Proof. We use standard identities for support functions. For convex bodies ${A,B\subset \mathbb R^2}$ and $t\ge 0$ , $ h_{A\oplus B}=h_A+h_B, h_{tA}=th_A, h_{R_\theta A}(\varphi )=h_A(\varphi -\theta ). $ These identities and the representation formulae below are classical (see Schneider [Reference Schneider17]). The Steiner point admits the representation

$$ \begin{align*} s(A)=\frac{1}{\pi}\int_0^{2\pi}\!u(\varphi)\,h_A(\varphi)\,d\varphi,\quad u(\varphi)=(\cos\varphi,\sin\varphi), \end{align*} $$

hence, it is Minkowski linear and rotation equivariant, and

$$ \begin{align*} s(A\oplus B)=s(A)+s(B),\quad s(tA)=t\,s(A),\quad s(R_\theta A)=R_\theta s(A). \end{align*} $$

Likewise, the mean width can be written as

$$ \begin{align*} w(A)=\frac{1}{\pi}\int_0^{\pi}(h_A(\varphi)+h_A(\varphi+\pi))\,d\varphi, \end{align*} $$

so w is Minkowski linear and rotation invariant:

$$ \begin{align*} w(A\oplus B)=w(A)+w(B),\quad w(tA)=t\,w(A),\quad w(R_\theta A)=w(A). \end{align*} $$

Apply these to $K'=(1-\lambda )K\oplus \lambda R_\theta K$ . Since $K\in \mathcal K_*$ has $s(K)=0$ and $w(K)=1$ ,

$$ \begin{align*} s(K')&=(1-\lambda)s(K)+\lambda\,s(R_\theta K)=(1-\lambda)\cdot 0+\lambda\,R_\theta s(K)=0,\\ w(K')&=(1-\lambda)w(K)+\lambda\,w(R_\theta K)=(1-\lambda)\cdot 1+\lambda\cdot 1=1. \end{align*} $$

Thus, $K'\in \mathcal K_*$ . Passing to support functions, the step is exactly $h\mapsto (1-\lambda )h+\lambda S_\theta h$ and the above computation shows that it preserves $w\equiv 1$ .

Proof. For $e_k(\varphi )=e^{ik\varphi }$ , we compute directly

$$ \begin{align*} S_\theta e_k(\varphi)=e_k(\varphi-\theta)=e^{ik(\varphi-\theta)}=e^{-ik\theta}e^{ik\varphi}=e^{-ik\theta}e_k(\varphi), \end{align*} $$

hence,

$$ \begin{align*} T e_k=(1-\lambda)e_k+\lambda S_\theta e_k=((1-\lambda)+\lambda e^{-ik\theta})e_k=\mu_k e_k. \end{align*} $$

By linearity and completeness of the Fourier basis in $L^2(\mathbb T)$ , if $h=\sum c_k e^{ik\varphi }$ , then $Th=\sum \mu _k c_k e^{ik\varphi }$ and iterating yields $T^n h=\sum \mu _k^{\,n} c_k e^{ik\varphi }$ .

It remains to characterise $|\mu _k|$ . For any real $\alpha $ ,

$$ \begin{align*} |(1-\lambda)+\lambda e^{-i\alpha}|^2=(1-\lambda)^2+\lambda^2+2\lambda(1-\lambda)\cos\alpha =1-2\lambda(1-\lambda)(1-\cos\alpha)\le 1, \end{align*} $$

with equality if and only if $\cos \alpha =1$ , that is, $\alpha \in 2\pi \mathbb Z$ . Taking $\alpha =k\theta $ gives $|\mu _k|<1$ exactly when $k\theta \notin 2\pi \mathbb Z$ , and $\mu _k=1$ exactly when $k\theta \in 2\pi \mathbb Z$ . Since $e^{0\cdot i\varphi }\equiv 1$ , we also have $\mu _0=(1-\lambda )+\lambda =1$ , which corresponds to preservation of the mean (and hence of the mean width).

