{\mathrm{d}}{\mathbf{y}}\bigg )^{1/2}\\&\quad \quad +2C_{\Psi }\bigg ( \int _{|{\mathbf{y}}|\le R, |{\mathbf{z}}|\le R} \big |F_{\delta }-F\big |^2 {\mathrm{d}}{\mathbf{z}}{\mathrm{d}}{\mathbf{y}}\bigg )^{1/2}\bigg ( \int _{|{\mathbf{y}}|\le R, |{\mathbf{z}}|\le R} \big |\Pi ({\mathbf{z}})\nabla _{{\mathbf{z}}}F \big |^2 {\mathrm{d}}{\mathbf{z}}{\mathrm{d}}{\mathbf{y}} \bigg )^{1/2}\\&\quad \rightarrow 0\quad (\delta \rightarrow 0^{+}). \end{aligned}$$

Convergence of r.h.s of (6.26):  We also have

$$\begin{aligned}&\bigg | \int _{Y\times {{{\mathbb {R}}}^{3}}} F^2_{\delta }\nabla _{{\mathbf{z}}}\cdot \Pi ({\mathbf{z}})\Psi {\mathrm{d}}{\mathbf{z}}{\mathrm{d}}{\mathbf{y}} - \int _{Y\times {{{\mathbb {R}}}^{3}}} F^2\nabla _{{\mathbf{z}}}\cdot \Pi ({\mathbf{z}})\Psi {\mathrm{d}}{\mathbf{z}}{\mathrm{d}}{\mathbf{y}}\bigg | \\&\quad \le C_{\Psi } \int _{|{\mathbf{y}}|\le R, |{\mathbf{z}}|\le R}\big | F^2_{\delta }-F^2\big | {\mathrm{d}}{\mathbf{z}}{\mathrm{d}}{\mathbf{y}}\\&\quad \le C_{\Psi }\bigg ( \int _{|{\mathbf{y}}|\le R, |{\mathbf{z}}|\le R}\big | F_{\delta }-F\big |^2{\mathrm{d}}{\mathbf{z}}{\mathrm{d}}{\mathbf{y}}\bigg )^{1/2}\bigg (\int _{|{\mathbf{y}}|\le R, |{\mathbf{z}}|\le R+1} \big |F\big |^2 {\mathrm{d}}{\mathbf{z}}{\mathrm{d}}{\mathbf{y}}\bigg )^{1/2}\\&\quad \rightarrow 0\quad (\delta \rightarrow 0^{+}). \end{aligned}$$

Thus from these two estimates, (6.26) yields that

$$\begin{aligned} \int _{Y\times {{{\mathbb {R}}}^{3}}} \Psi ({\mathbf{y}}, {\mathbf{z}})\cdot 2F({\mathbf{y}},{\mathbf{z}})\Pi ({\mathbf{z}})\nabla _{{\mathbf{z}}}F({\mathbf{y}},{\mathbf{z}}) {\mathrm{d}}{\mathbf{z}}{\mathrm{d}}{\mathbf{y}} =- \int _{Y\times {{{\mathbb {R}}}^{3}}} F^2({\mathbf{y}},{\mathbf{z}})\nabla _{{\mathbf{z}}}\cdot \Pi ({\mathbf{z}})\Psi ({\mathbf{y}},{\mathbf{z}}) {\mathrm{d}}{\mathbf{z}}{\mathrm{d}}{\mathbf{y}}. \end{aligned}$$

By Defintion 6.4, $F^2$ has the weak projection gradient $\Pi ({\mathbf{z}})\nabla _{{\mathbf{z}}}F^2({\mathbf{y}},{\mathbf{z}})$ in ${\mathbf{z}}\in {{{\mathbb {R}}}^{3}}{\setminus }\{\mathbf{0}\}$ and $\Pi ({\mathbf{z}})\nabla _{{\mathbf{z}}}F^2({\mathbf{y}},{\mathbf{z}})=2F({\mathbf{y}},{\mathbf{z}})\Pi ({\mathbf{z}})\nabla _{{\mathbf{z}}}F({\mathbf{y}},{\mathbf{z}}).$ $\quad \square$

Proof of Lemma 6.10

Suppose that F satisfies the assumption in Lemma 6.10 and let $F_{\delta }({\mathbf{y}},{\mathbf{z}})=(J_{\delta }*F({\mathbf{y}},\cdot ))({\mathbf{z}})=\int _{|\mathbf{u}\le 1} J(\mathbf{u})F({\mathbf{y}},{\mathbf{z}}-\delta \mathbf{u}){\mathrm{d}}{} \mathbf{u}$ be defined in Lemma 6.11. Then

$$\begin{aligned} c_{R}:=\inf _{{\mathbf{y}}\in Y, |{\mathbf{y}}|\le R, |{\mathbf{z}}|\le R+1}F({\mathbf{y}},{\mathbf{z}})>0,\quad \inf _{{\mathbf{y}}\in Y, |{\mathbf{y}}|\le R, |{\mathbf{z}}|\le R, 0<\delta <1}F_{\delta }({\mathbf{y}},{\mathbf{z}}) \ge c_{R}. \end{aligned}$$

(6.27)

Thus, as $\delta \rightarrow 0^{+}$,

$$\begin{aligned}&\int _{|{\mathbf{y}}|\le R,|{\mathbf{z}}|\le R}\Big |\sqrt{F_{\delta }}-\sqrt{F}\Big |{\mathrm{d}}{\mathbf{z}}{\mathrm{d}}{\mathbf{y}} \le \frac{1}{2\sqrt{c_R}}\int _{ |{\mathbf{y}}|\le R,|{\mathbf{z}}|\le R}|F_{\delta }-F| {\mathrm{d}}{\mathbf{z}}{\mathrm{d}}{\mathbf{y}}\rightarrow 0,\\&\int _{|{\mathbf{y}}|\le R, |{\mathbf{z}}|\le R} \Big |\frac{1}{\sqrt{F_{\delta }}}- \frac{1}{\sqrt{F}}\Big | {\mathrm{d}}{\mathbf{z}}{\mathrm{d}}{\mathbf{y}}\le \frac{1}{2c_R\sqrt{c_R}}\int _{ |{\mathbf{y}}|\le R, |{\mathbf{z}}|\le R} |F_{\delta }-F| {\mathrm{d}}{\mathbf{z}}{\mathrm{d}}{\mathbf{y}}\rightarrow 0. \end{aligned}$$

Since $F_{\delta }({\mathbf{y}},{\mathbf{z}})$ is smooth in ${\mathbf{z}}$ and has positive lower bound on any bounded set, it follows that

$$\begin{aligned}&\nabla _{\mathbf{z}} \sqrt{F_{\delta }({\mathbf{y}},{\mathbf{z}})} =\frac{\nabla _{\mathbf{z}} F_{\delta }({\mathbf{y}},{\mathbf{z}})}{ 2\sqrt{F_{\delta }({\mathbf{y}},{\mathbf{z}})} }. \end{aligned}$$

Thus for any $\Psi \in {{\mathcal {T}}}_{c}(Y\times {{{\mathbb {R}}}^{3}}, {{{\mathbb {R}}}^{3}})$

$$\begin{aligned} \int _{Y\times {{{\mathbb {R}}}^{3}}}\Psi ({\mathbf{y}},{\mathbf{z}})\cdot \frac{\Pi ({\mathbf{z}})\nabla _{\mathbf{z}} F_{\delta }({\mathbf{y}},{\mathbf{z}})}{2\sqrt{F_{\delta }({\mathbf{y}},{\mathbf{z}})}} {\mathrm{d}}{\mathbf{z}}{\mathrm{d}}{\mathbf{y}} =-\int _{Y\times {{{\mathbb {R}}}^{3}}}\sqrt{F_{\delta }({\mathbf{y}},{\mathbf{z}})}\, \nabla _{\mathbf{z}}\cdot \Pi ({\mathbf{z}})\Psi ({\mathbf{y}},{\mathbf{z}}) {\mathrm{d}}{\mathbf{z}}{\mathrm{d}}{\mathbf{y}}. \end{aligned}$$

(6.28)

