Proof. Since \(\overline{h_C}(t) = h_C(u_t)\), and by Schneider2014?*Thm. 1.7.2, the directional derivatives of \(h_C\) exist at \(u_t \in \mathbb{S}^{n-1}\) and are equal to \(h_{F(C,u_t)}\), the claim follows from the chain rule for directional derivatives. ◻

Proof. First, let \(t \in (-1,1)\). By 10, and choosing \(x\in F(C,u_t)\) appropriately, we obtain \[\begin{align} \partial_+\overline{h_C}(t) = h_{F(C, u_t)}\left(\frac{d}{dt}u_t\right) = \left\langle x, e_n - \frac{t}{\sqrt{1-t^2}}\bar{u}\right\rangle. \end{align}\] Consequently, since \(h_C(u_t) = \langle x,u_t \rangle\) as \(x \in F(C, u_t)\), \[\begin{align} R_C(t) &= \sqrt{1-t^2}(\overline{h_C}(t) - t\partial_- \overline{h_C}(t))\\ &=\sqrt{1-t^2}h_C(u_t)-t\sqrt{1-t^2}\langle x,e_n \rangle + t^2 \langle x,\bar u \rangle\\ &=\sqrt{1-t^2}\langle x,u_t \rangle-t\sqrt{1-t^2}\langle x,e_n \rangle + t^2 \langle x,\bar u \rangle\\ &=(1-t^2)\langle x,\bar u \rangle + t^2\langle x,\bar u \rangle = \langle x,\bar u \rangle, \end{align}\] which yields the first claim. The second claim follows either from the first claim or directly by a short computation from 10. ◻

Proof. It was shown in OrtegaMoreno2021?*Proof of Lem. 5.3 that the Hessian of \(h_C\) has eigenvalues \(\mathcal{A}_1^C(t) = \overline{h_C}(t) - t\overline{h_C}'(t)\) (with multiplicity \(n-2\)) and \(\mathcal{A}_2^C(t) = (1-t^2)\overline{h_C}''(t) + \overline{h_C}(t) - t\overline{h_C}'(t)\) (with multiplicity \(1\)) at every \(u \in \mathbb{S}^{n-1}\setminus\{\pm e_n\}\) such that \(t = \langle u,e_n \rangle\). As the density of \(S_{n-1}(C, \cdot)\) is the product of the eigenvalues, we obtain after applying cylindrical coordinates \[\begin{align} \label{eq:prfSurfAreaMeasSmooth} \int_{\mathbb{S}^{n-1}} f(\langle e_n,u \rangle) dS_{n-1}(C, u) = \omega_{n-1}\int_{-1}^{1} f(t) (1-t^2)^{\frac{n-3}{2}} (\mathcal{A}_1^C(t))^{n-2}\mathcal{A}_2^C(t) dt \end{align}\tag{11}\] for every \(f \in C[-1,1]\). Noting that \[\begin{align} R_C(t) = \sqrt{1-t^2}\mathcal{A}_1^C(t), \end{align}\] and since \[\begin{align} \label{eq:RCprimeSm} -\frac{R_C'(t)}{t} = (1-t^2)^{-\frac{1}{2}}\mathcal{A}_1^C(t) +\sqrt{1-t^2}\overline{h_C}''(t) = (1-t^2)^{-\frac{1}{2}}\mathcal{A}_2^C(t), \end{align}\tag{12}\] we conclude, via integration by parts, that \(R_C\) satisfies 8 for \(d \nu_{R_C}(t) = (1-t^2)^{-\frac{1}{2}}\mathcal{A}_2^C(t) dt\), that is, \(R_C \in \mathrm{BV}_0(-1,1)\). Moreover, by 11 and since \(R_C^{n-1}\) is again in \(\mathrm{BV}_0(-1,1)\) by 9[BV95095algebraic:product], we deduce the claim \[\begin{align} \int_{\mathbb{S}^{n-1}} f(\langle e_n,u \rangle) dS_{n-1}(C, u) = \frac{\omega_{n-1}}{n-1}\int_{(-1,1)} f(t) d\nu_{R_C^{n-1}}(t). \end{align}\] ◻

Proof of 16. First, let \(C\) be a body of revolution that does not contain a vertical segment in its boundary. Suppose that \((C_j)_{j \in \mathbb{N}}\) is a sequence of \(C^\infty_+\) bodies of revolution converging to \(C\) in the Hausdorff metric and fix a \(2\)-dimensional linear space \(E \subset \mathbb{R}^n\), containing \(e_n\). Then \(C_j|E \to C|E\), and, thus, \(S_1^{E}(C_j|E, \cdot) \rightharpoonup S_1^E(C|E, \cdot)\) by the continuity of the surface area measure. By 14 applied in \(E\), the measures \(\nu_{R_{C_j|E}}\) converge weakly to the pushforward of \(\frac{1}{\omega_{1}}S_{1}^E(C|E, \cdot)\) onto \((-1,1)\), denoted \(\nu\), as \(j \to \infty\). However, as the definition of \(R_C\) does not depend on the surrounding dimension and \(\overline{h_C} = \overline{h_{C|E}}\), we conclude that \(\nu_{R_{C_j}} \rightharpoonup \nu\).

Next, note that, by the Hausdorff convergence of \(C_j \to C\), \(\overline{h_{C_j}} \to \overline{h_C}\) uniformly on \([-1,1]\). Consequently, \(\mathcal{A}_1^{C_j} \to \mathcal{A}_1^C\) and \(\mathcal{A}_2^{C_j} \to \mathcal{A}_2^C\) converge in the sense of distributions, where \(\mathcal{A}_{1}^C\) and \(\mathcal{A}_{2}^C\) are defined as in the proof of 14, but in the distributional sense.

As, by 12 , \(d \nu_{R_{C_j}} (t) = (1-t^2)^{-\frac{1}{2}}\mathcal{A}_2^{C_j}(t) dt\), we conclude that \(\nu = (1-t^2)^{-\frac{1}{2}}\mathcal{A}_2^C\), as distributions. In particular, \(\mathcal{A}_2^C\) is a non-negative measure on \((-1,1)\).

Using 9 , the weak convergence thus implies \[\begin{align} \label{eq:prfSurfMeasRevContRj} R_{C_j}(t) - R_{C_j}(0) = -\int_{(0,t]} sd\nu_{R_{C_j}}(s) \to -\int_{(0,t]} sd\nu(s), \qquad j \to \infty, \end{align}\tag{13}\] for all \(t \in (-1,1)\), where \(\nu(\{t\}) = 0\), that is, as \(\nu\) is a Radon measure, for all except at most countably many \(t \in (-1,1)\). Moreover, 12 implies that \(R_{C_j}(0) = \overline{h_{C_j}}(0)\), and, therefore, \(R_{C_j}(0) \to R_C(0)\).

On the other hand, as \(\mathcal{A}_1^{C_j} \to \mathcal{A}_1^C\) in the sense of distributions, it follows that \(R_{C_j} \to R_C\). Together with 13 , the uniqueness of the limit, and the convergence of \(R_{C_j}(0)\), this implies that \[\begin{align} R_C(t) = R_C(0) - \int_{(0,t]} sd\nu(s), \end{align}\] and \(R_{C_j} \to R_C\) pointwise almost everywhere on \((-1,1)\).

Suppose now that \(f \in C^1_c(-1,1)\). Then by the weak convergence of \(\nu_{R_{C_j}}\), and 8 for \(R_{C_j}\), \[\begin{align} \label{eq:prfSurfMeasRevContLimN2} \int_{(-1,1)} f(t) t d\nu(t) &= \lim_{j \to \infty} \int_{(-1,1)} f(t) t d\nu_{R_{C_j}}(t) \nonumber \\ &= \lim_{j \to \infty} \int_{(-1,1)} f'(t) R_{C_j}(t) dt = \int_{(-1,1)} f'(t) R_{C}(t) dt, \end{align}\tag{14}\] where the last step follows by dominated convergence, using that the limit is bounded and thus integrable by 12. We conclude that \(R_C \in \mathrm{BV}_0(-1,1)\) since it satisfies property 8 for \(\nu_{R_{C}} = \nu\). Here, we used that the condition that \(C\) does not contain a vertical segment in its boundary implies that \(|\nu|(\{0\})=0\).

Returning to \(\mathbb{R}^n\), note that, by the continuity of the surface area measure, \(S_{n-1}(C_j, \cdot) \rightharpoonup S_{n-1}(C, \cdot)\), and by 14, the measures \(\nu_{R_{C_j}^{n-1}}\) converge weakly to the pushforward of \(\frac{n-1}{\omega_{n-1}}S_{n-1}(C, \cdot)\) onto \((-1,1)\), denoted \(\widehat{\nu}\), as \(j \to \infty\). Then, as also \(R_{C_j}^{n-1} \to R_C^{n-1}\) pointwise almost everywhere on \((-1,1)\), it follows by repeating the argument from 14 and the uniqueness of the measure \(\nu_{R_C^{n-1}}\) that \(\widehat{\nu} = \nu_{R_{C}^{n-1}}\). This implies ?? in the case where \(C\) does not contain a vertical segment in its boundary.

Finally, if \(C\) contains a vertical segment in its boundary, then, by 15, \(C = \widetilde{C} + \ell_C[0,e_n]\) and the boundary of \(\widetilde{C}\) does not contain a vertical segment. Note that \(R_C = R_{\widetilde{C}}\) and thus \(R_C \in \mathrm{BV}_0(-1,1)\) by the previous case. Moreover, by 5 , \[\begin{align} \int_{\mathbb{S}^{n-1}\setminus\{\pm e_n\}} f(\langle e_n,u \rangle) dS_{n-2}(\widetilde{C}, [0,e_n]; u) &= \int_{\mathbb{S}^{n-1}} f(\langle e_n,u \rangle) dS_{n-2}(\widetilde{C}, [0,e_n]; u)\\ =\frac{1}{n-1}\int_{\mathbb{S}^{n-2}(e_n^\perp)} f(\langle e_n,u \rangle) dS_{n-2}^{e_n^\perp}(\widetilde{C}|e_n^\perp, u) &=\frac{\omega_{n-1}}{n-1}R_{C}(0)^{n-2}f(0), \end{align}\] for all \(f \in C[-1,1]\), using that \(S_{n-2}(\widetilde{C}, [0,e_n]; \cdot)\) is concentrated on \(\mathbb{S}^{n-1} \setminus \{\pm e_n\}\) and, by 12, \(\widetilde{C}|e_n^\perp = \widetilde{C}|e_n^\perp = R_{C}(0) \mathbb{D}\). We can therefore use the multilinearity of the mixed area measure, \[\begin{align} \int_{\mathbb{S}^{n-1}\setminus\{\pm e_n\}} f(\langle e_n,u \rangle) dS_{n-1}(C; u) =& \int_{\mathbb{S}^{n-1}\setminus\{\pm e_n\}} f(\langle e_n,u \rangle) dS_{n-1}(\widetilde{C}; u) \\ &+ (n-1)\int_{\mathbb{S}^{n-1}\setminus\{\pm e_n\}} f(\langle e_n,u \rangle) dS_{n-2}(\widetilde{C}, [0,e_n]; u), \end{align}\] to deduce ?? in the general setting from the previous case. ◻

Proof. First, recall that for a convex body \(C\in\mathcal{K}(\mathbb{R}^n)\) and a Borel subset \(\beta\subseteq\mathbb{S}^{n-1}\), \(S_{n-1}(C,\beta)=\mathcal{H}^{n-1}(\tau(C,\beta))\) for the reverse spherical image \(\tau(C,\beta)\). Now, take a subspace \(E'\in\mathrm{Gr}_k(\mathbb{R}^n)\) and a Borel set \(\beta\subseteq \mathbb{S}^{k-1}(E')\). It is easy to see that \(F(C|E',v)=F(C,v)|E'\) for every \(v\in \mathbb{S}^{k-1}(E')\), and thus, \(\tau(C|E',\beta)=\tau(C,\beta)|E'\). Consequently, \[\label{eq:mixed95sph95lift95gnrl:proof1} S^{E'}_{k-1}(C|E',\beta) = \mathcal{H}^{k-1}(\tau(C|E',\beta)) = \mathcal{H}^{k-1}(\tau(C,\beta)|E') \leq \mathcal{H}^{k-1}(\tau(C,\beta)),\tag{16}\] where the final inequality is due to the fact that the Hausdorff measure decreases under a \(1\)-Lipschitz map.

