I prove the result only for strictly convex norms.
THEOREM If $\|\cdot\|$ is a strictly convex norm, then there exist two (one if $r=1$) continuous families $(x_u)_{u\in S}$ of maximisers of $\|\cdot\|$ constrained on $D_u$.
Let us start with a pictorial representation of the setting, for the case of the strictly convex norm $$\|(x,y)\| := \sqrt{x^2+0.5y^2 + xy} + 0.5|x| + |y|,$$ where the level curves of $\|\cdot\|$ are shown. This gives a visual idea of the line of proof.
The case $r = 1$ is trivially true, since $D_u = \{\mathbf{0}\}$ for all $u\in S$. The continuous family of maximisers is $\{\mathbf{0}\}_{u\in S}$. Let us now consider the case $r \in ]0,1[$.
LEMMA 1 If $\|\cdot\|$ is strictly convex, then $\partial (B + ru)$ and $\partial (B - ru)$ intersect at exactly two distinct points $\{P,Q\}$ that vary continuously on $u$.
Proof. Without loss of generality, fix $u$ and choose a reference system such that $u = \frac{[1,0]}{\|[1,0]\|} \in S$. From convexity, the curve $S$ is piecewise differentiable and the outward direction $\widehat{n}(x)$, $x\in S$, is well-defined even at corner points (in the non smooth case, we use the notion of tangent cone, and the proof still works). From strict convexity, the angle $\theta(x)$ between $u$ and $\widehat{n}(x)$ is strictly monotonic (and multivalued if the non-smooth case) as $x$ moves on $S$. Therefore, there exist only one point $P_1$ such that $\theta(x) = \frac{\pi}{2}$ and one point $P_2$ such that $\theta(x) = \frac{3\pi}{2}$. The points $P_1$ and $P_2$ split $S$ into a left and a right part, defined as $$S_R := \{x\in S \ | \ \widehat{n}(x)\cdot u > 0\};\\ S_L := \{x\in S \ | \ \widehat{n}(x)\cdot u < 0\}.$$
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From strict convexity, $S_R$ and $S_L$ can be viewed as the graphs of two continuous and piecewise differentiable functions $x = f_R(y)$ and $x = f_L(y)$, respectively, for $y \in [y_{P_2}, y_{P_1}]$. These functions fulfil:
- $f_R$ is a strictly concave function
- $f_L$ is a strictly convex function
- $f_R(y_{P_1}) = f_L(y_{P_1}) = x_{P_1}$
- $f_R(y_{P_2}) = f_L(y_{P_2}) = x_{P_2}$
The intersections between $\partial (B + ru)$ and $\partial (B - ru)$ are clearly given by the solutions of the equation $f_R(y)-r u_x = f_L(y) +r u_x$, i.e. $$f_R(y) - f_L(y) -2ru_x = 0. \label{inters_equation}\tag{1}$$
The function $g(y) := f_R(y) - f_L(y) - 2ru_x$ fulfils:
- $g$ is a strictly concave function
- $g(y_{P_1}) = g(y_{P_2}) = - 2ru_x < 0$
- there exists $\overline{y} \in [y_{P_2}, y_{P_1}]$ such that $g(\overline{y}) > 0$. This is true because $\mathring{D_u} \neq \emptyset$ since $r \in ]0,1[$.
These properties imply that \eqref{inters_equation} has exactly two solutions, say $y_Q < y_P$. Let $P := (f_R(y_P), y_P)$ and $Q := (f_R(y_Q), y_Q)$.
Now for the continuity w.r.t. $u$. Being in the reference system used in the proof entails a rotation that is continuous w.r.t. $u$ (even for non-smooth norms). Therefore, from strict convexity, $P_1$ and $P_2$ in turn vary continuously w.r.t. such rotation, and so do $f_L$, $f_R$, $g$, and the zeros of $g$. This completes the proof.
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LEMMA 2 The following sets are symmetric w.r.t. the origin:
- $D_u$ and $\partial D_u$
- $\partial (B+ru) \cap \partial (B-ru) = \{P,-P\}$, i.e. the points $P,Q$ of Lemma 1 are opposite to each other.
