Proof. Let $y\in V$ . Since $B_{1}\subset B_{2}(x)$ , there exists a closed hyperplane $\{f^{*}=1\}$ which is tangent at y to both $B_{1}$ and $B_{2}(x)$ simultaneously. Observe that this implies that $\|f^{*}\|=1$ , $f^{*}(y)=1$ , $\|y\|=1$ and $f^{*}(y-x)=2$ . Hence, we conclude that $f^{*}(-x)=1$ . Now let $z\in [-x,y]$ . Therefore, there is $\lambda \in [0,1]$ such that $z=\lambda (-x)+(1-\lambda )y$ . By the triangle inequality, we obtain that $\|z\|\leq 1$ and $\|z-x\|\leq 2$ . By linearity of $f^{*}$ , we get that $f^{*}(z)=1$ and $f^{*}(z-x)=2$ . Therefore, $z\in V$ .

Proof. First case. Let us start with $v\in \ker (u^{*})$ . By the differentiability of the norm, there exists a sequence $(\varepsilon _{n})_{n}\subset \mathbb {R}^{+}$ which tends to $0$ as n tends to infinity and such that the expression

$$ \begin{align*}\max\{|\|nu^{+}- v\|-\|nu^{+}\||,|\|nu^{-}- v\|-\|nu^{-}\||\}\leq \varepsilon_{n} \end{align*} $$

holds true for all $n\in \mathbb {N}$ . Now, if n is large enough, $n>f(v)$ and then

$$ \begin{align*}1\geq\dfrac{f(nu^+)-f(v)}{\|nu^+-v\|}\geq\dfrac{n-f(v)}{n+\varepsilon_n},\end{align*} $$

which implies $f(v)\geq -\varepsilon _{n}$ for all large n. Thus, $f(v)\geq 0$ . For the reverse inequality, observe that

$$ \begin{align*}1\geq \dfrac{|f(nu^-)-f(v)|}{\|nu^--v\|}\geq\dfrac{n+f(v)}{n+\varepsilon_n}\end{align*} $$

holds true for all $n>0$ . Finally, we arrive at $f(v)\leq \varepsilon _{n}$ , for all $n>0$ . Therefore, $f(v)\leq 0$ , implying that $f(v)=0$ .

Second case, Let $v\in X\setminus \ker (u^{*})$ . Without loss of generality, assume that $u^{*}(v)=\alpha >0$ . Let us consider the function $g:X\to \mathbb {R}$ defined by $g(x)=f(x+\alpha u^{+})-\alpha $ . Clearly, g is a $1$ -Lipschitz function such that $g(tu^{+})=t$ for all $t\geq 0$ . We claim that $g(tu^{-})= -t$ for all $t>0$ . Indeed, let us fix $t>0$ . For $s>0$ , we have that $u^{*}(-\alpha u^{+}+ su^{-})=-\alpha -s$ . Thus, $\|-\alpha u^{+}+ su^{-}\|=\alpha +s$ . Also, since $g(0)=0$ , $g(-\alpha u^{+}+ su^{-})=-\alpha -s$ and g is $1$ -Lipschitz, g must be linear along the segment $[0,-\alpha u^{+} +su^{-}]$ , that is,

$$ \begin{align*}g(\lambda (-\alpha u^{+}+ su^{-}))= -\lambda(\alpha+s),\quad\text{for all } \lambda\in [0,1], \quad\text{for all }s>0.\end{align*} $$

If $s>t$ , we can set $\lambda =t/s$ . Using the continuity of g, sending s to infinity, we get that $g(tu^{-})=-t$ . Finally, the function g satisfies the hypothesis to apply the first case at the vector $v-\alpha u^{+}\in \ker (u^{*})$ . Hence, we get that $g(v-\alpha u^{+})=0$ . Thus, by definition of g, $f(v)=\alpha $ , finishing the proof.

Proof. (i) $\Rightarrow $ (ii). This is due to the compactness of closed bounded subsets of $X^{*}$ .

(ii) $\Rightarrow $ (i). Reasoning by contradiction, if (i) does not hold true, then there are $\varepsilon>0$ and a sequence $(r_{j})_{j}$ , convergent to $0$ , such that

$$ \begin{align*}\max_{y\in \overline{B}_{r_{j}}(x)}\dfrac{|u(y)-u(x)-e^{*}(y-x)|}{r_{j}}\geq \varepsilon \quad\text{for all }j\in\mathbb{N}, \quad\text{for all } e^{*}\in X^{*},~\|e^{*}\|=S(x).\end{align*} $$

This clearly contradicts statement (ii).

Proof of Theorem 1.3

Let $x\in \Omega $ . We prove Lemma 3.5(ii). Let $(r_{j})_{j}\subset \mathbb {R}^{+}$ be a sequence that converges to $0$ . For each $j\in \mathbb {N}$ , let us define $v_{j}:r_{j}^{-1}(\Omega -x)\to \mathbb {R}^{*}$ by

$$ \begin{align*}v_{j}(y)=\dfrac{u(x+r_{j}y)-u(x)}{r_{j}}.\end{align*} $$

For each compact subset K of X, the functions $v_{j}$ are well defined on K for j large enough. Since u is a locally Lipschitz function, the $(v_{j})_{j}|_{K}$ form an equi-Lipschitz family vanishing at $0$ . So by the Arzelà–Ascoli theorem, up to a subsequence, we can assume that the sequence $(v_{j})$ converges uniformly on compact subsets of X towards a Lipschitz function v vanishing at $0$ , that is, $v(y)=\lim _{j} v_{j}(y)$ for any $y\in X$ . If v is linear, then we can take $e^{*}=v$ .

So, to prove Theorem 1.3, it remains only to show that v is necessarily linear. Let $S(x)$ be computed with the function u (see Corollary 2.3). Since a locally uniform limit of functions satisfying comparison with cones satisfies comparison with cones, we can apply Corollary 2.3 on v as well. From now on, we define the quantities

$$ \begin{align*}L^{+}(y,r):=\max_{z\in\partial B_{r}(y)}\dfrac{v(z)-v(y)}{r}, \quad L^{-}(y,r):=-\min_{z\in\partial B_{r}(y)}\dfrac{v(z)-v(y)}{r}, \end{align*} $$

for $y\in X$ and $r>0$ . Also, we define the corresponding values

$$ \begin{align*}L^{+}(y)=\lim_{r\to 0}L^{+}(y,r),\quad L^{-}(y)=\lim_{r\to 0}L^{-}(y,r),\end{align*} $$

where $L(y)=L^{+}(y)=L^{-}(y)$ , by Corollary 2.3.