Next, we take \(C\in\mathcal{K}(\mathbb{R}^n)\) to be a body of revolution that does not contain any vertical segment in its boundary. Note that \(C|e_n^{\perp}=r\mathbb{D}\) for some \(r\geq 0\), so for every \(u\in \mathbb{S}^{n-2}(e_n^{\perp})\), \[F(C,u)|e_n^{\perp} =F(C|e_n^\perp,u) =F(r\mathbb{D},u) = \{ru\}.\] which shows that the face \(F(C,u)\) is contained in the vertical line \(ru+\mathbb{R}e_n\). By our assumption on \(C\), we must have \(F(C,u)=\{ru+t e_n\}\) for some \(t\in\mathbb{R}\). Hence, the reverse spherical image \(\tau(C,\mathbb{S}^{n-i-2}(E^\perp))\) has the same Hausdorff dimension as \(\mathbb{S}^{n-i-2}(E^\perp)\), and thus, its \((n-i-1)\)-dimensional Hausdorff measure is zero. As an instance of 16 , for all \(u\in\mathbb{S}^{i}(E)\), \[S_{n-i-1}^{E^\perp \vee u}(C| (E^\perp \vee u), \mathbb{S}^{n-i-2}(E^\perp) ) \leq \mathcal{H}^{n-i-1}(\tau(C,\mathbb{S}^{n-i-2}(E^\perp))) = 0,\] so condition ?? is satisfied.

To establish the general case, let \(\mathcal{C}=(C_1,\ldots,C_{n-i-1})\) be a family of convex bodies of revolution in \(\mathbb{R}^n\) that do not contain any vertical segment in their respective boundaries. Then the body \(C=C_1+\cdots+C_{n-i-1}\), due to 15, also does not contain any vertical segment in its boundary. Note that for every Borel set \(\beta\subseteq \mathbb{S}^{n-i-1}(E^\perp\vee u)\), \[S_{n-i-1}^{E^\perp\vee u}(C|(E^\perp\vee u),\beta) = \sum_{j_1,\ldots,j_{n-i-1}=1}^{n-i-1} S^{E^\perp\vee u}((C_{j-1},\ldots,C_{j_{n-i-1}})|(E^\perp\vee u);\beta).\] For \(\beta=\mathbb{S}^{n-i-2}(E^\perp)\), by our previous argument, the left hand side vanishes. Since each term on the right hand side is non-negative, condition ?? is satisfied. Invoking 20 then concludes the proof. ◻

Proof. First, take some body of revolution \(C\in\mathcal{K}(\mathbb{R}^n)\) with no vertical segment in its boundary. by definition, for all \(u\in\mathbb{S}^{k-1}(E)\), \[(\pi_{E,C^{[n-k]}}f(\langle e_n,{}\cdot{} \rangle))(u) = \int_{\mathbb{H}^{n-k}(E,u)} f(\langle e_n,v \rangle)\, dS^{E^\perp \vee u}_{n-k-1}(C| (E^\perp \vee u),v).\] Next, note that whenever \(u\in\mathbb{S}^{k-1}(E)\) and \(v\in\mathbb{S}^{n-k}(E^\perp\vee u)\), then \(\langle e_n,v \rangle=\langle e_n,u \rangle\langle u,v \rangle\), and thus, \[h_{C|(E^\perp\vee u)}(v) = h_C(v) = \bar{h}_C(\langle e_n,v \rangle) = \bar{h}_C(\langle e_n,u \rangle\langle u,v \rangle).\] Hence, \(C|(E^\perp\vee u)\) is a body of revolution in \(E^\perp\vee u\) with axis \(u\), and denoting \(s=\langle e_n,u \rangle\), we have that \(\bar{h}_{C|(E^\perp\vee u)}(t)=\bar{h}_C(st)\). Hence, for \(t\in (-1,1)\), \[R_{C|(E^\perp\vee u)}(t) = (1-t^2)^{\frac{1}{2}}(\bar{h}_C(st)-st\partial_+\bar{h}_C(st)) = (1-t^2)^{\frac{1}{2}}(1-s^2t^2)^{-\frac{1}{2}}R_C(st).\] Suppose now that \(\mathcal{C}=(C_1,\ldots,C_{n-k})\) is a family of convex bodies with no vertical segment in their respective boundaries. By polarization, \(R_{\mathcal{C}|(E^\perp\vee u)}=R_{\mathcal{C},s}\), and thus, noting that \(\mathbb{H}^{n-k}(E^\perp\vee u)\subseteq \mathbb{S}^{n-k}(E^\perp\vee u)\setminus\{\pm u\}\), we obtain ?? from 18. ◻

Proof. We denote for \(s \in (0,1)\) and \(0 < t < 1/s\) and any tuple \(\mathcal{C}= (C_1, \dots, C_{n-i-1})\) \[\begin{align} Q_{\mathcal{C}}(s,t) = \left(\frac{(1-t^2)_+}{1-(st)^2}\right)^{\frac{n-i-1}{2}} \prod_{j=1}^{n-i-1}R_{C_j}(st), \end{align}\] where \((1-t^2)_+ = \max\{1-t^2, 0\}\) denotes the positive part.

Assume \(s > 0\), the claim for \(s < 0\) follows analogously. For every \(g \in C(-1,1)\), 8 , 10 , and letting \(u = st\) implies that for \(0<a<b<1\) \[\begin{align} -\int_{(a,b]} g(st) d\nu_{Q_{\mathcal{C}}(s,\cdot)}(t) &= \left.\frac{g(st)}{t}Q_{\mathcal{C}}(s,t)\right|_{t=a^+}^{b^+} - \int_{(a,b]}\frac{stg'(st)-g(st)}{t^2}Q_{\mathcal{C}}(s,t) dt\\ &= s \left(\left.\frac{g(u)}{u}Q_{\mathcal{C}}\left(s,\frac{u}{s}\right)\right|_{u=(sa)^+}^{(sb)^+} - \int_{(sa,sb]}\frac{ug'(u)-g(u)}{u^2}Q_{\mathcal{C}}\left(s,\frac{u}{s}\right) du\right)\\ &=-s\int_{(sa,sb]}g(u) d\nu_{Q_{\mathcal{C}}\left(s,\frac{(\cdot)}{s}\right)}(u). \end{align}\] By approximation, the same holds true for \(a=0\) and \(b=1\). Note also that \(Q_{\mathcal{C}}(s, \frac{(\cdot)}{s}) \in \mathrm{BV}_0(-1,1)\). Consequently, by 18 , \[\begin{align} \frac{n-i-1}{\omega_{n-i-1}}(\pi_{E,\mathbb{D}} g)(s) = T_{Q_{\mathbb{D}^{[n-i-1]}}(s,\frac{(\cdot)}{s})} g(s) - Q_{\mathbb{D}^{[n-i-1]}}(s,1) g(s) = T_{Q_{\mathbb{D}^{[n-i-1]}}(s,\frac{(\cdot)}{s})}g(s), \end{align}\] as \(Q_{\mathbb{D}^{[n-i-1]}}(s,1) = 0\). Observe now that, by 24[it:T95R95group95property] and since by assumption \(\widehat{T}_{\mathcal{C}} = T_{R_\mathcal{C}}\), \(T_{Q_{\mathbb{D}}(s,\frac{(\cdot)}{s})} \circ T_{R_\mathcal{C}} = T_{Q_{\mathbb{D}}(s,\frac{(\cdot)}{s}) R_{\mathcal{C}}}\). Hence, since \[\begin{align} Q_{\mathbb{D}}\left(s,\frac{t}{s}\right) R_{\mathcal{C}} (t) = \left(\frac{(1-(t/s)^2)_+}{(1-t^2)}\right)^{\frac{n-i-1}{2}}R_{\mathcal{C}}(t) = Q_{\mathcal{C}}\left(s,\frac{t}{s}\right), \quad t \in (-1,1), \end{align}\] the claim follows by reversing the steps taken above. ◻

Proof of 27. First, note that by 24[it:continuous95in95D95R], \(\bar{g} = \widehat{T}_\mathcal{C}\bar{f}\) and, thus, \(g\) is continuous.

Next, write \(C_j = \widetilde{C}_j + \ell_{C_j}[0,e_n]\), \(j = 1, \dots, n-i-1\), with \(\ell_{C_j} \geq 0\) chosen by 15 such that \(\widetilde{C}_j\) does not contain a vertical segment in its boundary. Then a similar argument as in the proof of 16 using the multilinearity of mixed area measures and 5 implies that \[\begin{align} \int_{\mathbb{S}^{n-1}} f \, dS_i(K,\mathcal{C};{}\cdot{}) = \int_{\mathbb{S}^{n-1}} f \, dS_i(K,\widetilde{\mathcal{C}};{}\cdot{}) + \frac{n-i-1}{(n-1)\omega_{n-i-1}}W_{\mathcal{C}} \int_{\mathbb{S}^{n-2}(e_n^\perp)}f(u) dS_i^{e_n^\perp}(K|e_n^\perp, u). \end{align}\] As \(f\) is zonal, \[\begin{align} \frac{1}{n-1}\int_{\mathbb{S}^{n-2}(e_n^\perp)}f(u) dS_i^{e_n^\perp}(K|e_n^\perp, u) = \bar{f}(0) V^{e_n^\perp}((K|e_n^\perp)^{[i]}, \mathbb{D}^{[n-i-1]}), \end{align}\] which, again by 5 , can be rewritten as \[\begin{align} &\bar{f}(0) V^{e_n^\perp}((K|e_n^\perp)^{[i]}, \mathbb{D}^{[n-i-1]}) = \bar{f}(0)\frac{n}{2} V(K^{[i]}, \mathbb{D}^{[n-i-1]},[-e_n,e_n])\\ &\qquad = \frac{\bar{f}(0)}{2} \int_{\mathbb{S}^{n-1}} |\langle e_n,u \rangle| dS_i(K, \mathbb{D}; u). \end{align}\] Consequently, as \(\widehat{T}_{\mathcal{C}}f = T_{\widetilde{\mathcal{C}}}f + \frac{n-i-1}{2\omega_{n-i-1}} W_{\mathcal{C}} f(0) |\cdot|\), it suffices to show the claim only for bodies of revolution which do not contain vertical segments in their respective boundary.

Next, note that both sides of ?? define continuous, translation-invariant and zonal valuations when seen as functionals in \(K \in \mathcal{K}(\mathbb{R}^n)\). By 4, it is therefore sufficient to show ?? for bodies \(K \in \mathcal{K}(\mathbb{R}^n)\) that are contained in some fixed \((i+1)\)-dimensional subspace \(E\) containing the axis \(e_n\).

