Proof of Theorem 1.1 The proof is long, and so we break it down into several steps. In Steps 1–3, we prove the result first for test functions f. Then, in Step 4, we deduce the result for general f.

Step 1: Maximum Principle. First suppose that $f\in C_c^{\infty }(A)$ is a test function. In particular, f is Lipschitz continuous on $\mathbb {R}^n$ . Then, $f^{\#}$ is also Lipschitz continuous on $\mathbb {R}^n$ ; see Section 7.5 of [Reference Baernstein10] or Lemma 6.7 in [Reference Sarvas35]. By standard regularity results, the solutions $u,v$ belong to $C^2(\overline {A})$ . See Theorem 6.31 in [Reference Gilbarg and Trudinger24] and the discussion following its proof on Fredholm alternatives or Theorem 3.2 of [Reference Ladyzenskaja and Ural’ceva27]. See also [Reference Nardi34] for an explicit discussion of the Neumann problem.

Put

$$ \begin{align*} w(r,\theta)=u^{\bigstar}(r,\theta)-Jv(r,\theta)-\varepsilon Q(r,\theta), \end{align*} $$

for $(r,\theta )\in A^{\bigstar }$ , where

$$ \begin{align*} Q(r,\theta)=(r-a)(r-b)+C\theta(\pi-\theta). \end{align*} $$

A direct computation shows

$$ \begin{align*} \Delta^{\bigstar}Q(r,\theta)=2+\frac{n-1}{r}(2r-a-b)-\frac{C}{r^{2}}\left(2+(n-2)(\pi-2\theta)\cot\theta\right). \end{align*} $$

Because $(\pi -2\theta )\cot \theta \geq 0$ on $(0,\pi )$ , we may choose $C>0$ sufficiently large to guarantee $\Delta ^{\bigstar }Q\leq 0$ on $A^{\bigstar }$ . We then compute in the weak sense

$$ \begin{align*} -\Delta^{\bigstar} w &= -\Delta^{\bigstar}u^{\bigstar} +\Delta^{\bigstar}Jv + \varepsilon \Delta^{\bigstar}Q\\ &\leq f^{\bigstar} + J\Delta v + \varepsilon \Delta^{\bigstar}Q\\ &=f^{\bigstar} - f^{\bigstar} + \varepsilon \Delta^{\bigstar}Q\\ &\leq 0, \end{align*} $$

where the first inequality follows from Theorems 2.4 and 2.5, the second equality follows from (2.5), and the second inequality follows from our choice of C. The Maximum Principle for weakly $\Delta ^{\bigstar }$ -subharmonic functions (see Theorem 1 of [Reference Littman33], for example) implies

(3.1) $$ \begin{align} \sup_{A^{\bigstar}}w \leq \max_{\partial A^{\bigstar}}w. \end{align} $$

If $\displaystyle \max _{\partial A^{\bigstar }}w \leq 0$ , then (3.1) implies

$$ \begin{align*} \sup_{A^{\bigstar}}(u^{\bigstar}-Jv)\leq \varepsilon \|Q\|_{L^{\infty}(A^{\bigstar})}, \end{align*} $$

and sending $\varepsilon \to 0^+$ gives $u^{\bigstar }\leq Jv$ in $A^{\bigstar }$ . In Steps 2 and 3 below, we therefore assume $\displaystyle \max _{\partial A^{\bigstar }}w> 0$ .

Step 2: Boundary analysis on vertical sides of $\boldsymbol \partial A^{\bigstar }$ . We first claim that $\displaystyle \max _{\partial A^{\bigstar }}w$ is not achieved at a point where $r=a$ or $r=b$ , i.e., on a vertical side of $\partial A^{\bigstar }$ . For $\theta \in [0,\pi ]$ and $h>0$ small, let $E(a+h,\theta ) \subseteq \mathbb {S}^{n-1}$ denote a measurable subset with $\sigma (E(a+h,\theta ))=\sigma (K(\theta ))$ for which the $\sup $ defining $u^{\bigstar }(a+h,\theta )$ is achieved. We estimate

$$ \begin{align*} & \frac{e^{-\alpha(a+h)}w(a+h,\theta)-e^{-\alpha a}w(a,\theta)}{h} \\ &\qquad\qquad =e^{-\alpha(a+h)}\int_{E(a+h,\theta)}\frac{u((a+h)\xi)}{h}\,d\sigma(\xi)- e^{-\alpha a}\int_{E(a,\theta)}\frac{u(a\xi)}{h}\,d\sigma(\xi)\\ &\qquad\qquad\quad -e^{-\alpha(a+h)}\int_{K(\theta)}\frac{v((a+h)\xi)}{h}\,d\sigma(\xi) +e^{-\alpha a}\int_{K(\theta)}\frac{v(a\xi)}{h}\,d\sigma(\xi) \\ &\qquad\qquad\quad -\varepsilon \frac{e^{-\alpha(a+h)}Q(a+h,\theta)-e^{-\alpha a}Q(a,\theta)}{h}\\ &\qquad\qquad \geq \int_{E(a,\theta)}\frac{e^{-\alpha(a+h)}u((a+h)\xi)-e^{-\alpha a}u(a\xi)}{h}\,d\sigma(\xi)\\ &\qquad\qquad\quad -\int_{K(\theta)}\frac{e^{-\alpha(a+h)}v((a+h)\xi)-e^{-\alpha a}v(a\xi)}{h}\,d\sigma(\xi)\\ &\qquad\qquad \quad -\varepsilon \frac{e^{-\alpha(a+h)}Q(a+h,\theta)-e^{-\alpha a}Q(a,\theta)}{h}.\\. \end{align*} $$

From the Dominated Convergence Theorem, we deduce

$$ \begin{align*} &\liminf_{h\to 0^+}\frac{e^{-\alpha(a+h)}w(a+h,\theta)-e^{-\alpha a}w(a,\theta)}{h} \\ &\qquad\qquad\qquad\geq -e^{-\alpha a}\int_{E(a,\theta)}(-u_r(a\xi)+\alpha u(a\xi))\,d\sigma(\xi)\\ &\qquad\qquad\qquad\quad +e^{-\alpha a}\int_{K(\theta)}(-v_r(a\xi)+\alpha v(a\xi))\,d\sigma(\xi)\\ &\qquad\qquad\qquad\quad +\varepsilon\left(e^{-\alpha a}(b-a)+\alpha e^{-\alpha a}C\theta(\pi-\theta)\right)\\ &\qquad\qquad\qquad=\varepsilon\left(e^{-\alpha a}(b-a)+\alpha e^{-\alpha a}C\theta(\pi-\theta)\right), \end{align*} $$

where the equality holds courtesy of the Robin boundary conditions. If $\displaystyle \max _{\partial A^{\bigstar }}w$ is achieved at $(a,\theta )$ , it follows that for all $h>0$ sufficiently small, we have

