Strict Subadditivity of Biharmonic Capacity for Externally Tangent Balls

1. Introduction

For an integer $m\ge 1$ and $n>2m$, the classical m-harmonic capacity of a compact set $K\subset {\mathbb{R}}^n$ is

$$\begin{array}{cc}\hfill {cap}_m\left(K\right):=inf\{& \sum _{\left|\alpha \right|=m}{\displaystyle \frac{m!}{\alpha !}}{\parallel D^{\alpha }u\parallel }_{L^2\left({\mathbb{R}}^n\right)}^2:u\in C_c^{\infty }\left({\mathbb{R}}^n\right),\hfill \\ \hfill & u=1\mathrm{in}\mathrm{a}\mathrm{neighbourhood}\mathrm{of}K\}.\hfill \end{array}$$

(1)

The capacity was introduced in the higher-order Dirichlet theory by Maz’ya; see [10,11,12]. It should be distinguished from the related higher-order capacities used in modern Wiener criteria, although the two theories share a common variational and polyharmonic background; see [9]. For $m\ge 2$, the elementary lattice argument available for Newtonian capacity breaks down because truncations and pointwise maxima do not interact monotonically with the full order-m energy.

Maz’ya asked whether

$${cap}_m(F\cup G)\le {cap}_m\left(F\right)+{cap}_m\left(G\right)$$

(2)

holds for arbitrary compact sets; see Problem 35 in [13]. The present paper treats a specific nontrivial configuration in the first higher-order case $m=2$.

Let $e_1=(1,0,\dots ,0)$ and let $R_{+},R_{-}>0$. Put

$$K_{+}:=\overline{B_{R_{+}}\left(R_{+}{e}_1\right)},K_{-}:=\overline{B_{R_{-}}(-R_{-}{e}_1)}.$$

(3)

Their interiors are disjoint and their boundaries are tangent at the origin.

Theorem 1.

Let $n>4$ and let $K_{\pm }$ be given by (3). Then

$${cap}_2(K_{+}\cup K_{-})<{cap}_2\left(K_{+}\right)+{cap}_2\left(K_{-}\right).$$

(4)

The proof uses three exact features of this geometry. First, inversion at the tangency point maps the exterior of $K_{+}\cup K_{-}$ to a slab and maps the exterior of a single ball to a half-space. Second, a reverse Kelvin construction identifies the capacity with the Robin regular part of the appropriate clamped biharmonic Green kernel. Third, after a partial Fourier transform, the comparison becomes one-dimensional.

The Green function literature for higher-order Dirichlet problems is substantially subtler than in the harmonic case because a general maximum principle is unavailable. Boggio’s formula gives an explicit positive Green kernel in a ball [3]; estimates and sign phenomena in general domains are developed, for example, in [4,5,7,8]. In the present slab geometry, no positivity theorem for arbitrary domains is invoked: the Green kernel is defined fiberwise, and the decisive inequality is proved directly for every Fourier frequency.

The result remains a special case of (2). It does not address separated balls, overlapping balls, general smooth sets, or $m\ge 3$.

2. The Variational Capacity and Its Far Field

2.1. The Homogeneous Space and the Capacitary Potential

For $n>4$, let ${\dot{H}}^2\left({\mathbb{R}}^n\right)$ be the completion of $C_c^{\infty }\left({\mathbb{R}}^n\right)$ under the norm

$${\parallel u\parallel }_{{\dot{H}}^2}:={\parallel \Delta u\parallel }_{L^2\left({\mathbb{R}}^n\right)}.$$

The Sobolev inequality gives a continuous embedding

$${\dot{H}}^2\left({\mathbb{R}}^n\right)\hookrightarrow L^{2n/(n-4)}\left({\mathbb{R}}^n\right),$$

(5)

so this completion has a canonical realization by functions rather than by classes modulo polynomials; see [11]. We shall use the following standard form of the homogeneous-space realization.

Lemma 1

(Realization of ${\dot{H}}^2$). Let $n>4$. A distribution u belongs to ${\dot{H}}^2\left({\mathbb{R}}^n\right)$ if and only if

$$u\in L^{2n/(n-4)}\left({\mathbb{R}}^n\right),\Delta u\in L^2\left({\mathbb{R}}^n\right),$$

(6)

in the distributional sense. Moreover,

$$\nabla u\in L^{2n/(n-2)}\left({\mathbb{R}}^n\right){,\parallel \nabla u\parallel }_{2n/(n-2)}+{\parallel u\parallel }_{2n/(n-4)}\le C{\parallel \Delta u\parallel }_2,$$

(7)

and, for every radial cutoff ${\chi }_R\left(x\right)=\chi (x/R)$ with $\chi =1$ on $B_1$ and $\chi =0$ outside $B_2$,

$${\chi }_{R}u⟶u\mathrm{in}{\dot{H}}^2\left({\mathbb{R}}^n\right)(R\to \infty ).$$

(8)

Proof. 

For $u\in C_c^{\infty }\left({\mathbb{R}}^n\right)$, Plancherel’s theorem gives

$${\displaystyle \sum _{i,j=1}^n}\parallel {\partial }_{ij}{u\parallel }_2^2={\parallel \Delta u\parallel }_2^2.$$

The homogeneous Sobolev inequalities of orders one and two then yield (7); passage to the completion proves the direct implication in (6).

Conversely, assume (6). Let $f=\Delta u\in L^2$ and define the Riesz potential

$$w:=-{(-\Delta )}^{-1}f,\widehat{w}\left(\xi \right)=-{\left|\xi \right|}^{-2}\widehat{f}\left(\xi \right).$$

The Hardy–Littlewood–Sobolev inequality gives $w\in L^{2n/(n-4)}$ and $\Delta w=f$. Hence $h:=u-w$ is a tempered harmonic distribution. Since $h\in L^{2n/(n-4)}$, Weyl’s lemma and the Liouville theorem imply $h=0$. Thus $u=w$, and standard mollification approximates u in the norm ${\parallel \Delta ·\parallel }_2$.

It remains to justify (8). On the annulus $A_R=\{R<|x|<2R\}$,

$$\Delta ({\chi }_{R}u-u)=({\chi }_R-1)\Delta u+2\nabla {\chi }_R·\nabla u+u\Delta {\chi }_R.$$

The first term tends to zero by absolute continuity of the $L^2$ norm. Using $|\nabla {\chi }_R|\le C{R}^{-1}$, $|\Delta {\chi }_R|\le C{R}^{-2}$, Hölder’s inequality, and (7),

$$\parallel \nabla {\chi }_R{·\nabla u\parallel }_{L^2\left(A_R\right)}\le C{\parallel \nabla u\parallel }_{L^{2n/(n-2)}\left(A_R\right)},$$

$$\parallel u\Delta {\chi }_R{\parallel }_{L^2\left(A_R\right)}\le C{\parallel u\parallel }_{L^{2n/(n-4)}\left(A_R\right)}.$$

Both right-hand sides vanish as $R\to \infty$. This proves the cutoff convergence and the characterization. See also [11]. □

For a compact set K, define

$${\mathcal{V}}_K:={\overline{\{\phi \in C_c^{\infty }\left({\mathbb{R}}^n\right):\phi =0\mathrm{in}\mathrm{a}\mathrm{neighbourhood}\mathrm{of}K\}}}^{{\dot{H}}^2}.$$

(9)

Choose $\eta \in C_c^{\infty }\left({\mathbb{R}}^n\right)$ with $\eta =1$ near K and set

$${\mathcal{A}}_K:=\eta +{\mathcal{V}}_K.$$

(10)

This affine space does not depend on the particular cutoff $\eta$. Moreover,

$${cap}_2\left(K\right)=\underset{u\in {\mathcal{A}}_K}{inf}{\parallel \Delta u\parallel }_2^2.$$

(11)

Indeed, every test function in the original definition belongs to ${\mathcal{A}}_K$, and conversely every element of ${\mathcal{A}}_K$ is the ${\dot{H}}^2$ limit of admissible smooth functions.

