Proof. We need to check that \(\widetilde{V}^{n \times n}_\varphi\) satisfies conditions (N2) and (N3) in Definition 1. The homogeneity property (N2) is obvious, in fact \(\widetilde{V}^{n \times n}_\varphi(av) = |a|\widetilde{V}^{n \times n}_\varphi(v)\) for every \(v \in \mathcal{M}(\Omega;\mathbb{R}^{n \times n}_{\rm sym})\) and any \(a \in \mathbb{R}\). In order to check (N3), namely the triangle inequality, let us consider \(v,w \in \mathcal{M}(\Omega;\mathbb{R}^{n \times n}_{\rm sym})\). We have \[\int_\Omega \psi : d(v+w) = \int_\Omega \psi : dv + \int_\Omega \psi : dw \le \widetilde{V}^{n \times n}_\varphi(v) + \widetilde{V}^{n \times n}_\varphi(w)\] for any \(\psi \in C^1_c(\Omega;\mathbb{R}^{n \times n}_{\rm sym})\) such that \(\|\psi\|_{\varphi^*} \le 1\). By taking the supremum over such \(\psi\) we have the thesis. ◻

Proof. We need to check that \(\varrho_{\widetilde{V}^{n \times n}_\varphi}\) satisfies the conditions of Definition 2. Conditions (i) and (iii) are obvious, respectively thanks to the properties of \(\varphi^*\) and that if \(\psi\) is an admissible test function, then also \(-\psi\) is. Condition (iv) is also easy to prove using the linearity of the first integral in the definition of \(\varrho_{\widetilde{V}^{n \times n}_\varphi}\) and the fact that the supremum of a sum is smaller than the sum of the suprema.
We are only left to prove condition (ii), namely that the function \(\lambda \mapsto \varrho_{\widetilde{V}^{n \times n}_\varphi}(\lambda v)\) on \([0, +\infty)\) is non-decreasing for any \(v \in \mathcal{M}(\Omega; \mathbb{R}^{n \times n}_{\rm sym})\). Indeed, if \(\lambda = 0\), then \(\varrho_{\widetilde{V}^{n \times n}_\varphi}(\lambda v) = 0\), but we know already that for any other \(\mu > \lambda = 0\) then \(\varrho_{\widetilde{V}^{n \times n}_\varphi}(\mu v) \ge 0\), since we can always choose \(\psi = 0\) as a test function. Let us instead fix \(0 < \lambda < \mu < +\infty\). We point out that since both \(\psi\) and \(-\psi\) are possible test functions, we can also only consider without loss of generality functions in the space \[\overline{T}^+ := \left\{ \psi \in C^1_c(\Omega; \mathbb{R}^{n \times n}_{\rm sym}) \ : \ \int_\Omega \psi : dv(x) \ge 0 \right\},\] since the second integral in the definition of the modular is always zero or negative. Then we have \[\begin{align} \varrho_{\widetilde{V}^{n \times n}_\varphi}(\lambda v) &= \sup \left\{ \int_\Omega \lambda\psi : dv(x) - \int_\Omega\varphi^*(x,|\psi|) \, dx \ : \ \psi \in C^1_c(\Omega; \mathbb{R}^{n \times n}_{\rm sym}) \right\} \\ &= \sup \left\{ \lambda \int_\Omega \psi : dv(x) - \int_\Omega\varphi^*(x,|\psi|) \, dx \ : \ \psi \in \overline{T}^+ \right\} \\ &\le \sup \left\{ \mu \int_\Omega \psi : dv(x) - \int_\Omega\varphi^*(x,|\psi|) \, dx \ : \ \psi \in \overline{T}^+ \right\} \\ &= \sup \left\{\int_\Omega \mu\psi : dv(x) - \int_\Omega\varphi^*(x,|\psi|) \, dx \ : \ \psi \in C^1_c(\Omega; \mathbb{R}^{n \times n}_{\rm sym}) \right\} = \varrho_{\widetilde{V}^{n \times n}_\varphi}(\mu v), \end{align}\] for any \(v \in \mathcal{M}(\Omega; \mathbb{R}^{n \times n}_{\rm sym})\), proving condition (ii).