(3.2) $$ \begin{align} w(a+h,\theta)>e^{\alpha h}w(a,\theta)+\frac{e^{\alpha h}h\varepsilon(b-a)}{2}>w(a,\theta), \end{align} $$

where the last inequality uses $w(a,\theta )>0$ and $\alpha \geq 0$ . When $\theta \in (0,\pi )$ , (3.2) says that $\displaystyle \max _{\partial A^{\bigstar }}w$ is not achieved at a point with $r=a$ , because otherwise the Maximum Principle (3.1) would be violated. Likewise, when $\theta =0$ or $\theta =\pi $ , $\displaystyle \max _{\partial A^{\bigstar }}w$ is not achieved at a point with $r=a$ , because according to (3.2), w assumes strictly larger values at other points $(r,\theta )\in \partial A^{\bigstar }$ with $r\in (a,b)$ .

We next do a similar analysis of the behavior of w at points along the boundary $\partial A^{\bigstar }$ with $r=b$ . To this end, say $\theta \in [0,\pi ]$ and $h>0$ is small. We estimate

$$ \begin{align*} &\frac{e^{\beta (b-h)}w(b-h,\theta)-e^{\beta b}w(b,\theta)}{h}\\ &\qquad\qquad\qquad = e^{\beta (b-h)}\int_{E(b-h,\theta)}\frac{u((b-h)\xi)}{h}\,d\sigma(\xi) - e^{\beta b}\int_{E(b,\theta)}\frac{u(b\xi)}{h}\,d\sigma(\xi)\\ &\qquad\qquad\qquad \quad -e^{\beta (b-h)}\int_{K(\theta)}\frac{v((b-h)\xi)}{h}\,d\sigma(\xi)+e^{\beta b}\int_{K(\theta)}\frac{v(b\xi)}{h}\,d\sigma(\xi)\\ &\qquad\qquad\qquad \quad -\varepsilon \frac{e^{\beta (b-h)}Q(b-h,\theta)-e^{\beta b}Q(b,\theta)}{h}\\ &\qquad\qquad\qquad \geq \int_{E(b,\theta)}\frac{e^{\beta (b-h)}u((b-h)\xi)-e^{\beta b}u(b\xi)}{h}\,d\sigma(\xi)\\ &\qquad\qquad\qquad \quad -\int_{K(\theta)}\frac{e^{\beta (b-h)}v((b-h)\xi)-e^{\beta b}v(b\xi)}{h}\,d\sigma(\xi)\\ &\qquad\qquad\qquad \quad -\varepsilon \frac{e^{\beta (b-h)}Q(b-h,\theta)-e^{\beta b}Q(b,\theta)}{h}. \end{align*} $$

As before, we use the Robin boundary conditions to deduce

$$ \begin{align*} \liminf_{h\to 0^+}\frac{e^{\beta (b-h)}w(b-h,\theta)-e^{\beta b}w(b,\theta)}{h} &\geq -e^{\beta b}\int_{E(b,\theta)}(u_r(b\xi)+\beta u(b\xi))\,d\sigma(\xi)\\ &\quad +e^{\beta b}\int_{K(\theta)}(v_r(b\xi)+\beta v(b\xi))\,d\sigma(\xi)\\ &\quad +\varepsilon\left(e^{\beta b}(b-a)+\beta e^{\beta b}C\theta(\pi-\theta)\right)\\ &=\varepsilon\left(e^{\beta b}(b-a)+\beta e^{\beta b}C\theta(\pi-\theta)\right). \end{align*} $$

If $\displaystyle \max _{\partial A^{\bigstar }}w$ is achieved at $(b,\theta )$ , then for all $h>0$ sufficiently small, we have

$$ \begin{align*} w(b-h,\theta)>e^{\beta h}w(b,\theta)+\frac{e^{\beta h}h\varepsilon(b-a)}{2}>w(b,\theta), \end{align*} $$

again because $w(b,\theta )>0$ and $\beta \geq 0$ . By the same argument as before, the string of strict inequalities immediately above shows that $\displaystyle \max _{\partial A^{\bigstar }}w$ cannot be achieved at a point with $r=b$ .

Step 3: Boundary analysis on horizontal sides of $\boldsymbol \partial A^{\bigstar }$ . We start by analyzing the behavior of $u^{\bigstar }(r,\pi )-Jv(r,\pi )$ in r. Define

$$ \begin{align*} \psi(r)=u^{\bigstar}(r,\pi)-Jv(r,\pi)=\int_{\mathbb{S}^{n-1}}(u(r\xi)-v(r\xi))\,d\sigma(\xi), \quad r\in[a,b]. \end{align*} $$

For $a\leq r_1<r_2\leq b$ , note that

$$ \begin{align*} r_2^{n-1}\psi'(r_2)-r_1^{n-1}\psi'(r_1)&= r_2^{n-1}\int_{\mathbb{S}^{n-1}}\left(u_r(r_2\xi)-v_r(r_2\xi)\right)\,d\sigma(\xi)\\ &\quad -r_1^{n-1}\int_{\mathbb{S}^{n-1}}\left(u_r(r_1\xi)-v_r(r_1\xi)\right)\,d\sigma(\xi)\\ &=\int_{\partial A(r_1,r_2)}\left(\frac{\partial u}{\partial \eta}-\frac{\partial v}{\partial \eta}\right)\,dS\\ &=\int_{A(r_1,r_2)}\left(\Delta u - \Delta v \right)\,dx\\ &=\int_{A(r_1,r_2)}\left(-f+f^{\#}\right)\,dx\\ &=0, \end{align*} $$

because f and $f^{\#}$ are rearrangements of each other on each sphere $\{|x|=r\}$ . It follows that

(3.3) $$ \begin{align} r^{n-1}\int_{\mathbb{S}^{n-1}}(u_r(r\xi)-v_r(r\xi))\,d\sigma(\xi)=k_1 \end{align} $$

is constant on $[a,b]$ . Integrating this equation shows that there exists a constant $k_2$ where