Convergence of l.h.s of (6.28):  Using (6.24) we have for all ${\mathbf{z}}\in {{{\mathbb {R}}}^{3}}{\setminus }\{\mathbf{0}\}, 0<\delta <1$ and ${\mathbf{y}}\in Y{\setminus } Z_{\delta ,{\mathbf{z}}}$ that

$$\begin{aligned}&\big |\Pi ({\mathbf{z}})\nabla _{\mathbf{z}} F_{\delta }({\mathbf{y}},{\mathbf{z}})-\Pi ({\mathbf{z}})\nabla _{\mathbf{z}} F({\mathbf{y}},{\mathbf{z}}) \big |\\&\quad \le \int _{|\mathbf{u}|\le 1}J(\mathbf{u})\big |\Pi ({\mathbf{z}}-\delta \mathbf{u})\nabla _{\mathbf{z}} F({\mathbf{y}},{\mathbf{z}}-\delta \mathbf{u}) -\Pi ({\mathbf{z}})\nabla _{\mathbf{z}} F({\mathbf{y}},{\mathbf{z}})\big | {\mathrm{d}}{} \mathbf{u}\\&\quad \quad +C \int _{|\mathbf{u}|\le 1}\bigg |\frac{F({\mathbf{y}},{\mathbf{z}}-\delta \mathbf{u})}{|{\mathbf{z}}-\delta \mathbf{u}|}- \frac{F({\mathbf{y}},{\mathbf{z}})}{|{\mathbf{z}}|}\bigg | {\mathrm{d}}{} \mathbf{u}. \end{aligned}$$

Then, thanks to Lemma 6.12,

$$\begin{aligned}&\int _{|{\mathbf{y}}|\le R, |{\mathbf{z}}|\le R}\big |\Pi ({\mathbf{z}})\nabla _{\mathbf{z}} F_{\delta }({\mathbf{y}},{\mathbf{z}})-\Pi ({\mathbf{z}})\nabla _{\mathbf{z}} F({\mathbf{y}},{\mathbf{z}}) \big |{\mathrm{d}}{\mathbf{z}}{\mathrm{d}}{\mathbf{y}} \\&\quad \le \int _{|\mathbf{u}|\le 1}J(\mathbf{u})\bigg (\int _{|{\mathbf{y}}|\le R, |{\mathbf{z}}|\le R} \big |\Pi ({\mathbf{z}}-\delta \mathbf{u})\nabla _{\mathbf{z}} F({\mathbf{y}},{\mathbf{z}}-\delta \mathbf{u}) -\Pi ({\mathbf{z}})\nabla _{\mathbf{z}} F({\mathbf{y}},{\mathbf{z}})\big |{\mathrm{d}}{\mathbf{z}}{\mathrm{d}}{\mathbf{y}}\bigg ) {\mathrm{d}}{} \mathbf{u} \\&\quad \quad +C\int _{|\mathbf{u}|\le 1}\bigg (\int _{|{\mathbf{y}}|\le R, |{\mathbf{z}}|\le R} \bigg |\frac{F({\mathbf{y}},{\mathbf{z}}-\delta \mathbf{u})}{|{\mathbf{z}}-\delta \mathbf{u}|}- \frac{F({\mathbf{y}},{\mathbf{z}})}{|{\mathbf{z}}|}\bigg | {\mathrm{d}}{\mathbf{z}}{\mathrm{d}}{\mathbf{y}}\bigg ){\mathrm{d}}{} \mathbf{u}\rightarrow 0\quad (\delta \rightarrow 0^{+}). \end{aligned}$$

Next we have

$$\begin{aligned}&\int _{|{\mathbf{y}}|\le R, |{\mathbf{z}}|\le R} \bigg |\frac{\Pi ({\mathbf{z}})\nabla _{\mathbf{z}} F_{\delta }({\mathbf{y}},{\mathbf{z}})}{2\sqrt{F_{\delta }({\mathbf{y}},{\mathbf{z}})}}- \frac{\Pi ({\mathbf{z}})\nabla _{\mathbf{z}} F({\mathbf{y}},{\mathbf{z}})}{2\sqrt{F({\mathbf{y}},{\mathbf{z}})}}\bigg | {\mathrm{d}}{\mathbf{z}}{\mathrm{d}}{\mathbf{y}}\\&\quad \le \frac{1}{2}\int _{|{\mathbf{y}}|\le R, |{\mathbf{z}}|\le R} \frac{|\Pi ({\mathbf{z}})\nabla _{\mathbf{z}} F_{\delta }({\mathbf{y}},{\mathbf{z}})-\Pi ({\mathbf{z}})\nabla _{\mathbf{z}} F({\mathbf{y}},{\mathbf{z}})|}{\sqrt{F_{\delta }({\mathbf{y}},{\mathbf{z}})}} {\mathrm{d}}{\mathbf{z}}{\mathrm{d}}{\mathbf{y}}\\&\quad \quad +\frac{1}{2}\int _{|{\mathbf{y}}|\le R, |{\mathbf{z}}|\le R}|\Pi ({\mathbf{z}})\nabla _{\mathbf{z}} F({\mathbf{y}},{\mathbf{z}})|\bigg |\frac{1}{\sqrt{F_{\delta }({\mathbf{y}},{\mathbf{z}})}}- \frac{1}{\sqrt{F({\mathbf{y}},{\mathbf{z}})}}\bigg | {\mathrm{d}}{\mathbf{z}}{\mathrm{d}}{\mathbf{y}} =: \frac{1}{2}A_{\delta }+\frac{1}{2}B_{\delta } \end{aligned}$$

and from the lower bound (6.27) we have

$$\begin{aligned} A_{\delta } \le \frac{1}{\sqrt{c_R}}\int _{ |{\mathbf{y}}|\le R, |{\mathbf{z}}\le R} |\Pi ({\mathbf{z}})\nabla _{\mathbf{z}} F_{\delta }({\mathbf{y}},{\mathbf{z}})-\Pi ({\mathbf{z}})\nabla _{\mathbf{z}} F({\mathbf{y}},{\mathbf{z}})| {\mathrm{d}}{\mathbf{z}}{\mathrm{d}}{\mathbf{y}}\rightarrow 0\quad (\delta \rightarrow 0^{+}). \end{aligned}$$

For $B_{\delta }$, we consider a decomposition: for any $0<M<\infty$,

$$\begin{aligned} B_{\delta }= & {} \int _{ |{\mathbf{y}}|\le R, |{\mathbf{z}}|\le R}|\Pi ({\mathbf{z}})\nabla _{\mathbf{z}} F|1_{\{|\Pi ({\mathbf{z}})\nabla _{\mathbf{z}} F|\le M\}}\bigg |\frac{1}{\sqrt{F_{\delta }}}- \frac{1}{\sqrt{F}}\bigg | {\mathrm{d}}{\mathbf{z}}{\mathrm{d}}{\mathbf{y}} \nonumber \\&+\int _{ |{\mathbf{y}}|\le R, |{\mathbf{z}}|\le R} |\Pi ({\mathbf{z}})\nabla _{\mathbf{z}} F({\mathbf{y}},{\mathbf{z}})| 1_{\{|\Pi ({\mathbf{z}})\nabla _{\mathbf{z}} F|> M\}} \bigg |\frac{1}{\sqrt{F_{\delta }}}- \frac{1}{\sqrt{F}}\bigg | {\mathrm{d}}{\mathbf{z}}{\mathrm{d}}{\mathbf{y}}\nonumber \\\le & {} M\int _{|{\mathbf{y}}|\le R, |{\mathbf{z}}|\le R}\bigg |\frac{1}{\sqrt{F_{\delta }}}- \frac{1}{\sqrt{F}}\bigg | {\mathrm{d}}{\mathbf{z}}{\mathrm{d}}{\mathbf{y}} + \frac{1}{\sqrt{c_R}}\int _{|{\mathbf{y}}|\le R, |{\mathbf{z}}|\le R} |\Pi ({\mathbf{z}})\nabla _{\mathbf{z}} F| 1_{\{|\Pi ({\mathbf{z}})\nabla _{\mathbf{z}} F|>M\}} {\mathrm{d}}{\mathbf{z}}{\mathrm{d}}{\mathbf{y}}.\nonumber \\ \end{aligned}$$

(6.29)

By definition of $F_{\delta }$, the first term in the righthand side of (6.29) tends to zero as $\delta \rightarrow 0^+$, while by the $L^1_{loc}$ integrability of $|\Pi ({\mathbf{z}})\nabla _{\mathbf{z}} F({\mathbf{y}},{\mathbf{z}})|$, the second term in the righthand side of (6.29) tends to zero as $M\rightarrow \infty$. Thus by first letting $\delta \rightarrow 0^{+}$ and then letting $M\rightarrow \infty$ we conclude $\lim \limits _{\delta \rightarrow 0^{+}}B_{\delta }=0$.