Since condition (iv) holds with the constant \(\beta = 1\), then \(\varrho_{\widetilde{V}^{n \times n}_\varphi}\) is a semimodular. ◻

Proof. Since \(u \in BD^\varphi(\Omega)\), the case \(\widetilde{V}^{n \times n}_\varphi(E u) = +\infty\) is excluded a priori. Moreover, if \(\widetilde{V}^{n \times n}_\varphi(Eu) = 0\), then \(\varrho_{\widetilde{V}^{n \times n}_\varphi}(\frac{Eu}{\lambda}) = 0\) for every \(\lambda > 0\), yielding \(\|Eu\|_{\varrho_{\widetilde{V}^{n \times n}_\varphi}} = 0\). Lastly, since the claim is homogeneous, the case \(\widetilde{V}^{n \times n}_\varphi(Eu) \in (0,+\infty)\) can always be reduced to \(\widetilde{V}^{n \times n}_\varphi(Eu) = 1\).
By the definition of \(\widetilde{V}^{n \times n}_\varphi\), see ?? , for any \(w \in C^1_c(\Omega;\mathbb{R}^{n \times n}_{\rm sym})\) such that \(\|w\|_{\varphi^*} \le 1\) it follows that \[\int_\Omega u \cdot E^*w \, dx \le \widetilde{V}^{n \times n}_\varphi(Eu) \, \|w\|_{\varphi^*} = \|w\|_{\varphi^*} \le 1 + \varrho_{\varphi^*}(|w|),\] where the last inequality comes from a general property of the Luxemburg norms, see [10]. Thus, by the definition of \(\varrho_{\widetilde{V}^{n \times n}_\varphi}\) given in 9 , \[\varrho_{\widetilde{V}^{n \times n}_\varphi}(Eu) \le \sup_{w \in C^1_c(\Omega; \mathbb{R}^{n \times n}_{\rm sym})} [1 + \varrho_{\varphi^*}(|w|) - \varrho_{\varphi^*}(|w|)] = 1,\] implying \(\|E u\|_{\varrho_{\widetilde{V}^{n \times n}_\varphi}} \le 1\). This concludes the proof of the first inequality.
We next establish the opposite inequality \(2 \, \|Eu\|_{\varrho_{\widetilde{V}^{n \times n}_\varphi}} \ge 1\), which is equivalent to \[\label{eq:mod95magg} \varrho_{\widetilde{V}^{n \times n}_\varphi} \left(\frac{2Eu}{\lambda} \right) > 1 \qquad \textrm{ for every } \lambda < 1.\tag{11}\] Indeed, on the one hand if \(\|v\|_{\varrho_{\widetilde{V}^{n \times n}_\varphi}} \ge 1\), by Definition 3 we have that \[\inf \left\{ k > 0 : \varrho_{\widetilde{V}^{n \times n}_\varphi}\left( \frac{v}{k} \right) \le 1 \right\} \ge 1,\] entailing that for any \(\lambda < 1\) it holds \(\|v\|_{\varrho_{\widetilde{V}^{n \times n}_\varphi}} > \lambda\). By [11] then this implies that \(\varrho_{\widetilde{V}^{n \times n}_\varphi}(\tfrac{v}{\lambda}) > 1\) for any \(\lambda < 1\). On the other hand, having \(\varrho_{\widetilde{V}^{n \times n}_\varphi}(\tfrac{v}{\lambda}) > 1\) for any \(\lambda < 1\) entails that, by Definition 3, \[\|v\|_{\varrho_{\widetilde{V}^{n \times n}_\varphi}} = \inf \left\{ \lambda > 0 \ : \ \varrho_{\widetilde{V}^{n \times n}_\varphi}\left( \frac{v}{\lambda} \right) \le 1 \right\} \ge 1\] directly from the definition of \(\inf\). Now, by [11] and thanks to the fact that \(\varphi^* \in \Phi_c(\Omega)\), see Lemma 1(iii), \(\varrho_{\varphi^*}(|w|) \le 1\) if and only if \(\|w\|_{\varphi^*} \le 1\), and we can immediately conclude that \[\begin{align} \varrho_{\widetilde{V}^{n \times n}_\varphi} \left(\frac{2Eu}{\lambda} \right) & \ge \sup \left\{ \int_\Omega \left[ \frac{2 u}{\lambda} \cdot E^* w - \varphi^*(x, |w|) \right] dx \ : \ w \in C^1_c(\Omega; \mathbb{R}^{n \times n}_{\rm sym}), \|w\|_{\varphi^*}\le 1 \right\} \\ & \ge \frac{2}{\lambda} \, \widetilde{V}^{n \times n}_\varphi(Eu) - 1 > 1. \end{align}\] This proves 11 and yields the thesis. ◻

Proof. Since \(u \in LD_{\rm loc}(\Omega)\), it follows from the definition of \(\widetilde{V}^{n \times n}_\varphi\) that \[\label{eq:integrationByParts} \widetilde{V}^{n \times n}_\varphi(Eu) = \sup \left\{ \int_\Omega \mathcal{E}u : w \, dx \ : \ w \in C^1_c(\Omega; \mathbb{R}^{n \times n}_{\rm sym}), \|w\|_{\varphi^*} \le 1 \right\}.\tag{12}\] The definition of the associate space norm (extended in a standard way to the vector valued setting), see Definition 10, implies that, chosen \(w \in C^1_c(\Omega; \mathbb{R}^{n \times n}_{\rm sym})\) such that \(\|w\|_{\varphi^*} \le 1\), \[\int_\Omega \mathcal{E}u : w \, dx \le \int_\Omega |\mathcal{E}u| |w| \, dx \le \| \mathcal{E}u\|_{(L^{\varphi^*}(\Omega;\mathbb{R}^{n \times n}))'} \|w\|_{\varphi^*}.\] Taking the supremum over those \(w\), we conclude that \(\widetilde{V}^{n \times n}_\varphi(\mathcal{E}u) \le \|\mathcal{E}u\|_{(L^{\varphi^*}(\Omega;\mathbb{R}^{n \times n}))'}\).