(3.4) $$ \begin{align} \int_{\mathbb{S}^{n-1}}(u(r\xi)-v(r\xi))\,d\sigma(\xi)=k_1\int_a^rt^{1-n}\,dt+k_2, \end{align} $$

on the interval $[a,b]$ . Using the Robin boundary conditions along $\{|x|=a\}$ and $\{|x|=b\}$ leads to the system of equations

(3.5) $$ \begin{align} 0&=\int_{\{|x|=a\}}\left(\left(\frac{\partial u}{\partial \eta}+\alpha u\right)-\left(\frac{\partial v}{\partial \eta}+\alpha v\right)\right)\,dS=-k_1+\alpha a^{n-1}k_2, \\ 0&=\int_{\{|x|=b\}}\left(\left(\frac{\partial u}{\partial \eta}+\beta u\right)-\left(\frac{\partial v}{\partial \eta}+\beta v\right)\right)\,dS=k_1+\beta b^{n-1}\left(k_1\int_a^bt^{1-n}\,dt+k_2\right).\nonumber \end{align} $$

Combining these equations shows that

$$ \begin{align*} 0=k_2\left(\alpha a^{n-1}+\beta b^{n-1}+\alpha a^{n-1}\beta b^{n-1}\int_a^bt^{1-n}\,dt\right). \end{align*} $$

If at least one of $\alpha ,\beta $ is nonzero, the equation above implies that $k_2=0$ and (3.5) implies that $k_1=0$ . Thus, (3.4) becomes

(3.6) $$ \begin{align} \int_{\mathbb{S}^{n-1}}\left(u(r\xi)-v(r\xi)\right)\,d\sigma(\xi)=0,\quad r\in [a,b]. \end{align} $$

If $\alpha =\beta =0$ , choosing either $r=a$ or $r=b$ in (3.3) shows that $k_1=0$ . Integrating (3.4) and using the normalization assumptions on $u,v$ gives

$$ \begin{align*} 0=\int_{A}(u-v)\,dx=k_2\int_a^br^{n-1}\,dr, \end{align*} $$

so $k_2=0$ and equation (3.6) holds in this case as well. Because $u^{\bigstar }-Jv=0$ when $\theta =0$ , we therefore deduce

$$ \begin{align*} w(r,0)=w(r,\pi)=-\varepsilon(r-a)(r-b). \end{align*} $$

Using (3.1), our work in Step 2 implies

$$ \begin{align*} \sup_{A^{\bigstar}}(u^{\bigstar}-Jv) &\leq \max_{r\in [a,b]}(-\varepsilon(r-a)(r-b))+ \varepsilon\|Q\|_{L^{\infty}(A^{\bigstar})} \\ &=\frac{\varepsilon}{4}(b-a)^2+\varepsilon \|Q\|_{L^{\infty}(A^{\bigstar})}. \end{align*} $$

Sending $\varepsilon \to 0^+$ gives $u^{\bigstar }\leq Jv$ in $A^{\bigstar }$ .

Step 4: Approximation for the general result. Now, say $f\in L^2(A)$ is a general function. Let $u,v$ be as in the statement of Theorem 1.1. Choose a sequence of test functions $f_k\in C^{\infty }_c(A)$ where $f_k\to f$ in $L^2(A)$ . If $\alpha =\beta =0$ , we assume $f,f_k$ have zero mean. Because cap symmetrization is a contraction in the $L^2$ -distance (see Theorem 3 or Corollary 1 of [Reference Baernstein8], for example), we also have $f_k^{\#} \to f^{\#}$ in $L^2(A)$ . Let $u_k$ and $v_k$ be solutions to

$$ \begin{align*} \begin{array}{r@{\;}c@{\;}lcc@{{\kern-4pt}}ccr@{\;}c@{\;}lcc} -\Delta u_k & = & f_k & \text{in} & A, & & & -\Delta v_k & = & f_k^{\#} & \text{in} & A,\\ \frac{\partial u_k}{\partial \eta} +\alpha u_k & = & 0 & \text{on} & \{|x|=a\}, & & & \frac{\partial v_k}{\partial \eta} + \alpha v_k & = & 0 & \text{on} & \{|x|=a\},\\ \frac{\partial u_k}{\partial \eta} +\beta u_k & = & 0 & \text{on} & \{|x|=b\}, & & & \frac{\partial v_k}{\partial \eta} + \beta v_k & = & 0 & \text{on} & \{|x|=b\}. \end{array} \end{align*} $$

If $\alpha =\beta =0$ , we also assume $u_k,v_k$ have zero mean. By our work in Steps 1–3, for each k,

(3.7) $$ \begin{align} u_k^{\bigstar}(r,\theta)\leq Jv_k(r,\theta), \end{align} $$

for $(r,\theta )\in A^{\bigstar }$ , and

(3.8) $$ \begin{align} \int_{\mathbb{S}^{n-1}}(u_k(r\xi)-v_k(r\xi))\,d\sigma(\xi)=0, \end{align} $$

for $r\in (a,b)$ . If either $\alpha $ or $\beta $ is nonzero, then (2.3) implies $u_k\to u$ in $L^2(A)$ . If $\alpha =\beta =0$ , we still have convergence in $L^2(A)$ ; see Corollary 1.29 of [Reference Langford28], for instance. In particular, we have convergence in $L^1(A)$ . Spelled out, we have

$$ \begin{align*} \lim_{k\to +\infty}\int_{a}^{b}\int_{\mathbb{S}^{n-1}}\big|u_{k}(r\xi)-u(r\xi)\big|\,d\sigma(\xi)\ r^{n-1}\,dr=0. \end{align*} $$

Thus, by passing to a subsequence, we may assume

(3.9) $$ \begin{align} \lim_{k\to +\infty}\int_{\mathbb{S}^{n-1}}\big|u_{k}(r\xi)-u(r\xi)\big|\,d\sigma(\xi)=0, \end{align} $$

for almost every $r\in (a,b)$ . Because cap symmetrization is a contraction in the $L^1$ -distance (again, see Theorem 3 of [Reference Baernstein8]), this last equality implies that for almost every $r\in (a,b)$ ,