Thus

$$\begin{aligned} \lim _{\delta \rightarrow 0^{+}}\int _{ |{\mathbf{y}}|\le R, |{\mathbf{z}}|\le R} \bigg |\frac{\Pi ({\mathbf{z}})\nabla _{\mathbf{z}} F_{\delta }({\mathbf{y}},{\mathbf{z}})}{2\sqrt{F_{\delta }({\mathbf{y}},{\mathbf{z}})}}- \frac{\Pi ({\mathbf{z}})\nabla _{\mathbf{z}} F({\mathbf{y}},{\mathbf{z}})}{2\sqrt{F({\mathbf{y}},{\mathbf{z}})}}\bigg | {\mathrm{d}}{\mathbf{z}}{\mathrm{d}}{\mathbf{y}}=0. \end{aligned}$$

From this we deduce the convergence:

$$\begin{aligned}&\bigg |\int _{Y\times {{{\mathbb {R}}}^{3}}}\Psi ({\mathbf{y}},{\mathbf{z}})\cdot \frac{\Pi ({\mathbf{z}})\nabla _{\mathbf{z}} F_{\delta }({\mathbf{y}},{\mathbf{z}})}{2\sqrt{F_{\delta }({\mathbf{y}},{\mathbf{z}})}} {\mathrm{d}}{\mathbf{z}}{\mathrm{d}}{\mathbf{y}} -\int _{Y\times {{{\mathbb {R}}}^{3}}}\Psi ({\mathbf{y}},{\mathbf{z}})\cdot \frac{\Pi ({\mathbf{z}})\nabla _{\mathbf{z}} F({\mathbf{y}},{\mathbf{z}})}{2\sqrt{F({\mathbf{y}},{\mathbf{z}})}} {\mathrm{d}}{\mathbf{z}}{\mathrm{d}}{\mathbf{y}}\bigg | \\&\quad \le C_{\Psi }\int _{ |{\mathbf{y}}|\le R, |{\mathbf{z}}|\le R}\bigg | \frac{\Pi ({\mathbf{z}})\nabla _{\mathbf{z}} F_{\delta }({\mathbf{y}},{\mathbf{z}})}{2\sqrt{F_{\delta }({\mathbf{y}},{\mathbf{z}})}}-\frac{\Pi ({\mathbf{z}})\nabla _{\mathbf{z}} F({\mathbf{y}},{\mathbf{z}})}{2\sqrt{F({\mathbf{y}},{\mathbf{z}})}}\bigg | {\mathrm{d}}{\mathbf{z}}{\mathrm{d}}{\mathbf{y}} \rightarrow 0\quad (\delta \rightarrow 0^{+}). \end{aligned}$$

Convergence of r.h.s of (6.28):  Since $\Psi \in {{\mathcal {T}}}_{c}(Y\times {{{\mathbb {R}}}^{3}}, {{{\mathbb {R}}}^{3}})$, we have

$$\begin{aligned}&\bigg |\int _{Y\times {{{\mathbb {R}}}^{3}}} \sqrt{F_{\delta }({\mathbf{y}},{\mathbf{z}})} \,\nabla _{\mathbf{z}}\cdot \Pi ({\mathbf{z}})\Psi ({\mathbf{y}},{\mathbf{z}}) {\mathrm{d}}{\mathbf{z}}{\mathrm{d}}{\mathbf{y}} -\int _{Y\times {{{\mathbb {R}}}^{3}}} \sqrt{F({\mathbf{y}},{\mathbf{z}})}\, \nabla _{\mathbf{z}}\cdot \Pi ({\mathbf{z}})\Psi ({\mathbf{y}},{\mathbf{z}}) {\mathrm{d}}{\mathbf{z}}{\mathrm{d}}{\mathbf{y}}\bigg |\\&\quad \le C_{\Psi }\int _{ |{\mathbf{y}}|\le R, |{\mathbf{z}}|\le R} \big |\sqrt{F_{\delta }({\mathbf{y}},{\mathbf{z}})}- \sqrt{F({\mathbf{y}},{\mathbf{z}})}\big | {\mathrm{d}}{\mathbf{z}}{\mathrm{d}}{\mathbf{y}}\rightarrow 0 \quad (\delta \rightarrow 0^{+}). \end{aligned}$$

Letting $\delta \rightarrow 0^{+}$ in (6.28) we finally obtain that

$$\begin{aligned} \int _{Y\times {{{\mathbb {R}}}^{3}}}\Psi ({\mathbf{y}},{\mathbf{z}})\cdot \frac{\Pi ({\mathbf{z}})\nabla _{\mathbf{z}} F({\mathbf{y}},{\mathbf{z}})}{2\sqrt{F({\mathbf{y}},{\mathbf{z}})}} {\mathrm{d}}{\mathbf{z}}{\mathrm{d}}{\mathbf{y}} =-\int _{Y\times {{{\mathbb {R}}}^{3}}} \sqrt{F({\mathbf{y}},{\mathbf{z}})} \,\nabla _{\mathbf{z}}\cdot \Pi ({\mathbf{z}})\Psi ({\mathbf{y}},{\mathbf{z}}) {\mathrm{d}}{\mathbf{z}}{\mathrm{d}}{\mathbf{y}}. \end{aligned}$$

Thus $\Pi ({\mathbf{z}})\nabla _{\mathbf{z}} \sqrt{F({\mathbf{y}},{\mathbf{z}})}$ exists and $\Pi ({\mathbf{z}})\nabla _{\mathbf{z}} \sqrt{F({\mathbf{y}},{\mathbf{z}})}=\frac{\Pi ({\mathbf{z}})\nabla _{\mathbf{z}} F({\mathbf{y}},{\mathbf{z}})}{2\sqrt{F({\mathbf{y}},{\mathbf{z}})}}.$ $\quad \square$

1.3.3 Application to Boltzmann collision operator

We begin with an algebraic lemma.

Lemma 6.13

Let ${\mathbf{n}}\in {{{\mathbb {S}}}^{2}}, {{{\mathbb {S}}}}^{1}({\mathbf{n}})=\{\sigma \in {{{\mathbb {S}}}^{2}}\,|\,\sigma \,\bot \, {\mathbf{n}}\,\}, A\in {{{\mathbb {R}}}}^{3\times 3}$, and $\mathbf{a}, \mathbf{b}\in {{{\mathbb {R}}}^{3}}$. Then

$$\begin{aligned}&\frac{1}{\pi }\int _{{{{\mathbb {S}}}}^1({\mathbf{n}})}\sigma ^{\tau } A\sigma {\mathrm{d}}\sigma =\mathrm{Tr}(A)-{\mathbf{n}}^{\tau }A{\mathbf{n}}, \end{aligned}$$

(6.30)

$$\begin{aligned}&\frac{1}{\pi }\int _{{{{\mathbb {S}}}}^1({\mathbf{n}})}(\mathbf{a}\cdot \sigma )\sigma {\mathrm{d}}\sigma = \Pi ({\mathbf{n}})\mathbf{a}, \quad \frac{1}{\pi }\int _{{{{\mathbb {S}}}}^1({\mathbf{n}})}( \mathbf{a}\cdot \sigma )(\mathbf{b}\cdot \sigma ){\mathrm{d}}\sigma = \mathbf{a}\cdot \Pi ({\mathbf{n}})\mathbf{b}. \end{aligned}$$

(6.31)