Under the density assumption, namely \(C^1_c(\Omega;\mathbb{R}^{n \times n})\) being dense in \(L^{\varphi^*}(\Omega;\mathbb{R}^{n \times n})\), we next show the opposite inequality, namely \(\|\mathcal{E}u\|_{(L^{\varphi^*}(\Omega;\mathbb{R}^{n \times n}))'} \le \widetilde{V}^{n \times n}_\varphi(\mathcal{E}u)\). Indeed, let \(w \in L^{\varphi^*}(\Omega; \mathbb{R}^{n \times n}_{\rm sym})\) with \(\|w\|_{\varphi^*} = 1\) and let \((w_j)_j \subset C^1_c(\Omega; \mathbb{R}^{n \times n}_{\rm sym})\) be a sequence such that \(w_j \to w\) in \(L^{\varphi^*}(\Omega; \mathbb{R}^{n \times n}_{\rm sym})\) and pointwise almost everywhere. Since we know that also \(w_j/\|w_j\|_{\varphi^*} \to w\) in \(L^{\varphi^*}(\Omega; \mathbb{R}^{n \times n}_{\rm sym})\), we may assume that \(\|w_j\|_{\varphi^*} = 1\) for every \(j \in \mathbb{N}\). By Fatou’s Lemma, we have \[\liminf_{j \to \infty} \int_\Omega \mathcal{E}u : w_j \, dx \ge \int_\Omega \mathcal{E}u : w \, dx.\] So, from 12 follows that \[\widetilde{V}^{n \times n}_\varphi(\mathcal{E}u) \ge \sup \left\{ \int_\Omega \mathcal{E}u : w \, dx \ : \ w \in L^{\varphi^*}(\Omega; \mathbb{R}^{n \times n}_{\rm sym}), \| w\|_{\varphi^*} \le 1 \right\}.\] Let \(h\in L^{\varphi^*}(\Omega)\). We can set \[w := \begin{dcases} \frac{\mathcal{E} u}{|\mathcal{E} u|} \, h & \textrm{ if } |\mathcal{E} u|\neq 0, \\ 0 & \textrm{ otherwise.} \end{dcases}\] This gives \[\widetilde{V}^{n \times n}_\varphi( \mathcal{E}u) \ge \sup \left\{ \int_\Omega |\mathcal{E}u| \, h \, dx \ : \ h \in L^{\varphi^*}(\Omega), \| h\|_{\varphi^*}\le 1 \right\} = \| \mathcal{E}u \|_{(L^{\varphi^*}(\Omega;\mathbb{R}^{n \times n}))'}.\] Hence \(\widetilde{V}^{n \times n}_\varphi(\mathcal{E}u) = \| \mathcal{E}u \|_{(L^{\varphi^*}(\Omega;\mathbb{R}^{n \times n}))'}\). This concludes the proof. ◻

Proof of Theorem 10. Since \(Eu = \mathcal{E}u + E^su\) and since \(u \in BD(\Omega)\), an integration by parts implies that \[\int_\Omega u \cdot E^*\psi \, dx = \int_\Omega \psi : dEu = \int_\Omega \mathcal{E}u : \psi \, dx + \int_\Omega \psi : dE^su\] for any \(\psi \in T^\varphi\). Hence, the claim follows from Propositions 11 and 12 once we prove that \[\begin{align} &\sup_{\psi \in T^\varphi} \left\{\int_\Omega [\mathcal{E}u : \psi - \varphi^*(x, |\psi|)] \, dx + \int_\Omega \psi : dE^su \right\} \\ = &\sup_{\psi \in T^\varphi} \int_\Omega [\mathcal{E}u : \psi - \varphi^*(x, |\psi|)] \, dx + \sup_{\psi \in T^\varphi} \int_\Omega \psi : dE^su. \end{align}\] The first inequality, namely “\(\le\)”, is obvious, so we only need to prove the opposite one.
