The normal-boundary Hodge estimate on four-dimensional balls

The whole-space first-order estimate. For a compactly supported smooth one-form $b$ on ${\mathbb{R}}^4$ put $f_{ij}={\mathrm{\partial }}_i{b}_j-{\mathrm{\partial }}_j{b}_i$ for all ordered indices and $g={\sum }_i{\mathrm{\partial }}_i{b}_i$. The identity $\mathrm{\Delta }{b}_j={\sum }_i{\mathrm{\partial }}_i{f}_{ij}+{\mathrm{\partial }}_{j}g$ and [F1] give ${\mathrm{\partial }}_k{b}_j=-{\sum }_i{R}_k{R}_i{f}_{ij}-R_k{R}_{j}g$: multiplication of the Fourier symbols proves equality away from zero frequency, hence almost everywhere in the $L^2$ transforms, and Plancherel injectivity identifies the functions. Riesz boundedness now gives $\mathrm{\parallel }\mathrm{\nabla }b{\mathrm{\parallel }}_p\le C_{p,V}(\mathrm{\parallel }db{\mathrm{\parallel }}_p+\mathrm{\parallel }{d}^{\ast }b{\mathrm{\parallel }}_p)$. Approximate compactly supported $W^{1,p}$ forms by compactly supported smooth ones using [F2]; all displayed first derivatives converge in $L^p$, so the estimate extends to them.

Reflection of a tangential one-form. For smooth $a$ on ${\bar{B}}_1$ with $a(\nu )=0$, put $\mathrm{ȷ}(x)=x\mathrm{/}\mathrm{\mid }x{\mathrm{\mid }}^2$ on $1<\mathrm{\mid }x\mathrm{\mid }<2$ and $b={\mathrm{ȷ}}^{\ast }a=D{\mathrm{ȷ}}^{T}a(\mathrm{ȷ}x)$ there, and put $b=a$ inside $B_1$. On the sphere $D\mathrm{ȷ}=I-2\nu \otimes \nu$, so the boundary values match. This continuous piecewise smooth form has its piecewise first derivatives as weak derivatives, by cancelling the value-jump terms in integration by parts. Outside the sphere, differentiating and cancelling the symmetric second derivatives of $\mathrm{ȷ}$ gives $db={\mathrm{ȷ}}^{\ast }(da)$. Also $D\mathrm{ȷ}D{\mathrm{ȷ}}^T=\mathrm{\mid }x{\mathrm{\mid }}^{-4}I$ and $\mathrm{\Delta }{\mathrm{ȷ}}_j=-4\mathrm{\mid }x{\mathrm{\mid }}^{-4}{x}_j$, whence $\mathrm{div}b=\mathrm{\mid }x{\mathrm{\mid }}^{-4}(\mathrm{div}a)(\mathrm{ȷ}x)+4\mathrm{\mid }x{\mathrm{\mid }}^{-2}x\cdot b$. The inversion Jacobian is $\mathrm{\mid }x{\mathrm{\mid }}^{-8}$; these formulas on the bounded annulus show that its $b,db,d^{\ast }b$ norms are controlled by the respective inside $a,da,d^{\ast }a$ norms, with an additional $C\mathrm{\parallel }a{\mathrm{\parallel }}_p$ in the codifferential estimate.

The estimate with a compact lower-order term. Multiply the reflected form by a fixed cutoff equal to one near ${\bar{B}}_1$ and supported in $B_2$. Its exterior derivative and codifferential acquire only bounded smooth multiples of $b$. Apply step 1.1 to the resulting compactly supported $W^{1,p}$ form and restrict to $B_1$, using step 1.2 and the product rule. This yields $\mathrm{\parallel }a{\mathrm{\parallel }}_{W^{1,p}}\le C_{p,V}(\mathrm{\parallel }da{\mathrm{\parallel }}_p+\mathrm{\parallel }{d}^{\ast }a{\mathrm{\parallel }}_p+\mathrm{\parallel }a{\mathrm{\parallel }}_p)$. It extends to $Y_p$ by its defining strong approximation. For the same approximants, integration by parts and zero normal value give $\int ⟨a_j,d\varphi ⟩=\int ⟨d^{\ast }{a}_j,\varphi ⟩$; passing to the limit proves the claimed identity for $a\in Y_p$.

The kernel has no nonzero elements. Suppose $a\in Y_p$ has $da=d^{\ast }a=0$. The identity used in step 1.1 makes each component of $a$ distributionally harmonic, hence smooth in the ball by [F4]. Closedness gives a smooth $V$-valued potential $u(x)={\int }_0^1{\sum }_i{x}_i{a}_i(tx)dt$: differentiating under the integral and using ${\mathrm{\partial }}_j{a}_i={\mathrm{\partial }}_i{a}_j$ makes ${\mathrm{\partial }}_{j}u$ the integral of $\frac{d}{dt}(t{a}_j(tx))$, so $du=a$. Normalize $u$ to have zero mean on $B_{1\mathrm{/}2}$. For $1\mathrm{/}2<R<1$, Poincare on $B_R$ gives $\mathrm{\parallel }u-u_{B_R}{\mathrm{\parallel }}_{L^2(B_R)}\le PR\mathrm{\parallel }a{\mathrm{\parallel }}_2$. Taking the mean over $B_{1\mathrm{/}2}$ bounds $\mathrm{\mid }{u}_{B_R}\mathrm{\mid }\le \mathrm{\mid }{B}_{1\mathrm{/}2}{\mathrm{\mid }}^{-1\mathrm{/}2}P\mathrm{\parallel }a{\mathrm{\parallel }}_2$. Consequently $\mathrm{\parallel }u{\mathrm{\parallel }}_{L^2(B_R)}\le C\mathrm{\parallel }a{\mathrm{\parallel }}_2$, uniformly in $R$. Here $a\in L^2$ since $p\ge 2$ and the ball has finite volume. Monotone convergence [F5] gives $u\in W^{1,2}(B_1)$ with $du=a$. The zero-normal identity of step 2.1 extends to every $W^{1,2}$ scalar test by [F5], since $a\in L^2$ and $d^{\ast }a=0$. Test with $u$ to obtain $\mathrm{\parallel }a{\mathrm{\parallel }}_2^2=0$. Thus $a=0$.