Proof. By Lemma 2.1, on $\mathcal K_*$ , the one-step map is $K\mapsto (1-\lambda )K\oplus \lambda R_\theta K$ and, on support functions, $h\mapsto Th=(1-\lambda )h+\lambda S_\theta h$ . Write the Fourier expansion $h_0(\varphi )=\sum _{k\in \mathbb Z}c_k e^{ik\varphi }$ in $L^2(\mathbb T)$ . Lemma 2.2 yields $ T e_k=\mu _k e_k,\; \mu _k=(1-\lambda )+\lambda e^{-ik\theta }, $ hence,

$$ \begin{align*} T^n h_0=\sum_{k\in\mathbb Z}\mu_k^{\,n} c_k e^{ik\varphi}\quad\text{in }L^2(\mathbb T). \end{align*} $$

For case (i), if $\theta /2\pi \notin \mathbb Q$ , then $k\theta \notin 2\pi \mathbb Z$ for all $k\ne 0$ , so $|\mu _k|<1$ by Lemma 2.2. Therefore, $\mu _k^{\,n}\to 0$ for $k\ne 0$ , while $\mu _0=1$ . It follows that $T^n h_0\to c_0 e^{i0\varphi }=\bar h$ in $L^2(\mathbb T)$ . Since the iteration preserves the mean width $w\equiv 1$ (Lemma 2.1), we have $\bar h=\frac 12$ , which is the support function of the unit mean width disk. This identifies the limit shape.

For case (ii), let $\theta =2\pi p/q$ with $(p,q)=1$ . Then, $\mu _k=1$ exactly when $k\equiv 0\pmod q$ , and $|\mu _k|<1$ otherwise (Lemma 2.2). Hence, $T^n h_0\to \sum _{m\in \mathbb Z}c_{mq} e^{i(mq)\varphi }$ in $L^2(\mathbb T)$ . The operator

$$ \begin{align*} P_q=\frac1q\sum_{j=0}^{q-1}S_{j\theta} \end{align*} $$

acts on Fourier modes by $P_q e_k=e_k$ if $k\equiv 0\pmod q$ and $P_q e_k=0$ otherwise, so indeed, $T^n h_0\to P_q h_0$ in $L^2(\mathbb T)$ . By Minkowski linearity of support functions and rotation covariance,

$$ \begin{align*} h_{(\frac1q\sum_{j=0}^{q-1}R_{j\theta}K_0)}=P_q h_0, \end{align*} $$

while s and w are preserved under this averaging (as in Lemma 2.1), so the geometric limit is exactly

$$ \begin{align*} K_\infty=\frac1q\sum_{j=0}^{q-1}R_{j\theta}K_0\in\mathcal K_*. \end{align*} $$

Uniqueness follows from the $L^2$ convergence of $T^n h_0$ to the indicated limit and linearity.

Proof. Write the Fourier expansion $h_0(\varphi )=\sum _{k\in \mathbb Z}c_k e^{ik\varphi }$ in $L^2(\mathbb T)$ . By Lemma 2.2, $ \mu _k=(1-\lambda )+\lambda e^{-ik\theta },\;T e_k=\mu _k e_k, $ hence, $T^n h_0=\sum _k \mu _k^{\,n} c_k e^{ik\varphi }$ .

1. (i) Resonant. If $\theta =2\pi p/q$ , then $\mu _k=1$ exactly when $k\equiv 0\pmod q$ and otherwise, $|\mu _k|\le \rho <1$ . Since $P_q e_k=e_k$ for $k\equiv 0\pmod q$ and $0$ otherwise,

   $$ \begin{align*} T^n h_0-P_q h_0=\sum_{k\not\equiv 0\,(q)}\mu_k^{\,n} c_k e^{ik\varphi}. \end{align*} $$

   Parseval gives the $L^2$ estimate. If $h_0\in C^\alpha $ , then $|c_k|\le C_\alpha \|h_0\|_{C^\alpha }|k|^{-(1+\alpha )}$ , so $\sum _k |c_k|<\infty $ and

   $$ \begin{align*} \|T^n h_0-P_q h_0\|_{L^\infty}\ \le\ \sum_{k\not\equiv 0\,(q)}|\mu_k|^{n}|c_k| \ \le\ \rho^{\,n}\sum_{k\not\equiv 0\,(q)}|c_k| \ \le\ C_\alpha\,\rho^{\,n}\,\|h_0\|_{C^\alpha}. \end{align*} $$

2. (ii) Hölder. For any real x, set $t:=2\lambda (1-\lambda )(1-\cos x)\in [0,4\lambda (1-\lambda )]$ . Then,

   $$ \begin{align*} |(1-\lambda)+\lambda e^{-ix}|^2=1-t,\quad |(1-\lambda)+\lambda e^{-ix}|=\sqrt{1-t}\le e^{-t/2}, \end{align*} $$
   and hence,
   $$ \begin{align*} |(1-\lambda)+\lambda e^{-ix}|\le \exp(-\lambda(1-\lambda)(1-\cos x)) \end{align*} $$
   and $1-\cos x\ge ({2}/{\pi ^2})\operatorname {dist}(x,2\pi \mathbb Z)^2$ . With the Diophantine bound, this yields
   $$ \begin{align*} |\mu_k|^n\ \le\ \exp(-a_0\,n\,|k|^{-2\tau}),\quad a_0=a_0(\lambda,c)>0. \end{align*} $$