Let \(K \in \mathcal{K}(E)\) be arbitrary, but fixed. Then, by 21 \[\begin{align} \label{eq:prfDictZonalCont} \int_{\mathbb{S}^{n-1}} f \, dS_i(K,\mathcal{C};{}\cdot{}) = \int_{\mathbb{S}^{i}(E)} \pi_{E,\mathcal{C}} f \, dS_i^E(K,\mathcal{C};{}\cdot{}) = \int_{\mathbb{S}^{i}(E)} (\bar\pi_{E,\mathcal{C}} \bar{f})(\langle e_n,\cdot \rangle) \, dS_i^E(K,\mathcal{C};{}\cdot{}). \end{align}\tag{19}\] Since, by 28, \(\bar{\pi}_{E, \mathcal{C}} \bar f = \bar{\pi}_{E, \mathbb{D}^{[n-i-1]}} \bar{g}\), repeating the steps in 19 for \(g\) and \(\mathbb{D}^{[n-i-1]}\) instead of \(f\) and \(\mathcal{C}\) yields the claim. ◻

Proof. Since the mixed area measure \(S_i(K,\mathcal{C};{}\cdot{})\) is locally determined by the convex bodies involved (see 2), we may assume that the reference bodies \(C_j\) do not contain any vertical segments in their respective boundaries (otherwise, replace them with \(\tilde{C}_j\)). Denoting by \(\mu\) and \(\sigma\) the pushforward of \(S_i(K,\mathcal{C};{}\cdot{})\) and \(S_i(K,\mathbb{D};{}\cdot{})\) under the map \(\langle e_n,{}\cdot{} \rangle\), respectively, 27 yields that \(\mu=T_{R_{\mathcal{C}}}^\ast\sigma\). The statement then follows from 53, which is postponed until 6.2. ◻

Proof. We give suitable estimates on the integral expressions in ?? . For the integral expression on the right-hand side, denoting \(t_+=\max\{0,t\}\), we have that for every \(t \in (0,1)\) \[\begin{align} \label{eq:mixedFrireyEst:proof} \begin{aligned} &\int_{\mathrm{Cap}(e_n,t)}\lvert \langle e_n,u \rangle \rvert\, S_i(K,\mathbb{D};du) \leq \int_{\mathbb{S}^{n-1}} {\langle e_n,u \rangle}_+ \, S_i(K,\mathbb{D};du) = nV(K^{[i]},\mathbb{D}^{n-i-1},[0, e_n]) \\ &\qquad = V^{e_n^\perp}((K|e_n^\perp)^{[i]},\mathbb{D}^{n-i-1}) = \frac{\kappa_{n-i-1}}{\binom{n-1}{i}} V_i(K|e_n^\perp), \end{aligned} \end{align}\tag{20}\] where we used the classical formulas 3 , 5 , and 4 . For the integral expression on the left-hand side of ?? and \(t \in (0,1)\), \[\int_{\mathrm{Cap}(e_n,t)} \lvert \langle e_n,u \rangle \rvert \, S_i(K,\mathcal{C};du) \geq t S_i(K,\mathcal{C};\mathrm{Cap}(e_n,t)).\] Finally, invoking 30 establishes ?? for the northern polar cap. The argument for the southern polar cap is analogous. ◻

Proof of 33. Fix a convex body \(K\in\mathcal{K}(\mathbb{R}^n)\) and some \(v\in\mathbb{S}^{n-1}\), and \(t \in (0,1)\). By applying 30 to the case where \(\mathcal{C}=(B^n)^{[n-i-1]}\), we have that \[\int_{\mathrm{Cap}(v,t)}\lvert \langle e_n,u \rangle \rvert\, S_i(K;du) = (1-t^2)^{\frac{n-i-1}{2}} \int_{\mathrm{Cap}(v,t)}\lvert \langle e_n,u \rangle \rvert\, S_i(K,\mathbb{D}(v^\perp);du),\] using the fact that \(R_{(B^n)^{[n-i-1]}}(t) = (1-t^2)^{\frac{n-i-1}{2}}\) and that Euclidean balls are symmetric around any axis. Note that for every Radon measure \(\mu\) on \(\mathbb{S}^{n-1}\), \[t \, \mu(\mathrm{Cap}(v,t)) \leq \int_{\mathrm{Cap}(v,t)}\lvert \langle e_n,u \rangle \rvert\, \mu(du) \leq \mu(\mathrm{Cap}(v,t))\] Applying this to \(S_i(K,{}\cdot{})\) and \(S_i(K,\mathbb{D},{}\cdot{})\) yields the following estimates. \[\begin{align} S_i(K; \mathrm{Cap}(v,t) ) &\leq \frac{1}{t}(1-t^2)^{\frac{n-i-1}{2}} S_i(K,\mathbb{D}(v^\perp); \mathrm{Cap}(v,t)).\\ S_i(K;\mathrm{Cap}(v,t)) &\geq t \,(1-t^2)^{\frac{n-i-1}{2}} S_i(K,\mathbb{D}(v^\perp);\mathrm{Cap}(v,t)). \end{align}\] Consequently, by dividing both sides by \((1-t^2)^{\frac{n-i-1}{2}}\) and passing to the limit \(t \to 1\), \[\lim_{t \to 1} \frac{S_i(K; \mathrm{Cap}(v,t) )}{(1-t^2)^{\frac{n-i-1}{2}}} = \lim_{t \to 1} S_i(K,\mathbb{D}(v^\perp),\mathrm{Cap}(v,t)) = S_i(K,\mathbb{D}(v^\perp);\{v\})\] Mixed area measures are locally determined by the convex bodies involved (see 2), so we may replace \(K\) in the final expression by any other convex body with the same face in the direction of \(v\). Hence, \[S_i(K,\mathbb{D}(v^\perp);\{v\}) = S_i(F(K,v),\mathbb{D}(v^\perp);\{v\}).\] Recall that \(S_{n-1}(C,{}\cdot{})=V_{n-1}(C)(\delta_v+\delta_{-v})\) for every convex body \(C\in\mathcal{K}(v^\perp)\). Hence, by polarizing this identity and by 4 , \[S_i(C,\mathbb{D}(v^\perp);{}\cdot{}) = V^{v^\perp}\!(C^{[i]},\mathbb{D}(v^\perp)^{[n-i-1]})(\delta_v + \delta_{-v}) = \frac{\kappa_{n-i-1}}{\binom{n-1}{i}} V_i(C)(\delta_v + \delta_{-v}).\] Combining the above steps, we obtain that \[\begin{align} \lim_{t \to 1} \frac{S_i(K; \mathrm{Cap}(v,t) )}{(1-t^2)^{\frac{n-i-1}{2}}} = S_i(F(K,v),\mathbb{D}(v^\perp),\{v\}) = \frac{\kappa_{n-i-1}}{\binom{n-1}{i}}V_i(F(K,v)), \end{align}\] which concludes the argument. ◻

Proof. By 24[it:T95R95group95property], the map \(T_R: C(-1,1)\to C(-1,1)\) is a bijection and \(T_R^{-1}=T_{\frac{1}{R}}\). Clearly, by 18 and the definition of \(\mathcal{D}_R\), \(T_R\) maps the subspace \(\mathcal{D}_R\) into \(C[-1,1]\), so it only remains to show that \(T_{\frac{1}{R}}\) maps \(C[-1,1]\) into \(\mathcal{D}_R\).

To this end, let \(g \in C[-1,1]\) and write \(f = T_{\frac{1}{R}} g\). By 63 in the appendix, to show \(f \in \mathcal{D}_R\), we only need to prove existence of the limits \(t \to 1\). Moreover, since \(g = T_R f\), by 18 , \[\begin{align} -\lim_{t\to 1} \int_{(0,t]} f(t)\, d\nu_R(t) = \lim_{t\to 1} \frac{1}{t}\left(R(t)f(t) - g(t)\right) = \lim_{t\to 1}\{R(t)f(t)\} - g(1), \end{align}\] and it suffices to consider the limit of \(R(t)f(t)\). Then for all \(0<t_0\leq t<1\), \[f(t) = \frac{g(t)}{R(t^+)} + t\int_{(0,t_0]}g \, d\nu_{\frac{1}{R}} + t\int_{(t_0,t]} g \, d\nu_{\frac{1}{R}}.\] By the mean value theorem for integrals, whenever \(0<t_0<t<1\), there exists some \(t_1\in (t_0,t)\) such that \[\begin{align} \int_{(t_0,t]} g\, d\nu_{\frac{1}{R}} = \int_{(t_0,t]} \frac{g(s)}{s} s \, d\nu_{\frac{1}{R}}(ds) = \frac{g(t_1)}{t_1} \int_{(t_0,t]} s \, d\nu_{\frac{1}{R}}(ds) = \frac{g(t_1)}{t_1} \left(\frac{1}{R(t_0^+)}-\frac{1}{R(t^+)}\right). \end{align}\]

In order to show that \(f\in\mathcal{D}_R\), we distinguish two cases. In the case when \(\lim_{t\to 1}R(t)\neq 0\), we let \(\varepsilon>0\) be arbitrary and choose \(t_0\in (0,1)\) such that \[\left| \frac{g(t)}{tR(t^+)} - \frac{g(t_0)}{t_0 R(t_0^+)}\right|<\varepsilon \qquad\text{and}\qquad \left|\frac{1}{R(t_0^+)}-\frac{1}{R(t^+)}\right|<\varepsilon \qquad\text{for all }t\in (t_0,1).\] By the previous step, for every \(t\in (t_0,1)\), there exists some \(t_1\in (t_0,t)\) such that \[\begin{align} &\frac{f(t)}{t} - \frac{f(t_0)}{t_0} = \frac{g(t)}{tR(t^+)} - \frac{g(t_0)}{t_0 R(t_0^+)} + \int_{(t_0,t]} g\, d\nu_{\frac{1}{R}} \\ &\qquad = \frac{g(t)}{tR(t^+)} - \frac{g(t_0)}{t_0 R(t_0^+)} - \frac{g(t_1)}{t_1} \left(\frac{1}{R(t_0^+)}-\frac{1}{R(t^+)}\right). \end{align}\] Consequently, we obtain that \[\left| \frac{f(t)}{t} - \frac{f(t_0)}{t_0} \right| \leq \varepsilon + \frac{\lVert g \rVert_\infty}{t_0} \varepsilon \qquad\text{for all }t\in (t_0,1).\] This shows that the limit \(\lim_{t\to 1}f(t)\), and thus, the limit \(\lim_{t\to 1}R(t)f(t)\), exists.