Since ${\dot{H}}^2$ is a Hilbert space, projection onto the closed affine space ${\mathcal{A}}_K$ gives a unique minimizer $U_K$. It satisfies

$${\int }_{{\mathbb{R}}^n}\Delta U_K\Delta \phi \mathrm{d}x=0(\phi \in {\mathcal{V}}_K)$$

(12)

and

$${cap}_2\left(K\right)={\int }_{{\mathbb{R}}^n}{|\Delta U_K|}^2\mathrm{d}x.$$

(13)

In particular, ${\Delta }^2{U}_K=0$ in ${\mathbb{R}}^n\setminus K$ and the distribution

$${\mu }_K:={\Delta }^2{U}_K$$

(14)

has compact support contained in K.

For the sets considered below, K is the closure of one ball or the union of two tangent closed balls. In this situation the condition $u\in {\mathcal{A}}_K$ is equivalent to $u=1$ quasi-everywhere on K. On the smooth portions of the boundary, a piecewise $H^2$ representative obtained by extending the constant value 1 through the ball has matching traces $u=1$ and ${\partial }_{\nu }u=0$. The single tangency point has zero ${\dot{H}}^2$ capacity when $n>4$, because a cutoff supported in a ball of radius $\epsilon$ has energy $O\left({\epsilon }^{n-4}\right)$.

2.2. The Fundamental Solution and the Far-Field Coefficient

Let

$${\Gamma }_n\left(x\right)={\gamma }_n{\left|x\right|}^{4-n},{\gamma }_n:={\displaystyle \frac{1}{2(n-2)(n-4){\omega }_{n-1}}},$$

(15)

where ${\omega }_{n-1}=\left|{\mathbb{S}}^{n-1}\right|$. With this normalization,

$${\Delta }^2{\Gamma }_n={\delta }_0$$

in distributions.

Lemma 2

(Total charge). Let $K\subset {\mathbb{R}}^n$ be compact and $n>4$. If $\eta \in C_c^{\infty }\left({\mathbb{R}}^n\right)$ is equal to 1 near K, then

$$\langle {\mu }_K,\eta \rangle ={cap}_2\left(K\right).$$

(16)

Consequently, the number $\langle {\mu }_K,1\rangle$, defined by the left-hand side for any such cutoff, is well defined and equals ${cap}_2\left(K\right)$.

Proof. 

Both $U_K$ and $\eta$ belong to ${\mathcal{A}}_K$, hence $U_K-\eta \in {\mathcal{V}}_K$. Taking $\phi =U_K-\eta$ in (12) gives

$$0=\int \Delta U_K\Delta (U_K-\eta )={cap}_2\left(K\right)-\int \Delta U_K\Delta \eta .$$

By the definition of ${\mu }_K$,

$$\int \Delta U_K\Delta \eta =\langle {\Delta }^2{U}_K,\eta \rangle =\langle {\mu }_K,\eta \rangle .$$

If two cutoffs are equal to 1 near K, their difference vanishes near $supp{\mu }_K$, so the value is independent of the cutoff. □

Lemma 3

(Potential representation). For every compact $K\subset {\mathbb{R}}^n$,

$$U_K={\Gamma }_n\ast {\mu }_K$$

(17)

in ${\mathcal{S}}^{\prime }\left({\mathbb{R}}^n\right)$ and almost everywhere.

Proof. 

Since ${\mu }_K={\Delta }^2{U}_K$, Fourier transformation gives ${\widehat{\mu }}_K={\left|\xi \right|}^4{\widehat{U}}_K$. Because the Fourier multiplier of ${\Gamma }_n$ is ${\left|\xi \right|}^{-4}$, the distribution

$$P:={\Gamma }_n\ast {\mu }_K-U_K$$

has Fourier transform supported at the origin; hence P is a polynomial. We show that this polynomial vanishes.

Choose R with $supp{\mu }_K\subset B_R$. If m is the order of the compactly supported distribution ${\mu }_K$, then for $\left|x\right|>2R$,

$$\begin{array}{cc}\hfill |({\Gamma }_n\ast {\mu }_K)\left(x\right)|& =\left|{\langle {\mu }_K\left(y\right),{\Gamma }_n(x-y)\rangle }_y\right|\hfill \\ \hfill & \le C\underset{\left|\alpha \right|\le m}{max}\underset{\left|y\right|\le R}{sup}\left|D_y^{\alpha }{\Gamma }_n(x-y)\right|\hfill \\ \hfill & \le {C\left|x\right|}^{4-n}.\hfill \end{array}$$

Thus ${\Gamma }_n\ast {\mu }_K$ tends to zero at infinity. If $P\ne 0$, there is an open cone on which $\left|P\left(x\right)\right|\ge {c\left|x\right|}^q$ for all sufficiently large $\left|x\right|$, where $q\ge 0$ is the degree of its leading homogeneous part. On that cone, $U_K={\Gamma }_n\ast {\mu }_K-P$ cannot belong to $L^{2n/(n-4)}$, contradicting (5). Therefore $P=0$. □

Lemma 4

(Far-field coefficient). Let $K\subset {\mathbb{R}}^n$ be compact and $n>4$. Then

$$U_K\left(x\right)={\gamma }_n{cap}_2{\left(K\right)\left|x\right|}^{4-n}+{O\left(\right|x|}^{3-n}\left)\right(\left|x\right|\to \infty ).$$

(18)

In particular, the remainder is $o\left(\right|x{|}^{4-n})$.

Proof. 

Let $R>0$ be such that $supp{\mu }_K\subset B_R$. By Lemma 3,

$$U_K\left(x\right)={\langle {\mu }_K\left(y\right),{\Gamma }_n(x-y)\rangle }_y.$$

For $\left|x\right|>2R$, Taylor expansion in y gives

$${\Gamma }_n(x-y)={\Gamma }_n\left(x\right)+\sum _{1\le \left|\alpha \right|\le M}{\displaystyle \frac{{(-y)}^{\alpha }}{\alpha !}}{D}^{\alpha }{\Gamma }_n\left(x\right)+{\mathcal{E}}_M(x,y).$$

A compactly supported distribution has finite order, so M can be chosen larger than that order. Since

$$D^{\alpha }{\Gamma }_n\left(x\right)={O\left(\right|x|}^{4-n-\left|\alpha \right|}),$$

all terms with $\left|\alpha \right|\ge 1$ are $O\left(\right|x{|}^{3-n})$, including the distributional remainder. The leading coefficient is

$${\Gamma }_n\left(x\right)\langle {\mu }_K,1\rangle ={\gamma }_n{\left|x\right|}^{4-n}{cap}_2\left(K\right)$$

by Lemma 2. □

3. Inversion and Green Kernels in the Transformed Domains

3.1. Geometry and Kelvin Transform

Let

$$I\left(y\right):={\displaystyle \frac{y}{{\left|y\right|}^2}}$$

be inversion about the unit sphere. The biharmonic Kelvin transform is

$$\left({\mathcal{K}}_{2}u\right)\left(y\right):={\left|y\right|}^{4-n}u\left(I\left(y\right)\right).$$

(19)

Away from the origin,

$${\Delta }_y^2\left({\mathcal{K}}_{2}u\right)\left(y\right)={\left|y\right|}^{-n-4}\left({\Delta }_x^{2}u\right)\left(I\left(y\right)\right);$$

(20)

see [2,5].

The two spherical boundaries in (3) become

$$y_1=b:={\displaystyle \frac{1}{2R_{+}}},y_1=-a:=-{\displaystyle \frac{1}{2R_{-}}}.$$

(21)

Therefore

$$I({\mathbb{R}}^n\setminus (K_{+}\cup K_{-}))=S_{a,b}:=\{(t,z)\in \mathbb{R}\times {\mathbb{R}}^{n-1}:-a<t<b\}.$$

(22)

The exterior of a single ball is mapped to a half-space with the pole at distance a or b from its boundary.

