Khovanov homology detects the unknot

Since $v$ is primitive, Bezout's identity gives $w\in H_1(T;\mathbb{Z})$ with $\det (v,w)=1$. The linear torus diffeomorphism taking the basis $(v,w)$ to the standard basis takes both curves to curves of class $(1,0)$. It is enough to prove the claim for that class.

Put the two curves in transverse position. If they intersect, lift them to the annular cover $A={\mathbb{R}}^2\mathrm{/}⟨(1,0)⟩$. Each lift is an essential embedded circle, and the deck translates of each one are disjoint. In the universal cover of $A$, a lift of either circle is a proper line invariant under horizontal unit translation and contained in a bounded horizontal strip; all such lines have the same two ends. The signed intersection sum in a fundamental period is zero because both curves have class $(1,0)$. Thus intersections have both signs. Two successive intersections of opposite sign cut off a bounded region between subarcs of the two lines. Choose among such regions one with the least number of interior intersections and, among those, least horizontal width modulo deck translation. If a lifted arc entered the region, its first entry and exit would cut off a region with fewer interior intersections; if a nontrivial deck translate met it, the overlap would yield a region of smaller horizontal width. Hence its interior is disjoint from all lifted arcs and translates, and its projection is an embedded bigon in $T$. Isotoping one arc across the bigon removes two intersections and creates none. Compactness and transversality give finitely many intersections, so finitely many such isotopies make the curves disjoint.

The disjoint curves still have primitive class $(1,0)$, so neither bounds a disk. Cutting the torus along the second curve gives a connected orientable surface with two boundary circles and Euler characteristic zero, hence an annulus. The first curve remains essential in that annulus. Cutting the annulus along it gives two annuli, so it is parallel to the second curve and isotopic to it. Composing this isotopy with the finite bigon-removal isotopies in step 2.1 proves the assertion for the original pair; undoing the linear diffeomorphism proves it for $v$.

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.

A finite harmonic decomposition. If $p$ is homogeneous of degree $d$ on ${\mathbb{R}}^4$, put $N=\lfloor d\mathrm{/}2\rfloor$, $c_0=1$, and $c_j=-c_{j-1}\mathrm{/}(4j(d-j+1))$ for $1\le j\le N$. Define $H={\sum }_{j=0}^N{c}_j\mathrm{\mid }x{\mathrm{\mid }}^{2j}{\mathrm{\Delta }}^{j}p$. For homogeneous $Q$ of degree $d-2j$, product differentiation and Euler's identity give $\mathrm{\Delta }(\mathrm{\mid }x{\mathrm{\mid }}^{2j}Q)=4j(d-j+1)\mathrm{\mid }x{\mathrm{\mid }}^{2j-2}Q+\mathrm{\mid }x{\mathrm{\mid }}^{2j}\mathrm{\Delta }Q$. Hence successive terms cancel in $\mathrm{\Delta }H$, and ${\mathrm{\Delta }}^{N+1}p=0$ gives $\mathrm{\Delta }H=0$. All denominators are positive in their stated range. Moreover $p-H$ is divisible by $\mathrm{\mid }x{\mathrm{\mid }}^2$. Induction on $d$, starting with degrees zero and one, therefore writes $p$ as a finite sum ${\sum }_k\mathrm{\mid }x{\mathrm{\mid }}^{2k}{H}_{d-2k}$ with homogeneous harmonic $H_{d-2k}$. Decomposing each homogeneous piece of an arbitrary polynomial and then restricting to the unit sphere writes its restriction as ${\sum }_{l=0}^L{G}_l{\mathrm{\mid }}_{\mathrm{\partial }{B}_1}$ with each $G_l$ homogeneous harmonic of degree $l$. All constructions apply coefficientwise to $V$.

A polynomial antiderivative of the Laplacian. On a monomial $x_1^a{x}_2^{b_2}{x}_3^{b_3}{x}_4^{b_4}$ define $I$ to multiply by $x_1^2\mathrm{/}((a+1)(a+2))$; extend linearly to polynomials. Then ${\mathrm{\partial }}_1^{2}I=\mathrm{i}\mathrm{d}$, and $I$ commutes with ${\mathrm{\Delta }}^{\mathrm{\prime }}={\mathrm{\partial }}_2^2+{\mathrm{\partial }}_3^2+{\mathrm{\partial }}_4^2$. For a polynomial $f$, choose $M$ so that $({\mathrm{\Delta }}^{\mathrm{\prime }}{)}^{M+1}f=0$ and set $p=-{\sum }_{k=0}^M(-1{)}^k{I}^{k+1}({\mathrm{\Delta }}^{\mathrm{\prime }}{)}^{k}f$. Applying ${\mathrm{\partial }}_1^2+{\mathrm{\Delta }}^{\mathrm{\prime }}$ telescopes the finite sum and gives $\mathrm{\Delta }p=-f$.

