Remark 2.4. The proof of existence is possible with a wide variety of regularity assumptions on the energy $\mathcal{E}$ , which can be tailored to a variety of different situations. For example, if the sequentially $\sigma$ -lower semicontinuous envelope of $|\partial \mathcal{E}|$

\begin{align*} |\partial ^- \mathcal{E}|\;:\!=\;\big \{ \liminf _{n\rightarrow \infty }|\partial \mathcal{E}|({x}^n)\,:\,{x}^n \stackrel {\sigma }{\rightharpoonup }{x},\sup _n{d({x}^n,{x}),\mathcal{E}({x}^n)}\lt +\infty \big \} \end{align*}

is a strong upper gradient, one can drop Assumption 2.b and instead prove existence of curves of maximal slope with respect to $|\partial ^- \mathcal{E}|$ . Further, if $|\partial \mathcal{E} |$ (or $|\partial ^- \mathcal{E}|$ respectively) is only a weak upper gradient (compare [Reference Ambrosio, Gigli and Savaré2, Theorem 2.3.3]), then Assumption 2.a has to be replaced by continuity of the energy.

Remark 2.6. The function defined in (2.3) is similarly employed in [3, 16, 17] and the proof strategy as displayed in Figure 2 resembles the max-ball arguments as in the previously mentioned works. The expression in (2.3) can also be seen as the infimal convolution [Reference Fenchel and Blackett39, Reference Hausdorff49] of $\mathcal{E}$ and $\chi _{\overline {B_\tau (0)}}$ , i.e., $\mathcal{E}_\tau = \chi _{\overline {B_\tau }}\, \square \, \mathcal{E}$ and can also be considered as the limit $p\to \infty$ of the Moreau envelope [72],

\begin{align*} \inf _{\tilde {{x}}} \left \{\mathcal{E}(\tilde {{x}}) + \frac {1}{p}\left \|{x} - \tilde {{x}}\right \|^p\right \} \end{align*}

which is typically defined for $p=2$ .

Remark 2.7. More recently, similar schemes to the one defined in (MinMove) have been introduced in an optimization context in [48]. Here, the operation on the right-hand side of (MinMove) was labelled the “ball-proximal” or “brox” operator.

Proof. Let ${x}\in \operatorname {dom}(\mathcal{E})$ , for any $\tau ^*\lt \infty$ we know that the mapping $\tau \mapsto \mathcal{E}_\tau ({x})$ is monotone decreasing on $[0,\tau ^*]$ and thus its variation can be bounded,

\begin{equation*} \mathcal{E}_0({x}) - \mathcal{E}_{\tau ^*}({x})= \mathcal{E}({x})-\mathcal{E}({x}_{\mathrm{min},\tau ^*})\lt \infty .\end{equation*}

Employing [Reference Saks90, Theorem 9.6, Chapter IV], this yields that the derivative exists for almost every $t\in (0,\tau ^*)$ and that (2.5) holds. To show (2.6), we observe that

\begin{equation*}B_r({x}_{\mathrm{min},\tau })\subset B_{\tau +r}({x}) \text{ and thus }\mathcal{E}_{\tau +r}({x})\leq \mathcal{E}_r({x}_{\mathrm{min},\tau }),\end{equation*}

see Figure 2, which yields

\begin{align*} -\left (\frac {\mathcal{E}({x}_{\mathrm{min},\tau })-\mathcal{E}_{\tau +r}({x})}{r}\right ) \leq -\left (\frac {\mathcal{E}({x}_{\mathrm{min},\tau })-\mathcal{E}_{r}({x}_{\mathrm{min},\tau })}{r}\right ). \end{align*}

It follows that

\begin{align*} \frac {\text{d}}{\text{d}\tau} \mathcal{E}_\tau ({x})&=\lim _{r\rightarrow 0} \frac {\mathcal{E}_{\tau +r}({x})-\mathcal{E}_{\tau }({x})}{r}\\ &= -\lim _{r\rightarrow 0} \frac {\mathcal{E}({x}_{\mathrm{min},\tau })-\mathcal{E}_{\tau +r}({x})}{r}\\ &\leq -\limsup _{r\rightarrow 0} \frac {\mathcal{E}({x}_{\mathrm{min},\tau })-\mathcal{E}_{r}({x}_{\mathrm{min},\tau })}{r}=-|\partial \mathcal{E}|({x}_{\mathrm{min},\tau }), \end{align*}

where we used the characterization of the slope from Lemma 2.8.

Proof. We consider the set of all possible iterates in the minimizing movement scheme $K=\{{x}^i_{\tau _n}\,:\, 0\leq i\leq n, n\in \mathbb{N}\}\subset \mathcal{S}$ . Recalling Definition 2.5, for every $n\in \mathbb{N}$ and $i,j \in \{0,\ldots ,n\}$ , we have the estimate

\begin{align*} d({x}_{\tau _n}^i,{x}_{\tau _n}^j)\leq \sum _{k=i}^{j-1} d \big({x}_{\tau _n}^k,{x}_{\tau _n}^{k+1} \big) \leq (j-i)\ {\tau _n} \leq T, \end{align*}

and therefore for every $n,m\in \mathbb{N}$ and $0\leq i\leq n,\ 0\leq j\leq m$ , we have

\begin{align*} d({x}_{\tau _n}^i,{x}_{\tau _m}^j)\leq d({x}_{\tau _n}^i,{x}^0) + d({x}^0,{x}_{\tau _m}^j)\leq 2T. \end{align*}

Furthermore, since ${x}^0\in \operatorname {dom}(\mathcal{E})$ , we also know that $\mathcal{E}({x}_{\tau _n}^i) \leq \mathcal{E}({x}^0) \lt \infty$ and thus $K$ is a $d$ -bounded set, contained in sublevels of $\mathcal{E}$ . Using relative compactness, i.e., Assumption 1.b, this ensures that $\overline {K}$ is a $\sigma$ -sequentially compact set and therefore fulfils 1 of Proposition B.1. In order to apply the latter, it remains to choose a function $\omega$ that fulfils 2. For this, we consider the sequence of curves $\left |{x}'_{\tau _n}\right |\;:\;[0,T]\to {\mathbb{R}}$ , which is by definition bounded in $L^\infty (0,T)$ , i.e.,

\begin{equation*}\left \|\left |{x}'_{\tau _n}\right |\right \|_{L^\infty (0,T)}\leq 1, \qquad \text{for every }n\in \mathbb{N}.\end{equation*}

For fixed $0\leq s \leq t \leq T$ , let us define

(2.11) \begin{align} s(n)\;:\!=\; \min _{k\in \{0,\ldots , n\}} \{ k\cdot \tau _n \,:\, s \leq k\cdot \tau _n \},\qquad t(n)\;:\!=\; \min _{k\in \{0,\ldots , n\}} \{ k\cdot \tau _n \,:\, t \leq k\cdot \tau _n \}. \end{align}

Using the triangle inequality and the fact that the distance between two consecutive iterates is bounded by $\tau$ , we obtain

(2.12) \begin{align} \limsup _{n\rightarrow +\infty } d(\bar {x}_{\tau _n}(s),\bar {x}_{\tau _n}(t)) &\leq \limsup _{n\rightarrow +\infty } \sum _{i=1}^{\frac {t(n)-s(n)}{\tau _n}} d(\bar {x}_{\tau _n}(s(n)+(i-1)\tau ),\bar {x}_{\tau _n}(s(n)+i\tau _n)) \end{align}

(2.13) \begin{align} &\leq \lim _{n\rightarrow +\infty } (t(n)-s(n))=|t-s |\;=\!:\; \omega (s,t). \end{align}

Therefore, 2 in Proposition B.1 is fulfilled, allowing us to apply [Reference Ambrosio, Gigli and Savaré2, Proposition 3.3.1] to extract another subsequence such that

\begin{align*} \bar {x}_{\tau _n}(t) \stackrel {\sigma }{\rightharpoonup }u(t) \text{ as } n\rightarrow \infty \quad \forall t\in [0,T], \quad u \text{ is } d\text{-continuous in } [0,T]. \end{align*}