Proof. The first assumption implies that $\mathrm {Lip}(v)\leq S(x)$ . Thanks to the monotonicity of $L^{\pm }(0,\cdot )$ (see Corollary 2.3) and the second assumption, we get that $S(x)\leq \min \{L^{-}(0,r), L^{+}(0,r)\}$ , and therefore, $S(x) =L^{+}(0,r)=L^{-}(0,r)$ . Further, this implies that $\mathrm {Lip}(v)\geq S(x)$ , and then $\mathrm {Lip}(v)= S(x)$ . By continuity of v and compactness of closed bounded sets, we can find $z_{r}^{+},z_{r}^{-}\in \partial B_{r}$ such that

$$ \begin{align*}L^{\pm}(0,r)=\pm\dfrac{v(z_{r}^{\pm})-v(0)}{r}=\pm\dfrac{v(z_{r}^{\pm})}{r}.\end{align*} $$

Therefore,

$$ \begin{align*}L^{+}(0,r)=L^{-}(0,r)=S(x)= \dfrac{v(z_{r}^{+})-v(z_{r}^{-})}{2r}.\end{align*} $$

Observe that the function v is an $S(x)$ -Lipschitz function such that

(3-1) $$ \begin{align} v(z_{r}^{+})-v(z_{r}^{-})=2S(x)r\leq S(x)\|z_{r}^{+}-z_{r}^{-}\|. \end{align} $$

Since $z_{r}^{+},z_{r}^{-}\in \partial B_{r}$ , $\|z_{r}^{+}-z_{r}^{-}\|\leq 2r$ , and applying (3-1), we get that $\|z_{r}^{+}-z_{r}^{-}\|=2r$ and that v is an affine function on $[z_{r}^{-},z_{r}^{+}]$ . Moreover, since $v(0)=0$ , we have that $v(z_{r}^{+})=S(x)r$ and $v(z_{r}^{-})=-S(x)r $ .

Let $u^{*}\in X^{*}$ be the linear form of norm $1$ such that $u^{*}(z^{+}_{1})=1$ . By Proposition 3.1 and Remark 3.2, we deduce that $u^{*}(z^{-}_{1})=-1$ . Indeed, if $u^{*}(z^{-}_{1})<-1$ , then $z^{-}_{1}\notin \partial B_{1}\cap \partial B_{2}(z^{+}_{1})$ . Thus, $\|z_{1}^{+}-z_{1}^{-}\|< 2$ , a contradiction since $\mathrm {Lip}(v)= S(x)$ . Let $r>1$ . Let us show that $u^{*}(z^{+}_{r})=r$ . Indeed, since $v(0)=0$ , $v(z^{+}_{r})=S(x)r$ and v is $S(x)$ -Lipschitz, v must be linear along the segment $[0,z^{+}_{r}]$ . Therefore, the vector $z^{+}_{r}/\|z^{+}_{r}\|$ may take the place of $z^{+}_{1}$ because $v(z^{+}_{r}/\|z^{+}_{r}\|)=S(x)$ . If $u^{*}(z^{+}_{r})<r$ , then $u^{*}(z^{+}_{r}/\|z^{+}_{r}\|)<1$ . By Proposition 3.1 and Remark 3.2, we get that

$$ \begin{align*}\bigg\|z_{1}^{-} - \dfrac{z^{+}_{r}}{\|z^{+}_{r}\|}\bigg\|<2,\end{align*} $$

a contradiction since $\mathrm {Lip}(v)= S(x)$ . As a direct consequence of $u^{*}(z^{+}_{r})=r$ we get that $u^{*}(z^{-}_{r})=-r$ .

Since X is a finite-dimensional space, there exist a sequence $(r(n))_{n}\subset \mathbb {R}^{+}$ which goes to infinity and two vectors $q^{+},q^{-}\in \partial B_{1}$ such that

$$ \begin{align*}\lim_{n\to\infty}\dfrac{z^{\pm}_{r(n)}}{\|z^{\pm}_{r(n)}\|}= q^{\pm}.\end{align*} $$

Clearly $u^{*}(q^{+})=1$ and $u^{*}(q^{-})=-1$ . As a consequence of the continuity of v and its linear behavior along the lines $[0,z^{+}_{r(n)}]$ , with slope $S(x)$ , we get that $v(tq^{+})=tS(x)$ for all $t\geq 0$ . Analogously, we get that $v(tq^{-})=-tS(x)$ . Finally, applying Proposition 3.3, we conclude that $v=S(x)u^{*}$ .

To finish the proof of Theorem 1.3, it remains only to prove the hypotheses of Proposition 3.6. We point out that this part of the proof follows as in the proof given in [Reference Crandall and Evans4], where X is a Euclidean space.

To this end, let us start with the case of the superscript $+$ . Let $y\in X$ and $z\in \partial B_{r}(y)$ be such that

(3-2) $$ \begin{align} L^{+}(y,r)=\dfrac{v(z)-v(y)}{r}=\lim_{j\to\infty}\dfrac{u(r_{j}z+x)-u(r_{j}y+x)}{r_{j}r}. \end{align} $$

Since $r_{j}z\in \partial B_{r_{j}r}(r_{j}y)$ , we get that

(3-3) $$ \begin{align} \dfrac{u(r_{j}z+x)-u(r_{j}y+x)}{r_{j}r}\leq S^{+}(r_{j}y+x,r_{j}r)\leq S^{+}(r_{j}y+x,R), \end{align} $$

for $r_{j}r<R<\mathrm {dist}(r_{j}y+x,\partial U)$ . Notice that in (3-3), the first and second inequalities are due to the definition of $S^{+}$ and to Corollary 2.3, respectively. Combining (3-2), (3-3) and using the continuity of the function $S^{+}(\cdot ,R)$ , we get that

$$ \begin{align*}L^{+}(y,r)\leq \lim_{j\to\infty} S^{+}(r_{j}y+x,R)\leq S^{+}(x,R).\end{align*} $$