3.2. One-Dimensional Fibers and Fourier Definition

Write $y=(t,z)\in \mathbb{R}\times {\mathbb{R}}^{n-1}$ and use

$$\widehat{f}\left(\xi \right)={\int }_{{\mathbb{R}}^{n-1}}{e}^{-iz·\xi }f\left(z\right)\mathrm{d}z,f\left(z\right)={\displaystyle \frac{1}{{\left(2\pi \right)}^{n-1}}}{\int }_{{\mathbb{R}}^{n-1}}{e}^{iz·\xi }\widehat{f}\left(\xi \right)\mathrm{d}\xi .$$

(23)

For $k=\left|\xi \right|$, the transformed operator is

$$L_k:={\left({\displaystyle \frac{{\mathrm{d}}^2}{\mathrm{d}{t}^2}}-k^2\right)}^2.$$

(24)

The full-line Green function with pole at the origin is

$$g_{\infty }(t;k)={\displaystyle \frac{1+k\left|t\right|}{4k^3}}{e}^{-k\left|t\right|},k>0.$$

(25)

For $k>0$, let $g_{a,b}(t,s;k)$ be the unique Green function of $L_k$ on $(-a,b)$ with

$$g_{a,b}={\partial }_t{g}_{a,b}=0\mathrm{at}t=-a,b.$$

For the half-line $(-d,\infty )$, let $g_{H_d}(t,s;k)$ be the unique Green function satisfying the same clamped conditions at $t=-d$ and decaying, together with its first three derivatives, as $t\to +\infty$. These conditions remove the polynomially growing homogeneous solutions that would otherwise destroy uniqueness in an unbounded domain.

When the pole is $s=0$ and $r=t+d>0$, solving the two boundary equations for the decaying correction gives

$$\begin{array}{cc}\hfill g_{H_d}(r-d,0;k)=& {\displaystyle \frac{1+k|r-d|}{4k^3}}{e}^{-k|r-d|}\hfill \\ \hfill & -{\displaystyle \frac{e^{-k(r+d)}}{4k^3}}\left(1+kd+kr+2k^{2}dr\right),r>0.\hfill \end{array}$$

(26)

Definition 1.

The Fourier-admissible clamped Green distributions with pole at the origin are

$$\begin{array}{cc}\hfill G_{S_{a,b}}(t,z;0)& :={\displaystyle \frac{1}{{\left(2\pi \right)}^{n-1}}}{\int }_{{\mathbb{R}}^{n-1}}{e}^{iz·\xi }{g}_{a,b}(t,0;|\xi \left|\right)\mathrm{d}\xi ,\hfill \end{array}$$

(27)

$$\begin{array}{cc}\hfill G_{H_d}(t,z;0)& :={\displaystyle \frac{1}{{\left(2\pi \right)}^{n-1}}}{\int }_{{\mathbb{R}}^{n-1}}{e}^{iz·\xi }{g}_{H_d}(t,0;|\xi \left|\right)\mathrm{d}\xi .\hfill \end{array}$$

(28)

The integrals are initially interpreted in ${\mathcal{S}}^{\prime }$. Their classical meaning away from the pole follows from the estimates below.

3.3. Low- and High-Frequency Estimates

The next two lemmas supply the estimates needed to justify Definition 1. They are stated separately because the mechanisms at $k=0$ and $k=\infty$ are different.

Lemma 5

(Analytic low-frequency fibers). There exists ${\lambda }_0>0$ such that, for $\left|\lambda \right|<{\lambda }_0$, the clamped Green function $g_{\lambda }$ of ${(D_t^2-\lambda )}^2$ on $(-a,b)$ with pole at 0 depends analytically on λ, piecewise with values in $C^{\infty }$. For real $\lambda =k^2\ge 0$ it satisfies $g_{\lambda }\left(t\right)=g_{a,b}(t,0;k)$. More precisely, for every $j,m\ge 0$,

$$\underset{-a\le t\le b}{sup}\left|{\partial }_{\lambda }^m{\partial }_t^j{g}_{\lambda }\left(t\right)\right|\le C_{j,m}$$

(29)

for real $0\le \lambda <{\lambda }_0$, where at $t=0$ the one-sided derivatives are used when $j\ge 3$. Consequently,

$$\xi ⟼{\partial }_t^j{g}_{a,b}(t,0;|\xi \left|\right),0\le j\le 2,$$

(30)

is $C^{\infty }$ near $\xi =0$, uniformly for $t\in [-a,b]$.

Proof. 

Introduce the entire functions

$$C_{\lambda }\left(x\right):={\displaystyle \sum _{m=0}^{\infty }}{\displaystyle \frac{{\lambda }^m{x}^{2m}}{\left(2m\right)!}},S_{\lambda }\left(x\right):={\displaystyle \sum _{m=0}^{\infty }}{\displaystyle \frac{{\lambda }^m{x}^{2m+1}}{(2m+1)!}}.$$

They satisfy $(D^2-\lambda )C_{\lambda }=(D^2-\lambda )S_{\lambda }=0$. Furthermore,

$$C_{\lambda },S_{\lambda },{\partial }_{\lambda }{C}_{\lambda },{\partial }_{\lambda }{S}_{\lambda }$$

form a fundamental system for ${(D^2-\lambda )}^{2}u=0$. At $\lambda =0$ they reduce to $1,x,x^2/2,x^3/6$.

Use this system, translated to the left and right endpoints, on the two intervals $(-a,0)$ and $(0,b)$. The endpoint conditions, continuity of $g,g^{\prime },g^{\prime \prime }$ at 0, and the fixed jump

$$g^{\prime \prime \prime }(0+)-g^{\prime \prime \prime }(0-)=1$$

form an $8\times 8$ linear system whose entries are entire in $\lambda$. At $\lambda =0$ the corresponding clamped $D^4$ problem has a unique solution: if $D^{4}u=0$ with $u=u^{\prime }=0$ at both endpoints, then

$$0={\int }_{-a}^{b}u{D}^{4}u\mathrm{d}t={\int }_{-a}^b{\left|u^{\prime \prime }\right|}^2\mathrm{d}t,$$

so $u=0$. Hence the coefficient matrix is invertible at $\lambda =0$ and its inverse is analytic for $\left|\lambda \right|<{\lambda }_0$. The coefficients, and therefore every fixed t-derivative on each closed half-interval, satisfy (29). The matching conditions control the common derivatives through order two at the pole. Finally, composition with $\lambda ={\left|\xi \right|}^2$ proves (30). □

Lemma 6

(Quantified high-frequency boundary system). Put $\ell =a+b$, $c=min\{a,b\}$, and

$$r_{a,b}(t;k):=g_{\infty }(t;k)-g_{a,b}(t,0;k).$$

For every $m\ge 0$ and $0\le j\le 2$, there are constants $C_{j,m}>0$ and integers $M_{j,m}\ge 0$ such that

$$\underset{-a\le t\le b}{sup}\left|{\partial }_k^m{\partial }_t^j{r}_{a,b}(t;k)\right|\le C_{j,m}{(1+k)}^{M_{j,m}}{e}^{-ck},k\ge 1.$$

(31)

The same conclusion holds after any finite number of ξ-derivatives of $r_{a,b}(t;|\xi \left|\right)$ on $\left|\xi \right|\ge 1$.

Proof. 

