Proof. Along the gradient orbit \(\gamma_w\), we have \[\frac{d}{dt} f(\gamma_w(t)) = \nabla f(\gamma_w(t))\cdot \dot{\gamma}_w(t) = -\|\nabla f(\gamma_w(t))\|^2 \leq 0.\] Thus \(f(\gamma_w(t))\leq f(w)\) for all \(t\geq 0\) and the curve \(\gamma_w\) lies in the sublevel set \(\{ f \leq f(w)\}\), which is compact, since \(f\) is coercive. Moreover, \[\int_0^\infty \|\nabla f(\gamma_w(t))\|^2\,dt \,\,\leq \,\, f(w)- \inf\,f \, <\, + \infty\,.\] Since \(t \mapsto \| \nabla f(\gamma_w(t))\|^2\) is integrable on \([0, +\infty)\), we infer that \(\lim_{t \to \infty} \| \nabla f(\gamma_w(t)) \| = 0\). Fix \(\{ t_n \}_{n \geq 1}\) with \(\lim_{n \to \infty} t_n = +\infty\). Since the set \(\{ f \leq f(w) \}\) is compact, there exist a subsequence \(\{ t_{n_k} \}\) and a vector \(p_w \in \mathbb{R}^d\) such that \(\lim_{k \to \infty} \gamma_w(t_{n_k}) = p_w\). Since \(\nabla f\) is continuous, we obtain \(\nabla f(p_w) = 0\) and so \(p_w \in \mathrm{Crit}(f)\), which completes the proof. ◻

Proof. Coercivity of \(f\) yields the nonemptiness of \(\mathrm{Crit}(f)\). Notice that the (non-negative) function \(V_2\) is convex and \(\mathrm{Crit}(f) = \{ V_2 = 0 \}\). Therefore \(\mathrm{Crit}(f)\) is convex. Take now any \(x,y\in \mathrm{Crit}(f)\). Then for every \(s \in (0, 1)\), we have \(x+s(y-x) \in \mathrm{Crit}(f)\) and \[\frac{d}{ds} f(x+s(y-x)) = \underbrace{\nabla f(x+s(y-x))}_{=0}\cdot (y-x) =0.\] It follows that \(f(x)=f(y)\) yielding that \(f\) is constant on \(\mathrm{Crit}(f)\). ◻

Proof. Fix \(x, y \in \mathbb{R}^d\). Denote by \(\gamma_x, \gamma_y : [0, + \infty) \to \mathbb{R}^d\) the gradient flows starting at \(x\) and \(y\), respectively, that is, \[\begin{align} \begin{dcases} \dot{\gamma_x}(t) = - \nabla f(\gamma_x(t)), \,\, t > 0 \\ \gamma_x(0) = x \end{dcases} \quad \text{ and } \quad \begin{dcases} \dot{\gamma_y}(t) = - \nabla f(\gamma_y(t)), \,\, t > 0 \\ \gamma_y(0) = y \end{dcases}. \end{align}\] Set \[\begin{align} \mu(t) := & \, \dfrac{\gamma_x(t) + \gamma_y(t)}{2}, \\ D(t) := & \, \dfrac{1}{2} \Big( f(\gamma_x(t)) + f(\gamma_y(t)) \Big) - f(\mu(t)). \end{align}\]

We first prove that the function \(t\mapsto D(t)\) is nonincreasing. First observe that \[\label{Gflow-id} \dfrac{d}{dt} f(\gamma_z(t)) = \langle \nabla f(\gamma_z(t)), \dot{\gamma}_z(t) \rangle = - \| \nabla f(\gamma_z(t)) \|^2 = - V_2(\gamma_z(t)) \quad \text{for z \in \{ x, y \} and t > 0.}\tag{1}\] It follows that \[\label{D1} D'(t) = - \dfrac{1}{2} \Big( V_2(\gamma_x(t)) + V_2(\gamma_y(t)) \Big) + \left\langle \nabla f(\mu(t)), \dfrac{\nabla f(\gamma_x(t)) + \nabla f(\gamma_y(t))}{2} \right\rangle.\tag{2}\] On the one hand, using the assumption that \(V_2\) is convex, we obtain \[\label{D2} \| \nabla f(\mu(t)) \| = \sqrt{V_2(\mu(t))} \leq \sqrt{\dfrac{V_2(\gamma_x(t)) + V_2(\gamma_y(t))}{2}}.\tag{3}\] On the other hand, using the convexity of the map \(w \mapsto \| w \|^2\), we have \[\label{D3} \left\| \dfrac{\nabla f(\gamma_x(t)) + \nabla f(\gamma_y(t))}{2} \right\| \leq \sqrt{\dfrac{V_2(\gamma_x(t)) + V_2(\gamma_y(t))}{2}}.\tag{4}\] Combining 2 , 3 and 4 , we conclude that \(D' \leq 0\) and hence it is nonincreasing. It follows that \[\label{descent95D} D(0) \geq D(t) \quad \text{ for every } t \geq 0.\tag{5}\]

We next prove that there exists a sequence \(\{ t_n \}\) such that \(\lim_{n \to \infty} D(t_n) = 0\). Applying Lemma 1 for \(w = x\) and subsequently, for \(w = y\), we obtain a sequence \(\{ t_n \}_{n\geq 1}\) and \(p_x, p_y \in \mathrm{Crit}(f) = \{ \nabla f = 0 \}\) such that \[\gamma_x(t_n) \underset{n \to \infty}{\longrightarrow} \,p_x \quad \text{ and } \quad \gamma_y(t_n) \underset{n \to \infty}{\longrightarrow} \,p_y.\] Consequently, thanks to the continuity of \(f\) and the fact that \(f\) is constant on the convex set \(\mathrm{Crit}(f)\) (see Lemma 2), we deduce \[\lim_{n \to \infty} D(t_n) = \dfrac{1}{2} (f(p_x) + f(p_y)) - f \left( \dfrac{p_x + p_y}{2} \right) = 0.\]