   Hence,

   $$ \begin{align*} \|T^n h_0-\bar h\|_{L^\infty}\ \le\ \sum_{k\ne 0}\! e^{-a_0 n |k|^{-2\tau}}\,|c_k|. \end{align*} $$

   Split at a cutoff $K\ge 1$ :

   $$ \begin{align*} \sum_{1\le |k|\le K}\! e^{-a_0 n |k|^{-2\tau}}|c_k|\ \le\ e^{-a_0 n K^{-2\tau}}\sum_{1\le |k|\le K}\! |c_k| \ \le\ C_1\,e^{-a_0 n K^{-2\tau}}\|h_0\|_{C^\alpha}, \end{align*} $$
   and using $|c_k|\le C_\alpha \|h_0\|_{C^\alpha }|k|^{-(1+\alpha )}$ ,
   $$ \begin{align*} \sum_{|k|>K}\!|c_k|\ \le\ C_2\,\|h_0\|_{C^\alpha}\,K^{-\alpha}. \end{align*} $$

   Choose $K=K(n)$ so that $e^{-a_0 n K^{-2\tau }}\approx K^{-\alpha }$ , that is, $a_0 n K^{-2\tau }\sim \alpha \log K$ . For $n\ge 2$ , one may take

   $$ \begin{align*} K\asymp \bigg(\frac{n}{\log n}\bigg)^{\!1/(2\tau)}, \end{align*} $$
   which gives
   $$ \begin{align*} \|T^n h_0-\bar h\|_{L^\infty}\ \le\ C\,\|h_0\|_{C^\alpha}\,K^{-\alpha} \ \le\ C\,\|h_0\|_{C^\alpha}\,n^{-{\alpha}/{2\tau}}(\log n)^{{\alpha}/{2\tau}}. \end{align*} $$

3. (iii) Analytic. If $|c_k|\le C_0 e^{-\sigma |k|}$ , then

   $$ \begin{align*} \|T^n h_0-\bar h\|_{L^\infty}\ \le\ C_0\sum_{k\ge1}\exp(-a_0 n k^{-2\tau}-\sigma k). \end{align*} $$

   Let $k_\ast \asymp (n)^{1/(2\tau +1)}$ be the minimiser of $a_0 n k^{-2\tau }+\sigma k$ . Then, for $k\le k_\ast $ , we have $a_0 n k^{-2\tau }\ge c\,n^{{1}/{(2\tau +1)}}$ , while for $k\ge k_\ast $ , $\sigma k\ge c\,n^{{1}/{(2\tau +1)}}$ . Hence, for all $k\ge 1$ ,

   $$ \begin{align*} \exp(-a_0 n k^{-2\tau}-\sigma k)\ \le\ \exp(-c\,n^{{1}/{(2\tau+1)}})\,e^{-{\sigma k}/{2}}, \end{align*} $$
   and summing the geometric tail $\sum _{k\ge 1}e^{-\sigma k/2}$ yields
   $$ \begin{align*} \|T^n h_0-\bar h\|_{L^\infty}\ \le\ C\,\exp(-\,c'\,n^{{1}/{(2\tau+1)}}). \\[-35pt] \end{align*} $$

Proof. Let $(q_\ell )$ be the convergents of $\theta /2\pi $ . Since $\theta $ is of bounded type, there exists $M\ge 1$ such that $q_{\ell +1}\le M q_\ell $ for all $\ell $ . Choose a sparse subsequence $(k_j)$ by setting $k_j:=q_{m_j}$ with $m_{j+1}$ large enough that $k_{j+1}\ge k_j^2$ and $k_1\ge 2$ . Define

$$ \begin{align*} h_0(\varphi)=\frac12+\sum_{j\ge1}b_j\cos(k_j\varphi),\quad b_j:=\min\bigg\{e^{-\sigma k_j},\ \frac{1}{4(k_j^2-1)}\bigg\}, \end{align*} $$

and fix $\sigma>0$ so that $\sum _j b_j k_j^2\le \frac 14$ . Then, $h_0$ is analytic in $\{|\Im \varphi |<\sigma \}$ , $h_0+h_0"\ge \frac 14>0$ , and since the first harmonic is absent and the mean is $\frac 12$ , we have $K_0\in \mathcal K_*$ .