The correction is homogeneous for $L_k$ and has at the endpoints the value and first derivative of $g_{\infty }$. Use the stable basis

$$u_1=e^{-k(t+a)},u_2=(t+a)e^{-k(t+a)},u_3=e^{-k(b-t)},u_4=(b-t)e^{-k(b-t)}.$$

If $r_{a,b}={\sum }_{j=1}^4{c}_j{u}_j$, the four endpoint traces give

$$B\left(k\right)c\left(k\right)=d\left(k\right),$$

(32)

where, with $\epsilon =e^{-k\ell }$,

$$B\left(k\right)=\left(\begin{array}{cccc}1& 0& \epsilon & \ell \epsilon \\ -k& 1& k\epsilon & (k\ell -1)\epsilon \\ \epsilon & \ell \epsilon & 1& 0\\ -k\epsilon & (1-k\ell )\epsilon & k& -1\end{array}\right)$$

(33)

and

$$d\left(k\right)=\left(\begin{array}{c}{g}_{\infty }(-a;k)\\ {\partial }_t{g}_{\infty }(-a;k)\\ g_{\infty }(b;k)\\ {\partial }_t{g}_{\infty }(b;k)\end{array}\right).$$

Write $B=B_0+E$, where

$$B_0\left(k\right)=\left(\begin{array}{cccc}1& 0& 0& 0\\ -k& 1& 0& 0\\ 0& 0& 1& 0\\ 0& 0& k& -1\end{array}\right),B_0{\left(k\right)}^{-1}=\left(\begin{array}{cccc}1& 0& 0& 0\\ k& 1& 0& 0\\ 0& 0& 1& 0\\ 0& 0& k& -1\end{array}\right).$$

For a fixed matrix norm,

$$\parallel B_0{\left(k\right)}^{-1}E\left(k\right)\parallel \le C_{\ell }{(1+k\ell )}^2{e}^{-k\ell }.$$

(34)

Choose $k_0\ge 1$ so large that the right-hand side is at most $1/2$ for all $k\ge k_0$. Then

$$B{\left(k\right)}^{-1}={(I+B_0^{-1}E)}^{-1}{B}_0^{-1},\parallel B{\left(k\right)}^{-1}\parallel \le C(1+k),k\ge k_0.$$

Differentiating either this identity or $B^{-1}B=I$ shows inductively that every k-derivative of $B^{-1}$ has at most polynomial growth. The boundary data satisfy, for every m,

$$|{\partial }_k^{m}d\left(k\right)|\le C_m{(1+k)}^{M_m}{e}^{-ck},k\ge 1,$$

(35)

because their two distances from the pole are a and b. Thus the coefficients and all their k-derivatives have the same form of bound. Derivatives of the four basis functions with respect to t and k add only polynomial factors on the bounded interval, proving (31) for $k\ge k_0$.

For $1\le k\le k_0$, invertibility follows from the coercive identity

$${\int }_{-a}^{b}u{L}_{k}u\mathrm{d}t={\int }_{-a}^b\left(|u^{\prime \prime }{|}^2+2k^2|u^{\prime }{|}^2+k^4{\left|u\right|}^2\right)\mathrm{d}t$$

for clamped u. Compactness of $[1,k_0]$ then gives the same estimate after increasing the constants. Finally, for a radial function $F\left(k\right)$ and $\left|\xi \right|\ge 1$,

$$|{\partial }_{\xi }^{\beta }F\left(\right|\xi \left|\right)|\le C_{\beta }{\displaystyle \sum _{r=1}^{\left|\beta \right|}}{\left|\xi \right|}^{r-\left|\beta \right|}|F^{\left(r\right)}\left(\right|\xi \left|\right)|$$

(with the evident interpretation for $\beta =0$). This transfers (31) to arbitrary finite $\xi$-derivatives. □

Proposition 1

(Decay of the inverse Fourier kernels). (i) Slab.For every multi-index α with $\left|\alpha \right|\le 2$ and every $N\ge 0$,

$$\underset{-a\le t\le b}{sup}|D_y^{\alpha }{G}_{S_{a,b}}(t,z;0)|\le C_{\alpha ,N}{(1+|z\left|\right)}^{-N},\left|z\right|\ge 1.$$

(36)

The estimate is uniform up to both boundary hyperplanes.

(ii) Half-space.Writing $H_d=\{(t,z):t>-d\}$, one has

$$\begin{array}{cc}\hfill G_{H_d}(t,z;0)=& {\Gamma }_n(t,z)-{\Gamma }_n(t+2d,z)\hfill \\ \hfill & -{\displaystyle \frac{d(t+d)}{(n-2){\omega }_{n-1}}}{\left({(t+2d)}^2+{\left|z\right|}^2\right)}^{(2-n)/2}.\hfill \end{array}$$

(37)

Consequently, for $\left|\alpha \right|\le 2$,

$$|D_y^{\alpha }{G}_{H_d}(y,0)|\le C_{\alpha }{(1+|y\left|\right)}^{2-n-\left|\alpha \right|}$$

(38)

for all sufficiently large $\left|y\right|$ in $H_d$.

Proof. 

Choose $\chi \in C_c^{\infty }\left({\mathbb{R}}^{n-1}\right)$ equal to one near the origin. By Lemma 5, $\chi \left(\xi \right){\partial }_t^j{g}_{a,b}(t,0;|\xi \left|\right)$ is a compactly supported smooth symbol, uniformly in t, for $j\le 2$. Its inverse transform is therefore a Schwartz function uniformly in t. By Lemma 6, the same is true of $(1-\chi ){\partial }_t^j{r}_{a,b}(t;|\xi \left|\right)$.

It remains to treat the full-line term. Direct differentiation of (25) gives, for $j\le 2$, every multi-index $\beta$, and $\left|\xi \right|\ge 1$,

$$\underset{t\in [-a,b]}{sup}|{\partial }_{\xi }^{\beta }{\partial }_t^j[(1-\chi \left(\xi \right))g_{\infty }(t;|\xi \left|\right)\left]\right|\le C_{j,\beta }{(1+|\xi \left|\right)}^{j-3-\left|\beta \right|}.$$

(39)

The same estimate, with the exponent increased by the number of tangential derivatives, holds after multiplication by the corresponding power of $\xi$. Regularize the inverse transform by $e^{-\epsilon \left|\xi \right|}$ and integrate repeatedly in the direction $z/\left|z\right|$. Once the number of integrations exceeds $n-1+j-3$ plus the number of tangential derivatives, the differentiated symbol is integrable by (39); no boundary term remains. Letting $\epsilon \downarrow 0$ gives an arbitrary power of ${\left|z\right|}^{-1}$, uniformly in t. This proves (36).

For the half-space, the tangential Fourier transforms of

$${\Gamma }_n(s,z),{\Phi }_n(s,z):={\displaystyle \frac{1}{(n-2){\omega }_{n-1}}}(s^2+{\left|z\right|}^2{)}^{(2-n)/2}$$

are respectively $(1+k|s\left|\right)e^{-k\left|s\right|}/\left(4k^3\right)$ and $e^{-k\left|s\right|}/\left(2k\right)$. Inverting (26) gives (37). Put $X=(t+2d,z)$. Taylor’s formula gives

$${\Gamma }_n(X-2d{e}_1)-{\Gamma }_n\left(X\right)=-2d{\partial }_1{\Gamma }_n\left(X\right)+{O\left(\right|X|}^{2-n}),$$

and

$$-2d{\partial }_1{\Gamma }_n\left(X\right)={\displaystyle \frac{d(t+2d)}{(n-2){\omega }_{n-1}}}{\left|X\right|}^{2-n}.$$

After subtracting the last term in (37), the remaining difference between $t+2d$ and $t+d$ is also $O\left(\right|X{|}^{2-n})$. Applying the differentiated Taylor formula through order two proves (38). □

3.4. The Green Equation and the Regular Part

Fiberwise uniqueness immediately gives uniqueness among tempered Green distributions whose partial Fourier transforms satisfy the clamped endpoint conditions and, in a half-space, the decay imposed above. No larger uniqueness class is used.

Proposition 2.

The kernels in Definition 1 are tempered distributions and satisfy

$${\Delta }^2{G}_D(·,0)={\delta }_0\mathit{in}D,G_D={\partial }_{\nu }{G}_D=0\mathit{on}\partial D.$$

(40)

Away from the pole they are classical. Moreover,

$$R_D\left(y\right):={\Gamma }_n\left(y\right)-G_D(y,0)$$

(41)

is biharmonic and smooth in all of D, including across $y=0$.

Proof. 

For the slab, Lemmas 5 and 6, together with (39), give a symbol of at most polynomial growth. For the half-space, the explicit formula (26) is bounded as $k\downarrow 0$ on compact t-intervals and has at most polynomial growth for $k\to \infty$. Hence both inverse transforms define tempered distributions.