To conclude, combining this with 5 , we get \(D(0) \geq \lim_{n \to \infty} D(t_n) = 0\). Therefore, \[\begin{align} f \left( \dfrac{x + y}{2} \right) \leq \dfrac{1}{2} (f(x) + f(y)). \end{align}\] Since \(f\) is continuous, we infer that \(f\) is convex, which completes the proof. ◻

Proof. First, consider the case in which \(\xi: [a, b] \to X\) a \(1\)–Lipschitz curve. In this case, the inequality 6 can be derived by combining [6] and [10], via the theory of eikonal equations in metric spaces. In the sequel, however, we give a self-contained and direct proof based on the lower Dini derivative, which avoids the machinery of viscosity solutions in general metric spaces.

Consider the function \[F(t) := f (\xi(t)) + \int_a^t s_f(\xi(\tau)) \, d\tau, \,\, \text{ for } t \in [a, b].\] Recall that its lower right Dini derivative at \(t \in (a, b)\) is defined by: \[D_+ F(t) := \liminf_{h \to 0_+} \dfrac{F(t + h) - F(t)}{h}.\] Fix \(t \in (a, b)\) and \(\varepsilon > 0\). By the definition of slope, there exists \(\delta > 0\) such that \[f(\xi(t)) - f(y) \leq (s_f(\xi(t)) + \varepsilon) \| \xi(t) - y \| \quad \text{ for every } y \in B_\delta(\xi(t)).\] Using the fact that \(\xi\) is 1–Lipschitz, we have, for every \(0 < h < \min \{\delta, b - t\}\), \[f(\xi(t + h)) - f(\xi(t)) \geq - (s_f(\xi(t)) + \varepsilon) \| \xi(t) - \xi(t + h) \| \geq - (s_f(\xi(t)) + \varepsilon)h.\] Therefore, we obtain \[\dfrac{F(t + h) - F(t)}{h} \geq - (s_f(\xi(t)) + \varepsilon) + \dfrac{1}{h} \int_t^{t + h} s_f(\xi(\tau)) \,d\tau \quad \text{ for every } 0 < h < \min \{ \delta, b - t \}.\] The continuity of \(s_f \circ \xi\) then yields \[D_+ F(t) \geq - (s_f(\xi(t)) + \varepsilon) + s_f(\xi(t)) = - \varepsilon.\] Since \(\varepsilon > 0\) and \(t \in (a, b)\) are chosen arbitrarily, we infer that \(D_+ F(t) \geq 0\) for every \(t \in (a, b)\). Therefore \(F\) is nondescreasing in \([a, b]\) (see e.g. [11] or [12]). It follows that, for any \(a \leq s \leq t \leq b\) \[f(\xi(s)) - f( \xi(t)) \leq \int_a^t s_f (\xi(\tau)) \,d\tau - \int_a^s s_f(\xi(\tau)) \,d\tau = \int_s^t s_f(\xi(\tau)) \,d\tau.\]

It remains to prove 6 for an arbitrary absolutely continuous curve \(\xi: [a, b] \to X\). Define the arc-length function \[\ell(r) := \int_a^r \|\dot{\xi}(\tau)\| \,d\tau, \quad \text{ for } r\in[a,b],\] and set \(L := \ell(b)\). The case \(L = 0\) is vacuous. Assume that \(L>0\). Since \(\xi\) is absolutely continuous, \(\ell\) is continuous, nondecreasing and satisfies \(\ell'(r)=\|\dot{\xi}(r)\|\) for a.e. \(r \in [a,b]\). Since \(\ell(r_1)=\ell(r_2)\) implies \(\xi(r_1)=\xi(r_2)\), the curve \(\eta:[0,L]\to X\) defined by \[\eta(\ell(r)) := \xi(r), \quad \text{ for } r\in[a,b],\] is well-defined. Furthermore, if \(0\le \rho_1\le \rho_2\le L\), we choose \(r_1,r_2\in[a,b]\) such that \(\ell(r_i) = \rho_i\) for \(i \in \{1,2\}\). Since \(\ell\) is nondecreasing, we may take \(r_1\le r_2\) and so \[\|\eta(\rho_2)-\eta(\rho_1)\| = \|\xi(r_2)-\xi(r_1)\| \le \int_{r_1}^{r_2}\|\dot{\xi}(\tau)\|\,d\tau = \rho_2-\rho_1.\] Thus \(\eta\) is \(1\)-Lipschitz. Applying the result already proved for \(1\)-Lipschitz curves, we obtain, for any \(a\le s\le t\le b\), \[\begin{align} f(\xi(s))-f(\xi(t)) = f(\eta(\ell(s))) - f(\eta(\ell(t)) ) \leq & \int_{\ell(s)}^{\ell(t)} s_f(\eta(\rho))\,d\rho \\ = & \int_s^t s_f(\eta(\ell(\tau)))\ell'(\tau)\,d\tau = \int_s^t s_f(\xi(\tau))\|\dot{\xi}(\tau)\|\,d\tau, \end{align}\] where we have used that \(\ell'(\tau) = \| \dot{\xi} (\tau) \|\) for a.e \(\tau \in [a, b]\). This completes the proof. ◻