Removing the lower-order term. If no constant bounded the $W^{1,p}$ norm by $\mathrm{\parallel }da{\mathrm{\parallel }}_p+\mathrm{\parallel }{d}^{\ast }a{\mathrm{\parallel }}_p$ on $Y_p$, choose $a_j\in Y_p$ with $\mathrm{\parallel }{a}_j{\mathrm{\parallel }}_{W^{1,p}}=1$ and $\mathrm{\parallel }d{a}_j{\mathrm{\parallel }}_p+\mathrm{\parallel }{d}^{\ast }{a}_j{\mathrm{\parallel }}_p<1\mathrm{/}j$. By [F3] a subsequence converges in $L^p$. Apply step 2.1 to its pairwise differences; their first-order data tend to zero, so the subsequence is Cauchy in $W^{1,p}$. Completeness follows componentwise from [F5] by passing the derivative test identities to the $L^p$ limits. The strong limit $a\in Y_p$ therefore has norm one, but $da=d^{\ast }a=0$, contrary to step 3.1. This proves $\mathrm{\parallel }a{\mathrm{\parallel }}_{W^{1,p}}\le C_{p,V}(\mathrm{\parallel }da{\mathrm{\parallel }}_p+\mathrm{\parallel }{d}^{\ast }a{\mathrm{\parallel }}_p)$ on the unit ball.

Critical nonlinear absorption. For $V=\mathfrak{s}\mathfrak{u}(2)$, the Frobenius matrix inequality gives $\mathrm{\mid }[X,Y]\mathrm{\mid }\le 2\mathrm{\mid }X\mathrm{\mid }\mathrm{\mid }Y\mathrm{\mid }$ and hence $\mathrm{\mid }a\wedge a\mathrm{\mid }\le 2\mathrm{\mid }a{\mathrm{\mid }}^2$. Fix $2\le p<4$ and let $p^{\ast }=4p\mathrm{/}(4-p)$. By [F3] there is $S_p$ with $\mathrm{\parallel }a{\mathrm{\parallel }}_{p^{\ast }}\le S_p\mathrm{\parallel }a{\mathrm{\parallel }}_{W^{1,p}}$. For $d^{\ast }a=0$, step 4.1 and Holder give $\mathrm{\parallel }a{\mathrm{\parallel }}_{W^{1,p}}\le C_{p,V}(\mathrm{\parallel }{F}_a{\mathrm{\parallel }}_p+2\mathrm{\parallel }a{\mathrm{\parallel }}_4\mathrm{\parallel }a{\mathrm{\parallel }}_{p^{\ast }})$, since $1\mathrm{/}p=1\mathrm{/}4+1\mathrm{/}{p}^{\ast }$. Choose ${\rho }_p=(4C_{p,V}{S}_p{)}^{-1}$. Then $\mathrm{\parallel }a{\mathrm{\parallel }}_4\le {\rho }_p$ absorbs the last term and gives $\mathrm{\parallel }a{\mathrm{\parallel }}_{W^{1,p}}\le 2C_{p,V}\mathrm{\parallel }{F}_a{\mathrm{\parallel }}_p$. All curvature products are in $L^p$, since $p^{\ast }\ge 2p$ for $p\ge 2$ on a finite-volume ball.

All radii and conclusion. Pull back by $y\mapsto ry$, so $b(y)=ra(ry)$. Smooth zero-normal forms pull back to smooth zero-normal forms, and strong approximation shows that $Y_p$ pulls back to $Y_p$. The $L^p$ norms of $\mathrm{\nabla }b,db,d^{\ast }b$ scale by $r^{2-4\mathrm{/}p}$, while $\mathrm{\parallel }b{\mathrm{\parallel }}_p=r^{1-4\mathrm{/}p}\mathrm{\parallel }a{\mathrm{\parallel }}_p$. Also $F_b(y)=r^2{F}_a(ry)$ and $\mathrm{\parallel }b{\mathrm{\parallel }}_4=\mathrm{\parallel }a{\mathrm{\parallel }}_4$. Dividing the unit-ball estimates of steps 4.1–5.1 by $r^{2-4\mathrm{/}p}$ proves both stated bounds. The integration identity rescales directly, and the kernel assertion follows from the estimate. Full AC is used through the exact Fourier, compactness and approximation suppliers [F1]–[F6], including the sequential choices in step 4.1.