(3.10) $$ \begin{align} \lim_{k\to +\infty}u_k^{\bigstar}(r,\theta)=u^{\bigstar}(r,\theta), \end{align} $$

for each $\theta \in (0,\pi )$ . By an identical argument, we may pass to another subsequence and assume that for almost every $r\in (a,b)$ ,

(3.11) $$ \begin{align} \lim_{k\to +\infty}\int_{\mathbb{S}^{n-1}}\big|v_{k}(r\xi)-v(r\xi)\big|\,d\sigma(\xi)=0, \end{align} $$

and

(3.12) $$ \begin{align} \lim_{k\to +\infty}Jv_k(r,\theta)=Jv(r,\theta), \end{align} $$

for each $\theta \in (0,\pi )$ . Sending $k\to +\infty $ in (3.8) and using (3.9) and (3.11) shows that

(3.13) $$ \begin{align} \int_{\mathbb{S}^{n-1}}\left(u(r\xi)-v(r\xi)\right)\,d\sigma(\xi)=0, \end{align} $$

for almost every $r\in (a,b)$ . Finally, sending $k\to +\infty $ in (3.7) and using (3.10) and (3.12) gives the theorem’s first conclusion. The theorem’s conclusion about convex means follows from (3.13) and Proposition 2.2.▪

Proof First suppose that $f\in C_c^{\infty }(A)$ is a test function. Then, as in the proof of Theorem 1.1, $f^{\#}$ is Lipschitz continuous and $v\in C^2(\overline {A})$ . Letting $f^{\#}$ play the role of f, Theorem 1.1 gives that $v^{\bigstar }(r,\theta )\leq Jv(r,\theta )$ on $A^{\bigstar }$ . The reverse inequality holds by definition of the star function, and so $v^{\bigstar }=Jv$ on $A^{\bigstar }$ . When the dimension $n=2$ , we use complex notation to write this equality as

$$ \begin{align*} \int_{-\theta}^{\theta}v(re^{i\phi})\,d\phi=\int_{-\theta}^{\theta}v^{\#}(re^{i\phi})\,d\phi, \end{align*} $$

for each $r\in (a,b)$ and $\theta \in (0,\pi )$ . Differentiating this equation with respect to $\theta $ gives

(3.14) $$ \begin{align} v(re^{i\theta})+v(re^{-i\theta})=2v^{\#}(re^{i\theta}). \end{align} $$

We claim that $v(re^{i\theta })=v(re^{-i\theta })$ for $\theta \in (0,\pi )$ . If this were not the case, then say $v(re^{i\theta })>v(re^{-i\theta })$ for some $\theta $ . We can then find $\varepsilon>0$ sufficiently small, so that

$$ \begin{align*} \int_{-\theta}^{\theta}v(re^{i\phi})\,d\phi<\int_{-\theta+\varepsilon}^{\theta+\varepsilon}v(re^{i\phi})\,d\phi, \end{align*} $$

and this contradicts the fact that $v^{\bigstar }(r,\theta )=\int _{-\theta }^{\theta }v(re^{i\phi })\,d\phi $ . Having verified the claim, (3.14) gives $v(re^{i\theta })=v^{\#}(re^{i\theta })$ .

If $n\geq 3$ , we write out the equation $v^{\bigstar }=Jv$ on $A^{\bigstar }$ using spherical coordinates. We have

$$ \begin{align*} &\int_0^\theta \int_{\mathbb{S}^{n-2}}v(r\cos t, ry\sin t)\,d\sigma(y)\sin^{n-2} t\,dt\\[-2pt] &\qquad \qquad \qquad \qquad =\int_0^\theta \int_{\mathbb{S}^{n-2}}v^{\#}(r\cos t, ry\sin t)\,d\sigma(y)\sin^{n-2} t\,dt, \end{align*} $$

for each $r\in (a,b)$ and $\theta \in (0,\pi )$ . Differentiating with respect to $\theta $ and dividing by $\sin ^{n-2}\theta $ gives

(3.15) $$ \begin{align} \int_{\mathbb{S}^{n-2}}v(r\cos \theta, ry\sin \theta)\,d\sigma(y)=\int_{\mathbb{S}^{n-2}}v^{\#}(r\cos \theta, ry\sin \theta)\,d\sigma(y), \end{align} $$

for each $r\in (a,b)$ and $\theta \in (0,\pi )$ . Note that $v^{\#}(r\cos \theta , ry\sin \theta )$ is constant in $y\in \mathbb {S}^{n-2}$ . We claim that $v(r\cos \theta , ry\sin \theta )$ is also constant for $y\in \mathbb {S}^{n-2}$ . Say this is not the case, and $v(r\cos \theta , ry_1\sin \theta )>v(r\cos \theta , ry_2\sin \theta )$ for $y_1,y_2\in \mathbb {S}^{n-2}$ . Then, we can find positive numbers $\varepsilon _1,\varepsilon _2>0$ where

(3.16) $$ \begin{align} \inf_{z_1\in B_1} v(rz_1)>\sup_{z_2\in B_2}v(rz_2); \end{align} $$

here $B_1= B((\cos \theta , y_1\sin \theta ),\varepsilon _1)$ and $B_2= B((\cos \theta , y_2\sin \theta ),\varepsilon _2)$ , where $B((\cos \theta , y_1\sin \theta ),\varepsilon _1)$ denotes the spherical cap (or geodesic ball) on $\mathbb {S}^{n-1}$ centered at $(\cos \theta , y_1\sin \theta )$ with radius $\varepsilon _1$ , and similarly for $B_2$ . We additionally assume

$$ \begin{align*} \sigma(B_1 \setminus K(\theta))=\sigma(B_2 \cap K(\theta)). \end{align*} $$

Write

$$ \begin{align*} K_1=(K(\theta)\setminus B_2))\cup (B_1\setminus K(\theta)). \end{align*} $$

Then, $\sigma (K_1)=\sigma (K(\theta ))$ , and using (3.16), we have

$$ \begin{align*} \int_{K_1}v(r\xi)\,d\sigma(\xi)&=\int_{K(\theta)\setminus B_2}v(r\xi)\,d\sigma(\xi)+\int_{B_1\setminus K(\theta)}v(r\xi)\,d\sigma(\xi)\\[-2pt] &>\int_{K(\theta)\setminus B_2}v(r\xi)\,d\sigma(\xi)+\int_{B_2\cap K(\theta)}v(r\xi)\,d\sigma(\xi)\\[-2pt] &=\int_{K(\theta)}v(r\xi)\,d\sigma(\xi), \end{align*} $$

which contradicts $v^{\bigstar }=Jv$ . We conclude that $v(r\cos \theta , ry\sin \theta )$ is constant for $y\in \mathbb {S}^{n-2}$ . Equation (3.15) implies

$$ \begin{align*} v(r\cos \theta, ry\sin \theta)=v^{\#}(r\cos \theta, ry\sin \theta), \end{align*} $$

for each $y\in \mathbb {S}^{n-2}$ , $r\in (a,b)$ , and $\theta \in (0,\pi )$ . That is, $v=v^{\#}$ .