This in particular ensures $u(0)={x}^0$ and (2.12) together with Assumption 1.a yields 1-Lipschitzness of $u$ , since for $s\leq t$ , we have

\begin{align*} d(u(s), u(t)) \leq \liminf _{n\to \infty } d(\bar {x}_{\tau _n}(s),\bar {x}_{\tau _n}(t))\leq t-s. \end{align*}

By construction, it holds that $d(\bar {x}_{\tau _n},\tilde {x}_{\tau _n})\leq \tau$ , which also yields

\begin{align*} \tilde {x}_{\tau _n}(t) \stackrel {\sigma }{\rightharpoonup }u(t) \text{ as } n\rightarrow \infty \quad \forall t\in [0,T]. \end{align*}

Observing that $\tilde {x}_{\tau _n}(0)=x^0=u(0)$ independent of $n$ , we take the limes inferior for (2.9) and use Assumption 2.a and Fatou’s lemma to obtain for all $t\in [0,T]$

\begin{align*} \mathcal{E}(u(0))&\geq \liminf _{n\rightarrow \infty } \left \{\mathcal{E}(\tilde {x}_{\tau _n}(t))-\int _0^t D_{\tau _n}(r)\,\text{d}r \right \} \geq \mathcal{E}(u(t))+\int _0^t \liminf _{n\rightarrow \infty } -D_{\tau _n}(r)\,\text{d}r\\ &\geq \mathcal{E}(u(t)) +\int _0^t |\partial \mathcal{E}(u(r))|\,\text{d}r. \end{align*}

The last inequality follows by the estimate

\begin{align*} |\partial \mathcal{E}|(u(t))\leq \liminf _{n\rightarrow \infty } |\partial \mathcal{E}|(\tilde {x}_{\tau _n}(t))\leq \liminf _{n\rightarrow \infty } -D_{\tau _n}(t)\quad \text{ for a.e.}\quad t\in (0,T), \end{align*}

which is a consequence of (2.8) and (2.6) and the $\sigma$ -lower semicontinuity of the slope. On the other hand, we know that $|\partial \mathcal{E}|$ is a strong upper gradient and $|u'|(r)\leq 1$ for a.e. $r \in [0,T]$ , such that

\begin{align*} \mathcal{E}(u(0))\leq \mathcal{E}(u(t))+\int _0^t |\partial \mathcal{E} |(u(r)) | u'|(r)\,\text{d}r\leq \mathcal{E}(u(t)) +\int _0^t |\partial \mathcal{E} |(u(r))\,\text{d}r. \end{align*}

In particular, the equality

\begin{align*} \mathcal{E}(u(0))= \mathcal{E}(u(t))+\int _0^t |\partial \mathcal{E} |(u(r)) | u'|(r)\,\text{d}r= \mathcal{E}(u(t)) +\int _0^t |\partial \mathcal{E} |(u(r))\,\text{d}r \end{align*}

must hold. It follows that $t\mapsto \mathcal{E}(u(t))$ is locally absolutely continuous and

\begin{align*} \frac {\text{d}}{\text{d}t}\mathcal{E}(u(t))=-|\partial \mathcal{E}|(u(t)) |u'|(t)=-|\partial \mathcal{E}|(u(t)) \text{ for a.e.}t\in (0,T). \end{align*}

Proof. We choose $\sigma$ to be the norm topology, such that Assumptions 1.a and 1.b are fulfilled and $ E$ fulfils Assumption 2.a. By Proposition 3.1, $\left |\partial E \right |$ is lower semicontinuous and a strong upper gradient. Therefore, also Assumption 2.b is fulfilled and the application of Theorem 2.11 yields the desired result.

Remark 3.3. Similar to [Reference Ambrosio, Gigli and Savaré2, section 3.1], we remark some properties of $E_\tau (x)=\inf _{\tilde {{x}}\in \overline {B_\tau }(x)}E(\tilde {{x}})$ in the case, when $E$ is convex. Since $E_\tau (x)$ is defined via the infimal convolution, see Remark 2.6, we can directly infer convexity in the $x$ argument, if $E$ was already convex. Furthermore, we also have convexity in $\tau$ , which can be seen as follows. Let $\tau _1,\tau _2\geq 0$ be arbitrary, where we also allow them to attain $0$ . For any $z_1\in \overline {B_{\tau _1}}(x), z_2\in \overline {B_{\tau _2}}(x)$ , we have that $\lambda z_1 + (1-\lambda ) z_2 \in \overline {B_{\tilde {\tau }}}(x)$ with $\tilde {\tau }=\lambda \tau _1 + (1-\lambda )\tau _2$ for any $\lambda \in [0,1]$ . The definition of $E_\tau$ and the convexity of $E$ yield

\begin{align*} E_{\tilde {\tau }}(x)\leq E (\lambda z_1 + (1-\lambda ) z_2) \leq \lambda E (z_1) + (1-\lambda ) E (z_2) \end{align*}

and since $z_1\in \overline {B_{\tau _1}}(x), z_2\in \overline {B_{\tau _2}}(x)$ were arbitrary, we obtain

\begin{align*} E_{\tilde {\tau }}(x)\leq \lambda E_{\tau _1}(x) + (1-\lambda ) E_{\tau _2}(x). \end{align*}

If $x\in \operatorname {dom}(E))$ , we have that $\operatorname {dom}(\tau \mapsto E_\tau (x))=[0,\infty )$ and thus $\tau \to E_\tau$ is continous on $(0,\infty )$ . If $ E$ is lower semicontinuous, we also obtain continuity at $0$ .

Proof. Let ${x}\in \operatorname {dom}(E)$ with $|\partial E({x})|\not = 0$ , then the mapping $\tau \mapsto E _\tau ({x})$ is non-increasing and not constant. Therefore, we can find an $\epsilon \gt 0$ such that $E_\tau ({x})\gt E_{\epsilon }({x})$ for all $0\lt \tau \lt \epsilon$ . Let $\left (\tilde {{x}}^{n}\right )_{n\in \mathbb{N}}$ be a sequence such that

\begin{align*} \lim _{n\to \infty } E (\tilde {{x}}_n) = E _\tau ({x}). \end{align*}

Since $ E_\tau ({x})\gt E _{\epsilon }({x})$ , we can find an element $\hat {{x}}$ that fulfils

\begin{align*} E (\tilde {{x}}_n)\gt E (\hat {{x}})\qquad \text{for every } n\in \mathbb{N}\qquad \text{and}\qquad \tau \lt \left \|{x}- \hat {{x}}\right \| \leq {\epsilon }. \end{align*}

Since $\hat {{x}}\notin \overline {B_\tau }({x})$ and $\tilde {{x}}_n\in \overline {B_\tau }({x})$ , the line between each pair $(\hat {{x}}, \tilde {{x}}_n)$ ,

\begin{align*} c_n\;:\;t\in [0,1]\mapsto t\hat {{x}}+(1-t)\tilde {{x}}_n \end{align*}

has to intersect the sphere $\partial B_\tau ({x})$ at some point $t_n\in [0,1)$ , where we define the intersection point as ${x}_n=c_n(t_n)\in \partial B_\tau ({x})$ . Due to convexity, we obtain

\begin{align*} E _\tau ({x}) \leq E ({x}_n) = E (t_n\hat {{x}}+(1-t_n)\tilde {{x}}_n)\leq t_n E (\hat {{x}}) + (1-t_n) E (\tilde {{x}}_n) \leq E (\tilde {{x}}_n). \end{align*}

Note that the last inequality would only be strict if $t_n\neq 0$ ; however, since $\tilde {{x}}_n$ might already be lying on the sphere, we only obtain the weak inequality. The sequence ${x}_n$ now is the desired sequence in (3.2).

