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Proof.
First, we prove Item (1).
By definition, Δ = d d ∗ + d ∗ d \Delta=dd^{*}+d^{*}d ν y = d y α \nu_{y}=dy_{\alpha} 1 ≤ α ≤ 3 1\leq\alpha\leq 3 1 ≤ p ≤ m − 3 1\leq p\leq m-3
First, we compute d ∗ d ω d^{*}d\omega
(3.168) d ω \displaystyle d\omega = f ( x ) ⋅ ∂ h ( y ) ∂ y β d y β ∧ d y α ∧ τ x + ∂ f ( x ) ∂ x j ⋅ h ( y ) ⋅ d x j ∧ ν y ∧ τ x , \displaystyle=f(x)\cdot\frac{\partial h(y)}{\partial y_{\beta}}dy_{\beta}\wedge dy_{\alpha}\wedge\tau_{x}+\frac{\partial f(x)}{\partial x_{j}}\cdot h(y)\cdot dx_{j}\wedge\nu_{y}\wedge\tau_{x},
which implies that
∗ d ω \displaystyle*d\omega = ( − 1 ) p f ( x ) ⋅ ∂ h ( y ) ∂ y β d y β α ^ ∧ ∗ T ( τ x ) + O ′ ( | y | k ) \displaystyle=(-1)^{p}f(x)\cdot\frac{\partial h(y)}{\partial y_{\beta}}dy_{\widehat{\beta\alpha}}\wedge*_{T}(\tau_{x})+O^{\prime}(|y|^{k})
(3.169) = ( − 1 ) p f ( x ) ( ∂ h ( y ) ∂ y α − 1 d y α + 1 − ∂ h ( y ) ∂ y α + 1 d y α − 1 ) ∧ ∗ T ( τ x ) + O ′ ( | y | k ) . \displaystyle=(-1)^{p}f(x)\Big(\frac{\partial h(y)}{\partial y_{\alpha-1}}dy_{\alpha+1}-\frac{\partial h(y)}{\partial y_{\alpha+1}}dy_{\alpha-1}\Big)\wedge*_{T}(\tau_{x})+O^{\prime}(|y|^{k}).
Differentiating the above equality,
d ∗ d ω \displaystyle d*d\omega = ( − 1 ) p f ⋅ ( ∂ 2 h ∂ y α ∂ y α − 1 d y α ∧ d y α + 1 − ∂ 2 h ∂ y α ∂ y α + 1 d y α ∧ d y α − 1 CLOSE \displaystyle=(-1)^{p}f\cdot\Big(\frac{\partial^{2}h}{\partial y_{\alpha}\partial y_{\alpha-1}}dy_{\alpha}\wedge dy_{\alpha+1}-\frac{\partial^{2}h}{\partial y_{\alpha}\partial y_{\alpha+1}}dy_{\alpha}\wedge dy_{\alpha-1}
(3.170) + ( ∂ 2 h ∂ y α − 1 2 + ∂ 2 h ∂ y α + 1 2 ) d y α − 1 ∧ d y α + 1 ) ∧ ∗ T ( τ x ) + O ′ ( | y | k − 1 ) . \displaystyle+\Big(\frac{\partial^{2}h}{\partial y_{\alpha-1}^{2}}+\frac{\partial^{2}h}{\partial y_{\alpha+1}^{2}}\Big)dy_{\alpha-1}\wedge dy_{\alpha+1}\Big)\wedge*_{T}(\tau_{x})+O^{\prime}(|y|^{k-1}).
Then it follows that
d ∗ d ω \displaystyle d^{*}d\omega = ( − 1 ) m p + m + 1 ∗ d ∗ d ω \displaystyle=(-1)^{mp+m+1}*d*d\omega
= f ⋅ ( ∂ 2 h ∂ y α ∂ y α − 1 d y α − 1 + ∂ 2 h ∂ y α ∂ y α + 1 d y α + 1 − ( ∂ 2 h ∂ y α − 1 2 + ∂ 2 h ∂ y α + 1 2 ) d y α ) ∧ τ x \displaystyle=f\cdot\Big(\frac{\partial^{2}h}{\partial y_{\alpha}\partial y_{\alpha-1}}dy_{\alpha-1}+\frac{\partial^{2}h}{\partial y_{\alpha}\partial y_{\alpha+1}}dy_{\alpha+1}-\Big(\frac{\partial^{2}h}{\partial y_{\alpha-1}^{2}}+\frac{\partial^{2}h}{\partial y_{\alpha+1}^{2}}\Big)dy_{\alpha}\Big)\wedge\tau_{x}
(3.171) + O ′ ( | y | k − 1 ) . \displaystyle+O^{\prime}(|y|^{k-1}).
On the other hand,
(3.172) ∗ ω = f ( x ) h ( y ) d y α ^ ∧ ∗ T ( τ x ) , \displaystyle*\omega=f(x)h(y)dy_{\widehat{\alpha}}\wedge*_{T}(\tau_{x}),
which implies
(3.173) d ∗ ω = f ( x ) ∂ h ∂ y α dvol N ∧ ∗ T ( τ x ) + O ′ ( | y | k ) . d*\omega=f(x)\frac{\partial h}{\partial y_{\alpha}}\dvol_{N}\wedge*_{T}(\tau_{x})+O^{\prime}(|y|^{k}).