Proof. Given \(x,y\in K\), apply 6 for the \(1\)-Lipschitz curve \(\xi(t):= y + t \,(\frac{x-y}{\|x-y\|})\),   \(t\in \big[\, 0, \|x-y\|\,\big]\). ◻

Proof. Since the functions \(f\) and \(s_f\) are continuous and \(s_f(x) > 0\), there exists \(\delta' > 0\) such that \[m_f(x):=\inf_{B_{\delta'}(x)}\,f > -\infty \qquad \text{and}\qquad 0 < \underbrace{\inf_{B_{\delta'}(x)} s_f}_{=: \, m_x} \leq \underbrace{\sup_{B_{\delta'}(x)} s_f}_{=: \, M_x} < + \infty.\] Consequently, thanks to Corollary 1, \(f\) is Lipschitz on \(B_{\delta'}(x)\) with the Lipschitz constant \(M_x\). Applying Proposition 2 to the case \(\Omega = B_{\delta'}(x)\), \(u = f\) and \(\ell = s_f\), there exists \(\delta \in (0, \delta')\) such that, for every \(\eta > 0\), there exists a \(1\)–Lipschitz curve \(\xi: [0, + \infty) \to X\) with \(\xi(0) = x\) satisfying \[\label{sig9501} f(x) \geq \int_0^r s_f(\xi(\rho)) \, d\rho + f(\xi(r)) - \eta(1 + r), \quad \text{ for every } 0 \leq r \leq R_\xi,\tag{7}\] where \(R_\xi > 0\) denotes the exit time of \(\xi\) from \(B_\delta(x)\). Since \(f\) is bounded from below, we get, for every \(\eta \leq \min\{1, m_x/2\}\), \[f(x) - m_f(x) + \eta \geq \int_0^{R_\xi} s_f(\xi(\rho)) \, d\rho - \eta R_\xi \geq (m_x - \eta) R_\xi \geq \dfrac{m_x R_\xi}{2}.\] Hence \[R_\xi \leq \dfrac{2(f(x) - m_f(x) + 1)}{m_x}.\] Since \(R_\xi\) is bounded by a quantity that depends only on \(f\) and \(x\), it follows from 7 that \[\label{sig9502} f(x) \geq \inf_\xi \left\{ \int_0^{R_\xi} s_f(\xi(\rho)) \, d\rho + f(\xi(R_\xi)) \right \},\tag{8}\] where the infimum is taken over all \(1\)–Lipschitz curves \(\xi: [0, + \infty) \to X\) with \(\xi(0) = x\). Fix \(\varepsilon > 0\). Then, by 8 , there exists a \(1\)–Lipschitz curve \(\sigma: [0, + \infty) \to X\) such that \(\sigma(0) = x\) and \[\label{sig9503} f(x) \geq \int_0^{R_\sigma} s_f(\sigma(\rho))\, d\rho + f(\sigma(R_\sigma))- \varepsilon.\tag{9}\] For any fixed \(r \in (0, R_\sigma)\), applying Lemma 3 to the case \(\xi = \sigma\), we have \[\label{sig9504} f(\sigma(r)) - f(\sigma(R_\sigma)) \leq \int_r^{R_\sigma} s_f(\sigma(\rho)) \, d\rho.\tag{10}\] Combining 9 and 10 , we obtain \[f(x) \geq \int_0^r s_f(\sigma(\rho)) \, d\rho + f(\sigma(r)) - \varepsilon, \quad \text{ for every } 0 \leq r \leq R_\sigma,\] which completes the proof. ◻

Proof. The case \(s_f(x) = 0\) is vacuous. Assume that \(s_f(x) > 0\). We split the proof into two steps.

Step 1: Local existence. Fix \(x \in X\) and \(\varepsilon > 0\). We prove that there exist \(T_{x} > 0\) depending only on \(x\) and a curve \(\gamma_x: [0, T_x) \to X\) such that \[\label{velo95local} \| \dot{\gamma}_x(t) \| \leq s_f^{m - 1}(\gamma_x(t)) \quad \text{ a.e t \in (0, T_x),}\tag{13}\] and \[\label{descent-local} f(x) - f(\gamma_x(t)) \geq \int_0^t s_f^m(\gamma_x(\tau)) \, d\tau - \varepsilon, \quad \text{ for every } t \in (0, T_x).\tag{14}\]