Proof. First, since every norm is an even function, then $B$ and $S$ are both symmetric w.r.t. the origin. Let $x \in D_u$. Then $x$ can be expressed as $x = y+tu = z-tu$, where $y,z\in B$. But from the symmetry of $B$, also $-y, -z \in B$. Therefore also $-y-tu = -z+tu = -x \in D_u$, and this proves that $D_u$ is symmetric. If, in addition, $x\in\partial D_u$, this means that either $\|y\| = 1$ or $\|z\| = 1$, therefore $\|-y\| = 1$ or $\|-z\| = 1$, therefore $-x\in\partial D_u$. If, in addition, $x\in \partial (B+ru) \cap \partial (B-ru)$, then $\|y\| = \|z\| = 1$, which implies $\|-y\| = \|-z\| = 1$, i.e. $-x\in \partial (B+ru) \cap \partial (B-ru)$, and this completes the proof.
LEMMA 3 It holds that $D_u \subset (B+tu) \cap (B-tu)$ for all $t \in [-r,r]$. Specifically, $D_u \subset B$.
Proof. Let $x \in D_u$. By definition of $D_u$, there exist $y,z\in B$ such that $x = y+ru = z-ru$, which yields $z = y+2ru$. From the convexity of $B$ we have that $y+2tu \in B$ for all $t\in [0,r]$. Analogously, we have $z-2tu \in B$ for all $t\in [0,r]$. This means that $x$ can be expressed as $x = \overline{y} + tu = \overline{z} - tu$, where $\overline{y} := y + (r-t)u \in B$ and $\overline{z} := z -(r-t)u \in B$. Therefore, $x\in (B+tu) \cap (B-tu)$ for all $t\in [0,r]$, and this completes the proof.
LEMMA 4 All maximisers of $\|x\|$ constrained to $D_u$ must be actually on $\partial D_u$.
Proof. If $x\in\mathring{D_u}$, there exists a ball $B_\ell (x) \subset D_u$, therefore $\overline{x} := \frac{\|x\| + \ell}{\|x\|}x$ fulfils $\|\overline{x}\| = \|x\|+\ell > \|x\|$. Since $D_u$ is closed, then $\overline{x} \in D_u$, and so $x$ does not maximise $\|\cdot\|$ over $D_u$. This completes the proof.
LEMMA 5 If $\|\cdot\|$ is strictly convex, there exists $0<\overline{\alpha} < 1$ such that $D_u\subset B_{\alpha}(\mathbf{0})$ if and only if $\alpha \ge \overline{\alpha}$.
Proof We start by proving that there exists $0<\alpha<1$ such that $D_u \subset B_\alpha(\mathbf{0})$. We will use the notations introduced in the proof of Lemma 1. By construction, it follows that $$S_L \cap (S_L+ru) = \emptyset;\\ S_R \cap (S_R-ru) = \emptyset.$$ The two above conditions imply that $$S_L \cap D_u = S_R \cap D_u = \emptyset.\label{empty1}\tag{2}$$
Now suppose by contradiction that $P_1 \in D_u$. Then there exists $x_1\in B$ such that $P_1 = x_1+tu$, which yields $x_1 = P_1-tu \in B$, which is not possible thanks to the strict convexity of $S$. We reason similarly for $P_2$ and we get $$P_1\notin D_u, \quad P_2\notin D_u. \label{empty2}\tag{3}$$ Since $S = S_L \cup S_R \cup \{P_1,P_2\}$, from \eqref{empty1} and \eqref{empty2} we finally have $$S \cap D_u = \emptyset.$$ Since $S=\partial B$ and $D_u$ are compact and disjoint, and $D_u \subset B$ from Lemma 3, then $\mathring{B}\setminus D_u$ is an open stripe of width $\delta > 0$ w.r.t. the norm $\|\cdot\|$ (this is true since all norms are equivalent in $\mathbb{R}^2$). Therefore, it is possible to shrink the radius of $B$ by $\delta$, i.e.: $$D_u \subset B_{1-\delta}(\mathbf{0}).$$ We are now allowed to define $$\overline{\alpha} := \inf \{0 < \alpha < 1 | D_u \subset B_\alpha(\mathbf{0})\}.$$ Since the $B_\alpha(\mathbf{0})$ are nested, then $$D_u \subset B_\alpha (\mathbf{0}) \qquad \forall \alpha > \overline{\alpha}.$$ We are left to show that $D_u \subset B_{\overline{\alpha}}(\mathbf{0})$. To this end, $$D_u \subset \bigcap_{\alpha > \overline{\alpha}} B_\alpha(\mathbf{0}) = B_{\overline{\alpha}}(\mathbf{0}),$$ and this completes the proof.