So it follows that
(3.174) d ∗ ω = ( − 1 ) m p + 1 ∗ d ∗ ω = − f ⋅ ∂ h ∂ y α ⋅ τ x + O ′ ( | y | k ) , d^{*}\omega=(-1)^{mp+1}*d*\omega=-f\cdot\frac{\partial h}{\partial y_{\alpha}}\cdot\tau_{x}+O^{\prime}(|y|^{k}),
and hence
(3.175) d d ∗ ω = − f ⋅ ( ∂ 2 h ∂ y α − 1 ∂ y α d y α − 1 + ∂ 2 h ∂ y α 2 d y α + ∂ 2 h ∂ y α + 1 ∂ y α d y α + 1 ) ⋅ d y α ∧ τ x + O ′ ( | y | k − 1 ) . dd^{*}\omega=-f\cdot\Big(\frac{\partial^{2}h}{\partial y_{\alpha-1}\partial y_{\alpha}}dy_{\alpha-1}+\frac{\partial^{2}h}{\partial y_{\alpha}^{2}}dy_{\alpha}+\frac{\partial^{2}h}{\partial y_{\alpha+1}\partial y_{\alpha}}dy_{\alpha+1}\Big)\cdot dy_{\alpha}\wedge\tau_{x}+O^{\prime}(|y|^{k-1}).
Therefore, combining (3.171 3.175
(3.176) Δ ω = ( d ∗ d + d d ∗ ) ω = − f ⋅ ( Δ 0 h ( y ) ) d y α ∧ τ x + O ′ ( | y | k − 1 ) , \Delta\omega=(d^{*}d+dd^{*})\omega=-f\cdot(\Delta_{0}h(y))dy_{\alpha}\wedge\tau_{x}+O^{\prime}(|y|^{k-1}),
where Δ 0 ( h ( y ) ) = − ∂ 2 h ∂ y 1 2 − ∂ 2 h ∂ y 2 2 − ∂ 2 h ∂ y 3 2 \Delta_{0}(h(y))=-\frac{\partial^{2}h}{\partial y_{1}^{2}}-\frac{\partial^{2}h}{\partial y_{2}^{2}}-\frac{\partial^{2}h}{\partial y_{3}^{2}}
Now we prove Item (2). Let ω = f ⋅ y α r d y 1 ∧ d y 2 ∧ d y 3 \omega=f\cdot\frac{y_{\alpha}}{r}dy_{1}\wedge dy_{2}\wedge dy_{3} d ∗ d ω d^{*}d\omega 3.2 d r = y γ d y γ r dr=\frac{y_{\gamma}dy_{\gamma}}{r}
(3.177) d ( y α r ) ∧ d y 1 ∧ d y 2 ∧ d y 3 = 0 . d(\frac{y_{\alpha}}{r})\wedge dy_{1}\wedge dy_{2}\wedge dy_{3}=0.
This implies that
(3.178) d ω = ∂ f ∂ x i ⋅ y α r d x i ∧ d y 1 ∧ d y 2 ∧ d y 3 , d\omega=\frac{\partial f}{\partial x_{i}}\cdot\frac{y_{\alpha}}{r}dx_{i}\wedge dy_{1}\wedge dy_{2}\wedge dy_{3},
and hence
(3.179) ∗ d ω = − ∂ f ∂ x i ⋅ y α r ∗ T ( d x i ) + O ′ ( r ) . *d\omega=-\frac{\partial f}{\partial x_{i}}\cdot\frac{y_{\alpha}}{r}*_{T}(dx_{i})+O^{\prime}(r).
Differentiating the above equality and applying Lemma 3.2
(3.180) d ∗ d ω = − ∂ f ∂ x i ( d y α r − y α y β ⋅ d y β r 3 ) ∗ ( d x i ) + O ′ ( 1 ) . \displaystyle d*d\omega=-\frac{\partial f}{\partial x_{i}}\Big(\frac{dy_{\alpha}}{r}-\frac{y_{\alpha}y_{\beta}\cdot dy_{\beta}}{r^{3}}\Big)*(dx_{i})+O^{\prime}(1).