For convergents, one has

$$ \begin{align*} \operatorname{dist}(q_\ell\theta,2\pi\mathbb Z)\le \frac{2\pi}{q_{\ell+1}}\le \frac{2\pi M}{q_\ell}, \quad 1-\cos(q_\ell\theta)\ \le\ \frac12 \operatorname{dist}(q_\ell\theta,2\pi\mathbb Z)^2\ \le\ \frac{C_1}{q_\ell^{2}}. \end{align*} $$

Set $t:=2\lambda (1-\lambda )(1-\cos x)$ . For $t\in [0,\frac 12]$ , we have $\sqrt {1-t}\ge e^{-t}$ . Thus, for j large enough,

$$ \begin{align*} |\mu_{k_j}|=\sqrt{1-2\lambda(1-\lambda)(1-\cos(k_j\theta))} \ \ge\ \exp\bigg(-\frac{C_2}{k_j^2}\bigg), \end{align*} $$

with $C_2=C_2(\lambda ,M)$ . Choose

$$ \begin{align*} n_j:=\bigg\lfloor \frac{\sigma}{2C_2}\,k_j^{3}\bigg\rfloor. \end{align*} $$

Then, $|\mu _{k_j}|^{\,n_j}\ge \exp (-\sigma k_j/2)$ , so the principal mode contributes at least $b_j e^{-\sigma k_j/2}$ at $n_j$ .

It remains to show that the contribution of the other modes is negligible. Since bounded type implies a Diophantine lower bound $\,\operatorname {dist}(k\theta ,2\pi \mathbb Z)\ge c_0/k\,$ for all $k\ge 1$ ,

$$ \begin{align*} |\mu_k| \le \exp(-c_1\,k^{-2}),\quad c_1=c_1(\lambda,c_0)>0. \end{align*} $$

For indices $i<j$ ,

$$ \begin{align*} |\mu_{k_i}|^{\,n_j} \le \exp\bigg(-c_1\,\frac{n_j}{k_i^2}\bigg) \le \exp\bigg(-c_3\,\frac{k_j^{3}}{k_{j-1}^{2}}\bigg) \le \exp(-c_4\,k_j^2), \end{align*} $$

using $k_{j-1}\le \sqrt {k_j}$ . For $i>j$ , we simply use $|\mu _{k_i}|^{n_j}\le 1$ and the tail of the analytic coefficients,

$$ \begin{align*} \sum_{i>j} b_i |\mu_{k_i}|^{\,n_j} &\le \sum_{i>j} e^{-\sigma k_i} \le C\,e^{-\sigma k_{j+1}} \le C\,e^{-\sigma k_j^2},\\ \sum_{i\ne j} b_i\,|\mu_{k_i}|^{\,n_j} &\le C\,e^{-\sigma k_j^2} \ll b_j e^{-\sigma k_j/2}\quad\text{for large } j,\\ \|T^{n_j}h_0&-\bar h\|_{L^\infty} \ge \tfrac12\,b_j e^{-\sigma k_j/2} \ge c\,e^{-C k_j}. \end{align*} $$

Since $n_j\asymp k_j^3$ , we conclude $ \|T^{n_j}h_0-\bar h\|_{L^\infty } \ge C \exp (-C'\,n_j^{1/3}), $ as claimed.

Proof. We work on support functions and use Lemma 2.2. Throughout, we ensure that the constructed $h_0$ is a valid support function in $\mathcal K_*$ by taking a small Fourier perturbation of the constant $\frac 12$ , which fixes the mean width and, if needed, avoids the first harmonic to keep the Steiner point at the origin.

1. (i) Resonant. Choose $r_*\in \{1,\ldots ,q-1\}$ attaining the maximum in the definition of  $\rho $ . Pick an integer $k\ge 2$ with $k\equiv r_*\pmod q$ . Let

   $$ \begin{align*} h_0(\varphi)=\frac12+\varepsilon\cos(k\varphi),\quad 0<\varepsilon\le \frac{1}{2\,(k^2-1)}. \end{align*} $$

   Then, $h_0\in C^\infty $ and $h_0+h_0"=\frac 12+\varepsilon (1-k^2)\cos (k\varphi )\ge 0$ , so $h_0$ is the support function of a convex body; its mean is $\frac 12$ and the $k\!=\!1$ mode is absent, and hence, $K_0\in \mathcal K_*$ . Since $k\not \equiv 0\pmod q$ , we have $P_q h_0=\frac 12$ . By Lemma 2.2,

   $$ \begin{align*} T^n h_0-P_q h_0=\varepsilon\,\Re (\mu_k^{\,n}e^{ik\varphi}), \quad \mu_k=(1-\lambda)+\lambda e^{-ik\theta}. \end{align*} $$