Finally, sending R to $0$ , we obtain that $L^{+}(y,r)\leq S(x)$ . To prove the second hypothesis, let us consider $y=0$ . Then we compute

$$ \begin{align*}L^{+}(0,r)= \max_{z\in \partial B_{r}}\dfrac{v(z)}{r}=\max_{z\in \partial B_{r}}\lim_{j\to \infty}\dfrac{u(r_{j}z+x)-u(x)}{r_{j}r}.\end{align*} $$

By compactness of $\partial B_{r}$ and continuity of u, for each j there exists $z_{j}\in \partial B_{r}$ satisfying

$$ \begin{align*}u(r_{j}z_{j}+x)=\max_{z\in \partial B_{r}}u(r_{j}z+x).\end{align*} $$

Let us consider any cluster point $\overline {z}$ of $(z_{j})_{j}\subset \partial B_{r}$ . Let $(j(n))$ be a subsequence such that $z_{j(n)}\to \overline {z}$ . Using the fact that u is Lipschitz in a neighborhood of x, we prove that

$$ \begin{align*}L^{+}(0,r)&\geq \lim_{n\to \infty}\dfrac{u(r_{j(n)}\overline{z}+x)-u(x)}{r_{j(n)}r}\\ &=\lim_{n\to \infty}\max_{z\in \partial B_{r}}\dfrac{u(r_{j(n)}z+x)-u(x)}{r_{j(n)}r}=\lim_{n\to \infty}S^{+}(x,r_{j(n)}r)=S(x).\end{align*} $$

Therefore, sending r to $0$ , we get that $L^{+}(0)\geq S(x)$ . Thus, $L^{+}(0)=S(x)$ . The case with superscript $-$ is analogous. This concludes the proof of Theorem 1.3.□

Proof. Let W be an open subset of $\Omega $ containing $0$ . Then there exists a segment $[z_{1},z_{2}]\subset W$ such that u is not an affine function on $[z_1,z_2]$ . Thus, there is an affine function $\ell $ on $[z_{1},z_{2}]$ and a point $y\in (z_{1},z_{2})$ such that $u(y)=\ell (y)$ and

$$ \begin{align*} u\geq \ell \text{ in }[z_{1},z_{2}] \text{ and } u(z_{i})> \ell(z_{i}), \text{ for }i=1,2,&\quad\text{or}\\ u\leq \ell \text{ in }[z_{1},z_{2}] \text{ and } u(z_{i})< \ell(z_{i}), \text{ for }i=1,2.& \end{align*} $$

We treat the first case; the second is similar. From Theorem 1.3, there exist vectors $e_{y,r}^{*}$ such that $\|e_{y,r}^{*}\|=S(y)$ and

$$ \begin{align*}\lim_{r\to 0}\max_{x\in B_{r}(y)} \dfrac{|u(x)-u(y)-e_{y,r}^{*}(x-y)|}{r}= 0.\end{align*} $$

By compactness, there is a sequence $(r_{i})_{i}$ , which converges to $0$ , such that $e_{y,r_{i}}^{*}\to e^{*}$ . Therefore, $\|e^{*}\|= S(y)$ and

(4-1) $$ \begin{align} \lim_{i\to\infty}\max_{x\in B_{r_{i}}(y)} \dfrac{|u(x)-g(x))|}{r_{i}}= 0, \end{align} $$

where g is the affine function defined by $g(x)=e^{*}(x)-e^{*}(y)+u(y)$ . Since $u\geq \ell $ in $[z_{1},z_{2}]$ and $u(y)=\ell (y)$ , the limit (4-1) implies that g coincides with $\ell $ in $[z_{1},z_{2}]$ , and then $z_{1},z_{2}\in \{u>g\}$ .

Reasoning by contradiction, suppose that there exists a polygonal line $\gamma \subset \{u>g\}$ connecting the points $z_{1}$ and $z_{2}$ . Let $\Gamma $ be the union of $\gamma $ with the segment $[z_{1},z_{2}]$ , and U be the union of all bounded connected components of $X\setminus \Gamma $ . Let $h^{*}\in X^{*}$ be a nonzero linear form such that $h^{*}(z_{2}-z_{1})=0$ . Using the fact that $y\notin \gamma $ and replacing $h^{*}$ by $-h^{*}$ if necessary, there exists $\delta>0$ such that $B_{\delta }(y)\cap \{h^{*}>0\}\subset U$ . Since $\gamma $ is compactly contained in $\{u> g\}$ , there exists $\varepsilon>0$ such that $u\geq g+\varepsilon h^{*}$ on $\gamma $ , hence also on $\Gamma $ . We have $u\geq g+\varepsilon h^{*}$ on $\partial U\subset \Gamma $ . Since u is an AML function, $u\geq g+\varepsilon h^{*}$ on U, so $u-g\geq \varepsilon h^{*}$ on $B_{\delta }(y)\cap \{h^{*}>0\}$ . This contradicts the limit (4-1), finishing the proof.

Proof. If $e_{1}^{*}=0$ , then u is identically $0$ in $\text {int}(B_{\rho })$ . Therefore, the second hypothesis cannot occur. So, without loss of generality, we assume that $e_{1}^{*}\neq 0$ . Let $R=C(\varepsilon ,X)>0$ be given by Lemma 4.3. Let us define $\displaystyle \lambda (\varepsilon ):={1}/{6C(\varepsilon ,X)}$ . If $w:\text {int}(B_{6R})\to \mathbb {R}$ is the function defined by

$$ \begin{align*}w(x):= \dfrac{6R}{\rho\|e^{*}_{1}\|}u\bigg( \dfrac{\rho x}{6R} \bigg),\end{align*} $$

and if $\lambda <\lambda (\varepsilon )$ , the function w satisfies the following assertions.

1. (a) $\sup \{|w(x)-{e_{1}^{*}(x)}/{\|e^{*}_{1}\|}|:x\in \text {int}(B_{6R})\}\leq 1$ .