We now consider the case when \(\lim_{t\to 1}R(t)= 0\), we let \(\varepsilon>0\) be arbitrary and choose \(t_0\in (0,1)\) such that \[\left| \frac{g(t)}{t} - \frac{g(t_1)}{t_1} \right| <\varepsilon \qquad\text{for all }t,t_1\in (t_0,1).\] By the first step of the proof, for every \(t\in (t_0,1)\), there exists some \(t_1\in (t_0,1)\) such that \[\begin{align} &\frac{1}{t}R(t^+)f(t) = \frac{g(t)}{t} + R(t^+)\int_{(0,t_0]} g \, d\nu_{\frac{1}{R}} + \frac{g(t_1)}{t_1} \left(\frac{R(t^+)}{R(t_0^+)}-1\right) \\ &\qquad = \left( \frac{g(t)}{t} - \frac{g(t_1)}{t_1} \right) + R(t^+)\left( \frac{1}{R(t_0^+)}\frac{g(t_1)}{t_1} + \int_{(0,t_0]} g \, d\nu_{\frac{1}{R}} \right). \end{align}\] By passing to the limit \(t\to 1\), we obtain that \(\limsup_{t\to 1} | \frac{1}{t}R(t^+)f(t) | \leq \varepsilon\). Since \(\varepsilon>0\) was arbitrary, this shows that \(\lim_{t\to 1} R(t)f(t)=0\). ◻

Proof. By symmetry, we may assume that \(j=1\). Next, note that \(h_{K_1}=h_{\{x\}}\) on \(\mathbb{S}^{n-1}\cap N(K_1,x)\). Hence, as mixed area measures are locally determined (see 2), \[S(K_1,\ldots,K_{n-1};\mathbb{S}^{n-1}\cap N(K_1,x)) = S(\{x\},K_2,\ldots,K_{n-1};\mathbb{S}^{n-1}\cap N(K_1,x)) = 0.\] Consequently, \(\mathop{\mathrm{spt}}S(K_1,K_2,\ldots,K_{n-1},{}\cdot{})\cap \mathrm{int}\, N(K_1,x) = \emptyset\), as claimed. ◻

Proof. Let \(\bar{f} \in \mathcal{D}_\mathcal{C}\) and write \(\bar{g} = \widehat{T}_\mathcal{C}\bar{f}\). By 38, \(\bar{g} \in C[-1,1]\), so \[\begin{align} \varphi(K) = \int_{\mathbb{S}^{n-1}} \bar{g}(\langle e_n,u \rangle) dS_i(K,\mathbb{D}; u), \quad K \in \mathcal{K}(\mathbb{R}^n), \end{align}\] well-defines a zonal valuation in \(\mathbf{Val}_i(\mathbb{R}^n)\). It remains to show that \(\varphi\) satisfies relation ?? .

To this end, let \(\varepsilon > 0\) and define \(I_{\mathcal{C}, \varepsilon} = (a_{\mathcal{C}, -} + \varepsilon, a_{\mathcal{C}, +} - \varepsilon)\) and \(\bar{f}_\varepsilon = \widehat{T}_{\mathbb{1}_{I_{\mathcal{C}, \varepsilon}}} \bar f\), that is, we cut \(\bar{f}\) at \(a_{\mathcal{C}, \pm} \mp \varepsilon\) and extend it linearly to \((-1,1)\) (see 25). If \(U_{\mathcal{C}, \varepsilon}^+ \neq \emptyset\), then, for \(f_\varepsilon = \bar{f}_\varepsilon(\langle e_n,\cdot \rangle)\), writing \(\mathbb{S}^{n-1}_+ = \{u \in \mathbb{S}^{n-1}:\, \langle e_n,u \rangle > 0\}\) for the upper hemisphere, \[\begin{align} \int_{\mathbb{S}^{n-1}_+} f_{\varepsilon} \, dS_i(K,\mathcal{C};{}\cdot{}) - \int_{\mathbb{S}^{n-1}_+\setminus U_{\mathcal{C},\varepsilon}^+} f \, dS_i(K,\mathcal{C};{}\cdot{}) = \int_{U_{\mathcal{C},\varepsilon}^+} f_\varepsilon \, dS_i(K,\mathcal{C};{}\cdot{}), \end{align}\] which tends to zero as \(\varepsilon \to 0^+\). Indeed, as \(\mathop{\mathrm{spt}}S_i(K,\mathcal{C};{}\cdot{}) \subseteq \mathbb{S}^{n-1} \setminus \mathrm{int} \, U_{\mathcal{C}, 0}^+\), and the maximum of \(|f_\varepsilon(t)|\) in \([a_{\mathcal{C},+}-\varepsilon, a_{\mathcal{C},+}]\) is attained at \(t = a_{\mathcal{C}, +}\), we can deduce from 31 that \[\begin{align} \left|\int_{U_{\mathcal{C}, \varepsilon}^+} f_\varepsilon \, dS_i(K,\mathcal{C};{}\cdot{})\right| \leq |f_\varepsilon(a_{\mathcal{C}, +})|S_i(K,\mathcal{C};U_{\mathcal{C}, \varepsilon}^+) \leq C_{n,i}\frac{|f_\varepsilon(a_{\mathcal{C}, +})| R_\mathcal{C}(a_{\mathcal{C}, +} - \varepsilon)}{a_{\mathcal{C}, +} - \varepsilon} V_i(K|e_n^\perp). \end{align}\] Inserting \(f_\varepsilon(a_{\mathcal{C}, +}) = \frac{f(a_{\mathcal{C},+} - \varepsilon)}{a_{\mathcal{C},+} - \varepsilon}a_{\mathcal{C}, +}\), we see that the right-hand side tends to zero as \(\varepsilon \to 0\), since by 39[it:D95R95endpoint95limits95zero], \(f(a_{\mathcal{C},+} - \varepsilon)R_\mathcal{C}(a_{\mathcal{C}, +} - \varepsilon) \to 0\). Here, we used that since \(U_{\mathcal{C}, \varepsilon}^+ \neq \emptyset\), we have \(\lim_{t \to a_{\mathcal{C}, +}}R_\mathcal{C}(t) = 0\).

If \(U_{\mathcal{C}, \varepsilon}^+ = \emptyset\), then both \(f\) and \(f_\varepsilon\) are continuous by 39[it:D95R95endpoint95limits95nzero] and \(f_\varepsilon \to f\) uniformly on \([0,1]\). Consequently, repeating the argument at \(- e_n\), we obtain \[\begin{align} \label{eq:prfGenDict1} \lim_{\varepsilon \to 0^+} \int_{\mathbb{S}^{n-1}} f_{\varepsilon} \, dS_i(K,\mathcal{C};{}\cdot{}) = \lim_{\varepsilon \to 0^+} \int_{\mathbb{S}^{n-1}\setminus U_{\mathcal{C},\varepsilon}} f \, dS_i(K,\mathcal{C};{}\cdot{}). \end{align}\tag{21}\] As \(\bar{f}_\varepsilon\) and \(\bar{g}_\varepsilon := \widehat{T}_\mathcal{C}\bar{f}_\varepsilon\) are continuous on \([-1,1]\) (by 38), 27 implies for \(g_\varepsilon = \bar{g}_\varepsilon(\langle e_n,\cdot \rangle)\) that \[\begin{align} \label{eq:prfGenDict2} \int_{\mathbb{S}^{n-1}} f_{\varepsilon} \, dS_i(K,\mathcal{C};{}\cdot{}) = \int_{\mathbb{S}^{n-1}} g_{\varepsilon} \, dS_i(K,\mathbb{D};{}\cdot{}), \quad K \in \mathcal{K}(\mathbb{R}^n). \end{align}\tag{22}\] Noting finally that, by 24[it:T95R95group95property], \(\bar{g}_\varepsilon = \widehat{T}_\mathcal{C}(\widehat{T}_{\mathbb{1}_{I_{\mathcal{C}, \varepsilon}}} \bar{f}) = \widehat{T}_{\mathbb{1}_{I_{\mathcal{C}, \varepsilon}}} \bar g\) and, therefore, \(\bar{g}_\varepsilon \to \bar{g}\) uniformly on \([-1,1]\), the claim follows by combining 21 and 22 . ◻

Proof. We may write \(I=(a_-,a_+)\), where \(-1\leq a_-<0<a_+\leq 1\).

First, assume that \(\mathop{\mathrm{spt}}\psi_{i,g} \subseteq \mathop{\mathrm{cls}}\{u\in\mathbb{S}^{n-1}: \langle e_n,u \rangle\in I\}\) and consider \[\begin{align} C_s = \mathrm{conv}\left(\mathbb{D}\cup \left\{\tfrac{\sqrt{1-s^2}}{s}e_n\right\}\right), \quad s \in [-1,1]\setminus\{0\}, \end{align}\] that is, the cone with basis \(\mathbb{D}\) and apex \(\frac{\sqrt{1-s^2}}{s}e_n\). In Brauner2024a?*Lemma 2.2, it is shown that \[\begin{align} \label{eq:prfSuppPsiConeEval} \psi_{i,g}(C_s)=\kappa_{n-1}\left(\bar{g}(\mathop{\mathrm{sgn}}s) + \frac{\bar{g}(s)}{\lvert s \rvert}\right). \end{align}\tag{23}\] If \(s\in (a_+,1]\), then \(\bar{h}_{C_{s}}(t)=\bar{h}_{\mathbb{D}}(t)\) for all \(t\in [-1,s] \supseteq I\), that is, \(h_{C_{s}}\) and \(h_{\mathbb{D}}\) coincide on some open neighborhood of \(\mathop{\mathrm{spt}}\psi_{i,g}\). Consequently, \(\psi_{i,g}(C_{s})=\psi_{i,g}(\mathbb{D})\), and by 23 \[\begin{align} \bar{g}(1) + \frac{\bar{g}(s)}{s} = 2 \bar{g}(1), \quad s\in (a_+,1], \end{align}\] which yields \(\bar{g}(s) = sg(1)\) on \((a_+, 1]\), and by continuity on \([a_+, 1]\). Repeating the argument for \([-1,a_-)\) implies that \(\bar{g}\) is linear outside \(I\), which is equivalent to \(\bar{g}=\widehat{T}_{\mathbb{1}_I}\bar{g}\), by 25.

Next, assume that \(\bar{g}=\widehat{T}_{\mathbb{1}_I}\bar{g}\). Let \(S_+:=\{u\in\mathbb{S}^{n-1}: \langle e_n,u \rangle\in [-1,a_+]\}\) and take \(K,K'\in \mathcal{K}(\mathbb{R}^n)\) such that \(h_K\) and \(h_{K'}\) coincide on some open neighborhood of \(S_+\). Since mixed area measures are locally determined by the respective convex bodies (see 2), the signed measure \(\nu:=S_i(K,\mathbb{D};{}\cdot{})-S_i(K',\mathbb{D};{}\cdot{})\) is supported in \(\mathbb{S}^{n-1}\setminus S_+\). Hence, as by assumption \(\bar{g}(t) = t\bar{g}(1)\) for \(t \in [a_+,1]\), \[\psi_{i,g}(K)-\psi_{i,g}(K') = \int_{\mathbb{S}^{n-1}} g(u)\, d\nu(u) \\ = \int_{\mathbb{S}^{n-1}} \langle e_n,u \rangle \bar{g}(1) \, d\nu(u) = 0,\] where the last equality is due to the fact that \(\nu\) is centered as difference of centered measures. We conclude that \(\mathop{\mathrm{spt}}\psi_{i,g}\subseteq S_+\). Repeating the argument for \(S_-:=\{u\in\mathbb{S}^{n-1}: \langle e_n,u \rangle\in [a_-,1]\}\) finishes the proof. ◻

Proof. First, suppose that \(\bar{f} \in \mathcal{D}_\mathcal{C}\) is given. Then 44 shows that ?? defines a zonal valuation \(\varphi \in \mathbf{Val}_i(\mathbb{R}^n)\), which is equal to \(\psi_{i,g}\), \(g=\bar{g}(\langle e_n,\cdot \rangle)\) and \(\bar{g} = \widehat{T}_\mathcal{C}\bar{f}\). By 38, \(\bar{g} = \widehat{T}_{\mathbb{1}_{I_\mathcal{C}}} \bar g\), so 45 implies the claim on \(\mathop{\mathrm{spt}}\varphi\).