Proof

Suppose $\mathbf{i}, \mathbf{j}\in {{\mathbb {S}}}^2$ and $\{\mathbf{i}, \mathbf{j},{\mathbf{n}}\}$ is an orthonormal base of ${{\mathbb {R}}}^3$. Then for the orthogonal matrix $T= ( \mathbf{i}\,\, \mathbf{j}\,\, {\mathbf{n}})^{\tau }\in {{\mathbb {R}}}^{3\times 3}$ we have $T\mathbf{i}=\mathbf{e}_1=(1,0,0)^{\tau },\, T\mathbf{j}=\mathbf{e}_2=(0,1,0)^{\tau },\, T{\mathbf{n}}=\mathbf{e}_3=(0,0,1)^{\tau }.$ It is easy to see that $\mathbf{i}^{\tau }A\mathbf{i}+\mathbf{j}^{\tau }A\mathbf{j}+{\mathbf{n}}^{\tau }A{\mathbf{n}} =\sum _{i=1}^3\mathbf{e}_i^{\tau }TAT^{-1}{} \mathbf{e}_i= \mathrm{Tr}(TAT^{-1})=\mathrm{Tr}(A)$. Hence $\frac{1}{\pi } \int _{{\mathbb S}^1({\mathbf{z}})} \sigma ^{\tau }A\sigma {\mathrm{d}}\sigma =\frac{1}{\pi } \int _{0}^{2\pi } (\cos (\theta )\mathbf{i}+\sin (\theta )\mathbf{j})^{\tau }A (\cos (\theta )\mathbf{i}+\sin (\theta )\mathbf{j}) {\mathrm{d}}\theta = \mathbf{i}^{\tau }A\mathbf{i}+\mathbf{j}^{\tau }A\mathbf{j} =\mathrm{Tr}(A)-{\mathbf{n}}^{\tau }A{\mathbf{n}}.$ It is not difficult to check that $\frac{1}{\pi } \int _{{{{\mathbb {S}}}}^1({\mathbf{n}})}(\mathbf{a}\cdot \sigma )\sigma {\mathrm{d}}\sigma =\frac{1}{\pi } \int _{0}^{2\pi }(\mathbf{a}\cdot (\cos (\theta )\mathbf{i}+\sin (\theta )\mathbf{j})) (\cos (\theta )\mathbf{i}+\sin (\theta )\mathbf{j}) {\mathrm{d}}\theta =(\mathbf{a}\cdot \mathbf{i}) \mathbf{i}+(\mathbf{a}\cdot \mathbf{j}) \mathbf{j} =\mathbf{a}-(\mathbf{a}\cdot {\mathbf{n}}){\mathbf{n}} =\Pi ({\mathbf{n}})\mathbf{a},$ which yields that $\frac{1}{\pi }\int _{{{{\mathbb {S}}}}^1({\mathbf{n}})}( \mathbf{a}\cdot \sigma )(\mathbf{b}\cdot \sigma ){\mathrm{d}}\sigma = \big (\frac{1}{\pi }\int _{{{{\mathbb {S}}}}^1({\mathbf{n}})}( \mathbf{a}\cdot \sigma )\sigma {\mathrm{d}}\sigma \big )\cdot \mathbf{b}=\Pi ({\mathbf{n}})\mathbf{a}\cdot \mathbf{b}=\mathbf{a}\cdot \Pi ({\mathbf{n}})\mathbf{b}.$ $\quad \square$

Since the Boltzmann collision operator involves integration on the unit sphere ${{{\mathbb {S}}}^{2}}$, to apply the definition of weak projection gradient to the weak coupling limit, we need to introduce the following space of test functions:

$$\begin{aligned}&{{\mathcal {T}}}_{c}(Y\times {{{{\mathbb {R}}}^{3}}\times {{\mathbb {S}}}^{2}})\nonumber \\&\quad =\bigg \{ \psi \in C_c(Y\times {{{\mathbb {R}}}^{3}}\times {{{\mathbb {S}}}^{2}})\,\,\Big | \,\,\psi ({\mathbf{y}},\cdot ,\cdot )\in C^1({{{\mathbb {R}}}^{3}}\times {{{\mathbb {S}}}^{2}}),\,\psi ({\mathbf{y}},\mathbf{0},\sigma )= 0\,\, \mathrm{for\,\,all\,\,} {\mathbf{y}}\in Y,\sigma \in {{{\mathbb {S}}}^{2}},\nonumber \\&\sup _{({\mathbf{y}},{\mathbf{z}},\sigma )\in Y\times {{{{\mathbb {R}}}^{3}}\times {{\mathbb {S}}}^{2}}, }|\nabla _{{\mathbf{z}}}\psi ({\mathbf{y}},{\mathbf{z}},\sigma )|<\infty ,\,\,\sup _{({\mathbf{y}}, {\mathbf{z}},\sigma )\in Y\times ({{{\mathbb {R}}}^{3}}{\setminus } \{\mathbf{0}\})\times {{{\mathbb {S}}}^{2}}}\frac{|\nabla _{\sigma }\psi ({\mathbf{y}},{\mathbf{z}},\sigma )|}{|{\mathbf{z}}| }<\infty \bigg \}. \end{aligned}$$

(6.32)

Here $\psi ({\mathbf{y}},\cdot ,\cdot )\in C^1({{{\mathbb {R}}}^{3}}\times {{{\mathbb {S}}}^{2}})$ can be understood as there is an open set $\Omega \supset {{{\mathbb {S}}}^{2}}$ (it may depend on ${\mathbf{y}}$) such that $\psi ({\mathbf{y}},\cdot ,\cdot )\in C^1({{{\mathbb {R}}}^{3}}\times \Omega )$, and so $\sigma \mapsto \nabla _{\sigma }\psi ({\mathbf{y}},{\mathbf{z}},\sigma )$ is the usual gradient taken in open set.

The following lemma is used to estimate bounds of difference quotients of functions on the sphere.

Lemma 6.14

Let $n\ge 2$ and $\psi \in C^1({{{\mathbb {S}}}}^{n-1})$ (which means that there is an open set $\Omega \supset {{{\mathbb {S}}}}^{n-1}$ such that $\psi \in C^1(\Omega )$). Then

$$\begin{aligned} |\psi (\sigma _1)-\psi (\sigma _2)|\le 4\sqrt{2}\Vert \nabla \psi \Vert _{{{{\mathbb {S}}}}^{n-1}}|\sigma _1-\sigma _2|\qquad \forall \, \sigma _1, \sigma _2\in {{{\mathbb {S}}}}^{n-1} \end{aligned}$$

where $\Vert \nabla \psi \Vert _{{{{\mathbb {S}}}}^{n-1}}=\max \limits _{\sigma \in {{{\mathbb {S}}}}^{n-1}}|\nabla \psi (\sigma )|$.

Proof

Denote $\Vert \nabla \psi \Vert =\Vert \nabla \psi \Vert _{{{{\mathbb {S}}}}^{n-1}}$. Take any $\sigma _1, \sigma _2\in {{{\mathbb {S}}}}^{n-1}$.

Case 1: $\sigma _1\cdot \sigma _2\ge 0.$ Consider $t\mapsto \gamma (t)=\mathbf{a}(t)/|\mathbf{a}(t)|\in {{{\mathbb {S}}}}^{n-1}$ with $\mathbf{a}(t)=t\sigma _1+(1-t)\sigma _2, t\in {[}0,1]$. We have $|\mathbf{a}(t)|\ge \frac{1}{\sqrt{2}}, \mathbf{a}'(t)=\sigma _1-\sigma _2,$ for all $t\in {[}0,1].$ This gives $|\gamma '(t)|\le 2|\mathbf{a}'(t)|/|\mathbf{a}(t)| \le 2\sqrt{2}|\sigma _1-\sigma _2|$ for all $t\in {[}0,1].$ So we have for some $t\in (0,1)$

$$\begin{aligned} |\psi (\sigma _1)-\psi (\sigma _2)|=|\psi \circ \gamma (1)-\psi \circ \gamma (0)| =|\nabla \psi (\sigma )\cdot \gamma '(t)|\big |_{\sigma =\gamma (t)}\le 2\sqrt{2}\Vert \nabla \psi \Vert |\sigma _1-\sigma _2|. \end{aligned}$$