2. (b) The set $\{w>{e^{*}}/{\|e^{*}_{1}\|}\}$ has at least two distinct connected components that intersect $B_{R}$ .

Therefore, Proposition 4.2 follows from Lemma 4.3.

Proof of Lemma 4.3

Let $f^{*}=e_{1}^{*}-e^{*}$ and let $\varepsilon>0$ . Without loss of generality, assume that $f^{*}\ne 0$ . By (H1), we have that

$$ \begin{align*} &\{f^{*}<-1\}\cap \text{int}(B_{6R})\subset \{u<e^{*}\},\\ &\{f^{*}>1\}\cap \text{int}(B_{6R})\subset \{u>e^{*}\}. \end{align*} $$

Thus, by hypothesis (H2), we can find a connected component $\mathcal {U}$ of $\{u>e^{*}\}$ that intersects $B_{R}$ and that is included in the strip $\mathcal {S}:=\{|f^{*}|\leq 1\}$ of width $2\|f^{*}\|^{-1}$ . If $R> \|f^{*}\|^{-1}$ , the set $\mathcal {S}\cap \partial B_{R}$ is the union of two disjoint arcs of $\partial B_{R}$ . Observe that $\mathcal {U}$ cannot be compactly contained in $\text {int}(B_{6R})$ . Otherwise, it would contradict the AML property of u (comparing against $e^{*}$ on $\overline {\mathcal {U}}$ ). Consider a polygonal line $\Gamma \subset \mathcal {U}\subset \mathcal {S}$ that starts in $B_{R}$ and exits $B_{6R}$ . Let $x_{0}\in X$ be a vector such that $\|x_{0}\|=3R$ and $f^{*}(x_{0})=0$ . Replacing $x_{0}$ by $-x_{0}$ if necessary, we can assume that $\Gamma $ intersects the two distinct arcs of $\mathcal {S}\cap \partial B_{R}(x_{0})$ . Let $v:B_{2R}\to \mathbb {R}$ be the function defined by $v(\cdot ):=u(\cdot +x_{0})-e_{1}^{*}(x_{0})$ . Observe that by (H1),

$$ \begin{align*}|v(x)-e_{1}^{*}(x)|\leq |u(x+x_{0})-e_{1}^{*}(x_{0}+x)|\leq 1,\quad\text{for all } x\in B_{2R},\end{align*} $$

and that, due to the fact that $f^{*}(x_{0})=0$ , $y\in \{v>e^{*}\}$ if and only if $y+x_{0}\in \{u>e^{*}\}$ . Therefore, replacing u by v (and $x_{0}$ by $-x_{0}$ if necessary), hypotheses (H1) and (H2) imply the following assertions.

1. (H1′) $\max \{|u(x)-e_{1}^{*}(x)|:x\in B_{2R}\}\leq 1$ for some $\|e_{1}^{*}\|=1$ .

2. (H2′) If $R>\|f^{*}\|^{-1}=\|e_{1}^{*}-e^{*}\|^{-1}$ , the set $\{u>e^{*}\}\cap B_{2R}$ has a connected component $\mathcal {U}$ , included in $\mathcal {S}=\{|f^{*}|\leq 1\}$ , that contains a polygonal line $\Gamma $ connecting the two distinct arcs of $\mathcal {S}\cap \partial B_{R}$ .

Lemma 4.3 follows from the next two lemmas.

Lemma 4.4 [Reference Savin12, Lemma 3]

Let $0<\gamma <1$ . If $R\geq C_{1}(\gamma ):=20\gamma ^{-2}$ , then

$$ \begin{align*}\|e^{*}\|\geq 1-\gamma.\end{align*} $$

Lemma 4.5. Let u be AML satisfying (H1 $^{\prime }$ ) and (H2 $^{\prime }$ ), let $\|e^{*}\|\geq \gamma>0$ and $\beta>0$ . There exists $C_{2}=C_{2}(\gamma ,\beta )$ such that if $R\geq C_{2}$ , then

$$ \begin{align*}\inf \bigg\{|f^{*}(h)|:h\in H\bigg(e^{*},\dfrac{\beta}{2}\bigg) \bigg\}< \gamma.\end{align*} $$

Let us finish the proof of Lemma 4.3. Since X is a finite-dimensional space with differential norm, $X^{*}$ is uniformly convex. Hence, there exists $\sigma (\varepsilon )>0$ such that for any two unit vectors $x^{*},y^{*}$ in $X^{*}$ satisfying $\| ({x^{*}+y^{*}})/{2}\|>\sigma (\varepsilon )$ , we have $\|x^{*}-y^{*}\| <\varepsilon $ . If $\gamma ,\beta>0$ are small, we have

(4-2) $$ \begin{align} \beta+\gamma<1-\sigma(\varepsilon/2) \end{align} $$

and

(4-3) $$ \begin{align} \dfrac{\beta}{2}+\dfrac{\gamma+\beta}{1-\beta}<\dfrac{\varepsilon}{2}. \end{align} $$

Let us fix $\beta =\beta (\varepsilon )$ and $\gamma =\gamma (\varepsilon )$ satisfying (4-2) and (4-3), and define $C(\varepsilon ,X):=\max \{C_{1}({\gamma }),C_{2}(\gamma ,\beta )\}$ . Assume that $R\geq C(\varepsilon ,X)$ . Lemma 4.4 implies that

(4-4) $$ \begin{align} \|e^{*}\| \geq 1-\gamma, \end{align} $$

and Lemma 4.5 implies the existence of a unit vector $h^{*}\in X^{*}$ satisfying the condition $\|h^{*}-{e^{*}}/{\|e^{*}\|} \| \leq \beta /2$ and a vector $h\in F_{h^{*}}$ such that $|f^{*}(h)|\leq \gamma $ . So $h\in H(e^{*},\beta /2)$ , and since Proposition 2.13 implies $\beta /2<\beta \leq \alpha ({e^{*}}/{\|e^{*}\|},\beta )$ , we have

(4-5) $$ \begin{align} (1-\beta)\|e^{*}\|<e^{*}(h). \end{align} $$

The condition $|f^{*}(h)|\leq \gamma $ implies that

(4-6) $$ \begin{align} e^{*}(h)\leq e^{*}_{1}(h)+\gamma\leq 1+\gamma. \end{align} $$