Let $\phi \in C_c^{\infty }\left(D\right)$ and let $\widehat{\phi }(t,\xi )$ be its partial Fourier transform. Insert cutoffs ${\mathbf{1}}_{\{\epsilon <|\xi |<R\}}$. For every retained frequency, the one-dimensional Green identity and the jump relation yield

$$\int g_D(t,0;k){(D_t^2-k^2)}^2\widehat{\phi }(t,-\xi )\mathrm{d}t=\widehat{\phi }(0,-\xi ).$$

The low-frequency bounds in Lemma 5 and the explicit half-line formula, the polynomial high-frequency bounds, and the rapid decay of $\widehat{\phi }$ furnish an integrable majorant independent of $\epsilon$ and R. Dominated convergence therefore gives

$$\langle G_D,{\Delta }^2\phi \rangle =\phi \left(0\right).$$

The frequency $k=0$ has measure zero; for the slab it may equivalently be filled in with the clamped Green function of $D_t^4$.

The value and first normal trace vanish identically at the endpoints in every fiber. The same symbol bounds justify inverse transformation of these traces, so the boundary conditions in (40) hold in the trace-distribution sense. Standard interior elliptic regularity then makes $G_D$ classical away from the pole. Since ${\Delta }^2{\Gamma }_n={\delta }_0$, the distribution $R_D$ is biharmonic through the pole; hypoellipticity of ${\Delta }^2$ yields $R_D\in C^{\infty }\left(D\right)$. □

The Robin regular part is therefore

$${\mathcal{R}}_D\left(0\right):=R_D\left(0\right)=\underset{y\to 0}{lim}\left({\Gamma }_n\left(y\right)-G_D(y,0)\right).$$

(42)

3.5. Reverse Kelvin Construction and the Capacity–Robin Identity

We first record a chain-rule estimate that makes the behavior at the inversion point explicit.

Lemma 7

(Derivatives under inversion). Let $F\in C^2$ on a region containing $I\left(x\right)$ and set $T_F\left(x\right)={\left|x\right|}^{4-n}F\left(I\left(x\right)\right)$. For $0<\left|x\right|<1$ and $\left|\alpha \right|\le 2$,

$$|D_x^{\alpha }{T}_F\left(x\right)|\le C_{\alpha }\sum _{\left|\beta \right|\le \left|\alpha \right|}{\left|x\right|}^{4-n-\left|\alpha \right|-\left|\beta \right|}\left|D_y^{\beta }F\left(I\left(x\right)\right)\right|.$$

(43)

Proof. 

For every $m\ge 1$, $|D^{m}I\left(x\right)|\le C_m{\left|x\right|}^{-m-1}$, while $|D^p{\left|x\right|}^{4-n}|\le C_p{\left|x\right|}^{4-n-p}$. In a term obtained by differentiating $F\left(I\right(x\left)\right)$ a total of q times, suppose that $\left|\beta \right|$ derivatives fall on F and that the associated derivatives of I have orders $m_1,\dots ,m_{\left|\beta \right|}$, with $m_1+\dots +m_{\left|\beta \right|}=q$. Their product is bounded by ${C\left|x\right|}^{-q-\left|\beta \right|}\left|D^{\beta }F\left(I\left(x\right)\right)\right|$. Combining this with the derivatives of the prefactor and summing the finitely many terms proves (43). □

Lemma 8

(Approximation at the tangent boundary). Let K be a closed ball or the union of the tangent balls in (3). Suppose that $v\in {\dot{H}}^2\left({\mathbb{R}}^n\right)$ vanishes in the interior of K and has zero value and zero first normal trace on every smooth piece of $\partial K$. In the two-ball case, assume additionally that, for every N,

$$|D^{\alpha }v\left(x\right)|\le C_{\alpha ,N}{\left|x\right|}^N,\left|\alpha \right|\le 2,x\to 0,x\notin K.$$

(44)

Then $v\in {\mathcal{V}}_K$.

Proof. 

By Lemma 1, radial truncation first reduces the problem to compactly supported v. Such a function belongs to $H^2\left({\mathbb{R}}^n\right)$: on its bounded support, the critical $L^p$ bounds give $v,\nabla v\in L^2$, and the Fourier identity ${\sum }_{i,j}\parallel {\partial }_{ij}{v\parallel }_2^2={\parallel \Delta v\parallel }_2^2$ supplies the second derivatives.

In the two-ball case, choose ${\chi }_{\epsilon }$ equal to zero on $B_{\epsilon }$, equal to one outside $B_{2\epsilon }$, and satisfying $|D^j{\chi }_{\epsilon }|\le C{\epsilon }^{-j}$ for $j\le 2$. On $A_{\epsilon }=B_{2\epsilon }\setminus B_{\epsilon }$,

$$\begin{array}{cc}\hfill \parallel v\Delta {\chi }_{\epsilon }{\parallel }_2& \le C{\epsilon }^{N-2+n/2},\hfill \\ \hfill \parallel \nabla v·\nabla {\chi }_{\epsilon }{\parallel }_2& \le C{\epsilon }^{N-1+n/2},\hfill \\ \hfill \parallel ({\chi }_{\epsilon }-1){\Delta v\parallel }_2& \le C{\epsilon }^{N+n/2}.\hfill \end{array}$$

Choosing N sufficiently large and expanding $\Delta ({\chi }_{\epsilon }v-v)$ gives ${\chi }_{\epsilon }v\to v$ in ${\dot{H}}^2$. Thus it is enough to approximate a compactly supported function whose support stays a positive distance from the tangency point.

Cover the remaining intersection of the support with $\partial K$ by finitely many open sets $U_j$ such that $U_j\cap \partial K$ is one smooth spherical patch and no other boundary component meets $U_j$. Choose a smooth partition of unity ${\theta }_0,{\theta }_1,\dots ,{\theta }_J$ on the support, where ${\theta }_0$ is supported a positive distance from K and $supp{\theta }_j⋐U_j$ for $j\ge 1$. For each $j\ge 1$ choose a bounded $C^{\infty }$ domain ${\mathcal{O}}_j\subset U_j\cap ({\mathbb{R}}^n\setminus K)$ such that $\overline{{\mathcal{O}}_j}\subset U_j\cap ({\mathbb{R}}^n\setminus intK)$, its boundary agrees with $\partial K$ near $supp{\theta }_j\cap \partial K$, and its artificial boundary lies where ${\theta }_j=0$. The localized function $w_j=\left({\theta }_{j}v\right){|}_{{\mathcal{O}}_j}$ belongs to $H^2\left({\mathcal{O}}_j\right)$ and has zero Dirichlet and first normal trace on the whole of $\partial {\mathcal{O}}_j$: on the spherical part this is the hypothesis, and on the artificial part it follows from the support of ${\theta }_j$.

For a smooth bounded domain, the trace characterization

$$H_0^2\left({\mathcal{O}}_j\right)=\{w\in H^2\left({\mathcal{O}}_j\right):Trw=0,Tr{\partial }_{\nu }w=0\}$$

(45)

therefore gives $w_j\in H_0^2\left({\mathcal{O}}_j\right)$. Hence there are $w_{j,m}\in C_c^{\infty }\left({\mathcal{O}}_j\right)$ converging to $w_j$ in $H^2\left({\mathcal{O}}_j\right)$. Their zero extensions converge in $H^2\left({\mathbb{R}}^n\right)$ and are smooth functions supported a positive distance from K. The interior term ${\theta }_{0}v$ is approximated by ordinary mollification with a radius smaller than its distance from K. Summing these approximations proves $v\in {\mathcal{V}}_K$. The one-ball case is identical without the preliminary $\epsilon$-cutoff. The trace characterization (45) is the standard density theorem for $H_0^2$ on smooth domains; see [1,6]. □

Proposition 3

(Capacity–Robin correspondence). Let K be one of $K_{+}$, $K_{-}$, or $K_{+}\cup K_{-}$, and let D be the corresponding half-space or slab under inversion. Then

$${cap}_2\left(K\right)={\gamma }_n^{-2}{\mathcal{R}}_D\left(0\right).$$

(46)

Proof. 