Case 2: $\sigma _1\cdot \sigma _2<0.$ In this case we have $|\sigma _1-\sigma _2|\ge \sqrt{2}$ and we can choose some $\sigma _0\in {{{\mathbb {S}}}}^{n-1}$ such that $\sigma _1\cdot \sigma _0\ge 0, \sigma _2\cdot \sigma _0\ge 0$. [For instance if $\sigma _1+\sigma _2=\mathbf{0}$, then take $\sigma _0\in {{{\mathbb {S}}}}^{n-1}$ such that $\sigma _0\cdot \sigma _1=0$; if $\sigma _1+\sigma _2\ne \mathbf{0}$, then choose $\sigma _0=(\sigma _1+\sigma _2)/|\sigma _1+\sigma _2|$.] Applying Case 1 to $\sigma _1, \sigma _0$ and $\sigma _0,\sigma _2$ respectively and noting that $|\sigma _1-\sigma _0|,|\sigma _0-\sigma _2|\le \sqrt{2}\le |\sigma _1-\sigma _2|$ we obrain

$$\begin{aligned} |\psi (\sigma _1)-\psi (\sigma _2)|\le |\psi (\sigma _1)-\psi (\sigma _0)|+|\psi (\sigma _0)-\psi (\sigma _2)| \le 4\sqrt{2}\Vert \nabla \psi \Vert |\sigma _1-\sigma _2|. \end{aligned}$$

$\square$

Lemma 6.15

For any $\Psi \in {{\mathcal {T}}}_{c}(Y\times {{{\mathbb {R}}}^{3}}, {{{\mathbb {R}}}^{3}})$, let $\psi ({\mathbf{y}},{\mathbf{z}},\sigma )=\Psi ({\mathbf{y}},{\mathbf{z}})\cdot \sigma$. Then $\psi \in {{\mathcal {T}}}_{c}(Y\times {{{{\mathbb {R}}}^{3}}\times {{\mathbb {S}}}^{2}})$ and

$$\begin{aligned}&\nabla _{{\mathbf{z}}}\psi ({\mathbf{y}},{\mathbf{z}},\sigma )\cdot \sigma = \sigma ^{\tau }\Psi '_{{\mathbf{z}}}({\mathbf{y}},{\mathbf{z}})\sigma ,\quad \nabla _{\sigma }\psi ({\mathbf{y}},{\mathbf{z}},\sigma ) =\Psi ({\mathbf{y}},{\mathbf{z}}),\\&\frac{1}{\pi }\int _{{{{\mathbb {S}}}}^1({\mathbf{n}})} \Big ( \nabla _{{\mathbf{z}}}\psi ({\mathbf{y}},{\mathbf{z}},\sigma )\cdot \sigma -\frac{\nabla _{\sigma }\psi ({\mathbf{y}},{\mathbf{z}},\sigma )}{|{\mathbf{z}}|}\cdot {\mathbf{n}} \Big ){\mathrm{d}}\sigma =\nabla _{\mathbf{z}} \cdot \Pi ({\mathbf{z}})\Psi ({\mathbf{y}},{\mathbf{z}}). \end{aligned}$$

Proof

That $\psi \in {{\mathcal {T}}}_{c}(Y\times {{{{\mathbb {R}}}^{3}}\times {{\mathbb {S}}}^{2}})$ and the first two equalities are obvious. Then applying Lemma 6.13, we have

$$\begin{aligned}&\frac{1}{\pi }\int _{{{\mathbb {S}}}^1({\mathbf{z}})} \sigma ^{\tau }\Psi '_{\mathbf{z}}({\mathbf{y}},{\mathbf{z}})\sigma {\mathrm{d}}\sigma =\mathrm{Tr}(\Psi '_{\mathbf{z}}({\mathbf{y}},{\mathbf{z}}))-{\mathbf{n}}^{\tau }\Psi '_{\mathbf{z}}({\mathbf{y}},{\mathbf{z}}){\mathbf{n}} \\&\quad =\nabla _{{\mathbf{z}}}\cdot \Psi ({\mathbf{y}},{\mathbf{z}})-{\mathbf{n}}^{\tau }\Psi '_{\mathbf{z}}({\mathbf{y}},{\mathbf{z}}){\mathbf{n}},\\ \\&\frac{1}{\pi } \int _{{{{\mathbb {S}}}}^1({\mathbf{n}})} \frac{ \nabla _{\sigma }\psi ({\mathbf{y}},{\mathbf{z}},\sigma )}{|{\mathbf{z}}|}\cdot {\mathbf{n}} {\mathrm{d}}\sigma =2\frac{\Psi ({\mathbf{y}},{\mathbf{z}})}{|{\mathbf{z}}|}\cdot {\mathbf{n}} \end{aligned}$$

and so using (6.14) yields the third equality. $\quad \square$

Let $\nu$ be a Borel measure on $Y\times {{{{\mathbb {R}}}^{3}}\times {{\mathbb {S}}}^{2}}$ defined by

$$\begin{aligned} \nu (E)=\int _{Y\times {{{\mathbb {R}}}^{3}}} \int _{{{{\mathbb {S}}}}^1({\mathbf{n}})} 1_{E}({\mathbf{y}},{\mathbf{z}},\sigma ){\mathrm{d}}\sigma {\mathrm{d}}{\mathbf{z}}{\mathrm{d}}{\mathbf{y}},\quad E\subset Y\times {{{{\mathbb {R}}}^{3}}\times {{\mathbb {S}}}^{2}} \end{aligned}$$

(6.33)

with ${\mathbf{n}}={\mathbf{z}}/|{\mathbf{z}}|$. It is easily seen that the following equality

$$\begin{aligned} \int _{Y\times {{{{\mathbb {R}}}^{3}}\times {{\mathbb {S}}}^{2}}}g({\mathbf{y}},{\mathbf{z}},\sigma ){\mathrm{d}}\nu ({\mathbf{y}},{\mathbf{z}},\sigma ) =\int _{Y\times {{{\mathbb {R}}}^{3}}} \int _{{{{\mathbb {S}}}}^1({\mathbf{n}})} g({\mathbf{y}},{\mathbf{z}},\sigma ){\mathrm{d}}\sigma {\mathrm{d}}{\mathbf{z}}{\mathrm{d}}{\mathbf{y}} \end{aligned}$$

(6.34)

holds for all nonnegative Borel measurable or $\nu$- integrable functions g on $Y\times {{{{\mathbb {R}}}^{3}}\times {{\mathbb {S}}}^{2}}$.

The last lemma below shows that the weak projection gradient $\Pi ({\mathbf{z}})\nabla _{{\mathbf{z}}}F({\mathbf{y}}, {\mathbf{z}})$ can be characterized by a vector-integration on the perpendicular circle ${{{\mathbb {S}}}}^1({\mathbf{n}})$, and plays an important role in the proof of Theorem 1.12 (From Eq. (MB) to Eq. (FPL)).

Lemma 6.16

A function $F\in L^1_{loc}(Y\times {{{\mathbb {R}}}^{3}})$ has the weak projection gradient $\Pi ({\mathbf{z}})\nabla _{{\mathbf{z}}}F({\mathbf{y}}, {\mathbf{z}})$ in ${\mathbf{z}}\in {{{\mathbb {R}}}^{3}}{\setminus }\{\mathbf{0}\}$, if and only if there is a Borel measurable function $D({\mathbf{y}},{\mathbf{z}},\sigma )$ on $Y\times {{{{\mathbb {R}}}^{3}}\times {{\mathbb {S}}}^{2}}$ such that $D\in L^1_{loc}(Y\times {{{{\mathbb {R}}}^{3}}\times {{\mathbb {S}}}^{2}},{\mathrm{d}}\nu )$ and it holds for all $\psi \in {{\mathcal {T}}}_{c}(Y\times {{{{\mathbb {R}}}^{3}}\times {{\mathbb {S}}}^{2}})$ that

$$\begin{aligned}&\int _{Y\times {{{\mathbb {R}}}^{3}}} \int _{{{{\mathbb {S}}}}^1({\mathbf{n}})} \psi ({\mathbf{y}},{\mathbf{z}},\sigma ) D({\mathbf{y}},{\mathbf{z}},\sigma ){\mathrm{d}}\sigma {\mathrm{d}}{\mathbf{z}}{\mathrm{d}}{\mathbf{y}} \nonumber \\&\quad = -\int _{Y\times {{{\mathbb {R}}}^{3}}}F({\mathbf{y}},{\mathbf{z}}) \int _{{{{\mathbb {S}}}}^1({\mathbf{n}})} \Big ( \nabla _{{\mathbf{z}}}\psi ({\mathbf{y}},{\mathbf{z}},\sigma )\cdot \sigma -\frac{\nabla _{\sigma }\psi ({\mathbf{y}},{\mathbf{z}},\sigma )}{|{\mathbf{z}}|}\cdot {\mathbf{n}} \Big ){\mathrm{d}}\sigma {\mathrm{d}}{\mathbf{z}}{\mathrm{d}}{\mathbf{y}}.\nonumber \\ \end{aligned}$$