If $f\in L^2(A)$ is a general function, pick a sequence of test functions $f_k\in C_c^{\infty }(A)$ where $f_k\to f$ in $L^2(A)$ . If $\alpha =\beta =0$ , we assume that each $f_k$ has zero mean. As cap symmetrization is a contraction in the $L^2$ -distance, $f_k^{\#} \to f^{\#}$ in $L^2(A)$ . As before, each $f_k^{\#}$ is Lipschitz continuous on $\overline {A}$ . Let $v_k$ solve

$$ \begin{align*} \begin{array}{r@{\;}c@{\;}lcc} -\Delta v_k & = & f_k^{\#} & \text{in} & A,\\ \frac{\partial v_k}{\partial \eta} + \alpha v_k & = & 0 & \text{on} & \{|x|=a\},\\ \frac{\partial v_k}{\partial \eta} + \beta v_k & = & 0 & \text{on} & \{|x|=b\}. \end{array} \end{align*} $$

If $\alpha =\beta =0$ , we assume that each $v_k$ has zero mean. As in the proof of Theorem 1.1, $v_k\to v$ in $L^2(A)$ . By the work above, $v_k=v_k^{\#}$ for each k. Again, cap symmetrization is a contraction in the $L^2$ -distance, and so $v_k^{\#}\to v^{\#}$ in $L^2(A)$ . We deduce that $v=v^{\#}$ a.e., as desired.▪

Proof The conclusions about slice functions follow from Theorem 1.1, equation (3.13), and Corollary 2.3; the remaining conclusions clearly follow.▪

Proof of Theorem 3.4 Let $\alpha _k$ denote a strictly increasing positive sequence with $\alpha _k\to +\infty $ . Let $u_k$ and $v_k$ denote solutions to the Poisson problems

$$ \begin{align*} \begin{array}{r@{\;}c@{\;}lcc@{{\kern-4pt}}ccr@{\;}c@{\;}lcc} -\Delta u_k & = & f & \text{in} & A, & & & -\Delta v_k & = & f^{\#} & \text{in} & A,\\ \frac{\partial u_k}{\partial \eta} +\alpha_k u_k & = & 0 & \text{on} & \{|x|=a\}, & & & \frac{\partial v_k}{\partial \eta} + \alpha_k v_k & = & 0 & \text{on} & \{|x|=a\},\\ \frac{\partial u_k}{\partial \eta} +\beta u_k & = & 0 & \text{on} & \{|x|=b\}, & & & \frac{\partial v_k}{\partial \eta} + \beta v_k & = & 0 & \text{on} & \{|x|=b\}. \end{array} \end{align*} $$

By Theorem 1.1 (and its proof), for almost every $r\in (a,b),$

(3.17) $$ \begin{align} \int_{\mathbb{S}^{n-1}}(u_k(r\xi)-v_k(r\xi))\,d\sigma(\xi)=0, \end{align} $$

for each k, and

(3.18) $$ \begin{align} u_k^{\bigstar}(r,\theta)\leq Jv_k(r,\theta), \end{align} $$

for each $\theta \in (0,\pi )$ . We claim that the functions $u_k$ are bounded in $H^1(A)$ . To back our claim, first note that

$$ \begin{align*} &\int_{A}|\nabla u_k|^2\,dx+\alpha_k\int_{\{|x|=a\}}u_k^2\,dS+\beta\int_{\{|x|=b\}}u_k^2\,dS\\ & \qquad \qquad \qquad \qquad \geq \int_{A}|\nabla u_k|^2\,dx+\alpha_1\int_{\{|x|=a\}}u_k^2\,dS+\beta\int_{\{|x|=b\}}u_k^2\,dS\\ & \qquad \qquad \qquad \qquad \geq c_2\|u_k\|_{H^1(A)}^2, \end{align*} $$

where $c_2$ denotes coercivity constant (see Section 2.1) of the bilinear form associated with the Robin problem with Robin parameter $\alpha _1$ on the inner sphere $\{|x|=a\}$ and Robin parameter $\beta $ on the outer sphere $\{|x|=b\}$ . Because

(3.19) $$ \begin{align} \int_{A}|\nabla u_k|^2\,dx+\alpha_k\int_{\{|x|=a\}}u_k^2\,dS+\beta\int_{\{|x|=b\}}u_k^2\,dS=\int_{A}fu_k\,dx, \end{align} $$

it follows that

$$ \begin{align*} \|u_k\|_{H^1(A)}\leq \frac{1}{c_2}\|f\|_{L^2(A)}, \end{align*} $$

for every k. The $u_k$ are thus bounded in $H^1(A)$ . By Banach–Alaoglu and Rellich–Kondrachov, we may pass to a subsequence and assume the existence of $u'\in H^1(A)$ where $u_k\to u'$ weakly in $H^1(A)$ and $u_k\to u'$ in $L^2(A)$ . By page 134 of [Reference Evans and Gariepy20],

$$ \begin{align*} \int_{\partial A}w^2\,dS\leq C\left(\int_{A}|\nabla w||w|\,dx+\int_{A}w^2\,dx\right), \end{align*} $$

for $w\in H^1(A)$ and for some universal constant C. Thus, $u_k\to u'$ in $L^2(\partial A)$ . Dividing (3.19) by $\alpha _k$ and sending $k\to +\infty $ shows $\int _{\{|x|=a\}}(u')^2\,dS=0$ , i.e., $u'\in H^1_a(A)$ .