Define

$$W_D\left(y\right):={\left|y\right|}^{4-n}-{\gamma }_n^{-1}{G}_D(y,0)={\gamma }_n^{-1}{R}_D\left(y\right).$$

(47)

By Proposition 2, $W_D$ is smooth and biharmonic in all of D, and

$$W_D\left(0\right)={\gamma }_n^{-1}{\mathcal{R}}_D\left(0\right).$$

(48)

Since the Green kernel is clamped, on every boundary hyperplane

$$W_D={\left|y\right|}^{4-n},{\partial }_{\nu }{W}_D={\partial }_{\nu }{\left|y\right|}^{4-n}.$$

(49)

Let $\Omega ={\mathbb{R}}^n\setminus K$. For $x\in \Omega$ set

$$U\left(x\right):={\left|x\right|}^{4-n}{W}_D\left(I\left(x\right)\right)=1-{\gamma }_n^{-1}{\left|x\right|}^{4-n}{G}_D(I\left(x\right),0),$$

(50)

and set $U=1$ in the interior of K. The Kelvin identity and (49) show that U is biharmonic in $\Omega$ and that its value and first normal derivative match those of the constant interior extension across every smooth spherical boundary point. Thus $U\in H_{\mathrm{loc}}^2$ there.

Consider $x\to 0$. In the two-ball case, $I\left(x\right)=(t,z)$ stays in the closed slab. Since $\left|I\left(x\right)\right|={\left|x\right|}^{-1}$ and t is bounded, for all sufficiently small $\left|x\right|$ one has $\left|z\right|\ge \frac{1}{2}{\left|x\right|}^{-1}$. Apply Lemma 7 with $F=G_{S_{a,b}}$. For every M and $\left|\alpha \right|\le 2$, choose the decay exponent in (36) larger than $M+n-4+2\left|\alpha \right|$. Then

$$|D_x^{\alpha }(U\left(x\right)-1)|\le C_{\alpha ,M}{\left|x\right|}^M,x\to 0,x\in \Omega .$$

(51)

For a single ball, (38) and (43) give term by term

$$|D_x^{\alpha }(U\left(x\right)-1)|\le C_{\alpha }{\left|x\right|}^{2-\left|\alpha \right|},\left|\alpha \right|\le 2.$$

(52)

Indeed, a term with $\left|\beta \right|$ derivatives of the Green function has exponent

$$4-n-\left|\alpha \right|-\left|\beta \right|+(n-2+|\beta \left|\right)=2-\left|\alpha \right|.$$

These bounds prove that the constant extension is $H^2$ near the inversion point. Hence $U\in H_{\mathrm{loc}}^2\left({\mathbb{R}}^n\right)$ in all cases.

As $\left|x\right|\to \infty$, $I\left(x\right)\to 0$. Taylor expansion of the smooth function $W_D$ at the origin and Lemma 7 yield, for $\left|\alpha \right|\le 2$,

$$D^{\alpha }U\left(x\right)=D^{\alpha }\left({\gamma }_n^{-1}{\mathcal{R}}_D\left(0\right){\left|x\right|}^{4-n}\right)+{O\left(\right|x|}^{3-n-\left|\alpha \right|}).$$

(53)

In particular,

$$U\left(x\right)={\gamma }_n^{-1}{\mathcal{R}}_D{\left(0\right)\left|x\right|}^{4-n}+{O\left(\right|x|}^{3-n}),$$

(54)

so $U\in L^{2n/(n-4)}$ and $\Delta U\in L^2$ at infinity. Together with the local $H^2$ result, Lemma 1 gives $U\in {\dot{H}}^2\left({\mathbb{R}}^n\right)$.

Choose $\eta \in C_c^{\infty }$ equal to one near K and, for the union, constant near the tangency point. Then $v=U-\eta$ vanishes in the interior of K, has the two zero traces on every smooth boundary piece, and in the union case satisfies (44) by (51). Lemma 8 gives $v\in {\mathcal{V}}_K$, hence $U\in {\mathcal{A}}_K$.

For every smooth test function vanishing near K, biharmonicity in $\Omega$ gives

$${\int }_{{\mathbb{R}}^n}\Delta U\Delta \phi \mathrm{d}x=0.$$

Passing to the closure in (9) extends this identity to all $\phi \in {\mathcal{V}}_K$. Thus U is the unique minimizer in ${\mathcal{A}}_K$, namely $U=U_K$. Comparing (54) with Lemma 4 gives

$${\gamma }_n{cap}_2\left(K\right)={\gamma }_n^{-1}{\mathcal{R}}_D\left(0\right),$$

which is (46). □

Remark 1.

No unrestricted uniqueness assertion is used for the clamped problem in a half-space. Such an assertion would be false: for a boundary $t=0$, the function $t^2$ satisfies ${\Delta }^2\left(t^2\right)=0$ and has zero value and normal derivative on the boundary. The Fourier-admissible decay condition in Definition 1 is essential.

4. Fourier Representation of the Robin Parts

Recall from (25) that $g_{\infty }(0;k)=1/\left(4k^3\right)$. Define the diagonal corrections

$$q_{a,b}\left(k\right):={\displaystyle \frac{1}{4k^3}}-g_{a,b}(0,0;k),q_H(k;d):={\displaystyle \frac{1}{4k^3}}-g_{H_d}(0,0;k).$$

(55)

Proposition 4

(Robin integrals). For $n>4$,

$$\begin{array}{cc}\hfill {\mathcal{R}}_{S_{a,b}}\left(0\right)& ={\displaystyle \frac{1}{{\left(2\pi \right)}^{n-1}}}{\int }_{{\mathbb{R}}^{n-1}}{q}_{a,b}\left(\right|\xi \left|\right)\mathrm{d}\xi ,\hfill \end{array}$$

(56)

$$\begin{array}{cc}\hfill {\mathcal{R}}_{H_d}\left(0\right)& ={\displaystyle \frac{1}{{\left(2\pi \right)}^{n-1}}}{\int }_{{\mathbb{R}}^{n-1}}{q}_H\left(\right|\xi |;d)\mathrm{d}\xi .\hfill \end{array}$$

(57)

Both integrals converge absolutely.

Proof. 

The partial Fourier transform of the fundamental solution is $g_{\infty }(t;k)$, so, first away from the pole and then distributionally,

$$R_D(t,z)={\displaystyle \frac{1}{{\left(2\pi \right)}^{n-1}}}{\int }_{{\mathbb{R}}^{n-1}}{e}^{iz·\xi }\left(g_{\infty }(t;k)-g_D(t,0;k)\right)\mathrm{d}\xi .$$

At $t=0$ the fiber difference is the correction in (55).

For the slab, analyticity of the clamped resolvent at $k=0$ gives

$$g_{a,b}(0,0;k)=O\left(1\right),q_{a,b}\left(k\right)={\displaystyle \frac{1}{4k^3}}+O\left(1\right).$$

For the half-space, either the physical formula (37) or the boundary correction in (26) gives the same bound. Thus

$$|q_{a,b}\left(k\right)|+|q_H(k;d)|\le C{k}^{-3},0<k\le 1.$$

(58)

At high frequency, the slab correction is precisely $r_{a,b}(0,0;k)$ from (31); the explicit half-space correction has the same exponential behavior. Therefore there are $c>0$ and an integer M such that

$$|q_{a,b}\left(k\right)|+|q_H{(k;d)|\le C(1+k)}^M{e}^{-ck},k\ge 1.$$

(59)

In polar coordinates on ${\mathbb{R}}^{n-1}$, the low-frequency majorant is $k^{n-2}{k}^{-3}=k^{n-5}$, which is integrable exactly for $n>4$; the high-frequency majorant is exponentially integrable. The inverse transform of the correction is therefore continuous at the origin, and dominated convergence gives (56) and (57). □

Scaling $x=kt$ gives

$$q_{a,b}\left(k\right)=k^{-3}Q(ka,kb),q_H(k;d)=k^{-3}h\left(kd\right).$$

(60)