In the reflexive case, the weak compactness of the unit ball guarantees weak convergence of a subsequence of $\left ({x}^{n}\right )_{n\in \mathbb{N}}$ to some ${x}_\tau \in \overline {B_\tau }({x})$ . Lower semicontinuity and convexity imply weak lower semicontinuity of $ E$ and thus

\begin{align*} E _\tau ({x})\leq E ({x}_\tau ) \leq \liminf _{n\to \infty } E ({x}_n) = E _\tau ({x}). \end{align*}

As above, we can choose an element $\hat {x}$ with $\left \|\hat {x} - x\right \|\gt \tau$ with $E(x_\tau ) \gt E (\hat {x})$ . Applying the same argument as above, there is some $t\in [0,1)$ such that $t\hat {x} + (1-t) x_\tau$ intersects $\partial B_\tau (x)$ . As above, if $t\neq 0$ , convexity yields

(3.3) \begin{align} E _\tau ({x})\leq E (t\hat {x} + (1-t) x_\tau ) \lt E (x_\tau ), \end{align}

which contradicts the fact that $ E_\tau ({x})= E (x_\tau )$ and thus $x_\tau$ must have already been on the boundary.

Proof. Step 1: The convex case.

We first assume that $E$ is convex. We choose $\tau$ small enough such that by Lemma 3.4, we obtain a sequence $\{{x}_n\}_n$ with $\|{x}-{x}_n\|=\tau$ and $\lim _{n\to \infty } E ({x}_n) = E _\tau ({x})$ . For each $n\in \mathbb{N}$ , we consider the line

\begin{align*} c_n(t)\;:\!=\; t\, {x}_n + (1-t)\, {x} \end{align*}

evaluated at $\tilde {t}=\tilde {\tau }/\tau$ for some $0\lt \tilde {\tau }\lt \tau$ , which yields $\left \|{x}-c_n(\tilde {t})\right \|=\tilde {\tau }/\tau \left \|{x}-{x}_n\right \| = \tilde {\tau }$ . Due to convexity, we obtain

\begin{align*} E (c_n(\tilde {t})) \leq \tilde {t}\, E ({x}_n)+\left (1-\tilde {t}\right )\, E ({x})\qquad \Rightarrow \qquad E ({x}) - E (c_n(\tilde {t})) \geq \tilde {t}\, \left ( E ({x}) - E ({x}_n)\right ). \end{align*}

Using the fact that $ E _{\tilde {\tau }}({x}) \leq E (c_n(\tilde {t}))$ and dividing by $\tilde {\tau }$ in the above inequality yields

\begin{align*} \frac {E ({x})- E _{\tilde {\tau }}({x})}{\tilde {\tau }}\geq \frac { E ({x})- E (c_n(\tilde {t}))}{\tilde {\tau }} \geq \frac { E ({x})- E ({x}_n)}{\tau }. \end{align*}

Considering the limit $n\to \infty$ , we obtain the following inequality,

\begin{align*} \frac {E({x})-E_{\tilde {\tau }}({x})}{\tilde {\tau }}\geq \limsup _{n\rightarrow \infty }\frac { E ({x})- E (c_n(\tilde {t}))}{\tilde {\tau }}\geq \lim _{n\rightarrow \infty }\frac {E({x})-E({x}_n)}{\tau }=\frac {E({x})-E_\tau ({x})}{\tau }. \end{align*}

This shows that $\tau \mapsto Q(\tau ) \;:\!=\; \frac {E({x})-E_\tau ({x})}{\tau }$ is decreasing in $\tau$ , and therefore, for a null sequence $\tau _n\to 0$ , $Q(\tau _n)$ is an increasing sequence. The monotone convergence theorem together with Lemma 2.8 shows (3.4).

Step 2: Extension to $C^1$ -perturbations.

We now assume that $E$ is a $C^1$ -perturbation of a convex function. By the definition of differentiability, we can write

\begin{align*} E (z)=\underbrace { E ^{\mathrm{c}}(z)+ E ^{\mathrm{d}}({x})-\langle D E ^{\mathrm{d}}({x}),{x}-z\rangle }_{\;=\!:\; F ({x})}+R({x},{x}-z), \end{align*}

with $R({x},{x}-z)\in o(|{x}-z|)$ for every $z\in \operatorname {dom}( E )$ . We observe that $ F$ is again a convex function. Let $\epsilon \gt 0$ , then we denote by ${x} ^{ E }_{\tau ,\epsilon },{x}^{ F }_{\tau ,\epsilon }\in \overline {B_\tau }({x})$ the quasi-minimizers that fulfil

\begin{align*} E ({x}^{ E }_{\tau ,\epsilon })- E_{\tau }({x})\leq \tau \epsilon \quad \text{and}\quad F ({x}^{F}_{\tau ,\epsilon })- F_{\tau }({x})\leq \tau \epsilon \quad \text{respectively}. \end{align*}

We use the estimate

\begin{gather*} E _{\tau }({x})- F _{\tau }({x})\leq E _{\tau }({x}) - F ({x}^{ F }_{\tau ,\epsilon }) + \tau \epsilon \\ =\underbrace {E _{\tau }({x})- E ({x}^{ F }_{\tau ,\epsilon })}_{\leq 0} + R({x},{x}-{x}^{ F }_{\tau ,\epsilon })+ \tau \epsilon \leq |R({x},{x}-{x}^{ F }_{\tau ,\epsilon })|+\tau \epsilon \end{gather*}

and analogously

\begin{gather*} F _{\tau }({x})- E _{\tau }({x})\leq F _{\tau }({x})- E ({x}^{E }_{\tau ,\epsilon }) + \tau \epsilon \\ =\underbrace {F _{\tau }({x})- F ({x}^{ E }_{\tau ,\epsilon })}_{\leq 0}-R({x},{x}-{x}^{ E }_{\tau ,\epsilon })+\tau \epsilon \leq |R({x},{x}-{x}^{E}_{\tau ,\epsilon })|+\tau \epsilon , \end{gather*}

to obtain

(3.5) \begin{align} |E_{\tau }({x})-F_{\tau }({x})|\leq \max \left \{ |R({x},{x}-{x}^{E}_{\tau ,\epsilon })|, |R({x},{x}-{x}^{F}_{\tau ,\epsilon })|\right \}+\tau \epsilon . \end{align}

Using that $ E ({x}) =F ({x})$ and dividing by $\tau$ in (3.5) yields the inequality

(3.6) \begin{gather} \left |\frac {E({x})-E_{\tau }({x})}{\tau }-\frac {F({x})-F_{\tau }({x})}{\tau } \right |\leq \underbrace {\frac {\max \{ |R({x},{x}-{x}^{ E }_{\tau }(\epsilon ))|, |R({x},{x}-{x}^{ F }_{\tau }(\epsilon ))|\}}{\tau }}_{\;:\!=\;r(\tau )}+\epsilon . \end{gather}

Since $\left |{x}-{x}^{E}_\tau (\epsilon )\right |= \left |{x}-{x}^{F}_\tau (\epsilon )\right |\leq \tau$ , it holds $\lim _{\tau \to 0} r(\tau ) = 0$ . Taking the $\limsup$ of (3.6) and sending $\epsilon$ to zero then yields,

\begin{align*} \lim _{\tau \to 0^+} \left |\frac {E({x})-E_{\tau }({x})}{\tau }-\frac {F({x})-F_{\tau }({x})}{\tau } \right | = 0. \end{align*}

Therefore, the limit in (3.4) exists,

\begin{align*} \lim _{\tau \to 0^+} \frac {E({x})-E_{\tau }({x})}{\tau } &=\lim _{\tau \to 0^+} \frac {E({x})-E_{\tau }({x})}{\tau }-\frac {F({x})-F_{\tau }({x})}{\tau } + \frac {F({x})-F_{\tau }({x})}{\tau }\\ &=\lim _{\tau \to 0^+} \frac {F({x})-F_{\tau }({x})}{\tau }=\left |\partial F\right |({x}), \end{align*}

where in the last step, we used that $F$ is convex together with Step 1.