If $v\in H_a^1(A)$ , then for each k,

$$ \begin{align*} \int_{A}\nabla u_k\cdot \nabla v\,dx+\beta\int_{\{|x|=b\}}u_kv\,dS=\int_{A}fv\,dx. \end{align*} $$

Thus, letting $k\to +\infty $ in the above equation and using weak convergence, we conclude

$$ \begin{align*} \int_{A}\nabla u'\cdot \nabla v\,dx+\beta\int_{\{|x|=b\}}u'v\,dS=\int_{A}fv\,dx. \end{align*} $$

By uniqueness, $u'=u$ , and so $u_k\to u$ in $L^2(A)$ . The argument in Step 4 in the proof of Theorem 1.1 shows that for almost every $r\in (a,b)$ ,

(3.20) $$ \begin{align} \int_{\mathbb{S}^{n-1}}u_k(r\xi)\,d\sigma(\xi) \to \int_{\mathbb{S}^{n-1}}u(r\xi)\,d\sigma(\xi) \end{align} $$

and

(3.21) $$ \begin{align} u^{\bigstar}_k(r,\theta)\to u^{\bigstar}(r,\theta), \end{align} $$

for each $\theta \in (0,\pi )$ . An analogous argument shows that we may assume $v_k\to v$ in $L^2(A)$ , and again, the proof of Theorem 1.1 shows that for almost every $r\in (a,b)$ ,

(3.22) $$ \begin{align} \int_{\mathbb{S}^{n-1}}v_k(r\xi)\,d\sigma(\xi) \to \int_{\mathbb{S}^{n-1}}v(r\xi)\,d\sigma(\xi) \end{align} $$

and

(3.23) $$ \begin{align} Jv_k(r,\theta)\to Jv(r,\theta), \end{align} $$

for each $\theta \in (0,\pi )$ . Combining (3.18), (3.21), and (3.23) gives the theorem’s first conclusion. Moreover, pairing (3.17) with (3.20) and (3.22) shows that for almost every $r\in (a,b)$ ,

$$ \begin{align*} \int_{\mathbb{S}^{n-1}}(u(r\xi)-v(r\xi))\,d\sigma(\xi)=0. \end{align*} $$

The equation above and Proposition 2.2 give the theorem’s conclusion about convex means.▪

Proof of Theorem 1.2 First, say $f\in C_c^{\infty }(B)$ is a test function. As before, $u,v\in C^2(\overline {B})$ . Say $\delta ,\varepsilon>0$ and let $A_{\delta }=A(\delta ,b)$ denote a spherical shell with inner radius $\delta $ and outer radius b. Let w be defined as in the proof of Theorem 1.1:

$$ \begin{align*} w(r,\theta)=u^{\bigstar}(r,\theta)-Jv(r,\theta)-\varepsilon Q(r,\theta), \end{align*} $$

where

$$ \begin{align*} Q(r,\theta)=r(r-b)+C\theta(\pi-\theta), \end{align*} $$

and again C is chosen, so that

$$ \begin{align*} \Delta^{\bigstar}Q(r,\theta)=2+\frac{n-1}{r}(2r-b)-\frac{C}{r^{2}}\left(2+(n-2)(\pi-2\theta)\cot\theta\right)\leq 0, \end{align*} $$

on $B^{\bigstar }$ . As before, we have

$$ \begin{align*} -\Delta ^{\bigstar}w\leq 0, \end{align*} $$

weakly in $A_{\delta }^{\bigstar }$ . By the Maximum Principle,

(3.24) $$ \begin{align} \sup_{A_{\delta}^{\bigstar}}w \leq \max_{\partial A_{\delta}^{\bigstar}}w, \end{align} $$

and letting $\delta \to 0^+$ , we see

(3.25) $$ \begin{align} \sup_{B^{\bigstar}}w \leq \max_{\partial B^{\bigstar}}w. \end{align} $$

As before, we may assume $\displaystyle \max _{\partial B^{\bigstar }}w>0$ .

Defining

$$ \begin{align*} \psi(r)=\int_{\mathbb{S}^{n-1}}(u(r\xi)-v(r\xi))\,d\sigma(\xi),\qquad r\in[0,b], \end{align*} $$

the proof of Theorem 1.1 shows that $r^{n-1}\psi '(r)=k_1$ is constant on $(0,b]$ . Letting $r\to 0^+$ and using $u,v\in C^2(\overline {B})$ , we conclude $k_1=0$ . Thus,

(3.26) $$ \begin{align} \int_{\mathbb{S}^{n-1}}(u(r\xi)-v(r\xi))\,d\sigma(\xi)=k_2 \end{align} $$

is constant on $[0,b]$ . We therefore have

$$ \begin{align*} b^{n-1}\int_{\mathbb{S}^{n-1}}\left(u_r(b\xi)-v_r(b\xi)\right)\,d\sigma(\xi)&=0,\\ b^{n-1}\beta \int_{\mathbb{S}^{n-1}}\left(u(b\xi)-v(b\xi)\right)\,d\sigma(\xi)&=b^{n-1}\beta k_2. \end{align*} $$

Adding these equations and using the Robin boundary conditions shows

$$ \begin{align*} 0=\int_{\{|x|=b\}}\left(\left(\frac{\partial u}{\partial \eta}+\beta u\right)-\left(\frac{\partial v}{\partial \eta}+\beta v\right)\right)\,dS=b^{n-1}\beta k_2. \end{align*} $$

If $\beta \neq 0$ , we conclude $k_2=0$ , and so

(3.27) $$ \begin{align} \int_{\mathbb{S}^{n-1}}(u(r\xi)-v(r\xi))\,d\sigma(\xi)=0,\qquad r\in[0,b]. \end{align} $$

If $\beta =0$ , we integrate (3.26) and use our normalization assumptions on $u,v$ :

$$ \begin{align*} 0=\int_{B}(u-v)\,dx=k_2\int_0^br^{n-1}\,dr, \end{align*} $$

so $k_2=0$ and again (3.27) holds.