(6.35)

Moreover, if (6.35) holds for such a function D (equivalently if $\Pi ({\mathbf{z}})\nabla _{{\mathbf{z}}}F({\mathbf{y}}, {\mathbf{z}})$ exists), then

$$\begin{aligned}&\Pi ({\mathbf{z}})\nabla _{{\mathbf{z}}}F({\mathbf{y}}, {\mathbf{z}}) =\frac{1}{\pi } \int _{{{{\mathbb {S}}}}^1({\mathbf{n}})} D({\mathbf{y}},{\mathbf{z}},\sigma )\sigma {\mathrm{d}}\sigma \quad \mathrm{a.e.}\quad on\quad Y\times {{{\mathbb {R}}}^{3}},\\&D({\mathbf{y}},{\mathbf{z}},\sigma )= \Pi ({\mathbf{z}})\nabla _{{\mathbf{z}}}F({\mathbf{y}}, {\mathbf{z}})\cdot \sigma \qquad \mathrm{a.e.}\quad [\nu ]\quad on\quad Y\times {{{{\mathbb {R}}}^{3}}\times {{\mathbb {S}}}^{2}}. \end{aligned}$$

Proof

First of all we note that the function $D({\mathbf{y}},{\mathbf{z}},\sigma )$ (up to a set of $\nu$-measure zero) is uniquely determined by the identity (6.35).

As before, we use smooth approximation. Let $j(t)=c\exp \big (\frac{-1}{1-t^2}\big )1_{\{|t|<1\}}$ with $c>0$ satisfy $\int _{{{{\mathbb {R}}}}}j(t){\mathrm{d}}t=1$, let $j_{\delta }(t)=\delta ^{-1} j(\delta ^{-1}t), \delta >0$, and let $\zeta _{\delta }(|{\mathbf{z}}|)=\zeta (|{\mathbf{z}}|/\delta )$ be the smooth function used in the proof of Lemma 6.8. Suppose that F has the weak projection gradient $\Pi ({\mathbf{z}})\nabla _{{\mathbf{z}}}F({\mathbf{y}}, {\mathbf{z}})$ in ${\mathbf{z}}\in {{{\mathbb {R}}}^{3}}{\setminus }\{\mathbf{0}\}$. Take any $\psi \in {{\mathcal {T}}}_{c}(Y\times {{{{\mathbb {R}}}^{3}}\times {{\mathbb {S}}}^{2}})$. Note that by definition of ${{\mathcal {T}}}_{c}(Y\times {{{{\mathbb {R}}}^{3}}\times {{\mathbb {S}}}^{2}})$ there are no problems of integrability for all integrands below. By continuity of $\sigma \mapsto \psi ({\mathbf{y}},{\mathbf{z}},\sigma ) \Pi ({\mathbf{z}})\nabla _{\mathbf{z}} F({\mathbf{y}},{\mathbf{z}})\cdot \sigma$ on ${{{\mathbb {S}}}^{2}}$, it is easily shown that (with ${\mathbf{n}}={\mathbf{z}}/|{\mathbf{z}}|$)

$$\begin{aligned}&\int _{Y\times {{{\mathbb {R}}}^{3}}} \int _{{{{\mathbb {S}}}}^1({\mathbf{n}})} \psi ({\mathbf{y}},{\mathbf{z}},\sigma ) \Pi ({\mathbf{z}})\nabla _{\mathbf{z}} F({\mathbf{y}},{\mathbf{z}})\cdot \sigma {\mathrm{d}}\sigma {\mathrm{d}}{\mathbf{z}}{\mathrm{d}}{\mathbf{y}} \\&\quad =\lim _{\delta \rightarrow 0^{+}} \int _{Y\times {{{\mathbb {R}}}^{3}}}\int _{{{{\mathbb {S}}}^{2}}}\zeta _{\delta }(|{\mathbf{z}}|) j_{\delta }({\mathbf{n}}\cdot \sigma )\psi ({\mathbf{y}},{\mathbf{z}},\sigma )\sigma \cdot \Pi ({\mathbf{z}})\nabla _{{\mathbf{z}}} F({\mathbf{y}},{\mathbf{z}}){\mathrm{d}}\sigma {\mathrm{d}}{\mathbf{z}}{\mathrm{d}}{\mathbf{y}} . \end{aligned}$$

Thanks to the smooth cutoff $\zeta _{\delta }(|{\mathbf{z}}|)$, the function $({\mathbf{y}},{\mathbf{z}})\mapsto \zeta _{\delta }(|{\mathbf{z}}|)j_{\delta }({\mathbf{n}}\cdot \sigma )\psi ({\mathbf{y}},{\mathbf{z}},\sigma )\sigma$ belongs to ${\mathcal T}_{c}(Y\times {{{\mathbb {R}}}^{3}}, {{{\mathbb {R}}}^{3}}).$ Hence, using also (6.19), we have

$$\begin{aligned}&\int _{{{{\mathbb {S}}}^{2}}} \int _{Y\times {{{\mathbb {R}}}^{3}}} \zeta _{\delta }(|{\mathbf{z}}|)j_{\delta }({\mathbf{n}}\cdot \sigma )\psi ({\mathbf{y}},{\mathbf{z}},\sigma )\sigma \cdot \Pi ({\mathbf{z}})\nabla _{{\mathbf{z}}} F({\mathbf{y}},{\mathbf{z}}) {\mathrm{d}}{\mathbf{z}}{\mathrm{d}}{\mathbf{y}} {\mathrm{d}}\sigma \\&\quad =-\int _{{{{\mathbb {S}}}^{2}}} \int _{Y\times {{{\mathbb {R}}}^{3}}}F({\mathbf{y}},{\mathbf{z}})\zeta _{\delta }(|{\mathbf{z}}|)\nabla _{{\mathbf{z}}} \cdot \big (\Pi ({\mathbf{z}}) j_{\delta }({\mathbf{n}}\cdot \sigma )\psi ({\mathbf{y}},{\mathbf{z}},\sigma )\sigma \big ){\mathrm{d}}{\mathbf{z}}{\mathrm{d}}{\mathbf{y}}{\mathrm{d}}\sigma . \end{aligned}$$

By elementary calculation we have (for ${\mathbf{z}}\in {{{\mathbb {R}}}^{3}}{\setminus }\{\mathbf{0}\}$)

$$\begin{aligned}&\nabla _{{\mathbf{z}}} \cdot \big (\Pi ({\mathbf{z}})j_{\delta }({\mathbf{n}}\cdot \sigma )\psi ({\mathbf{y}},{\mathbf{z}},\sigma )\sigma \big ) =j_{\delta }'({\mathbf{n}}\cdot \sigma ) \psi ({\mathbf{y}},{\mathbf{z}},\sigma )\frac{1-({\mathbf{n}}\cdot \sigma ) ^2}{|{\mathbf{z}}|}\\&\quad \quad +j_{\delta }({\mathbf{n}}\cdot \sigma )\Big (\nabla _{{\mathbf{z}}}\psi ({\mathbf{y}},{\mathbf{z}},\sigma )\cdot \sigma -( {\mathbf{n}}\cdot \nabla _{{\mathbf{z}}} \psi ({\mathbf{y}},{\mathbf{z}},\sigma )){\mathbf{n}}\cdot \sigma -2\frac{\psi ({\mathbf{y}},{\mathbf{z}},\sigma )}{|{\mathbf{z}}|}\sigma \cdot {\mathbf{n}} \Big ) \end{aligned}$$

