Appendix A
We provide the proof of a version of the envelope or Danskin’s theorem (see [8]), adapted to our purposes. First, we recall the notion of Hadamard differentiability. Given two Banach spaces \((\mathcal {X}, \Vert \cdot \Vert _{\mathcal {X}})\) and \((\mathcal {Z}, \Vert \cdot \Vert _{\mathcal {Z}})\) a map \(f: \mathcal {X}\rightarrow \mathcal {Z}\) is directionally differentiable at x if for all \(h\in \mathcal {X}\) the limit in \(\mathcal {Z}\)
$$\begin{aligned} Df(x,h):= \lim _{\tau \downarrow 0} \frac{f(x+\tau h)-f(x)}{\tau }, \end{aligned}$$
exists. If in addition, for all \(h\in \mathcal {X}\) the following equality in \(\mathcal {Z}\) holds
$$\begin{aligned} Df(x,h)= \lim _{\tau \downarrow 0, \; h'\rightarrow h} \frac{f(x+\tau h')-f(x)}{\tau } , \end{aligned}$$
then we say that f is directionally differentiable at x in the Hadamard sense. An important property of Hadamard differentiable functions is the chain rule. More precisely, if \((\mathcal {V}, \Vert \cdot \Vert _{\mathcal {V}})\) is another Banach space, \(g: \mathcal {V}\rightarrow \mathcal {X}\) is directionally differentiable at v and f is directionally differentiable at g(v) in the Hadamard sense, then the composition \(f\circ g\) is directionally differentiable at v (see e.g. [4, Proposition 2.47]) and \(D(f\circ g)(v, v')= Df(g(v),Dg(v,v'))\) for all \(v'\in \mathcal {V}\). If in addition, g is is also Hadamard directionally differentiable at v, then \(f\circ g\) is directionally differentiable at v in the Hadamard sense.
Now, suppose that \(K\subseteq \mathcal {X}\) is a weakly compact set. Let us consider the problem:
$$\begin{aligned} \qquad \qquad \qquad \qquad \qquad \qquad \qquad \quad \sup _{Z\in X} \langle d, Z \rangle \quad \text{ s.t. } \; Z\in K, \qquad \qquad \qquad \qquad \qquad \qquad \qquad {(AP_{d})} \end{aligned}$$
where \(d\in \mathcal {X}^{*}\) and \(\langle \cdot , \cdot \rangle \) denotes the bilinear pairing between \(\mathcal {X}\) and \(\mathcal {X}^{*}\). Let us define \(v: \mathcal {X}^{*} \rightarrow \mathbb {R}\) as the optimal value of problem \((AP_{d})\) and \(\mathcal {S}(d)\) the set of optimal solutions of \((AP_{d})\), i.e.
$$\begin{aligned} v(d):= \sup _{Z\in K} \langle d, Z \rangle , \quad \mathcal {S}(d):= \left\{ Z\in K \quad v(d)= \langle d, Z \rangle \right\} . \end{aligned}$$
Note that v is well defined, it is a Lipschitz function and \(\mathcal {S}(d)\ne \emptyset \). In fact,
$$\begin{aligned} | v(d_1)- v(d_2) | \le \Vert d_1- d_2\Vert _{\mathcal {X}^{*}}\sup _{Z\in K} \Vert Z\Vert _{\mathcal {X}}. \end{aligned}$$
(A.1)
The proof of the following result is a simple modification of the proof in [4, Theorem 4.13].
Lemma A.1
For any \(\bar{d}\in \mathcal {X}^{*}\), the following assertions hold true
(i) The set \(\mathcal {S}(\bar{d})\) is weakly compact.
(ii) The function v is directionally differentiable in the Hadamard sense and its directional derivative is
$$\begin{aligned} Dv(\bar{d}, \Delta d)= \sup _{Z\in \mathcal {S}(\bar{d})} \langle \Delta d, Z \rangle \quad \hbox {for} \,\hbox {all} \Delta d\in \mathcal {X}^{*}. \end{aligned}$$
(A.2)
Proof
The first assertion follows directly from the weak-continuity of \(\langle \bar{d}, \cdot \rangle \), which implies the weak closedness of \(\mathcal {S}(\bar{d})\). Now, in view of [4, Proposition 2.49] and (A.1) it suffices to show that v is directionally differentiable. Let \(\bar{Z}\in S(\bar{d})\) be such that \( \langle \Delta d, \bar{Z} \rangle = \sup _{Z\in \mathcal {S}(\bar{d})} \langle \Delta d, Z \rangle \) and for \(\tau >0\) set \(d_{\tau }:= \bar{d}+ \tau \Delta d\). By definition
$$\begin{aligned} v(d_{\tau })- v(\bar{d}) \ge \langle d_{\tau }-\bar{d}, \bar{Z}\rangle = \tau \langle \Delta d, \bar{Z} \rangle , \end{aligned}$$
which implies that
$$\begin{aligned} \textstyle \liminf _{\tau \rightarrow 0} \frac{ v(d_{\tau })- v(\bar{d})}{\tau } \ge \langle \Delta d, \bar{Z} \rangle = \sup _{Z\in \mathcal {S}(\bar{d})} \langle \Delta d, Z \rangle . \end{aligned}$$
(A.3)
Analogously, let \(Z_{\tau } \in S(d_{\tau })\). Then
$$\begin{aligned} \textstyle v(\bar{d})- v(d_\tau ) \ge - \langle d_{\tau }-\bar{d}, Z_{\tau }\rangle = - \tau \langle \Delta d, Z_\tau \rangle . \end{aligned}$$
(A.4)
On the other hand, using (A.1) we get that \(v(d_{\tau })\rightarrow v(\bar{d})\) as \(\tau \downarrow 0\), which implies, since \(d_{\tau }\rightarrow \bar{d}\) strongly in \(\mathcal {X}^*\), that any weak limit point of \(Z_{\tau }\) belongs to \(\mathcal {S}(\bar{d})\). Thus, (A.4) yields
$$\begin{aligned} \textstyle \limsup _{\tau \rightarrow 0} \frac{ v(d_{\tau })- v(\bar{d})}{\tau } \le \limsup _{\tau \rightarrow 0} \langle \Delta d, Z_{\tau } \rangle \le \sup _{Z\in \mathcal {S}(\bar{d})} \langle \Delta d, Z \rangle . \end{aligned}$$
(A.5)
Therefore, (A.2) is a consequence of (A.3) and (A.5).