Conditions (4-4), (4-5) and (4-6) imply

$$ \begin{align*}\|e_{1}^{*}+h^{*}\| &\geq(e_{1}^{*}+h^{*})(h) \geq e^{*}(h)-\gamma +1\\ &\geq (1-\beta)\|e^{*}\| +1-\gamma\geq (1-\beta)(1-\gamma)+1-\gamma.\end{align*} $$

Thus, $\|({e^{*}_{1}+h^{*}})/{2}\|\geq 1-\gamma -\beta \geq \sigma (\varepsilon /2)$ , and therefore, $\| e^{*}_{1}-h^{*}\|\leq \varepsilon /2$ . Conditions (4-4), (4-5) and (4-6) also imply

$$ \begin{align*}1-\gamma\leq \|e^*\|\leq \dfrac{1+\gamma}{1-\beta}.\end{align*} $$

So,

$$ \begin{align*}\|h^{*}-e^{*}\| \leq \bigg\|h^{*}-\dfrac{e^{*}}{\|e^{*}\|}\bigg\|+ |\|e^{*}\|-1|\leq \dfrac{\beta}{2}+\dfrac{\gamma+\beta}{1-\beta} \leq \varepsilon/2.\end{align*} $$

Finally, we get that $\|e^{*}-e_{1}^{*}\|\leq \varepsilon $ , finishing the proof of Lemma 4.3.

Proof. Reasoning by contradiction, let us assume that $\|e^{*}\|<1-\gamma $ . Since $f^{*}=e_{1}^{*}-e^{*}$ and $\|e_{1}^{*}\|=\|e_{1}\|=e_{1}^{*}(e_{1})=1$ , we have that $2\geq \|f^{*}\|\ge f^{*}(e_{1})>\gamma $ . Let $y_{0}:=-4\gamma ^{-1}e_{1}$ , and let $y_{1}$ be the point of intersection of $\{te_{1}:t\geq 0\}$ with the line $\{f^{*}=1\}$ . We have $\| y_{1}\|=f^{*}(e_{1})^{-1}<\gamma ^{-1}$ , so

$$ \begin{align*}4\gamma^{-1}<\|y_1-y_0\|<5\gamma^{-1}.\end{align*} $$

Since $R\ge C_{1}:=20\gamma ^{-2}>\| f^{*}\|^{-1}$ , we can apply (H2 $^{\prime }$ ), and we also have $y_{0}\in B_{R}$ and $y_{1}\in B_{R}(y_{0})\subset B_{2R}$ .

For $c\geq 0$ , let $V_{c}$ be the function defined on X by

$$ \begin{align*}V_{c}(x):=e_{1}^{*}(y_{0}) +1+c\|x-y_{0}\|.\end{align*} $$

Notice that, for $c> \|e^{*}\|$ , the set

$$ \begin{align*}E_{c}:=\{x\in X: V_{c}(x)\leq e^{*}(x)\}\end{align*} $$

is convex and compact. We claim that $u(y_{0})\le V_{c}(y_{0})<e^{*}(y_{0})$ . Indeed, $y_{0}\in B_{R}$ , so condition (H1 $^{\prime }$ ) implies the first inequality. On the other hand, $V_{c}(y_{0})=e_{1}^{*}(y_{0})+1=e^{*}(y_{0})+f^{*}(y_{0})+1=e^{*}(y_{0})+1-4f^{*}(e_{1})/\gamma $ , which implies the second inequality.

Let $m:=\max \{c>\|e^{*}\|:E_{c}\cap \partial (\{u<e^{*}\}\cap B_{2R})\neq \emptyset \}$ . The claim implies $y_{0}\in E_{c}$ for every $c>0$ . Since the diameter of $E_{c}$ tends to $0$ as c tends to infinity, we conclude that $E_{c}$ converges to $\{y_{0}\}$ in Hausdorff distance. The claim also implies $u(y_{0})<e^{*}(y_{0})$ , so $m<\infty $ . Now we set

$$ \begin{align*}c_{0}:=1-\dfrac{2}{\|y_{1}-y_{0}\|}> 1-\dfrac{\gamma}{2} >\|e^{*}\|.\end{align*} $$

The equality $\|y_{1}-y_{0}\|=e_{1}^{*}(y_{1}-y_{0})$ implies $V_{c_{0}}(y_{1})=e_{1}^{*}(y_{0})+\|y_{1}-y_{0}\|-1=e_{1}^{*}(y_{1})-f^{*}(y_{1}) =e^{*}(y_{1})$ . Therefore, $y_{1}\in E_{c_{0}}$ , and we know that $y_{0}\in E_{c_{0}}$ , so the segment $[y_{0},y_{1}]$ is included in the convex set $E_{c_{0}}$ . Since $[y_{0},y_{1}]$ crosses the strip $\mathcal {S}$ , it must intersect the polygonal line $\Gamma $ given by (H2 $^{\prime }$ ) which is included in $\{u>e^{*}\}$ . Therefore, $E_{c_{0}}\cap \partial (\{u<e^{*}\}\cap B_{2R})\neq \emptyset $ , which shows that $m\ge c_{0}$ . The compact $E_{m}$ (see Figure 1) is included in the interior of $B_{2R}$ . Indeed, let $x\in E_{m}$ . Since $\|e^{*}\|<1-\gamma $ , we get

Figure 1

Lemma 4.4: The set $E_{m}$ .

$$ \begin{align*}0\le e^*(x)-V_m(x)\le e^*(y_0)-V_m(y_0)+(1-\gamma-m)\|x-y_0\|.\end{align*} $$