Proof. For ${x}\in \mathcal{X}$ with $\|{x}\| \neq 1$ , the equality holds trivially. Therefore, we consider $\|{x}\|=1$ .

Step 1: $\mathcal{J}_\infty ({x})\subset \partial \chi _{\overline {B_1}}({x})$ .

Let $\xi \in \mathcal{J}_\infty ({x})$ , which means $\langle \xi , {x}\rangle = \left \|\xi \right \|_*$ , and consider an arbitrary $z\in \mathcal{X}$ . If $\left \|z\right \|\leq 1$ , we obtain

\begin{align*} \chi _{\overline {B_1}}(z)-\langle \xi , z-{x}\rangle =-\langle \xi , z\rangle +\|\xi \|_* \geq \|\xi \|_*\,( 1-\|z\|) \geq 0 =\chi _{\overline {B_1}}({x}), \end{align*}

while for $\|z\|\gt 1$ , the inequality holds trivially, thus we have $\xi \in \partial \chi _{\overline {B_1}}({x})$ .

Step 2: $\mathcal{J}_\infty ({x})\supset \partial \chi _{\overline {B_1}}({x})$ .

Let $\xi \in \partial \chi _{\overline {B_1}}({x})$ , then for all $z\in \overline {B_1}$ we get

\begin{align*} \underbrace {\partial \chi _{\overline {B_1}}(z)}_{=0}\geq \underbrace {\partial \chi _{\overline {B_1}}({x})}_{=0}+\langle \xi ,z-{x}\rangle \Longleftrightarrow \langle \xi , z\rangle \leq \langle \xi , {x}\rangle \leq \|\xi \|_*. \end{align*}

Taking the supremum over all $z\in \overline {B_1}$ yields the equality $\langle \xi , {x}\rangle = \|\xi \|_*$ and thus $\xi \in \mathcal{J}_\infty ({x})$ .

Proof. Let $t\in [0,T]$ be a point, where $u$ and $ E \circ u$ are differentiable, then we use the definition of the derivative to obtain

\begin{align*} \frac {\text{d}}{\text{d}t} E (u(t))-\langle \xi ,u'(t)\rangle &=\lim _{n\to \infty } \frac { E (u(t+h_n))- E (u(t))-\langle \xi , u(t+h_n)-u(t)\rangle }{h_n} \;=\!:\; (\spadesuit ), \end{align*}

where $\{h_n\}_n$ is a null sequence. We first consider only positive null sequences $h_n\gt 0$ , where we want to ensure that $u(t+h_n)\neq u(t)$ . If such a sequence does not exist, we infer that

\begin{equation*}\frac {\text{d}}{\text{d}t} E (u(t))= 0 =u'(t)\end{equation*}

and (3.8) holds. Now assuming that there exists a sequence with $u(t+h_n)\neq u(t)$ we continue,

\begin{align*} (\spadesuit )=\lim _{n\rightarrow \infty } \underbrace {\frac { E (u(t+h_n))- E (u(t))-\langle \xi , u(t+h_n)-u(t)\rangle }{\| u (t+h_n)-u(t)\|}}_{\;=\!:\;l_n}\cdot \underbrace {\frac {\| u (t+h_n)-u(t)\|}{h_n}}_{r_n}. \end{align*}

Note that $r_n\geq 0$ for all $n\in \mathbb{N}$ since we only allowed positive null sequences. Since $u$ is differentiable and in particular continuous at $t$ and since $\xi \in \partial E (u(t))$ (3.1) yields

\begin{align*} \liminf _{n\to \infty } l_n \geq 0, \end{align*}

i.e., for every null sequence $\{h_n\}_n$ , we can find a subsequence $\{h_n\}_n$ such that $l_n$ either converges to some limit $l\geq 0$ or diverges to $+\infty$ . In the convergent case, we obtain

\begin{align*} (\spadesuit ) = l \cdot \|u'(t)\| \geq 0. \end{align*}

In the divergent case, we also have $(\spadesuit )\geq 0$ , since we can find a $n_0$ such that $l_n$ is non-negative for all $n\geq N$ . Using the same arguments as above, but only allowing negative null sequences $h_n \lt 0$ , we instead obtain $(\spadesuit )\leq 0.$ This finally yields

\begin{align*} \frac {\text{d}}{\text{d}t} E (u(t))-\langle \xi ,u'(t)\rangle = 0. \end{align*}

Proof. Step 1: $(\text{i})\Leftrightarrow (\text{iii})$ .

Since $\psi$ is a monotone function, it is differentiable a.e., and thus we can find a Lebesgue null set $N\subset [0,T]$ , such that $u$ and $\psi$ are differentiable and $E (u(t))=\psi (t)$ for every $t\in [0,T]\setminus N$ . Using Lemma 3.7 and (3.7) for $t\in [0,1]\setminus N$ we obtain,

\begin{align*} \begin{aligned} \left .\begin{array}{cc} \psi '(t) \leq -|\partial E|(u(t)) \\ |u'|(t)\leq 1 \end{array}\right \} &\Leftrightarrow \left \{ \begin{array}{cc} \langle \xi , u'(t)\rangle =\psi '(t) \leq -\|\xi \|_*\quad \text{for all}\quad \xi \in \partial ^\circ E(u(t))\\ |u'|(t)\leq 1 \end{array} \right .\\ &\Leftrightarrow \langle \xi , u'(t)\rangle \leq - \|\xi \|_* - \chi _{\overline {B_1}}(u'(t))\quad \text{for all}\quad \xi \in \partial ^\circ E(u(t)). \end{aligned} \end{align*}

For each $\xi \in \partial ^\circ E(u(t))$ , the last statement is Item 2 with $f=\chi _{\overline {B_1}}$ and $f^*=\|\cdot \|_*$ , which is equivalent to Item 1, i.e.,

(3.9) \begin{align} \langle \xi , u'(t)\rangle \leq - \|\xi \|_* - \chi _{\overline {B_1}}(u'(t)) \quad &\Leftrightarrow \quad u'(t) \in \partial \|\cdot \|_{*}({-} \xi ), \end{align}

and thus we have shown $(\text{i})\Leftrightarrow (\text{iii})$ . The set identity in $(iii)$ ,

\begin{align*} \partial \left \|\cdot \right \|_*({-}\xi )\cap \mathcal{X} = {-}\operatorname*{arg\,max}_{{x}\in \overline {B_1}} \langle \xi , {x}\rangle , \end{align*}

follows from Corollary A.5.

Step 2: $(\text{i})\Leftrightarrow (\text{ii})$ .

Using the equivalence of Item 2 and Item 1 in (3.9), we also obtain that for a.e. $t\in [0,T]$ and all $\xi \in \partial E^\circ (u(t))$

\begin{align*} (i) \quad &\Leftrightarrow \quad -\xi \in \partial \chi _{\overline {B_1}}(u'(t)). \end{align*}

From Asplund’s theorem (Theorem 3.6), we have that

\begin{align*} -\xi \in \partial \chi _{\overline {B_1}}(u'(t)) \Longleftrightarrow -\xi \in \mathcal{J}_\infty (u'(t)) \end{align*}

which thus implies $(i)\Leftrightarrow (ii)$ .

Remark 3.9. In reflexive spaces, this is a very mild assumption, as the $\sigma$ -topology is often chosen to be the weak topology. In this case, we only need an assumption on the convex part $E^{\mathrm{c}}$ , namely lower semicontinuity, which together with convexity implies weak lower semicontinuity. The linearized part ${x}\mapsto E^{\mathrm{d}}(z)+\langle D E^{\mathrm{d}}(z),{x}-z\rangle$ is even weakly continuous and therefore, we do not need additional assumptions.