Next, suppose that \(\varphi\in\mathbf{Val}_i(\mathbb{R}^n)\) with \(\mathop{\mathrm{spt}}\varphi\subseteq \mathop{\mathrm{cls}}\mathbb{S}_{\mathcal{C}}\) is zonal. By 5, there exists \(g = \bar{g}(\langle e_n,\cdot \rangle) \in C(\mathbb{S}^{n-1})\) such that \(\varphi = \psi_{i,g}\). By 45, \(\bar g = T_{I_\mathcal{C}} \bar g\), that is \(\bar g \in T_{I_\mathcal{C}}(C[-1,1])\). Hence, by 38, there exists \(\bar f \in \mathcal{D}_\mathcal{C}\) such that \(\bar g = \widehat{T}_\mathcal{C}\bar f\). 44 then implies that the right-hand side of ?? defines a continuous valuation that coincides with \(\varphi\).

To show uniqueness of \(f\), assume that \(\varphi = 0\). Then 5 implies that \(\varphi = \psi_{i,g}\) and \(g\) is a zonal linear function restricted to \(\mathbb{S}^{n-1}\). As, by 38, \(\widehat{T}_\mathcal{C}\) is injective and direct computation shows that it maps linear functions to linear functions, we deduce that \(f\) is linear as well. ◻

Proof. By definition, \(U_{\mathcal{C}, \varepsilon}^+ = \emptyset\) if and only if \(a_+ = 1\) and \(\lim_{t \to 1}R_\mathcal{C}(t) > 0\). However, 12 shows that \(\lim_{t \to 1}R_\mathcal{C}(t) > 0\) if and only if all \(F(C_j, e_n)\) are non-degenerate disks for all \(j=1, \dots, n-i-1\). ◻

Proof. Without loss of generality, we may assume that \(E = \mathbb{R}^n\). Since \(\mu\) is a zonal measure, we have \[\int_{\mathbb{S}^{n-1}} f(u) \, \mu(du) = \int_{\mathbb{S}^{n-1}} f(\vartheta u) \, \mu(du), \qquad \vartheta \in \mathop{\mathrm{SO}}(n-1).\] Hence, averaging over all elements of \(\mathop{\mathrm{SO}}(n-1)\) yields \[\int_{\mathbb{S}^{n-1}} f(u) \, \mu(du) = \int_{\mathbb{S}^{n-1}} \breve{f}(u) \, \mu(du),\] where \[\breve{f}(u) := \int_{\mathop{\mathrm{SO}}(n-1)} f(\vartheta u)\, d\vartheta, \qquad u \in \mathbb{S}^{n-1}.\] Note that \(\breve{f}\) is a zonal function. If we show that \(\bar{\breve{f}} = \bar{f}\), then we reduce the problem to verifying (?? ) for zonal functions. To this end, for fixed \(t \in (-1,1)\) and any \(\vartheta \in \mathop{\mathrm{SO}}(n-1)\), we have \[\label{eq1:zonal95step} \int_{\mathbb{S}^{n-1} \cap H_t} f(\vartheta u)\, \sigma_t(du) = \int_{\mathbb{S}^{n-1} \cap H_t} f(u)\, \sigma_t(du),\tag{24}\] since \(\sigma_t\) is \(\mathop{\mathrm{SO}}(n-1)\)-invariant. Clearly, (24 ) and Fubini’s theorem yield \(\bar{\breve{f}} = \bar{f}\).

Now, to verify the statement for zonal functions, assume that \(f\) is zonal, and observe that \(f(u) = f(v)\) for all \(v \in \mathbb{S}^{n-1} \cap H_{\langle e_n,u \rangle}\), and so \(f(u) = \bar{f}(\langle e_n,u \rangle)\), since \(\bar{f}(\langle e_n,u \rangle)\) is precisely the average of all these values. Therefore, \[\int_{\mathbb{S}^{n-1}} f(u) \, \mu(du) = \int_{\mathbb{S}^{n-1}} \bar{f}(\langle e_n,u \rangle) \, \mu(du) = \int_{[-1,1]} \bar{f}(t) \, \bar{\mu}(dt),\] where the last equality follows from the fact that \(\bar{\mu}\) is the pushforward of \(\mu\) under the map \(u \mapsto \langle e_n,u \rangle\), and by the change of variables formula. ◻

Proof. First, note that [it:MinkChristDiskEx] clearly implies that \(\mu\) is centered. Moreover, from 25 , we see that the measure \(S_i(K, \mathbb{D}; \cdot)\) may be supported on \(\mathbb{S}^{n-2}(e_n^\perp)\) only if \(S_i^E(K|E, \cdot)\) is supported on \(\mathbb{S}^{i-1}(E \cap e_n^\perp)\) for all \(E \in \mathrm{Gr}_{i+1}(\mathbb{R}^n, e_n)\). However, as \(K|E\) is a body of revolution with axis \(e_n\), by Minkowski’s theorem, this is impossible for \(i > 1\). For \(i=1\), this implies that \(K|(e_n \vee u) \subset u^\perp\) for all \(u \in \mathbb{S}^{n-2}(e_n^\perp)\), which implies that \(K\) must be a segment – a contradiction. Hence, [it:MinkChristDiskEx] implies [it:MinkChristDiskCond].

Now assume that [it:MinkChristDiskCond] holds, and let \(E \in \mathrm{Gr}_{i+1}(\mathbb{R}^n, e_n)\) be arbitrary. If \(\mu\) is supported on \(\mathrm{span}\{e_n\}\), then choosing \[K = \left( \frac{\mu(\mathbb{S}^{n-1})}{2\kappa_{n-1}} \right)^{\frac{1}{i}} \mathbb{D}\] satisfies [it:MinkChristDiskEx], and we are done. Otherwise, by zonality, neither \(\mu\) nor \(\mu_E=(\bar{\mu})_E\) is concentrated on any great subsphere. Furthermore, by [it:MinkChristDiskCond] and ?? , the measure \(\mu_E\) is centered. Therefore, by the solution to the Minkowski problem in the subspace \(E\), there exists a full-dimensional convex body of revolution \(K_E \in \mathcal{K}(E)\) such that \(\mu_E = S_i^E(K_E, \cdot)\).

Note that \(\mu_E = \tau_\ast \mu_F\) whenever \(E, F \in \mathrm{Gr}_{i+1}(\mathbb{R}^n, e_n)\) and \(\tau \in \mathop{\mathrm{SO}}(n-1)\) satisfies \(\tau F = E\). Consequently, we have \(K_E = \tau K_F\), and so we define a convex body of revolution \(K \in \mathcal{K}(\mathbb{R}^n)\) by setting \(K|E = K_E\) for all \(E \in \mathrm{Gr}_{i+1}(\mathbb{R}^n, e_n)\). Using 25 and ?? , we obtain \[\begin{align} \int_{\mathbb{S}^{n-1}} f(u) \, d\mu(u) &= \int_{\mathbb{S}^{i}(E)} f(v) \, d\mu_E(v) = \int_{\mathbb{S}^{i}(E)} f(v) \, dS_i^E(K|E, v) \\ &= \int_{\mathrm{Gr}_{i+1}(\mathbb{R}^n,e_n)} \int_{\mathbb{S}^{i}(E)} f(u)\, dS^E_{i}(K|E,u) \, dE = \frac{\kappa_{i}}{\kappa_{n-1}} \int_{\mathbb{S}^{n-1}} f(u)\, dS_{i}(K,\mathbb{D};u), \end{align}\] for every zonal function \(f \in C(\mathbb{S}^{n-1})\). Scaling \(K\) appropriately, we have shown [it:MinkChristDiskEx].

For the uniqueness of \(K\), suppose that bodies of revolution \(K, L \in \mathcal{K}(\mathbb{R}^n)\) are given such that \(\mu = S_i(K, \mathbb{D}; {{}\cdot{}}) =S_i(L, \mathbb{D};{{}\cdot{}})\). Then, arguing as in the previous step of the proof, \[\begin{align} \int_{\mathbb{S}^{i}(E)} f(v) \, dS_i^E(K|E, v) = \frac{\kappa_{i}}{\kappa_{n-1}} \int_{\mathbb{S}^{n-1}} f(u)\, d\mu(u) = \int_{\mathbb{S}^{i}(E)} f(v) \, dS_i^E(L|E, v), \end{align}\] for all zonal \(f \in C(\mathbb{S}^{n-1})\) and \(E \in \mathrm{Gr}_{i+1}(\mathbb{R}^n, e_n)\). As \(f\) was arbitrary, we conclude that \(S_i^E(K|E, \cdot) = S_i^E(L|E, \cdot)\), and the uniqueness in the classical Minkowski problem (in \(E\)) implies that \(K|E\) and \(L|E\) are translates of each other. Since both bodies are bodies of revolution, we conclude \(K|E = L|E + c e_n\) for some \(c \in \mathbb{R}\), and, thus, again by symmetry, \(K = L + c e_n\), concluding the proof. ◻

Proof. First, let \(f \in C[a_-,a_+]\). Then, as \(R \in \mathrm{BV}_+(a_+,a_+)\), \(|\nu_R|((a_-,a_+)) < \infty\) and by 17 \[\begin{align} |T_R f(t)| \leq |R(0)f(t)| + \int_{(0,t]} |sf(t) - tf(s)| d|\nu_R|(s) \leq \left(|R(0)| + 2|\nu_R|((a_-,a_+))\right) \sup_{[a_-,a_+]}|f|, \end{align}\] whenever \(t \in (a_-,a_+)\). For \(t \geq a_+\) and \(t\leq a_-\), 25 shows that \(T_R f(t)\) is linear and, thus, can also be bounded in terms of \(\sup_{s \in [a_-,a_+]}|f(s)|\). Consequently, \(T_R:C[a_-,a_+] \to C[-1,1]\) is a bounded operator and has a well-defined adjoint \(T_R^\ast\).

Next, take some \(f\in C[a_-,a_+]\) and \(\sigma\in \mathcal{M}[-1,1]\). Then \[\begin{align} \int_{[a_-,a_+]} f(t)\, T_{\! R}^\ast\sigma(dt) = \int_{[-1,1]} T_{\! R} f(t)\, \sigma(dt). \end{align}\] We split the interval \([-1,1]=[-1,0)\cup [0,1]\). By Fubini’s theorem, we have that \[\begin{align} &\int_{[0,1]} T_{\! R} f(t)\, \sigma(dt) = \int_{[0,1]} R(t^+)f(t)\, \sigma(dt) + \int_{[0,1]} t \int_{[0,t]} f(s)\, \nu_R(ds)\, \sigma(dt) \\ &\qquad = \int_{[0,1]} R(t^+)f(t)\, \sigma(dt) + \int_{[0,1]} \int_{[s,1]} t\, \sigma(dt)\, f(s)\, \nu_R(ds), \end{align}\] and, since \(R\) and \(\nu_R\) vanish outside \([a_-,a_+]\), the desired formula holds on \([0,1]\). The argument for \([-1,0)\) is analogous.