LEMMA 6 The only points of $\partial D_u$ that are also on the optimal sphere $\partial B_{\overline{\alpha}}(\mathbf{0})$ are $\{P,-P\}$ in Lemma 2. Therefore, $P$ and $-P$ are the only maximisers of $\|\cdot\|$ constrained on $D_u$.
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Proof. We use the notation of Lemma 1. Since all level sets of $\|\cdot\|$ are rescalings of $S$, we can parametrise $\partial B_\overline{\alpha}(\mathbf{0})$ similarly as $S$, i.e. with two curves $$x = \widetilde{f}_L(y) := \overline{\alpha}f_L\left(\frac{y}{\overline{\alpha}}\right) \text{ and } x = \widetilde{f}_R(y) := \overline{\alpha}f_R\left(\frac{y}{\overline{\alpha}}\right), \quad y \in [\overline{\alpha}y_{P_2}, \overline{\alpha}y_{P_1}].\label{parametr_alpha}\tag{4}\\\\$$ Suppose by contradiction that there exists $T \in \partial B_{\overline{\alpha}}(\mathbf{0}) \cap (\partial D_u \setminus \{-P,P\})$. There are two cases:
- $T \in S_L+ru$;
- $T \in S_R-ru$.
We will address case 1) since case 2) is similar.
We start by proving that $\widetilde{f}'_L(y_T) = f'_L(y_T)$. Since $T\notin \{-P,P\}$ (i.e. $T$ is not the lowest nor the highest point of $D_u$), $\widetilde{f}'_L(y_T) \neq f'_L(y_T)$ would imply that $\partial B_{\overline{\alpha}}(\mathbf{0})$ crosses $\partial D_u$ at $T$ and therefore $ B_{\overline{\alpha}}(\mathbf{0}) \not\supseteq D_u$, a contradiction. Therefore $$\widetilde{f}'_L(y_T) = f'_L(y_T),\label{same_tangent}\tag{5}$$ i.e. both curves have the same tangent at $y_T$. By differentiating \eqref{parametr_alpha} and evaluating at $y = y_T$, we have $$\widetilde{f}'_L(y_T) = f'_L\left(\frac{y_T}{\overline{\alpha}}\right).\label{first_der}\tag{6}$$ By combining \eqref{same_tangent} and \eqref{first_der} we get $$f'_L(y_T) = f'_L\left(\frac{y_T}{\overline{\alpha}}\right).$$ From strict convexity, $f'_L$ is strictly increasing, so $y_T = \frac{y_T}{\overline{\alpha}}$, which using $\overline{\alpha} < 1$ gives $$y_T = 0.\label{yt0}\tag{7}$$ Differentiating \eqref{parametr_alpha} twice and using \eqref{yt0} gives $$\widetilde{f}''_L(y_T) = \frac{1}{\overline{\alpha}}f''_L\left(\frac{y_T}{\overline{\alpha}}\right) = \frac{1}{\overline{\alpha}} f''_L(0) = \frac{1}{\overline{\alpha}} f''_L(y_T).$$ So $\widetilde{f}_L$ has a higher (in absolute value) second derivative than $f_L + ru_x$ at $y_T$ (or, in the non-smooth case, limit of the second derivative as $y\rightarrow y_T$, which always exists for convex functions thanks to Alexandrov's theorem). So $\partial B_{\overline{\alpha}}(\mathbf{0})$ ventures inside $D_u$, so $B_{\overline{\alpha}}(\mathbf{0}) \not\supseteq D_u$, a contradiction. This completes the proof.
P.S. I believe the result holds also for non-strictly convex norms (such as the 1-norm or the infinity norm), but it is harder to prove since there can be more than two constrained maximisers.