Remark 3.11. The above scheme can also be recovered via the theory of doubly non-linear equations developed in [Reference Mielke, Rossi and Savaré69]. Namely, by considering the state-dependent dissipation potential

\begin{align*} \Psi _z(v) \;:\!=\; \chi _{\overline {B_1}}(v) + E^{\mathrm{d}}(z) + \langle D E^{\mathrm{d}}(z), v \rangle \end{align*}

the minimizing movement scheme defined in [Reference Mielke, Rossi and Savaré69, Eq. (4.9)] is given as

\begin{align*} x_{\mathrm {si},\tau }^{k+1}\in \operatorname*{arg\,min}_{x\in \mathcal{X}} \left \{\tau \Psi _{x_{\mathrm {si},\tau }^k}\left (\frac {x-x_{\mathrm {si},\tau }^k}{\tau }\right ) + E^{\mathrm{c}}(x)\right \} \end{align*}

which exactly recovers the scheme defined in Definition 3.10. The authors show convergence of this scheme towards solution of the equation

\begin{align*} \partial \Psi _{u(t)}(u'(t)) + \partial E^{\mathrm{c}}(u(t)) \ni 0 \end{align*}

which corresponds to the inclusion derived in Theorem 3.8. However, we cannot directly apply the results of [69] since the choice of dissipation potential as above violates condition (2. $\Psi _1$ ), since $\operatorname {dom}(\Psi ) \neq \mathcal{X}$ , (2. $\Psi _2$ ) since in general $\Psi _u(0)\neq 0$ and the growth condition on the Fenchel conjugate $\Psi _z^*(\xi ) = \left \|\xi - D E^{\mathrm{d}}(z)\right \|_* -E ^{\mathrm{d}}(z)$ is not fulfilled and also (2. $\Psi _3$ ). In fact, a more detailed study on how these assumptions could be relaxed would be very interesting, which we, however, leave for future work.

Proof. We compute

\begin{align*} x_{\mathrm {si},\tau }^{k+1}& \in \mathop{\mathrm{arg\,min}}\limits_{{x}\;:\; \|{x}-x_{\mathrm {si},\tau }^{k}\|\leq \tau } E(x_{\mathrm {si},\tau }^{k})+\langle D E(x_{\mathrm {si},\tau }^{k}) , {x}-x_{\mathrm {si},\tau }^{k} \rangle \\ &=\mathop{\mathrm{arg\,min}}\limits_{{x}\;:\; \|{x}-x_{\mathrm {si},\tau }^{k}\|\leq \tau } \langle D E(x_{\mathrm {si},\tau }^{k}) , {x}\rangle \\ &=-\mathop{\mathrm{arg\,max}}\limits_{{x}\;:\; \|{x}-x_{\mathrm {si},\tau }^{k}\|\leq \tau } \langle D E (x_{\mathrm {si},\tau }^{k}) , {x}\rangle \\ &=x_{\mathrm {si},\tau }^k -\tau \mathop{\mathrm{arg\,max}}\limits_{{x}\in \overline {B_1}} \langle D E(x_{\mathrm {si},\tau }^{k}), {x}\rangle \\ &=x_{\mathrm {si},\tau }^{k}-\tau \, \partial \|\cdot \|_*(D E(x_{\mathrm {si},\tau }^{k})), \end{align*}

where for the last identity, we used A.5.

Proof. Let $z, x\in \mathcal{X}$ , from Item 1 we know

\begin{align*} \partial E({x}) &= \partial E^{\mathrm{c}}({x}) + D E^{\mathrm{d}}({x}),\\ \partial E^{\mathrm{sl}}({x}; z) &= \partial E^{\mathrm{c}}({x}) + D E^{\mathrm{d}}(z), \end{align*}

and then Item 3 implies that there exists $\xi _1,\xi _2\in \partial E^{\mathrm{c}}({x})$ such that

\begin{align*} \left |\partial E\right |({x}) &= \min \left \{ \left \|\xi + D E^{\mathrm{d}}({x})\right \|_*\;:\;\xi \in \partial E^{\mathrm{c}}({x})\right \} = \left \|\xi _1 + D E^{\mathrm{d}}({x})\right \|_*,\\ \left |\partial E^{\mathrm{sl}}(\cdot ; z)\right |({x}) &= \min \left \{ \left \|\xi + D E^{\mathrm{d}}(z)\right \|_*\;:\;\xi \in \partial E^{\mathrm{c}}({x})\right \} = \left \|\xi _2 + D E^{\mathrm{d}}(z)\right \|_*. \end{align*}

We can then estimate

\begin{align*} \left |\partial E\right |({x})&\leq \left \|D E^{\mathrm{d}}({x})+ \xi _2\right \|_* \leq \|D E^{\mathrm{d}}({x})-D E^{\mathrm{d}}(z)\|_*+\|D E^{\mathrm{d}}(z)+\xi _2\|_*\\ &\leq \mathrm {Lip}(D E^{\mathrm{d}})\|{x}-z\|+\left |\partial E^{\mathrm{sl}}(\cdot ; z)\right |({x}), \end{align*}

and therefore

\begin{align*} \left |\partial E\right |({x}) - \left |\partial E^{\mathrm{sl}}(\cdot ;\ z)\right |({x})\leq \mathrm {Lip}(D E^{\mathrm{d}})\|{x}-z\|. \end{align*}

Analogously, we estimate

\begin{align*} \left |\partial E^{\mathrm{sl}}(\cdot ,z)\right |({x}) &\leq \| D E^{\mathrm{d}}(z) +\xi _1\|\leq \| D E^{\mathrm{d}}(z)-D E^{\mathrm{d}}({x})\|+\|D E({x})+\xi _1\|\\ &\leq \mathrm {Lip}(D E^{\mathrm{d}})\|{x}-z\|+|\partial E| ({x}). \end{align*}

and therefore

\begin{align*} \left |\partial E^{\mathrm{sl}}(\cdot ;\ z)\right |({x})-\left |\partial E\right |({x})\leq \mathrm {Lip}(D E^{\mathrm{d}})\|{x}-z\|. \end{align*}

This concludes the proof.

Proof. For (3.13), we apply Lemma 2.9 to the mapping ${x}\mapsto E^{\mathrm{sl}}({x};\ x_{\mathrm {si},\tau }^{k-1})$ to obtain

\begin{align*} \frac {\text{d}}{\text{d}t} E^{\mathrm{sl}}_{\big(t-t_\tau ^{k-1} \big)} \big({x};\ x_{\mathrm {si},\tau }^{k-1} \big) \leq |\partial E^{\mathrm{sl}} \big(\cdot ,x_{\mathrm {si},\tau }^{k-1} \big)|\left ({x}_{\text{min},t-t^{k-1}_\tau }\right ), \end{align*}

where ${x}_{\mathrm{min},t-t^{k-1}_\tau }\in \displaystyle\operatorname*{arg\,min}_{\tilde {{x}}} \{E^{\mathrm{sl}}(\tilde {{x}};\ x_{\mathrm {si},\tau }^{k-1})\,:\, \tilde {{x}}\in \overline {B_\tau }({x})\}$ . Choosing $v=x_{\mathrm {si},\tau }^{k-1}$ then yields

\begin{align*} \frac {\text{d}}{\text{d}t} E^{\mathrm{sl}}_{(t-t_\tau ^{k-1})}(x_{\mathrm {si},\tau }^{k-1}) \leq -|\partial E^{\mathrm{sl}}(\cdot ;\ x_{\mathrm {si},\tau }^{k-1})|(\tilde {x}_{\mathrm{si},\tau }(t)). \end{align*}