Let us first assume that \(\varrho_\varphi(|\mathcal{E}u|) + \int_\Omega \varphi^\infty \, d|E^su| < +\infty\) and let us fix \(\varepsilon> 0\). We can always choose \(\psi_1, \psi_2\in T^\varphi\) such that \[\label{eq95aux:choice951} \sup_{\psi \in T^\varphi} \int_\Omega [\mathcal{E}u : \psi - \varphi^*(x, |\psi|)] \, dx \le \int_\Omega [\mathcal{E}u : \psi_1 - \varphi^*(x, |\psi_1|)] \, dx + \varepsilon< +\infty\tag{13}\] and \[\label{eq95aux:choice952} \sup_{\psi \in T^\varphi} \int_\Omega \psi : dE^su \le \int_\Omega \psi_2 : dE^su + \varepsilon< +\infty.\tag{14}\] Since \(u \in BD(\Omega)\) and \(\psi_1, \psi_2 \in T^\varphi\), we have \(|\mathcal{E}u|\,|\psi_i| \in L^1(\Omega)\) and \(\varrho_{\varphi^*}(|\psi_i|) < +\infty\) for \(i \in \{1,2\}\). Thus, by the absolute continuity of the integral, we can find \(\delta > 0\) such that \[\label{eq95aux:stima951952} \left| \int_{\Omega \setminus \Omega_1} [\mathcal{E}u : \psi_i - \varphi^*(x, |\psi_i|)] \, dx \right| \le \int_{\Omega \setminus \Omega_1} [|\mathcal{E}u| \, |\psi_i| + \varphi^*(x, |\psi_i|)] \, dx \le \varepsilon\tag{15}\] for \(i \in \{1,2\}\) and any \(\Omega_1 \subset \Omega\) with \(|\Omega \setminus \Omega_1|< \delta\) and \[\left|\int_{\Omega \setminus \Omega_2} \psi_2 : dE^su \, \right| \le \int_{\Omega \setminus \Omega_2} \varphi^\infty \, d|E^su| \le \varepsilon\] for any \(\Omega_2 \subset \Omega\) with \(|E^su|(\Omega \setminus \Omega_2) < \delta\).
Since \({\rm supp}\, E^su\) is a set with zero Lebesgue measure, we can find a finite collection of open rectangles \(Q_i \subset \Omega\) such that \(|E^su|(\bigcup_{i \in \mathbb{N}} Q_i) > |E^su|(\Omega) - \delta\) and \(|\bigcup_{i \in \mathbb{N}} 2Q_i| < \delta\). Let us now choose \(\theta \in C^1_c(\Omega)\) such that \(0 \le \theta \le 1\), \(\theta=1\) in \(\Omega_2 := \bigcup_{i \in \mathbb{N}} Q_i\) and \(\theta = 0\) in \(\Omega_1 := \Omega \setminus \bigcup_{i \in \mathbb{N}} 2Q_i\), and let us fix \(\psi_\varepsilon:= \theta \, \psi_2 + (1 - \theta) \psi_1 \in C^1_c(\Omega; \mathbb{R}^{n\times n}_{\rm sym})\). Since \(\psi_\varepsilon\) is a pointwise convex combination, we get \[\varphi^*(\cdot, |\psi_\varepsilon|) \le \varphi^*(\cdot, \max\{|\psi_2|, |\psi_1|\}) \le \varphi^*(\cdot, |\psi_2|) + \varphi^*(\cdot, |\psi_1|).\] This yields that \(\varrho_{\varphi^*}(|\psi_\varepsilon|) \le \varrho_{\varphi^*}(|\psi_2|) + \varrho_{\varphi^*}(|\psi_1|) < +\infty\), yielding \(\psi_\varepsilon\in T^\varphi\). We then get that \[\begin{align} \varrho_{\widetilde{V}^{n \times n}_\varphi}(Eu) &= \sup_{\psi \in T^\varphi} \left\{\int_\Omega [\mathcal{E}u : \psi - \varphi^*(x, |\psi|)] \, dx + \int_\Omega \psi : dE^su \right\} \\ &\ge \int_\Omega [\mathcal{E}u : \psi_\varepsilon- \varphi^*(x, |\psi_\varepsilon|)] \, dx + \int_\Omega \psi_\varepsilon: dE^su \\ &\ge \int_{\Omega_1} [\mathcal{E}u : \psi_1 - \varphi^*(x, |\psi_1|)] \, dx + \int_{\Omega_2} \psi_2 : dE^su - c_\theta \\ &\ge \int_\Omega [\mathcal{E}u : \psi_1 - \varphi^*(x, |\psi_1|)] \, dx + \int_\Omega \psi_2 : dE^su - 5\varepsilon, \end{align}\] where, thanks to 15 , \[c_\theta := \int_{\Omega \setminus \Omega_1} [|\mathcal{E}u| \, |\psi_\varepsilon| + \varphi^*(x, |\psi_\varepsilon|)] \, dx + \int_{\Omega \setminus\Omega_2} |\psi_\varepsilon|\, d|E^su| \le 3\varepsilon\] by the absolute integrability assumptions and by Lemma 7. By the choices of \(\psi_1\) and \(\psi_2\), namely 13 and 14 , we then get \[\varrho_{\widetilde{V}^{n \times n}_\varphi}(Eu) \ge \sup_{\psi \in T^\varphi} \int_\Omega [\mathcal{E}u : \psi - \varphi^*(x, |\psi|)] \, dx + \sup_{\psi \in T^\varphi} \int_\Omega \psi : dE^su - 7 \varepsilon.\] The lower bound follows by letting \(\varepsilon\to 0^+\). This concludes the proof in the case \(\varrho_\varphi(|\mathcal{E}u|) + \int_\Omega \varphi^\infty \, d|E^su| < +\infty\).