Using the inequalities $m> 1-\gamma /2$ and $e^{*}(y_{0})-V_{m}(y_{0})\leq 8/\gamma $ , we obtain $\|x-y_{0}\|<16\gamma ^{-2}$ , and so $\|x\|<20\gamma ^{-2}\leq R$ . Therefore, if $x_{m}\in E_{m}\cap \partial (\{u<e^{*}\}\cap B_{2R})$ , then $\|x_{m}\|<R$ , and so $x_{m}\in \partial \{u<e^{*}\}$ , and by continuity of u, we have that $u(x_{m})=e^{*}(x_{m})$ . Notice now, by definition of m, that we have that $u\leq V_{m}$ in $\partial (E_{m}\setminus \{y_{0}\})$ . Hence, by comparison with cones, $u\leq V_{m}$ in $E_{m}$ . Since $V_{m}$ is an affine function restricted to $[x_{m}, y_{0}]$ and $u(x_{m})=V_{m}(x_{m})\geq V_{m}(y_{0})$ , we get

(4-7) $$ \begin{align} S(x_{m})=-\lim_{r\to 0}\min_{y\in\partial B_{r}(x_{m})}\dfrac{u(y)-u(x_{m})}{r} \geq-\lim_{r\to 0}\dfrac{V_{m}(y_{r})-V_{m}(x_{m})}{r}= m> c_{0}\geq 1-\dfrac{\gamma}{2}, \end{align} $$

where $y_{r}$ is the point of intersection of $\partial B_{r}(x_{m})$ with $[x_{m}, y_{0}]$ . However, we claim that $S(x_{m})\leq \|e^{*}\|+2R^{-1}$ . To this end, let $r>0$ be small and let $U^{\prime }$ be the open set relative to $B_{R}(x_{m})$ defined as the union of all connected components of $\{u>e^{*}\}\cap B_{R}(x_{m})$ that intersect $B_{r}(x_{m})$ . If $U^{\prime }=\emptyset $ , then $u\leq e^{*}$ in $B_{r}(x_{m})$ . Therefore, since $u(x_{m})=e^{*}(x_{m})$ , by Corollary 2.5 we get that $S(x_{m})\leq \|e^{*}\|$ , which proves the claim in this case. If $U^{\prime }\neq \emptyset $ , (H2 $^{\prime }$ ) implies that we have that $U^{\prime }\subset \mathcal {S}$ provided that $r<\mathrm {dist}(x_{m},\Gamma )$ . For ${x\in \partial U^{\prime }\cap \text{int}(B_R)(x_{m})}$ , we have that $u(x)=e^{*}(x)$ . For $x\in U^{\prime }\cap \partial B_{R}(x_{m})$ ,

$$ \begin{align*}u(x)\leq e^{*}(x)+2\leq e^{*}(x_{m})+ R\|e^{*}\|+2,\end{align*} $$

where the first inequality follows as in (4-9), in the proof of Lemma 4.5. Therefore, comparison with cones implies

(4-8) $$ \begin{align} u(x)\leq e^{*}(x_{m})+ (\|e^{*}\|+2R^{-1})\|x-x_{m}\|,\quad\text{for all } x\in U^{\prime} \cap B_{R}(x_{m}). \end{align} $$

Combining (4-8) and the fact that $u\leq e^{*}$ in $B_{r}(x_{m})\setminus U^{\prime }$ , we get that the inequality (4-8) holds in $B_{r}(x_{m})$ . By Corollary 2.5, we conclude that

$$ \begin{align*}S(x_{m})\leq \|e^{*}\|+2R^{-1}.\end{align*} $$

Since $R\geq C_{1}\geq 5\gamma ^{-1}$ , we arrive at $S(x_{m})\leq 1-\gamma /2$ . The last inequality contradicts (4-7), finishing the proof of Lemma 4.4.

Proof of Lemma 4.5

If $\|f^{*}\|\leq \gamma $ , the conclusion is direct. For this, let us assume that $\|f^{*}\|>\gamma $ . If we further assume that $C_{2}\geq 3/\gamma $ , since $R\geq C_{2}$ , the conclusion of hypothesis $(H^{\prime }_{2})$ is available for us. Reasoning by contradiction, we have

$$ \begin{align*} \inf \bigg\{|f^{*}(h)|:h\in H\bigg(e^{*},\dfrac{\beta}{2}\bigg) \bigg\}\geq \gamma. \end{align*} $$

Let e be a unit vector in X such that $e^{*}(e)=\|e^{*}\|$ , and let $x_{0}$ be the point of intersection of $\partial \mathcal {S}$ with the half line $\{-te:t>0\}$ . See Figure 2. We have that $x_{0}=-t_{0}e$ , where $t_{0}$ satisfies

$$ \begin{align*}1=t_{0}|f^{*}(e)|\geq t_{0}\gamma. \end{align*} $$

So, $\|x_{0}\|=t_{0}\leq 1/\gamma \leq C_{2}\leq R$ . Thus $-x_{0}\in B_{R}(x_{0})\subset B_{2R}$ . Hypotheses (H1 $^{\prime }$ ) and (H2 $^{\prime }$ ) imply

$$ \begin{align*} |u(x)-e^{*}(x)|\leq |u(x)-e^{*}_{1}(x)|+|e^{*}(x)-e_{1}^{*}(x)|\leq 2,& \quad\text{for all }x\in \mathcal{U}\cap B_{R}(x_{0}). \\ u(x)=e^{*}(x)& \quad\text{for all } x\in \partial \mathcal{U}\cap B_{R}(x_{0}). \end{align*} $$

Hence, if $x\in \mathcal {U}\cap \partial B_{R}(x_{0})$ , then

(4-9) $$ \begin{align} u(x)\leq e^{*}(x)+2\leq e^{*}(x_{0})+\sup_{x\in \mathcal{S}\cap \partial B_{R}(x_{0})} e^{*}(x-x_{0})+2. \end{align} $$

Figure 2

Lemma 4.5: Ball of radius R centered at $x_{0}$ .