First, let \(\delta > 0\) be defined as in Corollary 2. Notice that \[0 < \underbrace{\inf_{y \in B_\delta(x)} s_f(y)}_{:=\,m_x} \leq \underbrace{\sup_{y \in B_\delta(x)} s_f(y)}_{:=\,M_x} < + \infty.\] Therefore, thanks to Corollary 2, there exists a curve \(\sigma_x: [0, +\infty) \to X\) with \(\sigma_x(0) = x\) such that \(\| \dot{\sigma}_x(r) \| \leq 1\) a.e \(r > 0\) and \[\label{eq:24610} f(x) - f(\sigma_x(r)) \geq \int_0^r s_f(\sigma_x(\rho)) \, d\rho - \varepsilon \quad \text{ for every } r \in (0, R_x).\tag{15}\] Here, \(R_x\) denotes the exit time of \(\sigma_x\) from \(B_\delta(x)\): \[R_x := \inf\big\{ r \geq 0 : \sigma_x(r) \not\in B_\delta(x) \big\}.\] Since \(\sigma_x\) is \(1\)–Lipschitz, \(R_x \geq \delta\). Define \[\varphi(r) := \int_0^r \dfrac{1}{s_f^{m - 1}(\sigma_x(\rho))} \,d\rho, \qquad r \in [0,R_x).\] Since \(\inf_{r \in [0, R_x)} s_f(\sigma_x(r)) \geq m_x>0\), \(\varphi\) is strictly increasing and \(\mathcal{C}^1\) on \((0, R_x)\) with \[\varphi'(r) = \dfrac{1}{s_f^{m - 1}(\sigma_x(r))} \quad \text{ for every } r \in (0, R_x).\] Let \(\theta := \varphi^{-1} : [0, T_x) \to [0, R_x)\) be the inverse of \(\varphi\). Since \(R_x > \delta/2\), we have \[\label{T95hat} T_x = \lim_{r \nearrow R_x} \varphi(r) \geq \varphi(\delta/2) \geq \frac{\delta}{2 M_x^{m-1}} > 0 \quad \text{ where } M_x := \sup_{y \in B_\delta(x)} s_f(y) < + \infty.\tag{16}\] Set \(\gamma_x(t) := \sigma_x(\theta(t))\) for \(t \in [0, T_x)\). Then \(\gamma_x(0)=x\) and for every \(t \in (0, T_x)\), we have \[\theta'(t) = \frac{1}{\varphi'(\theta(t))} = s_f^{m-1}(\sigma_x(\theta(t))) = s_f^{m-1}(\gamma_x(t)).\] Therefore \[\|\dot{\gamma}_x(t)\| \leq \|\dot{\sigma}_x(\theta(t))\|\, \theta'(t) \leq s_f^{m-1}(\gamma_x(t)) \quad \text{for a.e. } t \in (0, T_x).\]

It remains to verify inequality 14 for \(t\in (0,T_x)\). To this end, let \(t \in (0, T_x)\) and apply the estimate 15 for \(\sigma_x\) at \(\theta(t)\), to obtain \[\begin{align} f(x)-f(\gamma_x(t)) = f(x)-f(\sigma_x(\theta(t))) \geq & ~ \int_0^{\theta(t)} s_f(\sigma_x(\rho))\,d\rho - \varepsilon \\ = & \,\int_0^t s_f(\sigma_x(\theta(\tau))) \theta'(\tau) \, d\tau - \varepsilon = \, \int_0^t s_f^m(\gamma_x(\tau)) \, d\tau - \varepsilon. \end{align}\]

Step 2: Global existence. We shall prove the existence of a curve \(\gamma_x: [0, + \infty) \to X\) with \(\gamma_x(0) = x\) satisfying 13 –14 for all \(t\in [0,+\infty)\) (that is, we can take \(T_x=+\infty\)).

Fix a continuous strictly increasing function \(\kappa : [0,+\infty]\to[0,\varepsilon]\) such that \[\kappa(0)=0, \quad \kappa(T) < \varepsilon\quad\text{for every }T<+\infty \quad \text{ and } \quad \kappa(+\infty) = \varepsilon .\] For instance, we can take \(\kappa(t) = \varepsilon(1 - e^{-t})\) or \(\kappa(t)=\frac{2\varepsilon}{\pi}\arctan t\). Let \(\EuScript A_x\) be the set of all pairs \((T, \gamma)\), where \(T\in(0,+\infty]\) and \(\gamma : [0, T) \to X\) is absolutely continuous such that \(\gamma(0) = x\), \[\label{velocity-T} \|\dot{\gamma}(t)\|\leq s_f(\gamma(t))^{m-1} \quad\text{for a.e. }t\in(0,T),\tag{17}\] and \[\label{descent-T} f(x) - f(\gamma(t)) \geq \int_0^t s_f^m(\gamma(s))\,ds - \kappa(T), \quad \text{ for every } t \in (0, T).\tag{18}\] Due to Step 1, we observe that \(\EuScript A_x\) is nonempty. Indeed, let \(T_x\) be defined as in Step 1. Applying the local existence result in Step 1 at \(x\) with any error \(\eta < \kappa(T_x)\) yields the existence of an admissible pair \((T_x, \gamma) \in \EuScript A_x\).

For any \((T, \gamma), (T', \gamma') \in \EuScript A_x\), we write

Observe that \((\EuScript A_x, \preceq)\) is partially ordered.

Proof of Claim 3. Let \(\{ (T_i, \gamma_i) \}_{i \in I}\) be a chain in \(\EuScript A_x\). Set \(T_\ast = \sup_{i \in I} T_i \in (0, + \infty]\). By total ordering, the curves are compatible: if \(T_i\le T_j\), then \(\gamma_j=\gamma_i\) on \([0, T_i)\). Thus we may define a curve \(\gamma_*\) on \([0,T_*)\) by setting \[\gamma_\ast(t):=\gamma_i(t) \quad \text{ whenever } t < T_i.\] It follows directly that \(\gamma_\ast\) is absolutely continuous, \(\gamma_\ast(0) = x\) and \[\| \dot{\gamma}_\ast(t) \| \leq s_f^{m - 1}(\gamma_\ast(t)) \quad \text{ for a.e } t \in (0, T_\ast).\] The curve \(\gamma_\ast\) also satisfies the inequality 18 for \(t \in (0, T_\ast)\). Indeed, for any fixed \(t < T_\ast\), there exists \(i \in I\) such that \(t < T_i\). Therefore, we have \[\begin{align} f(x) - f(\gamma_\ast(t)) = f(x) - f(\gamma_i(t)) \geq \int_0^t s_f^m(\gamma_\ast(\tau)) \, d\tau - \kappa(T_i) \geq \int_0^t s_f^m(\gamma_\ast(s)) \, ds - \kappa(T_\ast), \end{align}\] where we have used that \(\kappa(T_i) \leq \kappa(T_\ast)\). We have proved that \((T_\ast, \gamma_\ast) \in \EuScript A_x\). Therefore, every chain in \((\EuScript A_x, \preceq)\) admits an upper bound in \(\EuScript A_x\). \(\Diamond\)