Finally, recall that \(\nu_R\) is a signed Radon measure on \((-1,1)\) and \(|\nu_R|(\{0\})=0\), and thus, \(T_R^\ast\delta_0=R(0)\delta_0\). This immediately yields the “moreover” part of the statement. ◻

Proof. By 52 and Fubini’s theorem, for all \(t\in [0,a_+]\), \[\begin{align} &\int_{(t,a_{+}]} s\, \mu(ds) = \int_{(t,a_+]} s R(s^+)\, \sigma(ds) + \int_{(t,a_+]} \int_{[s,1]} x\, \sigma(dx)\, s\, \nu_R(ds) \\ &\qquad = \int_{(t,1]} s R(s^+)\, \sigma(ds) + \int_{(t,1]} \int_{(t,x]} s\, \nu_R(ds)\, x\, \sigma(dx). \end{align}\] Here, we used that \(R\) and \(\nu_R\) vanish on \((a_+, 1]\). Next, recall that, by 9 , \[\int_{(t,x]} s\, \nu_R(ds) = R(t^+)-R(x^+).\] Plugging this into the expression above yields the first part of the statement for \(t\in [0,a_+]\). The argument for \(t\in [a_-,0]\) is analogous.

For the “moreover” part of the lemma, observe that \[\lim_{t\nearrow a_+} \frac{\mu((t,a_+])}{R(t)} = \lim_{t\nearrow a_+} \frac{1}{R(t)} \int_{(t,a_+]} \frac{s}{a_+}\, \mu(ds) = \lim_{t\nearrow a_+} \int_{(t,1]} \frac{s}{a_+}\, \sigma(ds) = \int_{[a_+,1]} \frac{s}{a_+}\, \sigma(ds).\] The argument for the limit where \(t\searrow a_-\) is analogous. ◻

Proof. First, we take some function \(g\in C[a_-,a_+]\) and some \(a_-<a'<0<a<a_+\). Then \(\frac{1}{R} \in \mathrm{BV}_+(a',a)\) and, by 37 and 40, \(T_{\frac{1}{R}}g\in C[a',a]\), and thus, \[\begin{align} \int_{[0,a]} T_{\frac{1}{R}}g \,d\mu = \int_{[0,a]} \left[\frac{g(t)}{R(t^+)} + t\int_{(0,t]}g(s) \, \nu_{\frac{1}{R}}(ds)\right] \mu(dt). \end{align}\] By Fubini’s theorem and since \(\nu_{ \frac{1}{R}}(\{0\}) = 0\), we have that \[\begin{align} &\int_{[0,a]} t\int_{(0,t]}g(s) \, \nu_{\frac{1}{R}}(ds) \, \mu(dt) = \int_{[0,a]} g(t) \int_{[t,a]} s\, \mu(ds) \, \nu_{\frac{1}{R}}(dt) \\ &\qquad = \int_{[0,a]} g(t) \int_{[t,a_+]} s\, \mu(ds) \, \nu_{\frac{1}{R}}(dt) - \int_{(a,a_+]} s\, \mu(ds) \int_{[0,a]} g(t) \, \nu_{\frac{1}{R}}(dt). \end{align}\] Combining these equations, we obtain \[\begin{align} \label{eq:proof:lem:T95R95adjoint95inverse:I} \begin{aligned} \int_{[0,a]} T_{\frac{1}{R}}g \, d\mu &= \int_{[0,a]} g(t) \bigg[ \frac{1}{R(t^+)}\mu(dt) + \int_{[t,a_+]}s\, \mu(ds)\, \nu_{\frac{1}{R}}(dt) \bigg] \\ &\qquad - \int_{(a,a_+]} \! s\, \mu(ds) \int_{[0,a]} g(t) \, \nu_{ \frac{1}{R}}(dt). \end{aligned} \end{align}\tag{26}\]

Suppose now that \(\mu=T_R^\ast \sigma\) for some \(\sigma\in\mathcal{M}[-1,1]\). Take some function \(g\in C_c(a',a)\) with \(a_-<a'<0<a<a_+\). Then, \(T_{\frac{1}{R}}g \in C[a_-,a_+]\) and for \(t > a\) \[\begin{align} T_{\frac{1}{R}}g(t) = t \int_{(0,a]}g(s) d\nu_{ \frac{1}{R}}(s), \end{align}\] and similar for \(t<a'\), that is, \(T_{\frac{1}{R}}g\) is linear outside \([a',a]\). Consequently, \[\begin{align} \int_{[0,a]} T_{\frac{1}{R}}g \, d\mu + \int_{(a,a_+]} \! s\, \mu(ds) \int_{[0,a]} g(t) \, \nu_{ \frac{1}{R}}(dt) = \int_{[0,a]} T_{\frac{1}{R}}g \, d\mu + \int_{(a,a_+]}T_{\frac{1}{R}}g\, d\mu, \end{align}\] so repeating the argument for \([a',0]\), we obtain by 26 , \[\begin{align} \int_{[a_-,a_+]}T_{\frac{1}{R}}g\, d\mu = \int_{[a',a]} g(t) \bigg[ \frac{1}{R(t^\times)}\mu(dt) + \int_{[t,a_{\mathop{\mathrm{sgn}}t }]}s\, \mu(ds)\, \nu_{\frac{1}{R}}(dt) \bigg], \end{align}\] and \[\begin{align} \int_{[a_-,a_+]}T_{\frac{1}{R}}g\, d\mu = \int_{[-1,1]} T_R T_{\frac{1}{R}}g d\sigma = \int_{[-1,1]} g d\sigma. \end{align}\] Hence, letting \(a \nearrow a_+\) and \(a' \searrow a_-\), we obtain that \[\label{eq:proofTradjInvIntNec} \mathbb{1}_{(a_-,a_+)}(t)\sigma(dt) = \mathbb{1}_{(a_-,a_+)}(t) \left(\frac{1}{R(t^\times)} \mu(dt) + \bigg(\int_{[t,a_{\mathop{\mathrm{sgn}}t}]} \!s\, \mu(ds)\bigg) \nu_{\frac{1}{R}}(dt)\right).\tag{27}\] In particular, the expression on the right hand side is a finite signed Radon measure on \((a_-,a_+)\). In combination with 53, this shows that the conditions on \(\mu\) in the statement of the lemma are necessary. Moreover, 53 and 27 show that ?? defines a suitable preimage \(\sigma\).

Conversely, suppose now that \(\mu\) satisfies the conditions stated in the lemma and define \(\sigma\) to be the right hand side of ?? . In order to show that \(\mu=T_R^\ast \sigma\), we need to show that \[\begin{align} \int_{[a_-,a_+]} f d\mu = \int_{[-1,1]} T_R f d\sigma = \int_{[a_-,a_+]} T_R f d\sigma \end{align}\] holds for all \(f \in C[a_-,a_+]\). To this end, fix \(f \in C[a_-,a_+]\) and write \(g = T_R f\). Then, by 37 and 40, \(g \in C[a_-,a_+]\) and \(f = T_{\frac{1}{R}}g\). Hence, we need to show that \[\begin{align} \int_{[a_-,a_+]} T_{\frac{1}{R}}g d\mu = \int_{[a_-,a_+]} g d\sigma. \end{align}\] As by 25, \(T_{\mathbb{1}_{(-1,a)}}T_{\frac{1}{R}}g\) converges uniformly to \(T_{\frac{1}{R}}g\) on \([0,a_+]\) as \(a \nearrow a_+\), 26 implies that \[\begin{align} \int_{[0,a_+]} T_{\frac{1}{R}}g d\mu &= \lim_{a \nearrow a_+}\int_{[0,a_+]}T_{\mathbb{1}_{(-1,a)}}T_{\frac{1}{R}}g d\mu\\ &=\lim_{a \nearrow a_+} \left(\int_{[0,a]}T_{\frac{1}{R}}g d\mu + \int_{(a,a_+]}s d\mu(s) \frac{T_{\frac{1}{R}}g(a)}{a} \right)\\ &=\int_{[0,a_+)} g(t) d\sigma + \lim_{a \nearrow a_+}\int_{(a,a_+]}s d\mu(s) \left( -\int_{[0,a]} g(t) \, \nu_{ \frac{1}{R}}(dt) + \frac{T_{\frac{1}{R}}g(a)}{a}\right). \end{align}\] Plugging in the definition of \(T_{\frac{1}{R}}g\) inside the brackets, \[\begin{align} -\int_{[0,a]} g(t) \, \nu_{ \frac{1}{R}}(dt) + \frac{T_{\frac{1}{R}}g(a)}{a} = \frac{g(a)}{a R(a^+)}, \end{align}\] and using the definition of \(\sigma\) in ?? then yields \[\begin{align} \int_{[0,a_+]} T_{\frac{1}{R}}g d\mu &= \int_{[0,a_+)} g(t) d\sigma + \lim_{a \nearrow a_+}\frac{\int_{(a,a_+]}s d\mu(s)}{a R(a^+)} g(a)\\ &=\int_{[0,a_+)} g(t) d\sigma + \left(\lim_{a \nearrow a_+}\frac{\mu((a,a_+])}{R(a^+)}\right) g(a_+) = \int_{[0,a_+]} g(t) d\sigma. \end{align}\] Repeating the argument for \([a_-,0]\) then yields the claim \(\mu = T_R^\ast \sigma\). ◻

Proof. It follows directly from 52 that \(\widehat{T}_R\) is a bounded operator. To determine its adjoint, note that \[\begin{align} \int_{[a_-,a_+]} f d(\widehat{T}_R^\ast \sigma) = \int_{[-1,1]} (T_R f)(s) + c f(0) |s| d\sigma(s), \end{align}\] which immediately yields the remaining claims. ◻

Proof. By 55, using that the absolute value vanishes at \(t=0\), \[\begin{align} \int_{[a_-,a_+]} |\cdot | d \mu = \int_{[a_-,a_+]} |\cdot| d\bigg(T_R^\ast \sigma+ c \int_{[-1,1]} |s| d\sigma(s) \delta_0\bigg) = \int_{[-1,1]} T_R( |\cdot|) d\sigma. \end{align}\] Noting that by 17 , \(T_R(|\cdot|) = R(0) |\cdot|\) then yields the claim. ◻

Proof. By 55 and 56, \(\mu = \widehat{T}_R^\ast \sigma\), if and only if \[\begin{align} \mu = T_R^\ast \sigma + \left(c \int_{[-1,1]}|\cdot| d\sigma\right) \delta_0 = T_R^\ast \sigma + \left(\frac{c}{R(0)} \int_{[a_-,a_+]} |\cdot | d \mu\right) \delta_0. \end{align}\] 54 thus yields the claim, noting that the conditions remain essentially unchanged when adding mass at \(t=0\) to \(\mu\). ◻

Proof. Suppose that \(\mu=S_i(K,\mathcal{C};{}\cdot{})\), where \(K \in \mathcal{K}(\mathbb{R}^n)\) is either a full-dimensional body of revolution or \(K=\mathbb{D}\). Then 41 directly implies condition [thm:Christoffel95Minkowski95zonal:support]. To see [thm:Christoffel95Minkowski95zonal:concentr], note that by Schneider2014?*Thm. 5.1.8 \[\begin{align} \int_{\mathbb{S}^{n-1}} \lvert \langle e_n,u \rangle \rvert \, d\mu(u) = n V(K^{[i]},\mathcal{C},[-e_n,e_n]) > 0, \end{align}\] as we can find linearly independent segments in \(K, \dots, K, C_1, \dots, C_{n-i-1}\) spanning \(e_n^\perp\). Consequently, as \(\lvert \langle e_n,\cdot \rangle \rvert\) vanishes only on \(\mathbb{S}^{n-2}(e_n^\perp)\), \(\mu\) cannot be concentrated on \(\mathbb{S}^{n-2}(e_n^\perp)\).