If, instead, \(\varrho_\varphi(|\mathcal{E}u|) = +\infty\) and \(\int_\Omega \varphi^\infty \, d|E^su| < +\infty\), we simply estimate \[\begin{align} &\sup_{\psi \in T^\varphi} \left\{ \int_\Omega [\mathcal{E}u : \psi - \varphi^*(x, |\psi|)] \, dx + \int_\Omega \psi : dE^su \right\} \\ \ge &\sup_{\psi \in T^\varphi} \int_\Omega [\mathcal{E}u : \psi - \varphi^*(x, |\psi|)] \, dx - \sup_{\psi \in T^\varphi} \int_\Omega \psi : dE^su \\ = & \;\varrho_\varphi(|\mathcal{E}u|) - \int_\Omega \varphi^\infty \, d|E^su| = +\infty. \end{align}\] We are only left to consider the case \(\int_\Omega \varphi^\infty \, d|E^su| = +\infty\). By the proof of Proposition 11, see Appendix 5, there exists \(\psi_\varepsilon\in C^1_c(\Omega; \mathbb{R}^{n \times n}_{\rm sym})\) such that \[\int_\Omega \psi_\varepsilon: dE^su > \frac{1}{\varepsilon}\] and \(|\psi_\varepsilon| \le \frac{\varphi(\cdot, k)}{k}\) for some \(k = k_\varepsilon\). For any function \(\theta : \Omega \to [0,1]\), we have that \[\mathcal{E}u : (\theta \psi_\varepsilon) - \varphi^*(\cdot, |\theta \psi_\varepsilon|) \ge \mathcal{E}u : (\theta \psi_\varepsilon) - \varphi^* \left(\cdot, \frac{\varphi(\cdot, k)}{k} \right) \ge - \left(\frac{\varphi^+(k)}{k} \, |\mathcal{E}u| + \varphi^+(k) \right),\] see Lemma 1(iii). Since the function on the right-hand side is integrable, we can choose \(\delta_k > 0\) such that its integral over any measurable \(A\) with \(|A| < \delta_k\) is at least \(-1\). Furthermore, since \({\rm supp}\, E^su\) has measure zero, we can choose \(\theta \in C^\infty_c(\Omega)\) as before to have support with Lebesgue measure at most \(\delta_k\) and such that satisfies \[\int_\Omega (\theta \psi_\varepsilon) : dE^su > \frac{1}{2} \int_\Omega \psi_\varepsilon: dE^su > \frac{1}{2\varepsilon}.\] Then, \[\begin{align} &\sup_{\psi \in T^\varphi} \left\{\int_\Omega [\mathcal{E}u : \psi - \varphi^*(x, |\psi|)] \, dx + \int_\Omega \psi : dE^su \right\} \\ \ge & \int_\Omega (\theta \psi_\varepsilon) : dE^su + \int_\Omega [\mathcal{E}u : (\theta \psi_\varepsilon) - \varphi^*(x, |\theta \psi_\varepsilon|)] \, dx \ge \frac{1}{2\varepsilon} - 1. \end{align}\] Letting \(\varepsilon\to +\infty\), the claim follows. This concludes the proof of the theorem. ◻

Proof. First, let us observe that the claim of Lemma 7 is automatically satisfied in the case of \(\varphi\) given by 16 with the only requirement of strong \(\log\)-Hölder continuity in \(Y\) and the assumption \(p^+_\Omega\) is finite. Indeed, on the one hand, in view of 18 and Definition 8 (which is a limit in this specific case), it suffices to prove the inequality in the set \(Y\) given by 17 , since in \(\Omega \setminus Y\) the recession function is \(+\infty\). On the other hand though, whenever \(\psi \in C(\Omega)\) with \(\varrho_{\varphi^*}(|\psi|) < +\infty\), then by 18 we have that \(\varphi^*(x, |\psi|) = 0\) for almost every \(x \in Y\), implying that \(|\psi| \le 1=\varphi^\infty(x)\) in \(Y\), in view of Lemma 8, which proves our claim.