Using Claim 3, it follows from Zorn’s lemma that \((\EuScript A_x, \preceq)\) has a maximal element \((T_{\max}, \gamma)\). We prove that \(T_{\max} = + \infty\). Arguing by contradiction, assume that \(T_{\max} < + \infty\). Notice first that for every \(0 \leq s < t < T_{\max}\) we have: \[\| \gamma(s) - \gamma(t) \| \leq \int_s^t \| \dot{\gamma}(\tau) \| \, d\tau \underbrace{\,\leq\,}_{\eqref{velocity-T}} \int_s^t s_f^{m-1}(\gamma(\tau)) \, d\tau\,.\] Applying the Hölder inequality for \(p=m\) and \(q=\frac{m}{m-1}\) (where \(\frac{1}{p}+\frac{1}{q}=1\)) we obtain: \[\int_s^t s_f^{m-1}(\gamma(\tau)) \, d\tau \,\leq \, \left( \int_s^t 1^m d\tau \right)^{\frac{1}{m}} \, \left( \int_s^t s_f^{m-1}(\gamma(\tau))^{\frac{m}{m-1}} d\tau \right)^{\frac{m-1}{m}} = (t-s)^{\frac{1}{m}} \left( \int_s^t s_f^{m}(\gamma(\tau)) d\tau \right)^{\frac{m-1}{m}}.\] Finally, using an estimate similar to 18 we obtain: \[\int_s^t s_f^m(\gamma(\tau)) d\tau \leq f(\gamma(s)) - f(\gamma(t)) + \kappa(T_\ast) \leq f(x) - \inf \, f + \varepsilon\,.\] Combining the above, we obtain: \[\| \gamma(s) - \gamma(t) \| \, \leq \, C\,(t-s)^{\frac{1}{m}}\quad\text{where } \, C:=\left( f(x) - \inf \,f + \varepsilon \right)^{\frac{m-1}{m}}\,.\] Since \(X\) is complete, the limit \(\bar x := \gamma(T_{\max}) = \lim_{t \nearrow T_{\max} } \gamma(t)\) exists.

If \(s_f(\bar x) = 0\), take any \(\alpha > 0\) and define \(\widetilde{\gamma}:[0,T_{\max}+\alpha)\to X\) by \[\widetilde{\gamma}(t) := \begin{cases} \gamma(t), & 0\le t \leq T_{\max},\\ \phantom{t}\bar x\,, & T_{\max}\le t < T_{\max}+\alpha . \end{cases}\] One can directly check that \((T_{\max} + \alpha, \widetilde{\gamma}) \in \EuScript A_x\), which contradicts the maximality of \((T_{\max}, \gamma)\).

Consider the case \(s_f(\bar x) > 0\). Let \(T_{\bar x} > 0\) be defined as in Step 1 (at \(\bar x\)). Fix \(0 < \alpha < T_{\bar x}\). Since \(\kappa\) is strictly increasing, we may choose \[0 < \eta < \kappa(T_{\max} + \alpha) - \kappa(T_{\max}).\] Thanks to Step 1, there exists a curve \(\gamma_{\bar x} : [0, T_{\bar x}) \to X\) such that \(\gamma_{\bar x}(0) = \bar x\) and it satisfies 13 –14 with the error \(\eta\). Define \(\widetilde{\gamma}:[0,T_{\max} + \alpha)\to X\) by \[\widetilde{\gamma}(t):= \begin{dcases} \phantom{tri}\gamma(t), & 0\le t\le T_{\max},\\ \gamma_{\bar x}(t-T_{\max}), & T_{\max}\le t < T_{\max}+\alpha . \end{dcases}\]

We shall prove that \((T_{\max} + \alpha, \widetilde{\gamma}) \in \EuScript A_x\). Observe first that \(\widetilde{\gamma}\) satisfies the estimate 17 for a.e \(t \in (0, T_{\max} + \alpha)\). It remains to check that it satisfies 18 for every \(t \in (0, T_{\max} + \alpha)\). It suffices to consider the case \(t = T_{\max} + t^\prime\) for any \(t^\prime \in [0, \alpha)\). Using the monotone convergence theorem and the continuity of \(f\), we obtain \[\begin{align} f(x) - f(\widetilde{\gamma}(T_{\max})) = & \, f(x) - \lim_{t \nearrow T_{\max}} f(\gamma(t)) \\ \geq & \, \lim_{t \nearrow T_{\max}} \int_0^{t} s_f^m(\widetilde{\gamma}(\tau)) \, d\tau - \kappa(T_{\max}) \geq \int_0^{T_{\max}} s_f^m(\widetilde{\gamma}(\tau)) \, d\tau - \kappa(T_{\max} + \alpha). \end{align}\] Then, we have, for any \(t = T_{\max} + t^\prime \in (T_{\max}, T_{\max} + \alpha)\) \[\begin{align} f(x) - f(\widetilde{\gamma}(t)) = &\, f(x) - f(\gamma_{\bar x}(t^\prime)) \\ = & \, f(x) - f(\gamma(T_{\max})) + f(\bar x) - f(\gamma_{\bar x}(t^\prime)) \\ \geq & \, \int_0^{T_{\max}} s_f^{m}(\gamma(\tau)) \, d\tau - \kappa(T_{\max}) +\int_0^{t^\prime} s_f^{m}(\gamma_{\bar x}(\tau)) \, d\tau - \eta \\ = & \, \int_0^t s_f^{m}(\widetilde{\gamma}(\tau)) \, d\tau - (\eta + \kappa(T_{\max})) \\ \geq & \, \int_0^t s_f^{m}(\widetilde{\gamma}(\tau)) \, d\tau - \kappa(T_{\max} + \alpha), \end{align}\] where the last inequality follows from the choice of \(\eta\). Thus, we have \((T_{\max} + \alpha, \widetilde{\gamma}) \in \EuScript A_x\), again contradicting the maximality of \((T_{\max}, \gamma)\).