and then we compute using (2.2) that

$$\begin{aligned}&\int _{{{{\mathbb {S}}}^{2}}}\nabla _{{\mathbf{z}}} \cdot \big (\Pi ({\mathbf{z}}) j_{\delta }({\mathbf{n}}\cdot \sigma )\psi ({\mathbf{y}},{\mathbf{z}},\sigma )\sigma \big ){\mathrm{d}}\sigma \\&\quad =\delta ^{-1} \int _{-1/\delta }^{1/\delta } j'(s)\bigg (\int _{{{{\mathbb {S}}}}^1({\mathbf{n}})}\frac{ \psi \big ({\mathbf{y}},{\mathbf{z}},{\mathbf{n}}\delta s+\sqrt{1-(\delta s)^2}\sigma \big )}{|{\mathbf{z}}|}{\mathrm{d}}\sigma \bigg ) {\mathrm{d}}s \\&\quad \quad -\delta \int _{-1/\delta }^{1/\delta } s^2j'(s)\bigg (\int _{{{{\mathbb {S}}}}^1({\mathbf{n}})} \frac{\psi \big ({\mathbf{y}},{\mathbf{z}},{\mathbf{n}}\delta s+\sqrt{1-(\delta s)^2}\sigma \big )}{|{\mathbf{z}}|}{\mathrm{d}}\sigma \bigg ){\mathrm{d}}s \\&\quad \quad + \int _{-1/\delta }^{1/\delta }j(s)\bigg ( \int _{{{{\mathbb {S}}}}^1({\mathbf{n}})}\nabla _{{\mathbf{z}}}\psi \big ({\mathbf{y}},{\mathbf{z}},{\mathbf{n}}\delta s+\sqrt{1-(\delta s)^2}\sigma \big )\cdot \sigma {\mathrm{d}}\sigma \bigg ) \sqrt{1-(\delta s)^2}{\mathrm{d}}s \\&\quad \quad - 2\delta \int _{-1/\delta }^{1/\delta }sj(s)\bigg ( \int _{{{{\mathbb {S}}}}^1({\mathbf{n}})}\frac{\psi \Big ({\mathbf{y}},{\mathbf{z}},{\mathbf{n}}\delta s+\sqrt{1-(\delta s)^2}\sigma \bigg )}{|{\mathbf{z}}|} {\mathrm{d}}\sigma \Big ){\mathrm{d}}s \\&\quad =: I_{1,\delta }({\mathbf{y,z}})+I_{2,\delta }({\mathbf{y,z}})+I_{3,\delta }({\mathbf{y,z}})+I_{4,\delta }({\mathbf{y,z}}). \end{aligned}$$

Estimate of $I_{1,\delta }({\mathbf{y,z}})$:  Using $\int _{-1/\delta }^{1/\delta } j'(s){\mathrm{d}}s=0$, the first term becomes

$$\begin{aligned} I_{1,\delta }({\mathbf{y,z}}) = \int _{-1/\delta }^{1/\delta } j'(s)\bigg (\int _{{{{\mathbb {S}}}}^1({\mathbf{n}})}\frac{ \psi \Big ({\mathbf{y}},{\mathbf{z}},{\mathbf{n}}\delta s+\sqrt{1-(\delta s)^2}\sigma \Big )-\psi ({\mathbf{y}},{\mathbf{z}},\sigma )}{\delta |{\mathbf{z}}|}{\mathrm{d}}\sigma \bigg ) {\mathrm{d}}s. \end{aligned}$$

For any ${\mathbf{y}}\in Y,{\mathbf{z}}\in {{{\mathbb {R}}}^{3}}{\setminus }\{\mathbf{0}\},\sigma \in {{{\mathbb {S}}}}^1({\mathbf{n}})$, and $s\in {{{\mathbb {R}}}}$, we have

$$\begin{aligned} \lim _{\delta \rightarrow 0^+}\frac{ \psi \big ({\mathbf{y}},{\mathbf{z}},{\mathbf{n}}\delta s+\sqrt{1-(\delta s)^2}\sigma \big )-\psi ({\mathbf{y}},{\mathbf{z}},\sigma )}{\delta |{\mathbf{z}}|} =s\frac{ \nabla _{\sigma }\psi ({\mathbf{y}},{\mathbf{z}},\sigma )}{|{\mathbf{z}}|}\cdot {\mathbf{n}}. \end{aligned}$$

Also, applying Lemma 6.14 to ${{{\mathbb {S}}}^{2}}\ni \sigma \mapsto \psi ({\mathbf{y,z}},\sigma )$, there exists $0<C_{\psi }<\infty$ such that for all $\delta >0$

$$\begin{aligned} \sup _{y\in Y, {\mathbf{z}}\in {{{\mathbb {R}}}^{3}}{\setminus } \{\mathbf{0}\}, \sigma \in {{{\mathbb {S}}}}^1({\mathbf{n}})}\bigg | \frac{ \psi \big ({\mathbf{y}},{\mathbf{z}},{\mathbf{n}}\delta s+\sqrt{1-(\delta s)^2}\sigma \big ) -\psi ({\mathbf{y}},{\mathbf{z}},\sigma )}{\delta |{\mathbf{z}}|} \bigg | \le C_{\psi }|s|\quad \forall \, s\in [-1/\delta , 1/\delta ]. \end{aligned}$$

It follows from dominated convergence theorem and $\int _{-\infty }^{\infty }s j'(s){\mathrm{d}}s=-1$ that

$$\begin{aligned} \lim _{\delta \rightarrow 0^{+}}I_{1,\delta }({\mathbf{y,z}}) =\int _{-\infty }^{\infty }j'(s)\bigg (\int _{{{{\mathbb {S}}}}^1({\mathbf{n}})}s\frac{ \nabla _{\sigma }\psi }{|{\mathbf{z}}|}\cdot {\mathbf{n}}{\mathrm{d}}\sigma \bigg ){\mathrm{d}}s =-\int _{{{{\mathbb {S}}}}^1({\mathbf{n}})}\frac{ \nabla _{\sigma }\psi ({\mathbf{y}},{\mathbf{z}},\sigma )}{|{\mathbf{z}}|}\cdot {\mathbf{n}}{\mathrm{d}}\sigma . \end{aligned}$$

Moreover we have $\sup \limits _{{\mathbf{y}}\in Y,\,{\mathbf{z}}\in {{{\mathbb {R}}}^{3}}{\setminus }\{\mathbf{0}\},\delta >0} |I_{1,\delta }({\mathbf{y,z}})|<\infty$.

Estimates of $I_{2,\delta }({\mathbf{y,z}})$, $I_{3,\delta }({\mathbf{y,z}})$, $I_{4,\delta }({\mathbf{y,z}})$:  It is easily seen that

$$\begin{aligned}&\sup _{{\mathbf{y}}\in Y,\,{\mathbf{z}}\in {{{\mathbb {R}}}^{3}}{\setminus }\{\mathbf{0}\}}\big ( |I_{2,\delta }({\mathbf{y,z}})|+|I_{4,\delta }({\mathbf{y,z}})|\big )\le C_{\psi }\delta ,\quad \forall \,\delta>0;\\&\sup _{{\mathbf{y}}\in Y,\,{\mathbf{z}}\in {{{\mathbb {R}}}^{3}}{\setminus }\{\mathbf{0}\}, \delta >0}|I_{3,\delta }({\mathbf{y,z}})| <\infty ,\quad \lim _{\delta \rightarrow 0^{+}}I_{3,\delta }({\mathbf{y,z}}) =\int _{{{{\mathbb {S}}}}^1({\mathbf{n}})}\nabla _{{\mathbf{z}}}\psi ({\mathbf{y}},{\mathbf{z}},\sigma )\cdot \sigma {\mathrm{d}}\sigma . \end{aligned}$$

Thus it follows from dominated convergence theorem that

$$\begin{aligned}&\lim _{\delta \rightarrow 0^{+}} \int _{{{{\mathbb {S}}}^{2}}}\nabla _{{\mathbf{z}}} \cdot \Big (\Pi ({\mathbf{z}}) j_{\delta }({\mathbf{n}}\cdot \sigma )\psi ({\mathbf{y}},{\mathbf{z}},\sigma )\sigma \Big ){\mathrm{d}}\sigma \\&\quad = \int _{{{{\mathbb {S}}}}^1({\mathbf{n}})}\bigg ( -\frac{ \nabla _{\sigma }\psi ({\mathbf{y}},{\mathbf{z}},\sigma )}{|{\mathbf{z}}|}\cdot {\mathbf{n}} +\nabla _{{\mathbf{z}}}\psi ({\mathbf{y}},{\mathbf{z}},\sigma )\cdot \sigma \bigg ){\mathrm{d}}\sigma ,\\&\lim _{\delta \rightarrow 0^{+}} \int _{Y\times {{{\mathbb {R}}}^{3}}}\zeta _{\delta }(|{\mathbf{z}}|)F({\mathbf{y}},{\mathbf{z}})\bigg (\int _{{{{\mathbb {S}}}^{2}}}\nabla _{{\mathbf{z}}} \cdot \Big (\Pi ({\mathbf{z}}) j_{\delta }({\mathbf{n}}\cdot \sigma )\psi ({\mathbf{y}},{\mathbf{z}},\sigma )\sigma \Big ){\mathrm{d}}\sigma \bigg ){\mathrm{d}}{\mathbf{z}}{\mathrm{d}}{\mathbf{y}}\\&\quad =\int _{Y\times {{{\mathbb {R}}}^{3}}}F({\mathbf{y}},{\mathbf{z}}) \int _{{{{\mathbb {S}}}}^1({\mathbf{n}})}\Big ( \nabla _{{\mathbf{z}}}\psi ({\mathbf{y}},{\mathbf{z}},\sigma )\cdot \sigma -\frac{ \nabla _{\sigma }\psi ({\mathbf{y}},{\mathbf{z}},\sigma )}{|{\mathbf{z}}|}\cdot {\mathbf{n}} \Big ){\mathrm{d}}\sigma {\mathrm{d}}{\mathbf{z}}{\mathrm{d}}{\mathbf{y}}. \end{aligned}$$