The last equality of (3.13) follows by Lemma 3.13. To show (3.14), we again use Lemma 2.9 and get

\begin{align*} E^{\mathrm{sl}} \big(\tilde {x}_{\mathrm{si},\tau }(s);\ x_{\mathrm {si},\tau }^k \big)+\int _{s}^t \mathcal{D}_\tau (r) \,\text{d}r\geq E^{\mathrm{sl}} \big(\tilde {x}_{\mathrm{si},\tau }(t),x_{\mathrm {si},\tau }^k \big) \quad \text{for all}\quad t_\tau ^k\leq s\leq t\leq t_\tau ^{k+1}. \end{align*}

Due to Theorem C.1

\begin{align*} \left |E^{\mathrm{sl}}\big(\tilde {x}_{\mathrm{si},\tau }(s);\ x_{\mathrm {si},\tau }^k \big)- E(\tilde {x}_{\mathrm{si},\tau }(s))\right |&=\left | E^{\mathrm{d}} \big(x_{\mathrm {si},\tau }^k \big) + \langle D E^{\mathrm{d}}(x_{\mathrm {si},\tau }^k), \tilde {x}_{\mathrm{si},\tau }(s) - x_{\mathrm {si},\tau }^k\rangle - E^{\mathrm{d}}(\tilde {x}_{\mathrm{si},\tau }(s)) \right | \\ &\leq \bigg |\int _0^1 \left \langle D E^{\mathrm{d}}\left (x_{\mathrm {si},\tau }^k +r (\tilde {x}_{\mathrm{si},\tau }(s)-x_{\mathrm {si},\tau }^k)\right ),x_{\mathrm {si},\tau }^k -\tilde {x}_{\mathrm{si},\tau }(s) \right \rangle \\ &\quad -\langle D E^{\mathrm{d}}(x_{\mathrm {si},\tau }^k),\tilde {x}_{\mathrm{si},\tau }(t)-x_{\mathrm {si},\tau }^k \rangle \,\text{d}r\bigg |\\ &\leq \int _0^1 r \mathrm {Lip}(D E^{\mathrm{d}})\| \tilde {x}_{\mathrm{si},\tau }(s)-x_{\mathrm {si},\tau }^k\|^2\,\text{d}r \\ &\leq \frac {1}{2} \mathrm {Lip}(D E^{\mathrm{d}})\| \tilde {x}_{\mathrm{si},\tau }(s)-x_{\mathrm {si},\tau }^k\|^2 \leq \frac {1}{2} \mathrm {Lip}(D E^{\mathrm{d}}) \tau ^2 \end{align*}

and analogusly $|E^{\mathrm{sl}}(\tilde {x}_{\mathrm{si},\tau }(t),x_{\mathrm {si},\tau }^{k})- E(\tilde {x}_{\mathrm{si},\tau }(t))|\leq \frac {1}{2} \mathrm {Lip}(D E^{\mathrm{d}}) \tau ^2$ . Therefore, for all $t_\tau ^k\leq s\leq t\leq t_\tau ^{k+1}$ , we have that

(3.15) \begin{align} \begin{aligned} E(\tilde {x}_{\mathrm{si},\tau }(s))+\int _{s}^t \mathcal{D}_\tau (r) \,\text{d}r &\geq E^{\mathrm{sl}}(\tilde {x}_{\mathrm{si},\tau }(s);\ {x}_\tau ^k) +\int _{s}^t \mathcal{D}_\tau (r) \,\text{d}r -\frac {1}{2} \mathrm {Lip}(D E^{\mathrm{d}}) \tau ^2\\ &\geq E^{\mathrm{sl}}(\tilde {x}_{\mathrm{si},\tau }(t);\ {x}_\tau ^k) -\frac {1}{2} \mathrm {Lip}(D E^{\mathrm{d}}) \tau ^2\\ &\geq E(\tilde {x}_{\mathrm{si},\tau }(t)) - \mathrm {Lip}(D E^{\mathrm{d}}) \tau ^2. \end{aligned} \end{align}

Now for $s\in [t_\tau ^m,t_\tau ^{m+1}]$ and $t\in [t_\tau ^k,t_\tau ^{k+1}]$ with $m\leq k$ , we add up (3.15) to obtain

\begin{align*} E(\tilde {x}_{\mathrm{si},\tau }(s))&+ \int _s^{t_\tau ^{m+1}} \mathcal{D}_\tau (r) \,\text{d}r +\sum _{i=m+1}^{k-1} \int _{t_\tau ^{i}}^{t_\tau ^{i+1}} \mathcal{D}_\tau (r) \,\text{d}r+\int _{t_\tau ^k}^t \mathcal{D}_\tau (r) \,\text{d}r \\ &\geq E(\tilde {x}_{\mathrm{si},\tau }(t))-\sum _{i=m}^k \mathrm {Lip}(D E^{\mathrm{d}}) \tau ^2\\ &=E(\tilde {x}_{\mathrm{si},\tau }(t))- (k-m) \mathrm {Lip}(D E^{\mathrm{d}}) \tau ^2\\ &\geq E(\tilde {x}_{\mathrm{si},\tau }(t))- T \mathrm {Lip}(D E^{\mathrm{d}}) \tau \end{align*}

such that we finally obtain (3.14).

Proof. We simply replace the minimizing movement scheme in Definition 2.5 and De Giorgis variational interpolation (see Definition 2.10) by the semi-implicit scheme in Definition 3.10 and its corresponding variational interpolation of Lemma 3.15. Proceeding similarly as in the proof of Theorem 2.11, we use Proposition B.1 to show

\begin{align*} \bar {x}_{\mathrm{si},\tau _n}(t) \stackrel {\sigma }{\rightharpoonup }u(t) \text{ as } n\rightarrow \infty \quad \forall t\in [0,T] \end{align*}

for a subsequence $\tau _n$ , where $u$ is a $1$ -Lipschitz curve with $u(0)={x}^0$ . Then the same holds true for $\tilde {x}_{\mathrm{si},\tau _n}(t)$ .

Taking for $\tau _n$ the limes inferior for $n\rightarrow \infty$ of (3.14) and using Assumption 2.a, Assumption 2.b and (3.13) we again obtain

\begin{align*} E(u(0))\geq E(u(t))+\int _0^t |\partial E|(u(r)) \,\text{d}r \quad \text{ for all }t\in [0,T]. \end{align*}

Since on the other hand $|\partial E|$ is a strong upper gradient, equality in the above equation must hold.

Remark 3.17. Let $\tau _n$ be any sequence such that $\tau _n \rightarrow 0$ . If the $\infty$ -curve of maximal slope $u$ is unique, we can apply Theorem 3.16 to every subsequence of $\tau _n$ and find a further subsequence $\tilde {\tau }_n$ such that

\begin{align*} \bar {x}_{\mathrm{si},\tilde {\tau }_n}(t) \stackrel {\sigma }{\rightharpoonup }u(t) \text{ as } n\rightarrow \infty \quad \forall t\in [0,T]. \end{align*}

This implies that already for $\tau _n$

\begin{align*} \bar {x}_{\mathrm{si},\tau _n}(t) \stackrel {\sigma }{\rightharpoonup }u(t) \text{ as } n\rightarrow \infty \quad \forall t\in [0,T]. \end{align*}

and the semi-implicit scheme converges.