In conclusion, we have proved that the maximal element \((T_{\max}, \gamma)\) of \((\EuScript A_x, \preceq)\) satisfies \(T_{\max} = + \infty\). Using the fact that \(\kappa(+ \infty) = \varepsilon\), we conclude that the curve \(\gamma\) satisfies the desired estimates. This proves Lemma 4. ◻

Proof. Since \(f\) is continuous, writing \(s_f(x)\) as \[s_f(x) := \lim_{n\to \infty} \,\underbrace{\sup_{0 < \|h\|<\frac{1}{n}}\,\left(\dfrac{\max \{ f(x) - f(x+h), 0 \}}{\| h \|}\right)}_{:=\phi_n(x) \,\,\text{ (lsc function of x)}}\,, \quad\text{for every } x\in X,\] we deduce that \(s_f\) is a Baire-\(2\) function and consequently Baire-measurable ¹. The result follows directly from [13]. ◻

Proof of Theorem 6. Notice that if the function \(s_f\) is convex, then so is the function \(s_f^m\), for every \(m\geq 1\). Therefore, it suffices to establish the result for \(m > 1\).
Notice that according to Proposition 4, the convex function \(s_f^m\) is continuous and consequently, locally Lipschitz. Fix \(x, y \in X\) and \(\varepsilon > 0\). Thanks to Lemma 4, for each \(z \in \{x, y\}\), there exists a curve \(\gamma_z: [0, + \infty) \to X\) such that \(\gamma_z(0) = z\), \[\label{kmfdupje} \| \dot{\gamma_z}(t) \| \leq s_f^{m - 1}(\gamma_z(t)) \quad \text{ for a.e } t > 0,\tag{19}\] and \[\label{eq46gamma95epsi} f(z) - f(\gamma_z(t)) \geq \int_0^t V_m(\gamma_z(\tau)) \,d\tau - \varepsilon \quad \text{ for every } t > 0.\tag{20}\] Set \[\begin{align} \mu(t) := & \,\dfrac{\gamma_x(t) + \gamma_y(t)}{2} \qquad \text{and} \qquad D(t) := \, \dfrac{1}{2} \Big( f(\gamma_x(t)) + f(\gamma_y(t)) \Big) - f(\mu(t)). \end{align}\]

We first prove that for every \(t > 0\), one has \(D(t) - 2 \varepsilon \leq D(0)\). Indeed, it follows from 20 that \[D(t) - D(0) \leq - \dfrac{1}{2}\int_0^t \Big( V_m(\gamma_x(\tau)) + V_m(\gamma_y(\tau)) \Big) \, d\tau + 2 \varepsilon + f(\mu(0)) - f(\mu(t)).\] Applying Lemma 3 to the curve \(\mu: [0, t] \to X\), we obtain \[f(\mu(0)) - f(\mu(t)) \leq \int_0^t s_f(\mu(\tau)) \| \dot{\mu}(\tau) \| \, d\tau.\] Hölder’s inequality then yields \[\int_0^t s_f(\mu(\tau)) \| \dot{\mu}(\tau) \| \,d\tau \leq \left( \int_0^t V_m(\mu(\tau)) \,d\tau \right)^\frac{1}{m} \left( \int_0^t \| \dot{\mu}(\tau) \|^\frac{m}{m - 1} \,d\tau \right)^{\frac{m - 1}{m}}.\] On the one hand, it follows from the convexity of \(V_m\) that \[\int_0^t V_m(\mu(\tau)) \,d\tau \, \leq \, \dfrac{1}{2} \int_0^t V_m(\gamma_x(\tau)) + V_m(\gamma_y(\tau)) \, d\tau.\] On the other hand, recall that for each \(z \in \{ x, y \}\), one has \(\| \dot{\gamma}_z \| \leq s_f^{m - 1}(\gamma_z(t))\) for a.e \(t > 0\). Hence the convexity of the map \(w \mapsto \| w \|^\frac{m}{m - 1}\) yields \[\int_0^t \| \dot{\mu} (\tau) \|^\frac{m}{m - 1} \,d\tau \leq \dfrac{1}{2} \int_0^t \Big( \| \dot{\gamma}_x(\tau) \|^\frac{m}{m - 1} + \| \dot{\gamma}_x(\tau) \|^\frac{m}{m - 1} \Big) \, d\tau \leq \dfrac{1}{2} \int_0^t V_m(\gamma_x(\tau)) + V_m(\gamma_y(\tau)) \, d\tau.\] Therefore, \[f(\mu(0)) - f(\mu(t)) \leq \dfrac{1}{2} \int_0^t V_m(\gamma_x(\tau)) + V_m(\gamma_y(\tau)) \, d\tau\] which yields \[\label{D95differ95epsi} D(t) - D(0) \leq 2\varepsilon.\tag{21}\]