5. The One-Dimensional Comparison

Theorem 2

(Modewise slab inequality). For every $k,a,b>0$,

$$q_{a,b}\left(k\right)<q_H(k;a)+q_H(k;b).$$

(61)

5.1. The Finite-Interval Green Function

Let $A,B>0$, put $L=A+B$, and consider

$${(D^2-1)}^{2}g={\delta }_0\mathrm{on}(-A,B),g=g^{\prime }=0\mathrm{at}-A,B.$$

(62)

For $x<0$, use the clamped basis

$$\begin{array}{cc}\hfill {\varphi }_1\left(x\right)& =sinh(x+A)-(x+A)cosh(x+A),\hfill \\ \hfill {\varphi }_2\left(x\right)& =(x+A)sinh(x+A),\hfill \end{array}$$

(63)

and for $x>0$ use

$$\begin{array}{cc}\hfill {\psi }_1\left(x\right)& =sinh(B-x)-(B-x)cosh(B-x),\hfill \\ \hfill {\psi }_2\left(x\right)& =(B-x)sinh(B-x).\hfill \end{array}$$

(64)

Write $g=c_1{\varphi }_1+c_2{\varphi }_2$ on the left and $g=d_1{\psi }_1+d_2{\psi }_2$ on the right. Continuity of $g,g^{\prime },g^{\prime \prime }$ and the jump condition

$$g^{\prime \prime \prime }(0+)-g^{\prime \prime \prime }(0-)=1$$

(65)

lead to

$$M(A,B)\left(\begin{array}{c}{c}_1\\ c_2\\ d_1\\ d_2\end{array}\right)=\left(\begin{array}{c}0\\ 0\\ 0\\ 1\end{array}\right),$$

(66)

where

$$M(A,B)={\left.\left(\begin{array}{cccc}{\varphi }_1& {\varphi }_2& -{\psi }_1& -{\psi }_2\\ {\varphi }_1^{\prime }& {\varphi }_2^{\prime }& -{\psi }_1^{\prime }& -{\psi }_2^{\prime }\\ {\varphi }_1^{\prime \prime }& {\varphi }_2^{\prime \prime }& -{\psi }_1^{\prime \prime }& -{\psi }_2^{\prime \prime }\\ -{\varphi }_1^{\prime \prime \prime }& -{\varphi }_2^{\prime \prime \prime }& {\psi }_1^{\prime \prime \prime }& {\psi }_2^{\prime \prime \prime }\end{array}\right)\right|}_{x=0}.$$

(67)

The explicit determinant calculation is recorded in Appendix A. It gives

$$\begin{array}{cc}\hfill detM(A,B)& =2H\left(L\right),\hfill \\ \hfill H\left(L\right)& :=cosh\left(2L\right)-1-2L^2=2({sinh}^{2}L-L^2)>0.\hfill \end{array}$$

(68)

Thus the system is uniquely solvable. Put

$$v\left(A\right):=({\varphi }_1\left(0\right),{\varphi }_2\left(0\right),0,0),e_4={(0,0,0,1)}^T,$$

and introduce the bordered matrix

$$\tilde{M}(A,B):=\left(\begin{array}{cc}M(A,B)& e_4\\ v\left(A\right)& 0\end{array}\right).$$

(69)

Since $g_{A,B}\left(0\right)=v\left(A\right)M{(A,B)}^{-1}{e}_4$, the Schur-complement identity gives

$$g_{A,B}\left(0\right)=-{\displaystyle \frac{det\tilde{M}(A,B)}{detM(A,B)}}.$$

The Laurent-polynomial certificate in Appendix A, using only the addition formulas for sinh and cosh, gives the two exact determinant identities

$$detM(A,B)=2H\left(L\right),det\tilde{M}(A,B)=-{\displaystyle \frac{1}{2}}N(A,B).$$

(70)

Consequently,

$$g_{A,B}\left(0\right)={\displaystyle \frac{N(A,B)}{4H\left(L\right)}},$$

(71)

where

$$\begin{array}{cc}\hfill N(A,B):=& 4ABL-2A^{2}sinh\left(2B\right)-2B^{2}sinh\left(2A\right)\hfill \\ \hfill & -2A\left(cosh\left(2B\right)-1\right)-2B\left(cosh\left(2A\right)-1\right)\hfill \\ \hfill & -sinh\left(2A\right)-sinh\left(2B\right)+sinh\left(2L\right).\hfill \end{array}$$

(72)

Accordingly,

$$Q(A,B)={\displaystyle \frac{1}{4}}-{\displaystyle \frac{N(A,B)}{4H(A+B)}}.$$

(73)

The bordered determinant in (70) is also verified by exact symbolic arithmetic in the optional source-file script.

5.2. The Half-Line Correction

Shift the half-line so that its boundary is $r=0$ and its pole is $r=A$. The decaying homogeneous correction has the form

$$c\left(r\right)=(\alpha +\beta r)e^{-r}.$$

Matching c and $c^{\prime }$ to the full-line kernel at $r=0$ yields

$$\alpha ={\displaystyle \frac{1}{4}}{e}^{-A}(1+A),\beta ={\displaystyle \frac{1}{4}}{e}^{-A}(1+2A).$$

Therefore, at the pole,

$$h\left(A\right)=c\left(A\right)={\displaystyle \frac{1}{4}}{e}^{-2A}P\left(A\right),P\left(A\right):=1+2A+2A^2.$$

(74)

Restoring the original scale gives precisely the complete fiber (26) stated earlier.

Thus (61) is equivalent to

$$Q(A,B)<h\left(A\right)+h\left(B\right),A,B>0.$$

(75)

5.3. Factorization of the Gap

Substitution of (73) and (74), followed by clearing the denominator $H\left(L\right)$, gives

$$h\left(A\right)+h\left(B\right)-Q(A,B)={\displaystyle \frac{e^{-2L}}{8H\left(L\right)}}E(A,B),$$

(76)

where

$$\begin{array}{cc}\hfill E(A,B):=& e^{2L}\left[2P\left(L\right)+8ABL\right]\hfill \\ \hfill & -e^{2B}\left[8A(A+1)L^2+P(L+A)\right]\hfill \\ \hfill & -e^{2A}\left[8B(B+1)L^2+P(L+B)\right]\hfill \\ \hfill & +e^{-2A}P\left(A\right)+e^{-2B}P\left(B\right)-2.\hfill \end{array}$$

(77)

Since $H\left(L\right)>0$, it remains to prove $E(A,B)>0$.

5.4. Exact Coefficient Extraction

Expand

$$E(A,B)=\sum _{i,j\ge 0}{c}_{ij}{A}^i{B}^j.$$

(78)

For $i,j\ge 1$, define

$$d_{ij}:={\displaystyle \frac{i!j!}{2^{i+j}}}{c}_{ij}.$$

(79)

We now derive, rather than merely state, the coefficient formula.

For any polynomial monomial $A^p{B}^q$,

$$\left[A^i{B}^j\right]\left(e^{2(A+B)}{A}^p{B}^q\right)={\displaystyle \frac{2^{i+j-p-q}}{(i-p)!(j-q)!}},$$

(80)

with the convention that the right-hand side is zero if $i<p$ or $j<q$. The first positive term in (77) gives

$${\displaystyle \frac{i!j!}{2^{i+j}}}\left[A^i{B}^j\right]\left(e^{2L}2P\left(L\right)\right)={(i+j)}^2+(i+j)+2,$$

while

$${\displaystyle \frac{i!j!}{2^{i+j}}}\left[A^i{B}^j\right]\left(8ABL{e}^{2L}\right)=ij(i+j-2).$$

Thus the total positive contribution is

$$D_0(i,j):={(i+j)}^2+(i+j)+2+ij(i+j-2).$$

(81)

For the first negative term, expand its polynomial factor in powers of A:

$$\begin{array}{cc}\hfill 8A(A+1){(A+B)}^2+P(2A+B)=& 8A^4+(16B+8)A^3\hfill \\ \hfill & +(8B^2+16B+8)A^2\hfill \\ \hfill & +(8B^2+8B+4)A+(2B^2+2B+1).\hfill \end{array}$$