Next, note that by 27, \[\begin{align} \int_{[a_{\mathcal{C},-},a_{\mathcal{C},+}]}\bar f d\bar{\mu} = \int_{\mathbb{S}^{n-1}} f \, dS_i(K,\mathcal{C};{}\cdot{}) = \int_{\mathbb{S}^{n-1}} (\widehat{T}_\mathcal{C}\bar{f})(\langle e_n,\cdot \rangle) \, dS_i(K,\mathbb{D};{}\cdot{}), \end{align}\] for all zonal \(f \in C(\mathbb{S}^{n-1})\), that is, \(\bar{\mu} = \widehat{T}_\mathcal{C}^\ast \sigma\), where \(\sigma = (\langle e_n,\cdot \rangle)_\ast S_i(K,\mathbb{D},{}\cdot{}) \in \mathcal{M}[-1,1]\). Hence, 57 and 54 imply [thm:Christoffel95Minkowski95zonal:LimEx]. Moreover, by 54, \[\begin{align} \label{eq:prfNecCondChri} \mathbb{1}_{(a_{\mathcal{C},-},a_{\mathcal{C},+})}(t)\sigma(dt) = \frac{1}{R_\mathcal{C}(t)} \bar{\mu}(dt) + \bigg(\int_{[t,a_{\mathop{\mathrm{sgn}}(t)}]} \!s\, \bar{\mu}(ds)\bigg) \nu_{\frac{ 1}{R_\mathcal{C}}}(dt). \end{align}\tag{28}\] Since \(\sigma\) is non-negative and finite, this yields condition [mthm:Christoffel95Minkowski95zonal:measPos].

For condition [thm:Christoffel95Minkowski95zonal:AtEquat], note that by 3 and 5 , \[\int_{\mathbb{S}^{n-1}} \!\! \max\{0,\langle e_n,u \rangle\}\, \mu(du) = nV(K^{[i]},\mathcal{C},[0,e_n]) = V^{e_n^\perp}\!((K|e_n^\perp)^{[i]},\mathcal{C}|e_n^\perp) = \kappa_{n-1} R_K(0)^{i}R_{\mathcal{C}}(0).\] Moreover, by 15 and ?? , \[\begin{align} &\mu(\mathbb{S}^{n-1}\cap e_n^\perp) = S_i(K,\mathcal{C};\mathbb{S}^{n-1}\cap e_n^\perp) = W_{(K^{[i]},\mathcal{C})} \\ &\qquad = \frac{\omega_{n-1}}{n-1} \bigg( \frac{n-i-1}{\omega_{n-i-1}}R_K(0)^{i}W_{\mathcal{C}} + i \ell_KR_K(0)^{i-1}R_{\mathcal{C}}(0) \bigg) \\ &\qquad \geq \frac{\omega_{n-1}}{n-1} \frac{n-i-1}{\omega_{n-i-1}} R_K(0)^{i}W_{\mathcal{C}} = \frac{n-i-1}{\omega_{n-i-1}} \frac{W_{\mathcal{C}}}{R_{\mathcal{C}}(0)}\int_{\mathbb{S}^{n-1}} \max\{0,\langle e_n,u \rangle\}\, \mu(du). \end{align}\] Multiplying both sides with \(R_{\mathcal{C}}(0)\) yields condition [thm:Christoffel95Minkowski95zonal:AtEquat]. ◻

Proof. Let \(\mu \in \mathcal{M}(\mathbb{S}^{n-1})\) be a non-negative, centered, zonal Borel measure on \(\mathbb{S}^{n-1}\) satisfying conditions [thm:Christoffel95Minkowski95zonal:support] to [thm:Christoffel95Minkowski95zonal:AtEquat] in 58. Then \(\bar{\mu} = (\langle e_n,\cdot \rangle)_\ast \mu\) is a non-negative Borel measure concentrated on \([a_{\mathcal{C},-},a_{\mathcal{C},+}]\), that is, \(\bar{\mu} \in \mathcal{M}[-a_{\mathcal{C},-},a_{\mathcal{C},+}]\). The conditions [thm:Christoffel95Minkowski95zonal:measPos], [thm:Christoffel95Minkowski95zonal:LimEx] and [thm:Christoffel95Minkowski95zonal:AtEquat] assert that the conditions of 57 are satisfied so that there exists \(\sigma \in \mathcal{M}[-1,1]\) such that \(\bar{\mu} = \widehat{T}_\mathcal{C}^\ast \sigma\). Moreover, [thm:Christoffel95Minkowski95zonal:measPos] and [thm:Christoffel95Minkowski95zonal:AtEquat] imply that we can choose \(\sigma \geq 0\).

By 53 and since \(\mu\) is assumed to be centered, \[\begin{align} 0 = \int_{[-1,1]} s d\bar{\mu}(s) = R(0)\int_{[-1,1]} s d\sigma(s), \end{align}\] that is, \(\sigma\) is centered as well. Next, assume that \(\sigma\) is concentrated on \(\{0\}\). Then, by 52, \(\bar{\mu}\) is concentrated on \(\{0\}\), and thus \(\mu\) is concentrated on \(\mathbb{S}^{n-2}(e_n^\perp)\). This contradicts [thm:Christoffel95Minkowski95zonal:concentr], hence, \(\sigma\) is not concentrated on \(\{0\}\).

We conclude that the zonal measure \(\widetilde{\sigma}\) on \(\mathbb{S}^{n-1}\) uniquely determined by \((\langle e_n,\cdot \rangle)_\ast \widetilde{\sigma} = \sigma\) satisfies the conditions of 51, that is, there exists a body of revolution \(K \in \mathcal{K}(\mathbb{R}^n)\) such that \(\widetilde{\sigma} = S_i(K, \mathbb{D}, \cdot)\), and \(K\) is not a segment.

Finally, let \(f=\bar{f}(\langle e_n,\cdot \rangle) \in C(\mathbb{S}^{n-1})\) and write \(g = \bar{g}(\langle e_n,\cdot \rangle) \in C(\mathbb{S}^{n-1})\) with \(\bar g = \widehat{T}_\mathcal{C}\bar{f}\). Then, as \(\bar \mu = \widehat{T}_\mathcal{C}^\ast \sigma\) and by 27, \[\begin{align} \int_{\mathbb{S}^{n-1}} f d\mu = \int_{[a_{\mathcal{C},-},a_{\mathcal{C},+}]} \bar f d\bar{\mu} = \int_{[-1,1]} \bar{g} d\sigma = \int_{\mathbb{S}^{n-1}} g\,dS_i(K, \mathbb{D}; \cdot) = \int_{\mathbb{S}^{n-1}} f\,dS_i(K, \mathcal{C}; \cdot). \end{align}\] As \(f\) was arbitrary, we deduce that \(\mu = S_i(K, \mathcal{C}, \cdot)\), concluding the proof. ◻

Proof of 8. Observe that by 9 , it holds for \(t > 0\) that \[\begin{align} |R(t^+) - R((-t)^+)| \leq \int_{(-t,t]}|s|d\lvert \nu_R \rvert(s) \leq t \lvert \nu_R \rvert([-t,t]). \end{align}\] As \(\lvert \nu_R \rvert\) is a Radon measure, \(\lvert \nu_R \rvert([-t,t]) \to \lvert \nu_R \rvert(\{0\}) = 0\), as \(t\to 0^+\), that is, \(R(0^+) = R(0^-)\). Similarly, \[\begin{align} \left|\frac{R(t^+) - R(0^+)}{t}\right| + \left|\frac{R(0^+) - R((-t)^+)}{t}\right| \leq \frac{1}{t} \int_{(0,t] \cup (-t,0]} \lvert s \rvert\,d\lvert \nu_R \rvert(s) \leq \lvert \nu_R \rvert([-t,t]) \end{align}\] yields that both difference quotients on the left-hand side converge to zero, that is, \(R\) is differentiable at \(t=0\) with \(R'(0) =0\). ◻

Proof of 9. [BV95095algebraic:linear95comb] and [BV95095algebraic:unit] are clear from the definition. For [BV95095algebraic:product], first note that as \[\begin{align} (R(t)-R(0))(Q(t) - Q(0)) = R(t)Q(t) - R(0)Q(t) - Q(0)R(t) + R(0)Q(0), \end{align}\] by [BV95095algebraic:linear95comb] and [BV95095algebraic:unit], we can assume that \(R(0) = 0 = Q(0)\). Now, restricting first to integrals over the part \((0,1)\), 9 and Fubini’s theorem imply for every \(\phi \in C^1_c(-1,1)\) \[\begin{align} \int_{(0,1)} \phi'(t)R(t)Q(t)\, dt = -\int_{(0,1)} \phi'(t) Q(t) \int_{(0,t]}s \, d\nu_R(s)\, \, dt = -\int_{(0,1)} \int_{[s,1)} \phi'(t)Q(t)\, dt \, s \, d\nu_R(s). \end{align}\] Note here, that we can replace \(R(t)\) by \(R(t^+)\) in the first integral without changing its value. Applying 10 to the inner integral, we obtain, as \(\mathop{\mathrm{spt}}\phi \subset (-1,1)\) is compact, \[\begin{align} \int_{[s,1)} \phi'(t) Q(t) \, dt = -\phi(s)Q(s^-) + \int_{[s,1)} \phi(t) t\, d\nu_Q(t), \end{align}\] which yields in total \[\begin{align} \int_{(0,1)} \phi'(t)R(t)Q(t)\, dt = \int_{(0,1)} \phi(s) Q(s^-) s \, d\nu_R(s) - \int_{(0,1)} \int_{[s,1)} \phi(t) t\, d\nu_Q(t)\, s \, d\nu_R(s). \end{align}\] Exchanging again the order of integration in the last integral, we obtain, by 9 again, \[\begin{align} \int_{(0,1)} \int_{[s,1)} \phi(t) t\, d\nu_Q(t)\, s \, d\nu_R(s) = \int_{(0,1)} \int_{(0,t]}\, s \, d\nu_R(s)\, \phi(t) t\, d\nu_Q(t) = -\int_{(0,1)} \phi(t) t R(t^+)\, d\nu_Q(t), \end{align}\] yielding the claim for \((0,1)\). A similar argument for \((-1,0]\) then yields the first equality of [BV95095algebraic:product]. The second equality follows directly by exchanging \(R\) and \(Q\).

For [BV95095algebraic:inverse], note that if \(R\) is of locally bounded variation and \(\inf_I \lvert R \rvert > 0\) for every compact \(I \subset (-1,1)\) then \(1/R\) is also of locally bounded variation (and semi-continuous). By [BV95095algebraic:unit] and a similar argument as in the proof of [BV95095algebraic:product] for \(\mu_R\) instead of \(\nu_R\), \[\begin{align} 0 = d\mu_{R\cdot\frac{1}{R}}(t) = \frac{1}{R(t^-)} d\mu_R(t) + R(t^+)d\mu_{\frac{1}{R}}(t), \end{align}\] that is, \(d\mu_{\frac{1}{R}}(t) = \frac{-1}{R(t^-)R(t^+)}d\mu_R(t)\). Consequently, \(d\nu_{\frac{1}{R}}(t) = -\frac{1}{t}d\mu_{\frac{1}{R}}(t) = \frac{-1}{R(t^-)R(t^+)}d\nu_R(t)\) is a signed Radon measure, that is, \(1/R \in \mathrm{BV}_0(-1,1)\), concluding the proof. ◻

Proof of 24. For [it:T95R95cont95to95cont], let \(0\leq t' < t < 1\). By 17 , we obtain \[\begin{align} T_Rf(t)-T_Rf(t') &= R(0)(f(t)-f(t')) - \int_{(t',t]} (sf(t)-tf(s) )\, d\nu_R(s) \\ &\qquad - \int_{(0,t']} (s(f(t)-f(t'))-(t-t')f(s))\, d\nu_R(s). \end{align}\] As \(R\) is bounded, \(\nu_R\) is a Radon measure and \(f\) is continuous on \((-1,1)\) (whence \(f\) and \(s \mapsto sf(t)-tf(s)\) are bounded locally), the right-hand side becomes arbitrarily small whenever \(|t-t'|\) is small, showing continuity of \(T_R f\) on \([0,1)\). A similar argument shows continuity on \((-1,0]\), so \(T_Rf(0^-) = R(0)f(0) = T_Rf(0^+)\) yields the claim.