As a second step, let us remark that choosing \(\varphi\) as in 16 , with \(p(\cdot)\) continuous, , Proposition 12 is valid. In fact, by assumption \(\varphi\) is continuous, and by Remark 13, we know that (A0) and (aDec) are satisfied. Hence, we can conclude that \[\sup_{\psi \in T^\varphi} \int_\Omega \left[\mathcal{E}u : \psi - \varphi^*(x,|\psi|) \right] \, dx = \frac{1}{p(x)} \int_\Omega |\mathcal{E}u|^{p(x)} \, dx\] for any \(u \in BD(\Omega)\), where \(T^\varphi\) is defined as in ?? .
We are left to prove that Proposition 11 holds true under these hypotheses and choosing \(\varphi\) as in 16 , namely that \[\sup_{\psi \in T^\varphi} \int_\Omega \psi : dE^su = \int_\Omega \varphi^\infty \, d|E^su|.\] Note that the inequality "\(\le\)" is immediately verified thanks to the fact that Lemma 7 holds under our assumptions. So it suffices to prove that \[\int_\Omega \varphi^\infty d |E^s u|\leq \sup_{\psi \in T^\varphi}\int_\Omega \psi : d E^s u.\] Thanks to the continuity assumption on \(p\) in \(\Omega\), we obtain that 23 and 24 hold. In particular, the second one holds for any \(\beta \in [0,1]\), due to the non decreasing monotonicity of \((\cdot)^{p(x)}\). Observe that \(h_k\) is lower semicontinuous by definition (it is a lower semicontinuous envelope, see [41]) and increasing in \(k\). Consequently the same arguments as in the proof of Proposition 11 lead to the desired conclusion. Moreover, let us remark that for every \(u \in BD(\Omega)\) it holds \[\sup_{\psi \in T^\varphi} \int_\Omega \psi : d E^s u = \int_Y d|E^s u| = |E^s u|(Y),\] since \(\varphi^\infty(x) = 1\) in \(Y\) as observed before, which indeed says that \(E^s u\) is concentrated on \(Y\).
Finally, following the same steps of the proof of Theorem 10, since under our hypotheses the equivalent formulations of Lemma 7, Proposition 11 and Proposition 12 hold, we get the thesis. ◻

Proof. (i). Thanks to the definition of the functional \(TGV^{\varphi,2}_\alpha\) given in 1 , it is obvious to see that the two conditions (N2) and (N3) in Definition 1 hold, implying that \(TGV^{\varphi,2}_\alpha\) is a seminorm. Note that \(BGV^{\varphi,2}_\alpha(\Omega)\) is a normed space because \(\varphi\in \Phi_w(\Omega)\) and satisfies (A0), so \(\|\cdot\|_\varphi\) is a norm, see [11].
(ii). Let us denote the set of the test functions for the functional \(TGV^{\varphi,2}_\alpha\) by \[K^{\varphi,2}_\alpha(\Omega) := \{ \psi \in C^2_c (\Omega; \mathbb{R}^{n \times n}_{\rm sym}) \ : \ \|\psi\|_{\varphi^*} \le \alpha_1, \|\mathrm{div}\, \psi\|_{\varphi^*} \le \alpha_2 \}.\] Let us first assume that \(u\) is a polynomial of degree \(0\) or \(1\), implying that \(\nabla^2 u = 0\). Then, since the test function \(\psi\) is in \(K^{\varphi,2}_\alpha(\Omega)\) and thanks to two integration by parts, it is obvious that \(TGV^{\varphi,2}_\alpha(u) = 0\).
