A strong Lp Neumann inverse on four-dimensional balls

Reflection matching. Let $u$ be smooth near ${\bar{B}}_1$ with ${\mathrm{\partial }}_{\nu }u=0$. On $1<\mathrm{\mid }x\mathrm{\mid }<2$ put $\mathrm{ȷ}(x)=x\mathrm{/}\mathrm{\mid }x{\mathrm{\mid }}^2$ and $v(x)=u(\mathrm{ȷ}(x))$, and put $v=u$ on $B_1$. At $\mathrm{\mid }x\mathrm{\mid }=1$, $D\mathrm{ȷ}=I-2\nu \otimes \nu$; since $\mathrm{\nabla }u$ is tangential, the outside gradient $D{\mathrm{ȷ}}^T\mathrm{\nabla }u$ equals the inside gradient. The values match as well. Thus $v$ is $C^1$ across the sphere, smooth on each side, and has locally bounded piecewise second derivatives. Integration by parts on the two sides cancels the interface terms for $v$ and its first derivatives, so these piecewise derivatives are its weak derivatives through order two.

The reflected equation. Writing $s=\mathrm{\mid }x\mathrm{\mid }$, one has $D\mathrm{ȷ}=s^{-2}(I-2x\otimes x\mathrm{/}{s}^2)$, $D\mathrm{ȷ}D{\mathrm{ȷ}}^T=s^{-4}I$, and $\mathrm{\Delta }{\mathrm{ȷ}}_j=-4s^{-4}{x}_j$ in four dimensions. The chain rule therefore gives $\mathrm{\Delta }v=s^{-4}(\mathrm{\Delta }u)(\mathrm{ȷ}x)-4s^{-4}x\cdot (\mathrm{\nabla }u)(\mathrm{ȷ}x)$. Since $x\cdot \mathrm{\nabla }v=-s^{-2}x\cdot (\mathrm{\nabla }u)(\mathrm{ȷ}x)$, this is $\mathrm{\Delta }v=s^{-4}(\mathrm{\Delta }u)(\mathrm{ȷ}x)+4s^{-2}x\cdot \mathrm{\nabla }v$. All coefficients are bounded on the annulus. The inversion Jacobian has absolute value $s^{-8}$ and maps that annulus into $B_1\setminus {\bar{B}}_{1\mathrm{/}2}$, so change of variables bounds its reflected $u$, first-derivative and Laplacian norms by constant multiples of their norms on $B_1$.

A whole-space estimate for the extension. Choose the cutoff $\chi$ of [F5] equal to one near ${\bar{B}}_1$, and define $Eu=\chi v$ on $B_2$, zero elsewhere. Step 1.1 gives $Eu\in W^{2,p}({\mathbb{R}}^4)$ with compact support. Step 2.1 and $\mathrm{\Delta }(\chi v)=\chi \mathrm{\Delta }v+2\mathrm{\nabla }\chi \cdot \mathrm{\nabla }v+(\mathrm{\Delta }\chi )v$ give $\mathrm{\parallel }Eu{\mathrm{\parallel }}_p\le C\mathrm{\parallel }u{\mathrm{\parallel }}_p$ and $\mathrm{\parallel }\mathrm{\Delta }Eu{\mathrm{\parallel }}_p\le C_p(\mathrm{\parallel }\mathrm{\Delta }u{\mathrm{\parallel }}_p+\mathrm{\parallel }\mathrm{\nabla }u{\mathrm{\parallel }}_p+\mathrm{\parallel }u{\mathrm{\parallel }}_p)$. Since $Eu=u$ on $B_1$, $\mathrm{\parallel }\mathrm{\nabla }u{\mathrm{\parallel }}_p\le \mathrm{\parallel }\mathrm{\nabla }Eu{\mathrm{\parallel }}_p$. Apply [F2] to obtain $\mathrm{\parallel }{D}^{2}Eu{\mathrm{\parallel }}_p\le C_p(\mathrm{\parallel }\mathrm{\Delta }u{\mathrm{\parallel }}_p+\mathrm{\parallel }\mathrm{\nabla }Eu{\mathrm{\parallel }}_p+\mathrm{\parallel }u{\mathrm{\parallel }}_p)$. Interpolate the middle term by [F2] with $\eta >0$ chosen so $C_p\eta \le 1\mathrm{/}2$, and absorb it. Then interpolate once more for the first derivatives and restrict to $B_1$. The result is $\mathrm{\parallel }u{\mathrm{\parallel }}_{W^{2,p}(B_1)}\le C_p(\mathrm{\parallel }\mathrm{\Delta }u{\mathrm{\parallel }}_p+\mathrm{\parallel }u{\mathrm{\parallel }}_p)$. No estimate was applied before the reflected function's $W^{2,p}$ membership was proved.