Mean-zero polynomial density. Given an $L^2$ function on $B_1$, first approximate it in $L^2$ by a compactly supported smooth function using [F3]. Extend this smooth function by zero near the boundary; it is continuous on the compact closed ball. Apply the polynomial density of [F3] uniformly, which also gives $L^2$ approximation by [F4]. Thus polynomials are dense in $L^2(B_1)$, coefficientwise for $V$. For mean-zero $f$, take polynomial approximants $p_j\to f$ and replace $p_j$ by $f_j=p_j-(p_j{)}_{B_1}$. Holder gives $\mathrm{\parallel }(p_j{)}_{B_1}\mathrm{\parallel }\mathrm{\mid }{B}_1{\mathrm{\mid }}^{1\mathrm{/}2}\le \mathrm{\parallel }{p}_j-f{\mathrm{\parallel }}_2$, so these mean-zero polynomials still converge to $f$ in $L^2$. Full AC permits this sequential selection.

The normal correction for polynomial data. Suppose the polynomial $f$ has zero mean on $B_1$, choose $p$ from step 1.2, and let $g=-x\cdot \mathrm{\nabla }p$ restricted to $\mathrm{\partial }{B}_1$. By [F1], ${\int }_{\mathrm{\partial }{B}_1}g=-{\int }_{B_1}\mathrm{\Delta }p={\int }_{B_1}f=0$. Decompose $g={\sum }_l{G}_l{\mathrm{\mid }}_{\mathrm{\partial }{B}_1}$ by step 1.1. For $l\ge 1$, Euler's identity and [F1] give $l{\int }_{\mathrm{\partial }{B}_1}{G}_l={\int }_{\mathrm{\partial }{B}_1}{\mathrm{\partial }}_{\nu }{G}_l={\int }_{B_1}\mathrm{\Delta }{G}_l=0$. Thus the constant $G_0$ is zero: its sphere integral vanishes and the sphere has positive area, since [F1] applied to $x$ gives $\mathrm{\mid }\mathrm{\partial }{B}_1\mathrm{\mid }=4\mathrm{\mid }{B}_1\mathrm{\mid }>0$. The harmonic polynomial $h={\sum }_{l\ge 1}{G}_l\mathrm{/}l$ satisfies ${\mathrm{\partial }}_{\nu }h=g$ on the sphere. Hence $u=p+h$ satisfies $-\mathrm{\Delta }u=f$ and ${\mathrm{\partial }}_{\nu }u=0$. Subtracting its volume mean preserves these equations and gives a mean-zero polynomial solution.

The exact boundary Hessian estimate. For a scalar smooth $u$ with ${\mathrm{\partial }}_{\nu }u=0$ on $\mathrm{\partial }{B}_1$, consider $Z=(\mathrm{\Delta }u)\mathrm{\nabla }u-D^{2}u\mathrm{\nabla }u$. Direct differentiation cancels the third derivatives and gives $\mathrm{div}Z=(\mathrm{\Delta }u{)}^2-\mathrm{\mid }{D}^{2}u{\mathrm{\mid }}^2$. On the sphere $\mathrm{\nabla }u$ is tangential. Differentiating $\mathrm{\nabla }u\cdot \nu =0$ in that tangential direction gives $D^{2}u(\mathrm{\nabla }u,\nu )=-\mathrm{\mid }\mathrm{\nabla }u{\mathrm{\mid }}^2$, because ${\mathrm{\nabla }}_{\mathrm{\nabla }u}\nu =\mathrm{\nabla }u$ there. Thus $Z\cdot \nu =\mathrm{\mid }\mathrm{\nabla }u{\mathrm{\mid }}^2$. By [F1], $\mathrm{\parallel }\mathrm{\Delta }u{\mathrm{\parallel }}_2^2=\mathrm{\parallel }{D}^{2}u{\mathrm{\parallel }}_2^2+{\int }_{\mathrm{\partial }{B}_1}\mathrm{\mid }\mathrm{\nabla }u{\mathrm{\mid }}^2$. Sum the identity over an orthonormal basis of $V$. In particular every polynomial solution of step 2.1 satisfies $\mathrm{\parallel }{D}^{2}u{\mathrm{\parallel }}_2\le \mathrm{\parallel }f{\mathrm{\parallel }}_2$.