Remark 4.4. If $\mathcal{X}$ is separable and reflexive, then so is its dual. For a countable dense subset $\{x^*_n\}_{n\in \mathbb{N}}$ of $\overline {B_1^{\mathcal{X}*}}$ , we can define the norm

\begin{equation*}\| x \|_{\omega }=\sum _{n=1}^\infty \frac {1}{n^2} |\langle x^*_n,x\rangle |,\end{equation*}

which induces the weak topology $\sigma (\mathcal{X},\mathcal{X}^*)$ on bounded sets [Reference Morrison73, Lemma 3.2]. This norm is a so-called Kadec norm. In particular, we have that the Borel sigma algebra $\mathcal{B}(\mathcal{X})$ , generated by the norm topology, and the one generated by the weak topology $\mathcal{B}(\mathcal{X}_{\omega })$ coincide and thus $\mathcal{P}(\mathcal{X})=\mathcal{P}(\mathcal{X}_\omega )$ , see [Reference Edgar37, Theorem 1.1]. Now, let us assume that for a set $\mathcal{K}\subset \mathcal{P}(\mathcal{X})=\mathcal{P}(\mathcal{X}_\omega )$ , we have that

(4.5) \begin{align} \forall \epsilon \gt 0\quad \exists K_\epsilon \ \|\cdot \|\text{-bounded in } \mathcal{X}, \text{ such that } \mu (\mathcal{X}\setminus K_\epsilon )\leq \epsilon \quad \forall \mu \in \mathcal{K}. \end{align}

Since bounded sets are subsets of $\overline {B_\epsilon }$ for $\epsilon$ large enough and $\overline {B_\epsilon }$ is compact in the weak topology $\sigma (\mathcal{X},\mathcal{X}^*)$ , Prokhorov’s theorem can be applied for $\mathcal{X}_\omega$ . We obtain that there exists a subsequence $\left (\mu ^{n}\right )_{n\in \mathbb{N}}\subset \mathcal{K}$ and a limit $\mu \in \mathcal{P}(\mathcal{X})=\mathcal{P}(\mathcal{X}_\omega )$ such that

(4.6) \begin{align} \int _{\mathcal{X}} \varphi \,\text{d}\mu ^n \rightarrow \int _{\mathcal{X}} \varphi \,\text{d}\mu \quad \forall \varphi \in C^\omega _b(\mathcal{X}), \end{align}

where $C^\omega _b(\mathcal{X})$ now denotes the set of weakly continuous bounded functions.

Remark 4.6. For $1\leq p\lt \infty$ , convergence in $W_p$ is equivalent to narrow convergence and convergence of the $p$ -th moment [Reference Villani104, Theorem 6.9]. This equality is lost for the $\infty$ -Wasserstein distance, as convergence in the narrow topology ( $\mu ^n\stackrel {\sigma }{\rightarrow } \mu$ ) together with $\displaystyle\bigcup _n \operatorname{supp}(\mu ^n)$ being bounded or relatively compact no longer guarantees convergence in $W_\infty$ , as Example 3 demonstrates.

Remark 4.12 (Uniqueness of the velocity field). As mentioned before, if $\mathcal{X}$ satisfies the bounded approximation property, the velocity field obtained in Theorem 4.9 is minimal. For the case that $p\in (1,+\infty )$ and the norm of the underlying Banach space $\mathcal{X}$ is strictly convex, then $\| \cdot \|_{L^p(\mu _t;\mathcal{X})}$ is also strictly convex. Then the uniqueness of the minimal velocity field follows. In the other cases, the uniqueness is lost.

Remark 4.13. Whenever Theorem 4.11 is applicable, $\| \boldsymbol{v}_t\|_{L^\infty (\mu _t;\mathcal{X})}=|\mu '|(t)$ for a.e. $t\in (0,T)$ and thus (E.1) is actual an equality. For the Wasserstein spaces $p\in (1,+\infty )$ , we obtain

\begin{align*} \int _{\mathcal{X}} \left \|\int _{C(0,T;\;\mathcal{X})} u'(t) d\bar {\eta }_{x,t}\right \|^p \,\text{d}\mu _t= \int _{\mathcal{X}}\int _{C(0,T;\;\mathcal{X})} \left \|u'(t)\right \|^p d\bar {\eta }_{x,t} \,\text{d}\mu _t\quad \text{for a.e. } t\in (0,T) \end{align*}

or equivalently

\begin{align*} \left \|\int _{C(0,T;\;\mathcal{X})} u'(t) d\bar {\eta }_{x,t}\right \|^p=\int _{C(0,T;\;\mathcal{X})} \left \|u'(t)\right \|^p d\bar {\eta }_{x,t} \quad \text{ for } \bar {\mu }\text{-a.e. } (t,x)\in [0,T]\times \mathcal{X}. \end{align*}

from corresponding calculations [Reference Lisini62, Theorem 7]. Notice that this is the equality case of Jensen’s inequality. For a strictly convex norm $\|\cdot \|$ , this equality can only hold when $u'(t)$ is constant $\bar {\eta }_{x,t}$ -a.e. Thus, heuristically spoken, all curves passing through a point $x\in \mathcal{X}$ at time $t$ have the same derivative. This is in particular the reason why on an infinitesimal level optimal transport plans $\gamma _h\in \Gamma (\mu _t,\mu _{t+h})$ behave like classical optimal transport, i.e., for a.e. $t\in (0,T)$ (see [Reference Ambrosio, Gigli and Savaré2, Proposition 8.4.6]),

\begin{align*} \lim _{h \rightarrow 0} \left(\pi ^1,\frac {1}{h} (\pi ^2-\pi ^1)\right)_{\#} \gamma _h=({Id}\times \boldsymbol{v} _t)_{\#}\mu _t\quad \text{in } \mathcal{P}(\mathcal{X}\times \mathcal{X}). \end{align*}

This argument fails in the case $W_\infty$ or when the norm $\|\cdot \|$ is not strictly convex.

Proof. Every weakly open set $G\subset \mathcal{X}\cap \operatorname {dom}(E)$ is strongly open as well. And for strongly open sets $G$ , the lower inverse as defined in (F.1) is given as

\begin{equation*}\varphi _\tau ^l(G)=\{s\in \mathcal{X} |\ \exists \ x\in G \text{ with }\|s-x\|\leq \tau \}.\end{equation*}

Since $G$ is strongly open, this set is again strongly open, and thus in $\Sigma =\mathcal{B}(\mathcal{X})$ , yielding weak measurability of $\varphi _ \tau$ . To conclude, we observe that $\overline {B_\tau }(x)$ is nonempty and weakly compact.

Proof. As mentioned in remark Remark 4.4, $\mathcal{B}(\mathcal{X})$ and $\mathcal{B}(\mathcal{X}_{\omega })$ coincide in this particular setting. We choose the correspondence $\varphi _\tau$ from Lemma 4.14 and set $f(s,x)=-{E}(x)$ . Since Assumption 4.a guarantees that $f(s,x)=-E(x)$ is a Carathéodory function the application of Theorem F.3 yields this corollary, but only restricted to $\mathcal{X}\cap \operatorname {dom}_\tau (E)$ . However, we can extend ${E}_\tau (x)$ and $\boldsymbol{r}_\tau$ measurably by setting them to $+\infty$ and $\overline {B}_\tau (x)$ on $\operatorname {dom}_\tau (E)^c$ respectively.

Proof. Corollary 4.15 ensures the existence of measurable selectors of (4.11). We take $\tilde {\mu }$ , such that $W_\infty (\mu ,\tilde {\mu })\leq \tau$ and $\gamma \in \Gamma _0(\mu ,\tilde {\mu })$ , then by disintegration we get

\begin{align*} \int {E}(x) \text{d}\tilde {\mu }(x)=\int {E}({x})\text{d} \gamma (z,{x})= \int \int {E}({x}) \text{d} \rho _{z} ({x})\, \text{d} \mu (z) \end{align*}

with a Borel family of probability measures $\{ \rho _{z}\}_{z \in \mathcal{X}_1} \subset \mathcal{P}(\mathcal{X})$ and $\operatorname{supp}(\rho _{z}) \subset \overline {B_\tau } (z)$ . We further estimate,

\begin{align*} \int \int {E} ({x}) \text{d} \rho _{z} ({x}) \text{d} \mu (z) \geq \int {E}({r}_\tau (z)) \text{d} \mu (z)= \int {E}(z)\, \text{d} ({r}_\tau )_\# \mu (z), \end{align*}

and since $\tilde {\mu }$ was arbitrary, this concludes the proof.