Suppose now that \(TGV^{\varphi,2}_\alpha(u) = 0\) for some \(u \in L^1_{\rm loc}(\Omega)\). For any \(\psi \in C^2_c (\Omega; \mathbb{R}^{n \times n}_{\rm sym})\), one can find a \(\lambda > 0\) such that \[\|\lambda \psi\|_\infty \le \alpha_1 \quad \textrm{ and } \quad \|\mathrm{div}(\lambda \psi)\|_\infty \le \alpha_2.\] Since \(\varphi\) satisfies (A0), meaning that \(\varphi^*\) satisfies (A0) as well, see Lemma 1(iv), and since \(\Omega\) is bounded, thanks to Lemma 2 we have that \(\lambda\psi \in K^{\varphi,2}_\alpha(\Omega)\). Testing in 1 with \(\lambda\psi\), we then get \[\int_\Omega u \,\mathrm{div}^2 \psi \, dx = 0 \qquad \textrm{ for any } \psi \in C^2_c (\Omega; \mathbb{R}^{n \times n}_{\rm sym}),\] implying that \(\nabla^2 u = 0\) in the weak sense. Since \(\Omega\) is connected, this implies that \(u\) has to be a polynomial of degree less than \(2\).
(iii). The equivalence of the seminorms \(TGV^{\varphi,2}_\alpha\) and \(TGV^{\varphi,2}_{\tilde{\alpha}}\) can be proven noticing that, chosen \[c := \frac{\max\{\alpha_1, \alpha_2\}}{\min\{\tilde{\alpha}_1, \tilde{\alpha}_2\}},\] we have that \(c \, K^{\varphi,2}_\alpha(\Omega) \subset K^{\varphi,2}_{\tilde{\alpha}}(\Omega)\), meaning that \(c \, TGV^{\varphi,2}_\alpha(u) \le TGV^{\varphi,2}_{\tilde{\alpha}}(u)\) for any \(u \in L^1_{\rm loc}(\Omega)\). Interchanging the roles of \(\alpha\) and \(\tilde{\alpha}\) gives us the desired equivalence.
(iv). Thanks to [4], we know that given a matrix \(R \in SO(n)\), then \[\psi \in C^2_c (R^\textrm{T}\Omega; \mathbb{R}^{n \times n}_{\rm sym}) \quad \iff \quad \widetilde{\psi} := (\psi \circ R^\textrm{T})R^\textrm{T}\in C^2_c (\Omega; \mathbb{R}^{n \times n}_{\rm sym}),\] with \(\mathrm{div}^l \, \widetilde{\psi} = ((\mathrm{div}^l \, \psi) \circ R^\textrm{T})R^\textrm{T}\), with \(l = 0,1\) where \(\mathrm{div}^0 = id\). Hence, for \(l = 0,1\), \[\|\mathrm{div}^l \, \widetilde{\psi}\|_{\varphi^*} = \| ((\mathrm{div}^l \, \psi) \circ R^\textrm{T})R^\textrm{T}\|_{\varphi^*} = \| (\mathrm{div}^l \, \psi) \circ R^\textrm{T}\|_{\varphi^*} = \| \mathrm{div}^l \, \psi \|_{\varphi^*},\] implying that \(\psi \in K^{\varphi,2}_\alpha(R^\textrm{T}\Omega)\) if and only if \(\widetilde{\psi} \in K^{\varphi,2}_\alpha(\Omega)\). Lastly, for any \(\psi \in K^{\varphi,2}_\alpha(R^\textrm{T}\Omega)\) we have \[\int_{R^\textrm{T}\Omega} u(Rx) \,\mathrm{div}^2 \psi(x) \, dx = \int_\Omega u(x) \,\mathrm{div}^2 \psi(R^\textrm{T}x) \, dx = \int_\Omega u(x) \, \mathrm{div}^2 \widetilde{\psi}(x) \, dx,\] yielding \(TGV^{\varphi,2}_\alpha(u \circ R) = TGV^{\varphi,2}_\alpha(u)\).