We now quantify the difference between \(f(\mu(t))\) and \(f(\gamma_z(t))\) for each \(z \in \{x, y\}\). Fix \(z \in \{ x, y \}\) and \(t > 0\). Applying Lemma 3 to the straight-line segment connecting \(\gamma_z(t)\) and \(\mu(t)\), we get \[\label{f-differ} \begin{align} f(\gamma_z(t)) - f(\mu(t)) \leq & \, \int_0^1 s_f(\underbrace{\mu(t) + \tau(\gamma_z(t) - \mu(t))}_{\xi(\tau)}) \, \| \underbrace{\gamma_z(t) - \mu(t)}_{\dot{\xi}(\tau)} \| \, d\tau \\ = & \, \int_0^1V_m(\mu(t) + \tau(\gamma_z(t) - \mu(t))^{1/m } \| \gamma_z(t) - \mu(t) \| \, d\tau \\ \leq & \, \left( V_m(\gamma_x(t)) + V_m(\gamma_y(t)) \right)^{1/m}\| \gamma_z(t) - \mu(t) \|, \end{align}\tag{22}\] where convexity of \(V_m\) is used to obtain the last inequality. Furthermore, it follows from 11 and Hölder’s inequality that \[\| \gamma_z(t) - z \| \leq \int_0^t \| \dot{\gamma_z}(\tau) \| \, d\tau \leq t^{1/m} \left( \int_0^t s_f^m(\gamma_z(\tau)) \, d\tau \right)^{\frac{m - 1}{m}} \leq C_z t^{1/m},\] where \(C_z := \big( f(z) - \inf \, f + \varepsilon \big)^{(m - 1)/m}\). Therefore, there exists a constant \(C > 0\), depending on \(f, x, y\) and \(\varepsilon > 0\), such that \[\label{gama95mu} \| \gamma_z(t) - \mu(t) \| \leq C t^{1/m} \quad \text{ for every } t > 0.\tag{23}\] Combining 22 and 23 , we arrive at \[f(\gamma_z(t)) - f(\mu(t)) \leq C t^{1/m}\left( V_m(\gamma_x(t)) + V_m(\gamma_y(t)) \right)^{1/m}.\] By a similar argument, we obtain \[\label{f95differ95abs} \big| f(\gamma_z(t)) - f(\mu(t)) \big| \leq C t^{1/m}\left( V_m(\gamma_x(t)) + V_m(\gamma_y(t)) \right)^{1/m} \quad \text{ for every } t > 0 \text{ and } z \in \{x, y\}.\tag{24}\]

To continue, we need the following fundamental fact.

Proof of Claim 8. Arguing by contradiction, assume that there exists \(\varepsilon > 0\) and \(T > 0\) such that \[\begin{align} t h(t) > \varepsilon \,\, \text{ for every } t > T. \end{align}\] It follows that \[\int_T^\infty h(t) \, dt \geq \varepsilon \int_T^\infty \dfrac{dt}{t} = + \infty,\] which contradicts the fact that \(h\) is integrable. \(\Diamond\)

Thanks to 20 , we observe that \[\int_0^\infty V_m(\gamma_x(t)) + V_m(\gamma_y(t)) \, dt< + \infty.\] Therefore, applying Claim 8 to the case \(h(t) = V_m(\gamma_x(t)) + V_m(\gamma_y(t))\), there exists \(\{ t_n \}\) such that \(t_n \underset{n \to \infty}{\longrightarrow} \infty\) and \[t_n \Big( V_m(\gamma_x(t_n)) + V_m(\gamma_y(t_n)) \Big) \to 0 \text{ as } n \to \infty.\] Substituting \(t = t_n\) into 24 , we obtain \[|D(t_n)| \leq \dfrac{1}{2} \Big( |f( \gamma_x(t_n) ) - f( \mu(t_n) ) | + |f( \gamma_y(t_n) ) - f( \mu(t_n) ) | \Big) \to 0 \quad \text{ as } n \to \infty.\] We have proved that there exists a sequence \(\{t_n\}\) such that \[\label{limD95zero} \lim_{n \to \infty} D(t_n) = 0.\tag{25}\]

To conclude, combining 21 and 25 , we get \(D(0) \geq - 2\varepsilon.\) Since \(\varepsilon > 0\) is chosen arbitrarily, we deduce \(D(0) \geq 0\), equivalently \[f \left( \dfrac{x + y}{2} \right) \leq \dfrac{1}{2} (f(x) + f(y)).\] Therefore, \(f\) is convex, which completes the proof. ◻

Proof. We first prove that the convex subdifferential \(\partial f\) is a singleton at every point in \(\mathcal{H}\). Indeed, arguing by contradiction, we assume that there exists \(\bar x \in \mathcal{H}\) such that \(\partial f(\bar x)\) is not a singleton. Then, there exists \(\xi_0 \in \partial f(\bar x)\) with \(\xi_0 \neq \xi_{\bar x}\), where \(\{ \xi_{\bar x} \} = \mathrm{argmin} \big\{ \| \xi \| : \xi \in \partial f(\bar x) \big\}\). Consequently, \[\label{eq46contra} \| \xi_0 \| > \|\xi_{\bar x}\| = \min_{\xi \in \partial f(\bar x)} \| \xi \| = s_f(\bar x).\tag{26}\] Denote \(e := \xi_0 / \| \xi_0 \|\). For any fixed \(t > 0\), it follows from the monotonicity of \(\partial f\) that \[\langle \eta - \xi_0, te \rangle \geq 0 \,\, \text{ for every } \eta \in \partial f(\bar x + te).\] Hence, using the Cauchy–Schwarz inequality and the fact that \(\| e \| = 1\), we get \[\| \eta \| \geq \langle \eta, e \rangle \geq \langle \xi_0, e \rangle = \| \xi_0 \| \,\, \text{ for every } \eta \in \partial f(\bar x + te).\] Taking the infimum with respect to \(\eta\), we obtain \[s_f(\bar x + te) \geq \| \xi_0 \| \,\, \text{ for every } t > 0,\] which, thanks to the continuity of slope, leads to \[s_f(\bar x) = \lim_{t \to 0^+} s_f(\bar x + te) \geq \| \xi_0 \|,\] which contradicts 26 . Therefore, \(\partial f\) is singleton at every point in \(\mathcal{H}\).