Next, the second part of [it:T95R95group95property] is an immediate consequence of ?? , 9[BV95095algebraic:inverse], and the fact that \(T_1 = \mathrm{id}\). To show ?? , let \(f\in C(-1,1)\) and note first that, by 9 , \[\begin{align} \label{eq:prfTRgroup95firstid} sT_Qf(t) - tT_Qf(s) =& Q(0)(sf(t) - tf(s)) - \int_{(0,s]} sxf(t) - stf(x) -(txf(s)-tsf(x)) d\nu_Q(x)\nonumber\\& - s\int_{(s,t]} xf(t) - tf(x) d\nu_Q(x)\nonumber\\ =& Q(x^+)(sf(t) - tf(s)) - s\int_{(s,t]} xf(t) - tf(x) d\nu_Q(x) \end{align}\tag{29}\] for all \(0 < s \leq t < 1\). Moreover, Fubini’s theorem and 9 imply for \(t \in (0,1]\) that \[\begin{align} \label{eq:prfTRgroup95secid} \int_{(0,t]}s\int_{(s,t]} xf(t) - tf(x) d\nu_Q(x) d\nu_R(s) &= \int_{(0,t]} (xf(t) - tf(x)) \int_{(0,x)} s d\nu_R(s) d\nu_Q(x)\nonumber\\ &=-\int_{(0,t]} (xf(t) - tf(x)) (R(x^-) - R(0))d\nu_Q(x). \end{align}\tag{30}\] Combining 29 and 30 with the definition of \(T_R\), then yields for all \(t\in [0,1)\), \[\begin{align} T_R (T_Q f)(t) &= R(0) T_Qf(t) - \int_{(0,t]} (sT_Qf(t) - t T_Q f(s)) d\nu_R(s)\\ &=R(0) Q(0)f(t) - \int_{(0,t]} (xf(t) - tf(x)) \left(Q(x^+) d\nu_R(x) + R(x^-)d\nu_Q(x)\right). \end{align}\] Thus, by 9[BV95095algebraic:product], \(T_R(T_Q f)(t)=T_{RQ}f(t)\) for all \(t \in [0,1)\). By 63, the claim also follows for \(t \in (-1,0)\). For [it:continuous95in95D95R], finally, let \(f\in C[-1,1]\). Then clearly, the limits \(\lim_{t\to\pm 1} R(t)f(t)\) exist. Moreover, for any given \(\varepsilon>0\), there exists \(t_0\in (0,1)\) such that \(\lvert f(t)-f(1) t \rvert<\varepsilon\) for all \(t\in (t_0,1)\). Due to the fact that \(R\) is of bounded variation, the signed measure \(\nu_R\) is finite, so for all \(t\in (t_0,1)\), \[\begin{align} \bigg| \int_{(t_0,t]} f\, d\nu_R - f(1)(R(t_0^+)-R(t^+)) \bigg| = \bigg| \int_{(t_0,t]} (f(t)-f(1)t) \, \nu_R(dt) \bigg| \leq \lvert \nu_R \rvert(-1,1) \, \varepsilon. \end{align}\] This argument shows that whenever \((t_n)_n\) is a sequence in \((-1,1)\) that tends to \(1\), then \(\int_{(0,t_n]}f\, d\nu_R + f(1)R(t_n^+)\) yields a Cauchy sequence. By our assumption on \(R\), this implies that the limit \(\lim_{t\to 1} \int_{(0,t]} f\, d\nu_R\) exists. The argument for the corresponding limit at \(t=-1\) is completely analogous, concluding the proof. ◻

Proof of 64. Clearly, \(R \circ z\) is again a function of bounded variation. Moreover, a short calculation substituting \(u=z(s)\) yields \[\begin{align} \int_{(-1,1)} \phi'(s) (R \circ z)(s) ds = \int_{(a_{-},a_+)}(\phi \circ z^{-1})'(u)R(u) du = \int_{(a_-,a_+)} (\phi \circ z^{-1})(u) u d\nu_R(u) \end{align}\] for every \(\phi \in C^1_c(-1,1)\). Consequently, \(R \circ z \in \mathrm{BV}_0(-1,1)\) with \[\begin{align} \label{eq:prfTransfNuRz} \nu_{R \circ z} = \left(|a_{-}| \mathbb{1}_{(-1,0)} + a_+ \mathbb{1}_{(0,1)}\right)(z^{-1})_\ast \nu_R, \end{align}\tag{31}\] which directly implies that \(R \circ z \in \mathrm{BV}_+(-1,1)\) as well. Next, 17 implies that for \(t > 0\) \[\begin{align} T_{R \circ z}(f\circ z)(t) &= (R \circ z)(0) (f\circ z)(t) - \int_{(0,t]} s (f\circ z)(t) - t (f\circ z)(s) d\nu_{R \circ z}(s) \\ &=R(0)f(z(t)) - \int_{(0,z(t)]} u f(z(t)) - z(t) f(u) d\nu_{R}(u) = (T_R f)(z(t)). \end{align}\] A similar argument for \(t < 0\) then yields the second claim. The last claim, finally, follows directly from 31 and the definition of \(\mathcal{D}_R\). ◻

Proof of 37. First, observe that the map \(J_I : C[a_-,a_+]\to T_{\mathbb{1}[I]}(C[-1,1])\) is bijective. Indeed, if \(\tilde{g}\in C[a_-,a_+]\), then \(\left. (J_I\tilde{g}) \right|_{\mathop{\mathrm{cls}}I}=\tilde{g}\), and conversely, if \(g\in T_{\mathbb{1}[I]}(C[-1,1])\), then \(J_I(\left. g \right|_{\mathop{\mathrm{cls}}I})=T_{\mathbb{1}[I]}g=g\). Hence, it suffices to show that the map \(\left. T_R \right|_{I}: \mathcal{D}_R\to C[a_-,a_+]\) is bijective.

Thus, the map \(\left. T_R \right|_{I}: \mathcal{D}_R\to C[a_-,a_+]\) is bijective, which concludes the argument. ◻

Proof of 39. We show claim [it:D95R95endpoint95limits95zero] first for \(I = (-1,1)\). By 63, we can assume without loss of generality that \(a = 1\). Moreover, if \(-\nu_R \geq 0\), \(R\) is positive and monotonically increasing on \((0,1)\) and so \(\lim_{t \to 1}R(t)\) can never be zero. Hence, the assumptions imply that we must have \(\nu_R \geq 0\) in [it:D95R95endpoint95limits95zero].

For [it:D95R95endpoint95limits95zero], let \(0 < t_0 < t < 1\), where we pick \(t_0\) to be a continuity point of \(R\), that is, \(R(t_0^-) = R(t_0^+) = R(t_0)\). Since by assumption \(R\) is positive and \(\nu_R \geq 0\) on \([0,1)\), we can estimate \[\int_{[t_0,t)} f\, d\nu_{R} = \int_{[t_0,t)} \frac{R(s)f(s)}{s} \frac{s}{R(s)}\, \nu_{R}(ds) \geq \frac{1}{t_0}\, \inf_{s \in [t_0,t)} \left\{R(s)f(s)\right\} \, \int_{[t_0,t)}\frac{s}{R(s)}\, \nu_{R}(ds).\] Again, as \(\nu_R\) is non-negative and, thus, \(R\) is decreasing on \([0,1)\), by 9 , \[\begin{align} \int_{[t_0,t)}\frac{s}{R(s)}\, \nu_{R}(ds) \geq \frac{1}{R(t_0)} \int_{[t_0,t)}s\, \nu_{R}(ds) = \frac{R(t_0^-) - R(t^-)}{R(t_0)} \geq 1 - \frac{R(t^-)}{R(t_0)}, \end{align}\] where, for fixed \(t_0\), the last term tends to one as \(t\) tends to zero, using that \(\lim_{t\to 1} R(t) = 0\). Consequently, we obtain for \(t < 1\) large enough \[\begin{align} \inf_{s \in [t_0,1)} \left\{R(s)f(s)\right\} \leq \inf_{s \in [t_0,t)} \left\{R(s)f(s)\right\} \leq 2t_0 \int_{[t_0,t)} f\, d\nu_{R}. \end{align}\] As \(f \in \mathcal{D}_R\) and, thus, \(|\int_{[0,1)} f\, d\nu_{R}| < \infty\), this implies that \[\begin{align} \liminf_{s\to 1}\left\{R(s)f(s)\right\} \leq 2 \liminf_{s\to 1} \int_{[s,1)} f\, d\nu_{R} = 0. \end{align}\] Since actually the limit of the left-hand side exists, we conclude that \(\lim_{s \to 1} R(s)f(s) \leq 0\). Noting that with \(f \in \mathcal{D}_R\), we also have \(-f \in \mathcal{D}_R\), we can repeat the argument to obtain \(\lim_{s \to 1} R(s)f(s) = 0\), concluding the proof of [it:D95R95endpoint95limits95zero] for \(I = (-1,1)\).

For general \(I \subset (-1,1)\), take \(f\in\mathcal{D}_R\). By 64, we have that \(R\circ z\in\mathrm{BV}_+(-1,1)\) and \(f\circ z\in\mathcal{D}_{R\circ z}\). Since \(\lim_{t\to \mathrm{sgn} (a)}(R\circ z)(t)=\lim_{t\to a}R(t)=0\), by the first part, \[\lim_{t\to a} R(t)f(t) = \lim_{t\to \mathrm{sgn}(a) } (R\circ z)(t)(f\circ z)(t) = 0,\] which concludes the argument.

For [it:D95R95endpoint95limits95nzero], finally, we note that as \(f \in \mathcal{D}_R\), by definition, \[\begin{align} \lim\limits_{t \to a} R(t) f(t) \end{align}\] exists. As \(R(t) > 0\) and \(\lim_{t \to a}R(t) > 0\), however, this directly implies that \[\begin{align} \lim\limits_{t \to a} f(t) = \frac{\lim\limits_{t \to a} R(t) f(t)}{\lim\limits_{t \to a} R(t)} \end{align}\] exists, concluding the proof. ◻

Proof of 38. First note that \(\widehat{T}_R\) clearly is well-defined. Moreover, since by 17 , \[\begin{align} (T_R f)(0) = R(0) f(0), \end{align}\] by 37, the inverse of \(\widehat{T}_R\) is given by \[\begin{align} (\widehat{T}_R)^{-1}g = T_R^{-1}\left(g - c\frac{g(0)}{R(0)} |\cdot|\right) = T_R^{-1}g - c\frac{g(0)}{R(0)^2}|\cdot|. \end{align}\] Here, we used that, by assumption, \(R\) is strictly positive on \(I\) and that \(T_{\mathbb{1}_I}|\cdot| = |\cdot|\), that is, \(|\cdot| \in T_{\mathbb{1}_I}(C[-1,1])\). ◻