Since $R\geq C_{2}$ , $|f^{*}(Rh)|\geq 3$ for every $h\in H(e^{*},\beta /2)$ . Therefore, $|f^{*}(Rh-x_{0})|\geq 2>1$ for every $h\in H(e^{*},\beta /2)$ , that is,

$$ \begin{align*}(\mathcal{S}\cap \partial B_{R}(x_{0}) )\cap (RH(e^{*},\beta/2)-x_{0}) = \emptyset.\end{align*} $$

By Proposition 2.14, with $\sigma =\beta /2$ , we obtain $\rho \in (0,1)$ depending on $\beta $ , such that

(4-10) $$ \begin{align} e^{*}(x-x_{0})\leq(1-\rho)\|e^{*}\|\, \|x-x_{0}\|, \quad\text{for all }x\in \mathcal{S}\cap \partial B_{R}(x_{0}). \end{align} $$

Let us assume now that $C_{2}\geq {3}/{(\gamma \rho )}> {3}/{\gamma }$ . Since $R\geq C_{2}$ and $\|e^{*}\|\geq \gamma $ , we get $ R\|e^{*}\|\rho \geq 2$ . Combining (4-9) and (4-10), we get

$$ \begin{align*}u(x)\leq e^{*}(x_{0})+\|e^{*}\|\,\|x-x_{0}\|, \quad\text{for all }x\in \mathcal{U}\cap \partial B_{R}(x_{0}).\end{align*} $$

From comparison with cones, we obtain

$$ \begin{align*}u(x)\leq e^{*}(x_{0})+\|e^{*}\|\,\|x-x_{0}\|,\quad\text{for all }x\in \mathcal{U}\cap B_{R}(x_{0}). \end{align*} $$

In particular,

$$ \begin{align*}u(x)\leq e^{*}(x),\quad\text{for all }x\in \mathcal{U}\cap B_{R}(x_{0})\cap \{x_{0}+te:t>0\}. \end{align*} $$

This is a contradiction with $(H_{2}^{\prime })$ since $\mathcal {U}\cap B_{R}(x_{0})\cap \{x_{0}+te:t>0\}$ necessarily intersects $\Gamma $ .□

Proof of Theorem 1.5

If $\|e_{1}^{*}\|=0$ , there is nothing to prove. If $\|e_{1}^{*}\|\ne 0$ , by homogeneity, we can assume $\|e_{1}^{*}\|=1$ . By Theorem 1.3, there exists $(e^{*}_{0,r})_{r}\subset X^{*}$ such that $\|e^{*}_{0,r}\|=S(0)$ for every r and

(4-11) $$ \begin{align} |u(x)-u(0)-e_{0,r}^{*}(x)|\leq r\sigma(r),\quad\text{for all }x\in B_{r}, \end{align} $$

where $\sigma :\mathbb {R}_{+}\to \mathbb {R}$ is a positive function such that $\sigma (r)$ tends to $0$ as r tends to $0$ .

Let us fix $\varepsilon>0$ . We need to find $\delta :=\delta (\varepsilon )>0$ such that, if $|u(x)-e_{1}^{*}(x)|\leq \delta $ in $\text {int}(B_{1})$ , then

(4-12) $$ \begin{align} \limsup_{r\to 0}\|e_{0,r}^{*}-e_{1}^{*}\|\leq \varepsilon. \end{align} $$

First case. Suppose that u is not identical to an affine function in any neighborhood of $0$ . We show that if $\delta \leq \delta _{1}(\varepsilon )=\min \{\lambda (\varepsilon /4)/4, 1/2\}$ , where $\lambda $ is the function given in Proposition 4.2, then (4-12) holds. Let $r\in (0,1/2)$ be such that $\sigma (r)\le \lambda (\varepsilon /4)S(0)/4$ . Thanks to Proposition 4.1, replacing u by $-u$ if necessary, there exist $y\in \text {int}(B_{r/24})$ and a linear form $e^{*}\in X^{*}$ satisfying $\| e^{*}\|=S(y)$ , such that the set

$$ \begin{align*}\mathcal{O}=\{u>e^{*}+u(y)-e^{*}(y)\}\cap \text{ int }(B_{1})\end{align*} $$

has at least two distinct connected components intersecting $\text {int}(B_{r/24})$ . The function $v(\cdot ):=u(\cdot +y)-u(y)$ is well defined on int $(B_{1/2})$ . Let us check that v satisfies the hypotheses of Proposition 4.2. The set $\{v>e^{*}\}=(\mathcal {O}-y)\cap \text {int}(B_{1/2})$ has at least two distinct connected components intersecting $B_{r/12}\subset B_{1/12}$ . On the other hand, for $x\in B_{1/2}$ we have

$$ \begin{align*}|v(x)-e_{1}^{*}(x)|\leq |u(x+y)-e^{*}_{1}(x+y)|+|u(y)-e^{*}_{1}(y)|\leq 2\delta.\end{align*} $$

Since $2\delta \le \lambda (\varepsilon /4)/2$ , thanks to Proposition 4.2 applied with $\rho =1/2$ , we get

(4-13) $$ \begin{align} \|e^{*}-e^{*}_{1}\|\leq \dfrac{\varepsilon}{4}. \end{align} $$

Since $|u(x)-e^{*}_{1}(x)|\leq \delta $ in $\text {int}(B_{1})$ and $\|e_{1}^{*}\|=1$ , we obtain

$$ \begin{align*} \|e^{*}_{0,r}\|=S(0)\leq S^{+}\big(0,\tfrac{1}{2}\big)= 2\max_{x\in \partial B_{1/2}} {u(x)-u(0)}\leq 1+4\delta. \end{align*} $$

We now apply Proposition 4.2 to the function v on $B_{r/2}$ . The set $\{v>e^{*}\}\cap \text {int}(B_{r/2})$ has at least two distinct connected components that intersect $B_{r/12}$ . On the other hand, thanks to (4-11), for $x\in B_{r/2}$ we have that

$$ \begin{align*} |v(x)-e_{0,r}^{*}(x)|\leq |u(x+y)-u(0)-e_{0,r}^{*}(x+y)|+ |u(y)-u(0)-e_{0,r}^{*}(y)|\leq 2r\sigma(r). \end{align*} $$

Since $2\sigma (r)\le \lambda (\varepsilon /4)\|e_{0,r}^{*}\|/2$ , we get that

$$ \begin{align*}|v(x)-e_{0,r}^{*}(x)|\leq \dfrac{r}{2}\lambda(\varepsilon/4)\|e_{0,r}^{*}\|,\quad\text{for all }x\in B_{r/2}.\end{align*} $$

Finally, we can apply Proposition 4.2 with $\rho =r/2$ to obtain

(4-14) $$ \begin{align} \|e^{*}-e_{0,r}^{*}\|\leq\dfrac{\varepsilon\|e_{0,r}^{*}\|}{4}\leq\dfrac{(1+4\delta)\varepsilon}{4}\leq\dfrac{3\varepsilon}{4}. \end{align} $$

Combining (4-13) with (4-14), we get that $\|e_{1}^{*}-e_{0,r}^{*}\| \leq \varepsilon $ . Thus (4-12) is satisfied in this case whenever $\delta \le \delta _{1}(\varepsilon )$ .