Proof. Since $\frac {{E}(x)-{E}_\tau (x)}{\tau }\geq 0$ we can use Fatou’s lemma to show

\begin{align*} \int _{\mathcal{X}} |\partial {E}|(x) \,\text{d}\mu (x)&=\int _{\mathcal{X}} \lim _{\tau \rightarrow 0} \frac {{E}(x)-{E}_\tau (x)}{\tau } \,\text{d}\mu (x)\\&\leq \liminf _{\tau \rightarrow 0} \int _{\mathcal{X}} \frac {{E}(x)- {E}_\tau (x)}{\tau } \,\text{d}\mu (x)\\& \leq \limsup _{\tau \rightarrow 0} \int _{\mathcal{X}} \frac {{E}(x)-{E}({r}_\tau (x))}{\tau } \,\text{d}\mu (x)\\& = \limsup _{\tau \rightarrow 0} \frac {\mathcal{E}(\mu ) - \mathcal{E}(\mu _\tau )}{\tau } = |\partial \mathcal{E}|(\mu ), \end{align*}

where in the last step, we employ Lemma 2.8 . This implies that when $\int _{\mathcal{X}} |\partial {E}|(x) \,\text{d}\mu (x)=+\infty$ then $|\partial \mathcal{E}|(\mu )=+\infty$ . In the case $\int _{\mathcal{X}} |\partial {E}|(x) \,\text{d}\mu (x)\lt +\infty$ , we use Lemma 2.8 (for $\mathcal{E}$ ) and Lemma 3.5 (for $E$ ) to calculate

\begin{align*} \int _{\mathcal{X}} |\partial {E}|(x) \,\text{d}\mu (x)&=\int _{\mathcal{X}} \lim _{\tau \rightarrow 0} \frac {{E}(x)- {E}_\tau (x)}{\tau }\,\text{d}\mu (x) \\ &=\lim _{\tau \rightarrow 0} \frac {\int _{\mathcal{X}} {E}(x)-{E}({r}_\tau (x)) \,\text{d}\mu (x)}{\tau }\\ &=\lim _{\tau \rightarrow 0} \frac {\mathcal{E}(\mu )-\mathcal{E}(\mu _\tau )}{\tau }=|\partial \mathcal{E}|(\mu ), \end{align*}

where the dominated convergence theorem was used to draw the limit into the integral. For the upper bound, we observe

\begin{align*} |\partial {E}|(x)=&\limsup _{z\rightarrow x}\frac {({E}^{\text{c}}(x)-{E}^{\text{c}}(z)+{E}^{\text{d}}(x)-{E}^{\text{d}}(z))^+}{\|x-z\|}\\ \geq &\limsup _{z\rightarrow x} \frac {({E}^{\text{c}}(x)-{E}^{\text{c}}(z))^+}{\|x-z\|} -\frac {|{E}^{\text{d}}(x)-{E}^{\text{d}}(z)|}{\|x-z\|}\\ \geq &\limsup _{z\rightarrow x} \frac {({E}^{\text{c}}(x)-{E}^{\text{c}}(z))^+}{\|x-z\|}-\mathrm {Lip}({E}^{\text{d}})=|\partial {E}^{\text{c}}|(x)-\mathrm {Lip}({E}^{\text{d}}) \end{align*}

and by [Reference Ambrosio, Gigli and Savaré2, Theorem 2.4.9]

\begin{equation*} \sup _{z\neq {x}}\frac {({E}^{\text{c}}({x})-{E}^{\text{c}}(z))^+}{\|{x}-z\|} =| \partial {E}^{\text{c}}|({x}).\end{equation*}

Then we can give an upper bound by

\begin{align*} \frac {{E}(x)-{E}(r_\tau (x))}{\tau }&\leq \frac {{E}^{\text{c}}(x)-{E}^{\text{c}}(r_\tau (x))}{\|x-r_\tau (x)\|}+\frac {{E}^{\text{d}}(x)-{E}^{\text{d}}(r_\tau (x))}{\|x-r_\tau (x)\|} \\ &\leq \sup _{z\neq {x}}\frac {({E}^{\text{c}}(x)-{E}^{\text{c}}(z))^+}{\|x-z\|} +\mathrm {Lip}({E}^{\text{d}})=|\partial {E}^{\text{c}}|(x)+\mathrm {Lip}({E}^{\text{d}}) \\&\leq |\partial {E}|(x)+2\mathrm {Lip}({E}^{\text{d}}). \end{align*}

To prove that $|\partial \mathcal{E}|$ is a strong upper gradient, let $\mu _t$ be an absolutely continuous curve in $\mathcal{W}_\infty (\mathcal{X})$ . Since $\left |\partial \mathcal{E}\right |(\mu )= \int _{\mathcal{X}} |\partial {E}|(x) \,\text{d}\mu ({x})$ and by Item 4 the slope $|\partial {E}|(x)$ is lower semicontinuous, it follows from [Reference Ambrosio, Gigli and Savaré2, Lemma 5.1.7] that $\left |\partial \mathcal{E}\right |(\mu )$ is lower semicontinuous w.r.t. narrow convergence and in particular $t\mapsto |\partial \mathcal{E}|(\mu _t)$ is lower semicontinuous and thus Borel. Assume that $\int _s^t \int _{\mathcal{X}} |\partial {E}|(x) \,\text{d}\mu _r({x}) |\mu '|(r) \text{d}r = \int _s^t \left |\partial \mathcal{E}\right |(\mu _r) |\mu '|(r)\,\text{d}r\lt +\infty$ , otherwise (1.4) holds trivially. We can estimate

(4.13) \begin{align} \begin{split} |\mathcal{E}(\mu _t)-\mathcal{E}(\mu _s)|&=\left |\int _{\mathcal{X}} {E}(x)\,\text{d}\mu _t(x)-\int _{\mathcal{X}} {E}(x)\,\text{d}\mu _s(x)\right |\\ &=\left |\int _{\mathcal{X}} {E}({x})\text{d}{e_t}_\# \eta ({x})-\int _{\mathcal{X}} {E}({x})\text{d}{e_s}_\# \eta ({x})\right |\\ &=\left |\int _{C(0,T;\;\mathcal{X})} {E}(u(t))\,\text{d}\eta (u)-\int _{C(0,T;\;\mathcal{X})} {E}(u(s))\text{d} \eta (u)\right | \\&\leq \int _{C(0,T;\;\mathcal{X})} |{E}(u(t))-{E}(u(s))| \,\text{d}\eta (u)\\ &\overset {(i)}{\leq }\int _{C(0,T;\;\mathcal{X})} \int _s^t | \partial {E}|(u(r))\, |u'|(r)\,\text{d}r \,\text{d}\eta (u)\\ &\overset {(ii)}{=}\int _s^t\int _{C(0,T;\;\mathcal{X})} | \partial {E}|(u(r))\, |u'|(r) \,\text{d}\eta (u)\,\text{d}r\\ &\overset {(iii)}{\leq } \int _s^t\int _{C(0,T;\;\mathcal{X})} | \partial {E}|(u(r)) \,\text{d}\eta (u) |\mu '|(r)\,\text{d}r\\ &= \int _s^t\int _{\mathcal{X}} | \partial {E}|(x) \,\text{d}\mu _r |\mu '|(r)\,\text{d}r\lt +\infty , \end{split} \end{align}

where $(t,u)\mapsto | \partial {E}|(u(t))$ is $\bar {\eta }$ -measurable since it is lower semicontinuous on $[0,T]\times C(0,T,\mathcal{X})$ and measurability of $\left |u'\right |$ follows as in the proof of [Reference Lisini62, Theorem 7] and Theorem 4.9. For $(ii)$ , we use the theorem of Fubini–Tonelli, while for $(i)$ , we observe that $\eta$ from Theorem 4.7 is concentrated on ${\mathrm{AC}}^\infty (0,T;\;\mathcal{X})$ and $|\partial {E}|$ is a strong upper gradient (c.f. Definition 1.4) and for (iii) we use $|\mu '|(t)=\| |u'|(t)\|_{L_p(\eta )}$ .