Collecting the above results we conclude that

$$\begin{aligned}&\int _{Y\times {{{\mathbb {R}}}^{3}}} \int _{{{{\mathbb {S}}}}^1({\mathbf{n}})} \psi ({\mathbf{y}},{\mathbf{z}},\sigma ) \Pi ({\mathbf{z}})\nabla _{\mathbf{z}} F({\mathbf{y}},{\mathbf{z}})\cdot \sigma {\mathrm{d}}\sigma {\mathrm{d}}{\mathbf{z}}{\mathrm{d}}{\mathbf{y}} \\&\quad =-\int _{Y\times {{{\mathbb {R}}}^{3}}}F({\mathbf{y}},{\mathbf{z}}) \int _{{{{\mathbb {S}}}}^1({\mathbf{n}})}\Big ( \nabla _{{\mathbf{z}}}\psi ({\mathbf{y}},{\mathbf{z}},\sigma )\cdot \sigma -\frac{ \nabla _{\sigma }\psi ({\mathbf{y}},{\mathbf{z}},\sigma )}{|{\mathbf{z}}|}\cdot {\mathbf{n}} \Big ){\mathrm{d}}\sigma {\mathrm{d}}{\mathbf{z}}{\mathrm{d}}{\mathbf{y}}. \end{aligned}$$

This shows that the function $D({\mathbf{y}},{\mathbf{z}},\sigma ):= \Pi ({\mathbf{z}})\nabla _{{\mathbf{z}}}F({\mathbf{y}}, {\mathbf{z}})\cdot \sigma$ satisfies (6.35).

Conversely, suppose that there is a Borel measurable function $D({\mathbf{y}},{\mathbf{z}},\sigma )$ on $Y\times {{{{\mathbb {R}}}^{3}}\times {{\mathbb {S}}}^{2}}$ such that $D\in L^1_{loc}(Y\times {{{{\mathbb {R}}}^{3}}\times {{\mathbb {S}}}^{2}},{\mathrm{d}}\nu )$ and (6.35) holds. Let

$$\begin{aligned} \mathbf{D}({\mathbf{y}},{\mathbf{z}})=\frac{1}{\pi } \int _{{{{\mathbb {S}}}}^1({\mathbf{n}})} D({\mathbf{y}},{\mathbf{z}},\sigma )\sigma {\mathrm{d}}\sigma ,\quad {\mathbf{n}}={\mathbf{z}}/|{\mathbf{z}}|. \end{aligned}$$

Then $\mathbf{D}({\mathbf{y}},{\mathbf{z}})\in {{{\mathbb {R}}}}^2({\mathbf{z}})\,({\mathbf{z}}\ne \mathbf{0})$ and $\mathbf{D}\in L^1_{loc}(Y\times {{{\mathbb {R}}}^{3}}, {{{\mathbb {R}}}^{3}})$. For any $\Psi \in {{\mathcal {T}}}_{c}(Y\times {{{\mathbb {R}}}^{3}}, {{{\mathbb {R}}}^{3}})$, let $\psi ({\mathbf{y}},{\mathbf{z}},\sigma )= \Psi ({\mathbf{y}},{\mathbf{z}})\cdot \sigma$, then $\psi \in {{\mathcal {T}}}_{c}(Y\times {{{{\mathbb {R}}}^{3}}\times {{\mathbb {S}}}^{2}})$ and by (6.35) and Lemma 6.15 we have

$$\begin{aligned}&\int _{Y\times {{{\mathbb {R}}}^{3}}} \Psi ({\mathbf{y}}, {\mathbf{z}})\cdot \mathbf{D}({\mathbf{y}}, {\mathbf{z}}) {\mathrm{d}}{\mathbf{z}}{\mathrm{d}}{\mathbf{y}} =\frac{1}{\pi }\int _{Y\times {{{\mathbb {R}}}^{3}}}\int _{{{{\mathbb {S}}}}^1({\mathbf{n}})}\psi ({\mathbf{y}}, {\mathbf{z}},\sigma ) D({\mathbf{y}},{\mathbf{z}},\sigma ){\mathrm{d}}\sigma {\mathrm{d}}{\mathbf{z}}{\mathrm{d}}{\mathbf{y}} \\&\quad = -\frac{1}{\pi }\int _{Y\times {{{\mathbb {R}}}^{3}}}F({\mathbf{y}},{\mathbf{z}}) \int _{{{{\mathbb {S}}}}^1({\mathbf{n}})} \Big ( \nabla _{{\mathbf{z}}}\psi ({\mathbf{y}},{\mathbf{z}},\sigma )\cdot \sigma -\frac{ \nabla _{\sigma }\psi ({\mathbf{y}},{\mathbf{z}},\sigma )}{|{\mathbf{z}}|}\cdot {\mathbf{n}} \Big ){\mathrm{d}}\sigma {\mathrm{d}}{\mathbf{z}}{\mathrm{d}}{\mathbf{y}}\\&\quad =-\int _{Y\times {{{\mathbb {R}}}^{3}}}F({\mathbf{y}},{\mathbf{z}}) \nabla _{\mathbf{z}} \cdot \Pi ({\mathbf{z}})\Psi ({\mathbf{y}},{\mathbf{z}}) {\mathrm{d}}{\mathbf{z}}{\mathrm{d}}{\mathbf{y}}. \end{aligned}$$

Thus, by definition, F has the weak projection gradient $\Pi ({\mathbf{z}})\nabla _{{\mathbf{z}}}F({\mathbf{y}}, {\mathbf{z}})$ in ${\mathbf{z}}\in {{{\mathbb {R}}}^{3}}{\setminus }\{\mathbf{0}\}$ and

$$\begin{aligned} \Pi ({\mathbf{z}})\nabla _{{\mathbf{z}}}F({\mathbf{y}}, {\mathbf{z}})= \mathbf{D}({\mathbf{y}},{\mathbf{z}})=\frac{1}{\pi } \int _{{{{\mathbb {S}}}}^1({\mathbf{n}})} D({\mathbf{y}},{\mathbf{z}},\sigma )\sigma {\mathrm{d}}\sigma . \end{aligned}$$

Finally using (6.31) we also have

$$\begin{aligned} \frac{1}{\pi } \int _{{{{\mathbb {S}}}}^1({\mathbf{n}})}\big (\Pi ({\mathbf{z}})\nabla _{{\mathbf{z}}}F({\mathbf{y}}, {\mathbf{z}}) \cdot \sigma \big )\sigma {\mathrm{d}}\sigma =\Pi ({\mathbf{z}})\nabla _{{\mathbf{z}}}F({\mathbf{y}}, {\mathbf{z}}). \end{aligned}$$

Since $D({\mathbf{y}},{\mathbf{z}},\sigma )$ is unique, it follows that $D({\mathbf{y}},{\mathbf{z}},\sigma )= \Pi ({\mathbf{z}}) \nabla _{{\mathbf{z}}} F({\mathbf{y}}, {\mathbf{z}})\cdot \sigma$  a.e. $[\nu ]$ on $Y\times {{{{\mathbb {R}}}^{3}}\times {{\mathbb {S}}}^{2}}.$ This completes the proof. $\quad \square$