   Because $k\equiv r_*\pmod q$ and $\theta =2\pi p/q$ , we have $e^{-ik\theta }=e^{-2\pi i r_*/q}$ and thus, $|\mu _k|=\rho $ . Taking $L^\infty $ norms gives $ \|T^n h_0-P_q h_0\|_{L^\infty }=\varepsilon \,|\mu _k|^{\,n}=\varepsilon \,\rho ^{\,n}, $ so the claim holds with $c=\varepsilon $ . In particular, no smaller base than $\rho $ can work uniformly.

2. (ii) Nonresonant. Fix $\rho _0\in (0,1)$ . Since $\theta /2\pi $ is irrational, the set $\{e^{-ik\theta }:k\in \mathbb Z\}$ is dense on the unit circle, and hence there exists $k\ge 2$ such that

   $$ \begin{align*} |\mu_k|=|(1-\lambda)+\lambda e^{-ik\theta}|> \rho_0. \end{align*} $$

   Define $h_0(\varphi )=\frac 12+\varepsilon \cos (k\varphi )$ with $0<\varepsilon \le 1/(2(k^2-1))$ as above, so $h_0$ is the support function of a body in $\mathcal K_*$ . Then, exactly as in part (i), $ \|T^n h_0-\bar h\|_{L^\infty }=\varepsilon \,|\mu _k|^{\,n}. $ Given any $C>0$ , since $|\mu _k|/\rho _0>1$ , the ratio

   $$ \begin{align*} \frac{\varepsilon |\mu_k|^{\,n}}{C \rho_0^{\,n}}=\frac{\varepsilon}{C}\bigg(\frac{|\mu_k|}{\rho_0}\bigg)^{\!n}\ \to\ \infty\quad(n\to\infty), \end{align*} $$
   so the putative bound $\|T^n h_0-\bar h\|_{L^\infty }\le C\rho _0^{\,n}$ fails for all large n. Therefore, no uniform exponential estimate can hold over all admissible initial data in the irrational case.

Proof.

1. (i) For any normed space X among $L^2$ , $L^\infty $ , $C^\alpha $ , the shift $S_\theta $ is an isometry: $\|S_\theta f\|_X=\|f\|_X$ . Hence,

   $$ \begin{align*} \|Th-Tg\|_X&=\|(1-\lambda)(h-g)+\lambda(S_\theta h-S_\theta g)\|_X\\ & \le (1-\lambda)\|h-g\|_X +\lambda\|S_\theta(h-g)\|_X=\|h-g\|_X, \end{align*} $$
   showing $1$ -Lipschitz. For convex bodies, $\|h_{K}-h_{L}\|_{L^\infty }$ equals the Hausdorff distance $d_H(K,L)$ , so nonexpansiveness transfers to the geometric map.

2. (ii) Let $K\subset X$ be compact and fix $\varepsilon>0$ . If $\{x_1,\ldots ,x_M\}\subset K$ is $(n,\varepsilon )$ -separated (in Bowen’s sense), then for some $0\le j<n$ , we must have $\|T^j x_a-T^j x_b\|_X\ge \varepsilon $ whenever $a\ne b$ . Since T is $1$ -Lipschitz, $\|x_a-x_b\|_X\ge \varepsilon $ already at time $0$ . Thus, M is bounded by the $\varepsilon $ -packing number of K, which depends on $K,\varepsilon $ but not on n. Therefore, the exponential growth rate of $(n,\varepsilon )$ -separated sets is zero, that is, $h_{\mathrm {top}}(T|_K)=0$ . Taking the supremum over compact K gives $h_{\mathrm {top}}(T)=0$ .

3. (iii) The description of $\mathrm {Fix}(T)$ follows from Lemma 2.2: $T e_k=\mu _k e_k$ with $\mu _k=1$ if and only if $k=0$ in the irrational case, and if and only if $k\equiv 0\pmod q$ in the resonant case. The convergence statements are exactly Theorem 2.3. In particular, every orbit converges to a single point in the relevant fixed set: to $\bar h$ when $\theta /2\pi \notin \mathbb Q$ and to $P_q h_0$ when $\theta =2\pi p/q$ . This proves the stated attractor properties.