Step 2 in the proof of Theorem 1.1 shows that $\displaystyle \max _{\partial B^{\bigstar }}w$ cannot be achieved at a point with $r=b$ . Now, $u^{\bigstar }-Jv=0$ by definition when $\theta =0$ and (3.27) implies that $u^{\bigstar }-Jv=0$ when $\theta =\pi $ . Moreover, (3.27) implies $u(0)=v(0)$ , and so $u^{\bigstar }-Jv=0$ when $r=0$ . Thus, (3.25) implies

$$ \begin{align*} \sup_{B^{\bigstar}}(u^{\bigstar}-Jv) \leq \max_{\partial B^{\bigstar}}(-\varepsilon Q)+\varepsilon\|Q\|_{L^{\infty}(B^{\bigstar})}. \end{align*} $$

Sending $\varepsilon \to 0^+$ shows $u^{\bigstar }-Jv\leq 0$ in $B^{\bigstar }$ . The approximation argument from the proof of Theorem 1.1 applies almost verbatim to our new setting, and shows that for a given function $f\in L^2(B)$ , for almost every $r\in (0,b)$ , the inequality $u^{\bigstar }(r,\theta )\leq Jv(r,\theta )$ holds for each $\theta \in (0,\pi )$ . Equation (3.27) and the approximation argument in the proof of Theorem 1.1 also show that for general $f\in L^2(B)$ ,

(3.28) $$ \begin{align} \int_{\mathbb{S}^{n-1}}(u(r\xi)-v(r\xi))\,d\sigma(\xi)=0, \end{align} $$

for almost every $r\in (0,b)$ . Thus, the theorem’s conclusion about convex means follows from Proposition 2.2.▪

Proof of Theorem 1.3 Again, the proof is rather long, so we break it down into manageable steps.

Step 1: Maximum Principle. If $\alpha>0$ , let $Q_1$ denote the solution to the Robin problem

$$ \begin{align*} \begin{array}{r@{\;}c@{\;}lcc} \Delta Q_1 & = & 0 & \text{in} & D,\\ \frac{\partial Q_1}{\partial \eta} +\alpha Q_1 & = & 1 & \text{on} & \partial D. \end{array} \end{align*} $$

If $\alpha =0$ , let $Q_1$ be the solution to the Neumann problem

$$ \begin{align*} \begin{array}{r@{\;}c@{\;}lcc} \Delta Q_1 & = & \frac{\mathcal H^{n-2}(\partial D)}{\mathcal L^{n-1}(D)} & \text{in} & D,\\ \frac{\partial Q_1}{\partial \eta} & = & 1 & \text{on} & \partial D, \end{array} \end{align*} $$

where $Q_1$ is additively normalized to have zero mean. Here, we have written $\mathcal H^{n-2}$ for $(n-2)$ -dimensional Hausdorff measure and $\mathcal L^{n-1}$ for $(n-1)$ -dimensional Lebesgue measure. By Theorem 6.31 in [Reference Gilbarg and Trudinger24], Theorem 3.2 in [Reference Ladyzenskaja and Ural’ceva27], and [Reference Nardi34] (see also [Reference Bers and Nirenberg12]), $Q_1$ belongs to $C^2(\overline {D})$ . We are led to define a function Q by

$$ \begin{align*} Q(x,y)=Q_1(x)+Cy(\ell-y), \end{align*} $$

for $(x,y)\in \Omega $ , where $C>0$ is chosen large enough to guarantee

$$ \begin{align*} \Delta Q=\Delta_x Q_1-2C\leq 0, \end{align*} $$

on $\Omega $ . Define for $\varepsilon>0$ ,

$$ \begin{align*} w(x,y)=u^{\bigstar}(x,y)-Jv(x,y)-\varepsilon Q(x,y), \end{align*} $$

for $(x,y)\in \Omega $ .

In the weak sense, we have

$$ \begin{align*} -\Delta w &= -\Delta u^{\bigstar} +\Delta Jv + \varepsilon \Delta Q\\ &\leq f^{\bigstar} + J\Delta v + \varepsilon \Delta Q\\ &=f^{\bigstar} - f^{\bigstar} + \varepsilon \Delta Q\\ &\leq 0, \end{align*} $$

where the first inequality follows from Proposition 2.6 and Theorem 2.7, the second equality follows from (2.8), and the second inequality follows from our choice of C. Thus, w is weakly subharmonic in $\Omega $ , so by the Maximum Principle (Theorem 1 in [Reference Littman33]),

(3.29) $$ \begin{align} \sup_{\Omega}w\leq \max_{\partial \Omega}w. \end{align} $$

As before, we assume $\displaystyle \max _{\partial \Omega }w>0$ , because otherwise we are done.

Step 2: Boundary analysis on $\boldsymbol \partial \Omega _2$ . We first show that $\displaystyle \max _{\partial \Omega }w$ is not achieved at a point of $\partial \Omega _2$ . Fix $(x,y)\in \partial \Omega _2$ , so that $x\in \partial D$ and $y\in [0,\ell ]$ . Let $n=n(x)$ denote the inner normal vector at x to $\partial D$ . Let $E(x)$ denote a measurable subset of $(0,\ell )$ of length y for which the $\sup $ defining $u^{\bigstar }(x,y)$ is achieved. By translating, we assume $0\in D$ . Write $r=|x|$ . We then have for $h>0$ ,

$$ \begin{align*} &\frac{e^{-\alpha(r+h)}w(x+hn,y)-e^{-\alpha r}w(x,y)}{h} \qquad \qquad \qquad \qquad \qquad \qquad \\ &\qquad \qquad \qquad = e^{-\alpha(r+h)}\int_{E(x+hn)}\frac{u(x+hn,t)}{h}\,d t - e^{-\alpha r}\int_{E(x)}\frac{u(x,t)}{h}\,d t \\ &\qquad \qquad \qquad \quad -e^{-\alpha(r+h)}\int_0^y\frac{v(x+hn,t)}{h}\,d t +e^{-\alpha r}\int_0^y\frac{v(x,t)}{h}\,d t\\ &\qquad \qquad \qquad \quad -\varepsilon \frac{e^{-\alpha(r+h)}Q(x+hn,y)-e^{-\alpha r}Q(x,y)}{h}\\ &\qquad \qquad \qquad \geq \int_{E(x)}\frac{e^{-\alpha(r+h)}u(x+hn,t)-e^{-\alpha r}u(x,t)}{h}\,d t\\ &\qquad \qquad \qquad \quad -\int_0^y\frac{e^{-\alpha(r+h)}v(x+hn,t)-e^{-\alpha r}v(x,t)}{h}\,d t\\ &\qquad \qquad \qquad \quad -\varepsilon \frac{e^{-\alpha(r+h)}Q(x+hn,y)-e^{-\alpha r}Q(x,y)}{h}. \end{align*} $$