The remaining terms and the weak identity. Integration by parts [F1] for a mean-zero polynomial solution gives $\mathrm{\parallel }\mathrm{\nabla }u{\mathrm{\parallel }}_2^2=\int ⟨f,u⟩\le \mathrm{\parallel }f{\mathrm{\parallel }}_2\mathrm{\parallel }u{\mathrm{\parallel }}_2\le P\mathrm{\parallel }f{\mathrm{\parallel }}_2\mathrm{\parallel }\mathrm{\nabla }u{\mathrm{\parallel }}_2$ by [F2] and [F4]. Dividing when the gradient norm is nonzero, and treating zero separately, gives $\mathrm{\parallel }\mathrm{\nabla }u{\mathrm{\parallel }}_2\le P\mathrm{\parallel }f{\mathrm{\parallel }}_2$ and $\mathrm{\parallel }u{\mathrm{\parallel }}_2\le P^2\mathrm{\parallel }f{\mathrm{\parallel }}_2$. Together with step 3.1 this yields $\mathrm{\parallel }u{\mathrm{\parallel }}_2+\mathrm{\parallel }\mathrm{\nabla }u{\mathrm{\parallel }}_2+\mathrm{\parallel }{D}^{2}u{\mathrm{\parallel }}_2\le (1+P+P^2)\mathrm{\parallel }f{\mathrm{\parallel }}_2$. The all-test-function weak identity first holds for tests smooth near the closed ball by [F1], since the normal term vanishes; it then holds for every $W^{1,2}$ test by [F4]–[F5].

Completion constructs the strong inverse. For each $f_j$ choose the polynomial mean-zero solution $u_j$ of step 2.1. Their differences are polynomial mean-zero solutions for $f_j-f_k$, so steps 3.1–4.1 make $u_j$, all its first derivatives and all its second derivatives Cauchy in $L^2$. Completeness [F4] gives limits for each of these finitely many components. Passing in the weak derivative identities against a compactly supported smooth test identifies the first-derivative limits as the weak derivatives of the limit $u$, and then identifies the second-derivative limits as derivatives of those first derivatives. Hence $u\in W^{2,2}(B_1)$ and $u_j\to u$ strongly in that space. Its mean is zero by Holder. Passing to the limit in the weak identity of step 4.1 gives the displayed equation for every $W^{1,2}$ test. Norm continuity passes both estimates of steps 3.1–4.1 to $u$, including the constant-one Hessian bound.

Uniqueness, linearity and compatibility. If two mean-zero $W^{1,2}$ solutions have the same datum, test their difference equation with the difference itself. Its gradient has zero $L^2$ norm; [F2] makes its $L^2$ norm zero, so the solutions agree. This proves uniqueness in the claimed $W^{2,2}$ class as well. Linear combinations satisfy the same linear weak identity and mean-zero normalization; uniqueness proves linearity of the inverse. Testing any weak solution with each constant vector in $V$ gives $\int f=0$, proving necessity. The polynomial solution of step 2.1 equals the completed solution by this uniqueness.

Every radius. For data $f$ on $B_r$, put $f_r(y)=r^{2}f(ry)$ on $B_1$ and solve there for $v$ by step 5.1. Set $u(x)=v(x\mathrm{/}r)$. Changing variables in the weak identity proves the required identity and mean-zero normalization on $B_r$. In four dimensions, $\mathrm{\parallel }{f}_r{\mathrm{\parallel }}_{L^2(B_1)}=\mathrm{\parallel }f{\mathrm{\parallel }}_{L^2(B_r)}$, $\mathrm{\parallel }{D}^{2}v{\mathrm{\parallel }}_2=\mathrm{\parallel }{D}^{2}u{\mathrm{\parallel }}_2$, $\mathrm{\parallel }\mathrm{\nabla }v{\mathrm{\parallel }}_2=r^{-1}\mathrm{\parallel }\mathrm{\nabla }u{\mathrm{\parallel }}_2$, and $\mathrm{\parallel }v{\mathrm{\parallel }}_2=r^{-2}\mathrm{\parallel }u{\mathrm{\parallel }}_2$. Thus the bounds follow with $C=1+P+P^2$, independent of $r$. The construction preserves polynomiality for polynomial data.

Given finite equivariance and conclusion. For $\gamma \in \mathrm{\Gamma }$ let $T_{\gamma }u(x)=\sigma (\gamma {)}^{-1}u(\gamma x)$. Orthogonal changes of variables preserve the ball, volume, the gradient pairing and the zero mean; orthogonality of $\sigma (\gamma )$ preserves the target pairing. Therefore $T_{\gamma }u$ solves the same weak equation as $u$ when $T_{\gamma }f=f$. Uniqueness from step 6.1, rescaled as in step 7.1, gives $T_{\gamma }u=u$. This proves the equivariance for every element of the given finite group, including the stated Lie-algebra case. All existence, uniqueness and estimate claims are proved; full AC enters only through the analytic suppliers and the sequential approximation choices [F2]–[F6].

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.