It follows that
(3.181) ∗ d ∗ d ω = ( − 1 ) m + 1 ⋅ ∂ f ∂ x i ⋅ d y α ^ ∧ d x i r + ( − 1 ) m ∂ f ∂ x i ⋅ y α y β r 3 ⋅ d y β ^ ∧ d x i + O ′ ( 1 ) \displaystyle*d*d\omega=(-1)^{m+1}\cdot\frac{\partial f}{\partial x_{i}}\cdot\frac{dy_{\widehat{\alpha}}\wedge dx_{i}}{r}+(-1)^{m}\frac{\partial f}{\partial x_{i}}\cdot\frac{y_{\alpha}y_{\beta}}{r^{3}}\cdot dy_{\widehat{\beta}}\wedge dx_{i}+O^{\prime}(1)
Therefore,
d ∗ d ω \displaystyle d^{*}d\omega = ( − 1 ) m + 1 ∗ d ∗ d ω \displaystyle=(-1)^{m+1}*d*d\omega
(3.182) = ∂ f ∂ x i ⋅ d y α ^ ∧ d x i r − ∂ f ∂ x i ⋅ y α y β r 3 ⋅ d y β ^ ∧ d x i + O ′ ( 1 ) . \displaystyle=\frac{\partial f}{\partial x_{i}}\cdot\frac{dy_{\widehat{\alpha}}\wedge dx_{i}}{r}-\frac{\partial f}{\partial x_{i}}\cdot\frac{y_{\alpha}y_{\beta}}{r^{3}}\cdot dy_{\widehat{\beta}}\wedge dx_{i}+O^{\prime}(1).
Now we compute d d ∗ ω dd^{*}\omega 3.13 dvol N \dvol_{N} 3.92
∗ ( d y 1 ∧ d y 2 ∧ d y 3 ) \displaystyle*(dy_{1}\wedge dy_{2}\wedge dy_{3}) = ∗ ( dvol N + A i γ β y γ d y γ ^ ∧ d x i ) + O ~ ( r 2 ) \displaystyle=*(\dvol_{N}+A_{i\gamma\beta}y_{\gamma}dy_{\widehat{\gamma}}\wedge dx_{i})+\widetilde{O}(r^{2})
(3.183) = ( 1 + H β y β ) dvol T + A i γ β y γ d y γ ∧ ∗ T ( d x i ) + O ~ ( r 2 ) , \displaystyle=(1+H^{\beta}y_{\beta})\dvol_{T}+A_{i\gamma\beta}y_{\gamma}dy_{\gamma}\wedge*_{T}(dx_{i})+\widetilde{O}(r^{2}),
so we have
∗ ω \displaystyle*\omega = f ⋅ y α r ⋅ dvol T + f ⋅ H β ⋅ y α y β r ⋅ dvol T + f ⋅ A i γ β y α y γ r d y γ ∧ ∗ T ( d x i ) + O ′ ( r 2 ) \displaystyle=f\cdot\frac{y_{\alpha}}{r}\cdot\dvol_{T}+f\cdot H^{\beta}\cdot\frac{y_{\alpha}y_{\beta}}{r}\cdot\dvol_{T}+f\cdot A_{i\gamma\beta}\frac{y_{\alpha}y_{\gamma}}{r}dy_{\gamma}\wedge*_{T}(dx_{i})+O^{\prime}(r^{2})
(3.184) ≡ 𝔗 1 + 𝔗 2 + 𝔗 3 + O ′ ( r 2 ) . \displaystyle\equiv\FT_{1}+\FT_{2}+\FT_{3}+O^{\prime}(r^{2}).
By collecting the leading terms, it is easy to compute the leading term in the above equality,
(3.185) d ∗ d ( 𝔗 1 ) \displaystyle d*d(\FT_{1}) = ∂ f ∂ x i ( d x i ∧ d y α ^ r − y α y β d x i ∧ d y β ^ r 3 ) − 2 ω + O ′ ( 1 ) \displaystyle=\frac{\partial f}{\partial x_{i}}\Big(\frac{dx_{i}\wedge dy_{\widehat{\alpha}}}{r}-\frac{y_{\alpha}y_{\beta}dx_{i}\wedge dy_{\widehat{\beta}}}{r^{3}}\Big)-2\omega+O^{\prime}(1)
(3.186) d ∗ d ( 𝔗 2 ) \displaystyle d*d(\FT_{2}) = r − 5 Π 3 ( 4 ) , d ∗ d ( 𝔗 3 ) = r − 5 Π 3 ( 4 ) . \displaystyle=r^{-5}\Pi_{3}^{(4)},\ d*d(\FT_{3})=r^{-5}\Pi_{3}^{(4)}.
Therefore,
(3.187) d d ∗ ω = − ∂ f ∂ x i ⋅ d x i ∧ d y α ^ r + ∂ f ∂ x i ⋅ y α y β r 3 ⋅ d x i ∧ d y β ^ + 2 ω + r − 5 Π 3 ( 4 ) + O ′ ( 1 ) . dd^{*}\omega=-\frac{\partial f}{\partial x_{i}}\cdot\frac{dx_{i}\wedge dy_{\widehat{\alpha}}}{r}+\frac{\partial f}{\partial x_{i}}\cdot\frac{y_{\alpha}y_{\beta}}{r^{3}}\cdot dx_{i}\wedge dy_{\widehat{\beta}}+2\omega+r^{-5}\Pi_{3}^{(4)}+O^{\prime}(1).
By (3.182 3.187
(3.188) Δ ω = ( d ∗ d + d d ∗ ) ω = 2 ω + r − 5 Π 3 ( 4 ) + O ′ ( 1 ) . \displaystyle\Delta\omega=(d^{*}d+dd^{*})\omega=2\omega+r^{-5}\Pi_{3}^{(4)}+O^{\prime}(1).
So the proof is done.