Using the Dominated Convergence Theorem, we therefore have

$$ \begin{align*} &\underset{h\rightarrow 0^+}{\text{lim inf}}\frac{e^{-\alpha(r+h)}w(x+hn,y)-e^{-\alpha r}w(x,y)}{h}\\ &\qquad \qquad\qquad\qquad\qquad\geq -e^{-\alpha r}\int_{E(x)}\left(\frac{\partial u}{\partial \eta}(x,t)+\alpha u(x,t)\right)\,d t \\ &\qquad \qquad\qquad\qquad\qquad\quad+e^{-\alpha r}\int_0^y\left(\frac{\partial v}{\partial \eta}(x,t)+\alpha v(x,t)\right)\,d t \\ &\qquad\qquad\qquad\qquad\qquad \quad+\varepsilon e^{-\alpha r}\left(1+\alpha Cy(\ell-y)\right)\\ &\qquad\qquad\qquad\qquad\qquad = \varepsilon e^{-\alpha r}\left(1+\alpha Cy(\ell-y)\right), \end{align*} $$

where the last equality holds courtesy of the Robin boundary conditions on $\partial \Omega _2$ . If $\displaystyle \max _{\partial \Omega }w$ is achieved at $(x,y)$ , then for all $h>0$ sufficiently small, we have

(3.30) $$ \begin{align} w(x+hn,y)>e^{\alpha h}w(x,y)+\frac{e^{\alpha h}h\varepsilon}{2}>w(x,y), \end{align} $$

because $w(x,y)>0$ and $\alpha \geq 0$ . When $y\in (0,\ell )$ , inequality (3.30) says that $\displaystyle \max _{\partial \Omega }w$ is not achieved at a point $(x,y)\in \partial \Omega _2$ , because otherwise the Maximum Principle (3.29) would be violated. Similarly, when $y=0$ or $y=\ell $ , $\displaystyle \max _{\partial \Omega }w$ is not achieved at a point $(x,y)\in \partial \Omega _2$ , because by (3.30), w assumes strictly larger values at points of $\partial \Omega _1$ .

Step 3: Boundary analysis on $\boldsymbol \partial \Omega _1$ . We start by defining

$$ \begin{align*} \psi(x)=\int_0^{\ell}(u(x,t)-v(x,t))\,d t,\qquad x\in D. \end{align*} $$

Using the Dominated Convergence Theorem and our assumption that $u,v\in C^2(\overline \Omega )$ , we see that $\psi $ is harmonic in D:

$$ \begin{align*} \Delta \psi(x)&=\int_0^{\ell}(\Delta_x u(x,t)-\Delta_x v(x,t))\,d t\\ &=\int_0^{\ell}(-u_{yy}(x,t)-f(x,t)+v_{yy}(x,t)+f^{\#}(x,t))\,d t\\ &= \int_0^{\ell}(-f(x,t)+f^{\#}(x,t))\,d t-u_y(x,\ell)+u_y(x,0)+v_y(x,\ell)-v_y(x,0)\\ &=0. \end{align*} $$

This last equality holds by the Neumann boundary conditions of $u,v$ along $\partial \Omega _1$ and because f and $f^{\#}$ are rearrangements of each other.

Taking $x\in \partial D$ , we next investigate the Robin boundary condition of $\psi $ :

$$ \begin{align*} \frac{\partial \psi}{\partial \eta}(x)+\alpha\psi(x)=\int_0^{\ell}\left(\left(\frac{\partial u}{\partial \eta}(x,t)+\alpha u(x,t)\right)-\left(\frac{\partial v}{\partial \eta}(x,t)+\alpha v(x,t)\right)\right)\,d t=0, \end{align*} $$

because u and v themselves satisfy Robin boundary conditions on $\partial \Omega _2$ . If $\alpha =0$ , note that $\psi $ has zero mean by our normalization assumption on $u,v$ :

$$ \begin{align*} \int_D\psi(x)\,d x=\int_{\Omega}(u-v)\,d x\,dy=0. \end{align*} $$

The function $\psi $ therefore solves the problem

$$ \begin{align*} \begin{array}{r@{\;}c@{\;}lcc} \Delta \psi & = & 0 & \text{in} & D,\\ \frac{\partial \psi}{\partial \eta} +\alpha \psi & = & 0 & \text{on} & \partial D, \end{array} \end{align*} $$

and when $\alpha =0$ , $\int _{D}\psi \,dx=0$ . Integrating by parts, $\psi $ satisfies the equation

$$ \begin{align*} \int_{D}|\nabla \psi|^2\,dx=-\alpha\int_{\partial D}\psi^2\,dS, \end{align*} $$

so we deduce $\psi \equiv 0$ in D. Because $u^{\bigstar }-Jv=0$ when $y=0$ , we conclude that

$$ \begin{align*} w(x,0)=w(x,\ell)=-\varepsilon Q_1(x), \end{align*} $$

for $x\in D$ .

Our work from Steps 2 and 3 and the Maximum Principle (3.29) together give

$$ \begin{align*} \sup_{\Omega} (u^{\bigstar}-Jv)\leq \varepsilon \|Q\|_{L^{\infty}(\Omega)}+\sup_{D}(-\varepsilon Q_1). \end{align*} $$

Sending $\varepsilon \to 0^+$ gives $u^{\bigstar }-Jv\leq 0$ in $\Omega $ . Note that because $\psi (x)\equiv 0$ on D, for each $x\in D$ , the slice functions $u(x,\cdot )$ and $v(x,\cdot )$ have the same mean on $(0,\ell )$ . Thus, Proposition 2.2 gives the theorem’s claims on the convex means of u and v.▪

Proof Letting $f=f^{\#}$ , Theorem 1.3 gives that $v^{\bigstar }\leq Jv$ on $\Omega $ . Because the reverse inequality holds by definition, we have $v^{\bigstar }= Jv$ on $\Omega $ . Spelled out, this means that for each $x\in D$ ,

$$ \begin{align*} \int_0^yv^{\#}(x,t)\,dt=\int_0^yv(x,t)\,dt, \end{align*} $$

for each $y\in (0,\ell )$ . Differentiating the above equation with respect to y shows that $v(x,y)=v^{\#}(x,y)$ on $\Omega $ , as desired.▪

Proof The proof of Theorem 1.3 shows that for each $x\in D$ ,

$$ \begin{align*} \int_0^{\ell}u(x,t)\,dt=\int_0^{\ell}v(x,t)\,dt. \end{align*} $$

This equality together with Theorem 1.3 and Corollary 2.3 gives the claimed inequalities on the slice functions of u and v, and as before, the subsequent inequalities follow.▪