To conclude, we will prove that \(\nabla f(\cdot)\) is continuous. Let \(\{ x_n \} \subset \mathcal{H}\) and \(x \in \mathcal{H}\) be such that \(x_n \to x\) as \(n \to \infty\). It follows that \(\| \nabla f (x_n) \| \to \| \nabla f (x)\|\) as \(n \to \infty\). Hence \(\{\nabla f (x_n) \}_n\) is bounded and so there exist \(\xi \in \mathcal{H}\) such that, up to a subsequence, \(\nabla f(x_n) \rightharpoonup \xi\) as \(n \to \infty\). One can show that \(\xi \in \partial f(x)\) and hence \(\xi = \nabla f(x)\). Since \(\{ \nabla f(x_n) \}\) converges weakly and in norm to \(\nabla f(x)\), we infer that it strongly converge to \(\nabla f(x)\). Proposition 9 is proven. ◻

Proof. \((i)\) Fix \(\bar x \in \mathbb{R}\) and choose \(\delta >0\) sufficiently small. Set \(I := [\bar x-\delta,\bar x+\delta]\). Since \(|f'|\) is locally \(\beta\)–Hölder, there exists \(C_I>0\) such that \[\big||f'|(x)-|f'|(y)\big| \leq C_I |x-y|^\beta \qquad \text{for every } x,y\in I.\] We claim that \(f'\) is \(\beta\)–Hölder on \(I\). Fix \(x,y\in I\). The case \(f'(x)\cdot f'(y) = 0\) is immediate. If \(f'(x)\) and \(f'(y)\) have the same sign, then \[|f'(x)-f'(y)| = \big||f'|(x)-|f'|(y)\big| \leq C_I |x-y|^\beta .\]

It remains to consider the case in which \(f'(x)\) and \(f'(y)\) have opposite signs. By the continuity of \(f'\), there exists \(z\) between \(x\) and \(y\) such that \(f'(z)=0\). Hence \[|f'(x)| \leq C_I |x-z|^\beta \text{ and } |f'(y)| \leq C_I |y-z|^\beta\] The concavity of the map \(r \mapsto r^\beta\) implies that \[|x-z|^\beta+|y-z|^\beta \leq 2^{1 - \beta} (|x - z| + |y - z|)^\beta = 2^{1-\beta}|x-y|^\beta.\] Therefore, \[|f'(x)-f'(y)| \leq |f'(x)|+|f'(y)| \leq2^{1 -\beta} C_I |x - y|^\beta.\] Thus \(f'\in \mathcal{C}^{0,\beta}(I)\). Since \(\bar x\) was arbitrary, we conclude that \(f\in \mathcal{C}^{1,\beta}_{\mathrm{loc}}(\mathbb{R})\).

\((ii)\) Fix \(0<\beta\leq 1\) and define \[f(x,y)=e^x+\frac{1}{1+\beta/2}|y|^{1+\beta/2}, \qquad \text{ for } (x,y)\in \mathbb{R}^2 .\] Observe first that \(f\geq 0\), \(f\in \mathcal{C}^1(\mathbb{R}^2)\) and is convex. Also a direct computation yields \[\nabla f(x,y) = \big(e^x,\operatorname{sgn}(y)|y|^{\beta/2}\big) \quad \text{ and } \quad V(x, y) := \| \nabla f(x, y) \| = \big(e^{2x}+|y|^\beta\big)^{1/2}.\] It is straightforward to see that \(\nabla f\) is locally \((\beta/2)\)–Hölder and fails to be \(\beta\)–Hölder around the origin.

We claim that \(V\) is locally \(\beta\)–Hölder. Fix \((\bar x,\bar y)\in\mathbb{R}^2\) and let \(K\subset\mathbb{R}^2\) be a compact neighborhood of \((\bar x,\bar y)\). For \(z=(x,y)\) and \(z'=(x',y')\in K\), we have \[\begin{align} |V(z)-V(z')| &= \Big| \big(e^{2x}+|y|^\beta\big)^{1/2} - \big(e^{2x'}+|y'|^\beta\big)^{1/2} \Big| = \frac{\big|e^{2x}-e^{2x'}+|y|^\beta-|y'|^\beta\big|}{ \big(e^{2x}+|y|^\beta\big)^{1/2} + \big(e^{2x'}+|y'|^\beta\big)^{1/2} }. \end{align}\] Since \(e^{2x}\geq 1\) for all \(x\in\mathbb{R}\), we get \[|V(z)-V(z')| \leq \frac{1}{2} |e^{2x}-e^{2x'}| + \frac{1}{2} \bigl||y|^\beta-|y'|^\beta\bigr|.\] Note that \(x \mapsto e^{2x}\) is smooth in \(\mathbb{R}\) and for \(0<\beta\leq 1\), the map \(t\mapsto |t|^\beta\) is globally \(\beta\)–Hölder on \(\mathbb{R}\). Therefore, there exists \(C_K > 0\) such that \[|V(z)-V(z')| \leq C_K\big(|x-x'|^\beta+|y-y'|^\beta\big) \leq C_K |z-z'|^\beta \qquad\text{for all } z,z'\in K.\] Hence \(V=\|\nabla f\|\) is locally \(\beta\)–Hölder. This completes the proof. ◻