Removing the zeroth-order term for Neumann solutions. For mean-zero polynomial $f$, [F1] gives a polynomial mean-zero solution $u$ with classical zero normal derivative and $\mathrm{\parallel }u{\mathrm{\parallel }}_{W^{2,2}}\le C\mathrm{\parallel }f{\mathrm{\parallel }}_2$. By [F3], $\mathrm{\parallel }u{\mathrm{\parallel }}_p\le C_p\mathrm{\parallel }f{\mathrm{\parallel }}_2\le C_p\mathrm{\mid }{B}_1{\mathrm{\mid }}^{1\mathrm{/}2-1\mathrm{/}p}\mathrm{\parallel }f{\mathrm{\parallel }}_p$, where the final Holder bound uses $p\ge 2$. Substitution in step 3.1, with $-\mathrm{\Delta }u=f$, proves $\mathrm{\parallel }u{\mathrm{\parallel }}_{W^{2,p}(B_1)}\le C_{p,V}\mathrm{\parallel }f{\mathrm{\parallel }}_p$ for these polynomial solutions, uniformly in their degrees.

Construction for all mean-zero Lp data. Extend $f$ by zero to ${\mathbb{R}}^4$, approximate in $L^p$ by smooth functions using [F4], and restrict to the closed ball. Each approximant is continuous there and hence uniformly approximable by polynomials by [F4]. A choice of uniform error tending to zero gives polynomial approximants in $L^p$; subtract their means to obtain mean-zero $f_j\to f$ in $L^p$, since $\mathrm{\parallel }(f_j-f{)}_{B_1}\mathrm{\parallel }\mathrm{\mid }{B}_1{\mathrm{\mid }}^{1\mathrm{/}p}\le \mathrm{\parallel }{f}_j-f{\mathrm{\parallel }}_p$ before normalization. Let $u_j$ be the polynomial solutions from [F1]. Applying step 4.1 to their differences makes all derivative components through order two Cauchy in $L^p$. Completeness [F4] and passage in the weak-derivative test identities give a limit $u\in W^{2,p}$ with $u_j\to u$ strongly and the same bound. Finite volume and $p\ge 2$ imply $u_j\to u$ in $W^{1,2}$ and $f_j\to f$ in $L^2$; their weak Neumann identities pass to the limit for every $W^{1,2}$ test. Thus the solution is the unique $L^2$ Neumann solution of [F1], now with the stronger regularity and norm estimate.

The complete normalized domain. By step 5.1 the inverse maps into $X_p^0$ and is bounded there. Conversely let $u\in X_p^0$ and choose polynomial zero-normal approximants $u_j\to u$ in $W^{2,p}$. Subtracting their means preserves the normal condition and convergence. The sources $f_j=-\mathrm{\Delta }{u}_j$ have zero mean, by the polynomial weak identity of [F1], and converge to $f=-\mathrm{\Delta }u\in L_0^p$. Each $u_j$ is the normalized solution for $f_j$, so the bounded inverse from step 5.1 sends $f$ to $u$. This proves bijectivity. Also $\mathrm{\parallel }\mathrm{\Delta }u{\mathrm{\parallel }}_p\le C\mathrm{\parallel }u{\mathrm{\parallel }}_{W^{2,p}}$ and the inverse bound gives the reverse norm comparison. The zero-mean subspace of $L^p$ is closed, since its mean functional is bounded by Holder, hence complete by [F4]. Its isomorphism with $X_p^0$ gives completeness for $\mathrm{\parallel }\mathrm{\Delta }u{\mathrm{\parallel }}_p$.

Scaling, uniqueness and symmetry. For $B_r$ put $f_r(y)=r^{2}f(ry)$, solve on $B_1$ and set $u(x)=u_r(x\mathrm{/}r)$. Direct change of variables gives $\mathrm{\parallel }{f}_r{\mathrm{\parallel }}_p=r^{2-4\mathrm{/}p}\mathrm{\parallel }f{\mathrm{\parallel }}_p$, $\mathrm{\parallel }{D}^2{u}_r{\mathrm{\parallel }}_p=r^{2-4\mathrm{/}p}\mathrm{\parallel }{D}^{2}u{\mathrm{\parallel }}_p$, $\mathrm{\parallel }\mathrm{\nabla }{u}_r{\mathrm{\parallel }}_p=r^{1-4\mathrm{/}p}\mathrm{\parallel }\mathrm{\nabla }u{\mathrm{\parallel }}_p$, and $\mathrm{\parallel }{u}_r{\mathrm{\parallel }}_p=r^{-4\mathrm{/}p}\mathrm{\parallel }u{\mathrm{\parallel }}_p$. Divide the unit-ball bound of step 5.1 by $r^{2-4\mathrm{/}p}$ to obtain the displayed estimate; polynomial approximations rescale as well. The weak identity passes by change of variables. Uniqueness, linearity and preservation of prescribed finite orthogonal equivariance follow from [F1], since all these solutions belong to its unique mean-zero $W^{1,2}$ solution class. Full AC is inherited through [F2]–[F5] and the sequential approximation choices.