Second case. Suppose that there exists $e_{0}^{*}\in X^{*}$ such that $u=e_{0}^{*}$ in a neighborhood of $0$ . Let

$$ \begin{align*}R=\max\{r\in(0,1];\,\{u=e_{0}^{*}\}\subset B_{r}\}.\end{align*} $$

If $R\ge 1/2$ , notice that $e_{0,r}^{*}=e_{0}^{*}$ satisfies (4-11) whenever $r\leq 1/2$ . Assume $\delta \le \varepsilon /2$ and $|u(x)-e^{*}_{1}(x)|<\delta $ in $\text {int}(B_{1})$ . Since $u=e_{0}^{*}$ in $B_{1/2}$ , we get $\limsup _{r\to 0}\|e_{0,r}^{*}-e_{1}^{*}\|=\|e_{0}^{*}-e_{1}^{*}\|\leq \varepsilon $ .

If $R< 1/2$ , there exists $x_{0}\in \partial B_{R}$ such that u is not identical to an affine function in any neighborhood of $x_{0}$ . Let us define the AML function $v:B_{1}\to \mathbb {R}$ by

$$ \begin{align*}v(\cdot):= u\bigg(\dfrac{\cdot}{2}+x_{0}\bigg)-u(x_{0}).\end{align*} $$

Since v is not affine in any neighborhood of $0$ , we wish to apply the first case to the function v. According to Theorem 1.3, there exists $(e^{*}_{x_{0},r})_{r}\subset X^{*}$ such that $\|e^{*}_{0,r}\|=S(x_{0})$ for every $r\in (0,1)$ and

$$ \begin{align*}|u(x)-u(x_{0})-e_{x_{0},r}^{*}(x-x_{0})|\leq r\widetilde{\sigma}(r),\quad\text{for all }x\in B_{r}(x_{0}),\end{align*} $$

where $\widetilde {\sigma }:\mathbb {R}_{+}\to \mathbb {R}$ is a positive function such that $\widetilde {\sigma }(r)$ tends to $0$ as r tends to $0$ . So, for $r\in (0,1/2)$ , we have

(4-15) $$ \begin{align} \bigg|v(x)-v(0)-\dfrac{e_{x_{0},r}^{*}}{2}(x)\bigg|=\bigg|u\bigg(\dfrac{x}{2}+x_{0}\bigg)-u(x_{0}) -e_{x_{0},r}^{*}\bigg(\dfrac{x}{2}\bigg)\bigg|\leq r\widetilde{\sigma}(r),\quad\text{for all }x\in B_{r}. \end{align} $$

Let us suppose that $\delta \le \delta _{1}(\varepsilon /2)/2$ . Since $|u(x)-e_{1}^{*}(x)|\leq \delta $ in $\text {int}(B_{1})$ , we have, for every $x\in B_{1}$ ,

(4-16) $$ \begin{align} \bigg|v(x)-\dfrac{e_{1}^{*}}{2}(x)\bigg|\leq \bigg|u\bigg(\dfrac{x}{2}+x_{0}\bigg)-e_{1}^{*}\bigg(\dfrac{x}{2}+x_{0}\bigg)\bigg|+|u(x_{0})-e_{1}^{*}(x_{0})|\leq \delta_{1}\bigg(\frac{\varepsilon}{2}\bigg). \end{align} $$

Let us check that $S_{v}(0)>0$ . We know that $\|e_{0}^{*}\|=S_{u}(0)\ne 0$ . Since $\|x_{0}\|=R$ and $u=e_{0}^{*}$ on $B_{R}$ , we can apply Corollary 2.4 to get

$$ \begin{align*}S_{v}(0)=\dfrac{S_{u}(x_{0})}{2}>0. \end{align*} $$

Conditions (4-15) and (4-16) allow us to apply the first case to v. We get

(4-17) $$ \begin{align} \limsup_{r\to 0}\bigg\| \dfrac{e_{1}^{*}}{2}-\dfrac{e_{x_{0},r}^{*}}{2} \bigg\|\leq \dfrac{\varepsilon}{2}. \end{align} $$

Let us now show that $e^{*}_{x_{0},r}$ tends to $e_{0}^{*}$ as r tends to $0$ . Reasoning by contradiction, assume that there exists a null sequence $(r_{i})_{i}$ such that $e_{x_{0},r_{i}}^{*}$ converges to some $h^{*}\neq e_{0}^{*}$ . So, there exist $z\in \partial B_{1}$ and $t>0$ such that the open segment $(x_{0},x_{0}+t z)$ is included in U and ${(e_{0}^{*}-h^{*})(z)\neq 0}$ . Finally, we compute

$$ \begin{align*}\lim_{i\to\infty}\dfrac{u(x_{0}+r_{i}z)-u(x_{0})}{r_{i}}-e_{x_{0},r_{i}}^{*}(z)=e_{0}^{*}(z)-h^{*}(z) \neq 0, \end{align*} $$

which contradicts Theorem 1.3. Hence, $e_{x_{0},r}^{*}$ converges to $e_{0}^{*}$ , and from (4-17) we get

$$ \begin{align*}\limsup_{r\to 0} \|e_{0,r}^{*}-e_{1}^{*} \|=\|e_{0}^{*}-e_{1}^{*} \| = \limsup_{r\to 0} \|e^{*}_{x_{0},r}-e_{1}^{*} \|\leq \varepsilon.\end{align*} $$

Thus, (4-12) is satisfied whenever

$$ \begin{align*}\delta (\epsilon )=\min \{\delta _{1}(\varepsilon /2)/2,\varepsilon /2\}=\min \{\lambda (\varepsilon /8)/8, 1/4, \varepsilon /2\}.\end{align*} $$

This completes the proof of Theorem 1.5.