(82)

Multiplication by $e^{2B}$ and use of (80) give the normalized contribution

$$\begin{array}{cc}\hfill T_1\left(j\right)& =j^2+j+2,\hfill \\ \hfill T_2\left(j\right)& =j^2+3j+4,\hfill \\ \hfill T_3\left(j\right)& =6j+6,\hfill \\ \hfill T_4\left(j\right)& =12,\hfill \\ \hfill T_i\left(j\right)& =0(i\ge 5).\hfill \end{array}$$

(83)

The second negative term is obtained by interchanging i and j. The remaining three terms in (77) contain only one variable and do not contribute when $i,j\ge 1$. Hence

$$d_{ij}=D_0(i,j)-T_i\left(j\right)-T_j\left(i\right).$$

(84)

If $i,j\ge 5$, then $d_{ij}=D_0(i,j)>0$. If $1\le i\le 4$ and $j\ge 5$, then

$$\begin{array}{cc}\hfill d_{1j}& =j^2+j+2,\hfill \\ \hfill d_{2j}& =2j^2+2j+4,\hfill \\ \hfill d_{3j}& =4j^2+4j+8,\hfill \\ \hfill d_{4j}& =5j^2+17j+10,\hfill \end{array}$$

(85)

all strictly positive. The symmetric range is identical. For $1\le i,j\le 4$,

$$\left(d_{ij}\right)=\left(\begin{array}{cccc}0& 0& 2& 10\\ 0& 2& 10& 32\\ 2& 10& 32& 76\\ 10& 32& 76& 146\end{array}\right),$$

(86)

which is entrywise nonnegative.

On the coordinate axes, direct substitution into (77) gives

$$E(A,0)=4P\left(A\right)({sinh}^{2}A-A^2),$$

(87)

and symmetrically for $E(0,B)$. Since

$${sinh}^{2}A-A^2={\displaystyle \sum _{r=2}^{\infty }}{\displaystyle \frac{2^{2r-1}}{\left(2r\right)!}}{A}^{2r},$$

(88)

all axis coefficients are nonnegative. Therefore every coefficient $c_{ij}$ in (78) is nonnegative. Since, for example, $d_{13}=2$, at least one mixed coefficient is positive, and hence

$$E(A,B)>0(A,B>0).$$

(89)

Combining (89) with (76) proves (75), and scaling proves 2.

6. Proof of the Capacity Inequality

Proof 

(Proof of 1). Let $a={\left(2R_{-}\right)}^{-1}$ and $b={\left(2R_{+}\right)}^{-1}$. By Proposition 3,

$$\begin{array}{cc}\hfill {cap}_2(K_{+}\cup K_{-})& ={\gamma }_n^{-2}{\mathcal{R}}_{S_{a,b}}\left(0\right),\hfill \\ \hfill {cap}_2\left(K_{-}\right)& ={\gamma }_n^{-2}{\mathcal{R}}_{H_a}\left(0\right),\hfill \\ \hfill {cap}_2\left(K_{+}\right)& ={\gamma }_n^{-2}{\mathcal{R}}_{H_b}\left(0\right).\hfill \end{array}$$

Using Proposition 4 and 2, the integrable difference

$$\xi ⟼q_H\left(\right|\xi |;a)+q_H\left(\right|\xi |;b)-q_{a,b}\left(\right|\xi \left|\right)$$

is continuous and strictly positive for every $\xi \ne 0$. It is therefore positive on every compact annulus and has a strictly positive integral. Hence

$$\begin{array}{cc}\hfill {\mathcal{R}}_{S_{a,b}}\left(0\right)& ={\displaystyle \frac{1}{{\left(2\pi \right)}^{n-1}}}{\int }_{{\mathbb{R}}^{n-1}}{q}_{a,b}\left(\right|\xi \left|\right)\mathrm{d}\xi \hfill \\ \hfill & <{\displaystyle \frac{1}{{\left(2\pi \right)}^{n-1}}}{\int }_{{\mathbb{R}}^{n-1}}\left(q_H\left(\right|\xi |;a)+q_H\left(\right|\xi |;b)\right)\mathrm{d}\xi \hfill \\ \hfill & ={\mathcal{R}}_{H_a}\left(0\right)+{\mathcal{R}}_{H_b}\left(0\right).\hfill \end{array}$$

Multiplication by ${\gamma }_n^{-2}>0$ gives (4). □

7. Normalization Check: The Capacity of a Ball

Corollary 1.

For $R>0$,

$${cap}_2\left(\overline{B_R}\right)={(n-2)}^2(n-4){\omega }_{n-1}{R}^{n-4}.$$

(90)

Consequently,

$${cap}_2(K_{+}\cup K_{-})<{(n-2)}^2(n-4){\omega }_{n-1}\left(R_{+}^{n-4}+R_{-}^{n-4}\right).$$

(91)

Proof. 

By scaling it is enough to take $R=1$. The exterior biharmonic function matching the constant interior value and its first derivative at $r=1$ is

$$U\left(r\right)={\displaystyle \frac{n-2}{2}}{r}^{4-n}-{\displaystyle \frac{n-4}{2}}{r}^{2-n},r\ge 1,$$

(92)

with $U=1$ for $r\le 1$. Indeed, $U\left(1\right)=1$ and $U^{\prime }\left(1\right)=0$. The constant interior extension is in $H_{\mathrm{loc}}^2$, belongs to ${\mathcal{A}}_{\overline{B_1}}$, and is biharmonic outside the ball. For every $\phi \in {\mathcal{V}}_{\overline{B_1}}$, approximation by smooth functions vanishing near the ball gives

$${\int }_{{\mathbb{R}}^n}\Delta U\Delta \phi \mathrm{d}x=0.$$

Thus U satisfies the Euler–Lagrange equation and, by strict convexity of the energy on the affine space, is the capacitary minimizer. Since $r^{2-n}$ is harmonic away from the origin,

$$\Delta U\left(r\right)=-(n-2)(n-4)r^{2-n}(r>1).$$

Therefore

$$\begin{array}{cc}\hfill {cap}_2\left(\overline{B_1}\right)& ={\omega }_{n-1}{(n-2)}^2{(n-4)}^2{\int }_1^{\infty }{r}^{4-2n}{r}^{n-1}\mathrm{d}r\hfill \\ \hfill & ={\omega }_{n-1}{(n-2)}^2{(n-4)}^2{\int }_1^{\infty }{r}^{3-n}\mathrm{d}r\hfill \\ \hfill & ={(n-2)}^2(n-4){\omega }_{n-1}.\hfill \end{array}$$

Scaling gives (90).

As an independent check of all Fourier constants, inserting

$$q_H(k;d)={\displaystyle \frac{e^{-2kd}}{4k^3}}\left(1+2kd+2k^2{d}^2\right)$$

into (57), evaluating the three Gamma integrals, and using $R={\left(2d\right)}^{-1}$ reproduces

$${\gamma }_n^{-2}{\mathcal{R}}_{H_d}\left(0\right)={(n-2)}^2(n-4){\omega }_{n-1}{R}^{n-4}.$$

□

8. Discussion and Limitations

The proof establishes a stronger statement than the integrated capacity inequality: the slab correction is strictly smaller than the sum of the two half-space corrections at every nonzero tangential Fourier frequency. The two radii remain independent through $A=ka$ and $B=kb$.

External tangency is essential to the present method. Inversion sends both spherical boundaries to parallel hyperplanes, so the transformed problem separates after a tangential Fourier transform. For separated balls, this one-dimensional reduction is lost. For $m\ge 3$, the transformed fiber operator has order $2m$, and neither the finite-interval formula nor the coefficient argument used here persists without substantial new work.

No general positivity-preserving principle for biharmonic Green functions is assumed. This is important because such principles fail in general domains; see [5,7,8]. Positivity enters only through the exact two-parameter entire function $E(A,B)$.

Accordingly, 1 resolves a concrete special configuration in Maz’ya’s upper-subadditivity problem, but not the general problem.