(v). Note that \[\mathrm{div}(v \circ r^{-1} \mathbb{I}_n) = r^{-1} (\mathrm{div}\, v) \circ r^{-1} \mathbb{I}_n,\] implying that, chosen \(\psi \in C^l_c(r^{-1}\Omega;\mathbb{R}^{2 \times 2}_{\rm sym})\) and \(\bar\psi := r^2 \, \psi \circ r^{-1} \mathbb{I}_n\), we have \[\| \mathrm{div}^l \bar\psi \|_{\varphi^*} = r^{2-l} \| \mathrm{div}^l \psi \|_{\varphi^*},\] for \(l = 0,1\). As a consequence, \(\psi \in K^{\varphi,2}_\alpha(r^{-1} \Omega)\) if and only if \(\bar\psi \in K^{\varphi,2}_{\bar\alpha}(\Omega)\) with \(\bar\alpha = (\bar\alpha_1, \bar\alpha_2) := (\alpha_1 r^2, \alpha_2 r)\). Lastly, this yields \[\int_{r^{-1} \Omega} u(rx) \,\mathrm{div}^2 \psi(x) \, dx = r^{-n} \int_\Omega u(x) \,\mathrm{div}^2 \psi(r^{-1} x) \, dx = \int_\Omega u(x) \, \mathrm{div}^2 \bar\psi(x) \, dx,\] implying \(TGV^{\varphi,2}_\alpha(u \circ r \, \mathbb{I}_n) = r^{-n} \, TGV^{\varphi,2}_{\bar\alpha}(u)\). ◻

Proof. We only prove the lemma for \(p \in [1, +\infty)\), as the case \(p = +\infty\) is analogous.
Chosen \(\psi \in C^2_c(\Omega; \mathbb{R}^{n \times n}_{\rm sym})\) with \(\|\psi\|_{\varphi^*} \le \alpha_1\) and \(\|\mathrm{div}\, \psi\|_{\varphi^*} \le \alpha_2\) as in 1 , then \(\mathrm{div}^2 \psi \in L^\infty(\Omega)\). Then, chosen \((u_k)_k \subset L^p(\Omega)\) such that \(u_k \rightharpoonup u\) weakly in \(L^p(\Omega)\) we get \[\int_\Omega u \, \mathrm{div}^2\psi \, dx = \liminf_{k \to +\infty} \int_\Omega u_k \, \mathrm{div}^2\psi \, dx \le \liminf_{k \to +\infty} TGV^{\varphi,2}_\alpha(u_k).\] The thesis follows by taking the supremum over all such \(\psi\). ◻

Proof. Thanks to Proposition 17(iii), we know that all the seminorms \(TGV^{\varphi,2}_\alpha\), respectively norms \(\|\cdot\|_{BGV^{\varphi,2}_\alpha}\), are equivalent for any \(\alpha \in \mathbb{R}^+ \times \mathbb{R}^+\), so we only need to prove that \(BGV^{\varphi,2}_\alpha(\Omega)\) is complete for some \(\alpha \in \mathbb{R}^+ \times \mathbb{R}^+\).
Note that, since \(\Omega\) is bounded and \(\varphi\) satisfies (A0), then by Lemma 2 we have \(L^\varphi(\Omega) \hookrightarrow L^1(\Omega)\). This, together with Lemma 9, implies that \(TGV^{\varphi,2}_\alpha\) is a lower semicontinuous functional with respect to \(L^\varphi(\Omega)\) for any \(\alpha\) and \(\varphi\) chosen as in the hypotheses.
Now let \((u_n)_n\) be a Cauchy sequence in \(BGV^{\varphi,2}_\alpha(\Omega)\). This yields that \((u_n)_n\) is a Cauchy sequence also in \(L^\varphi(\Omega)\), implying the existence of a limit \(u \in L^\varphi(\Omega)\) such that \[TGV^{\varphi,2}_\alpha(u) \le \liminf_{n \to +\infty} TGV^{\varphi,2}_\alpha(u_n),\] thanks to the lower semicontinuity of the functional. As a consequence, \(u \in BGV^{\varphi,2}_\alpha(\Omega)\) and the only thing left to be shown is that \(u\) is the limit in the corresponding norm as well. In order to do so, let us choose \(\varepsilon> 0\) and \(n \in \mathbb{N}\) such that for any \(\bar{n} \ge n\) it holds that \(TGV^{\varphi,2}_\alpha(u_n - u_{\bar{n}}) \le \varepsilon\). Thanks again to the lower semicontinuity of the functional (with respect to the \(L^1\) weak convergence), letting \(\bar{n} \to +\infty\) we then get \[TGV^{\varphi,2}_\alpha(u_n - u) \le \liminf_{\bar{n} \to +\infty} TGV^{\varphi,2}_\alpha(u_n - u_{\bar{n}}) \le \varepsilon,\] implying that \(u_n \to u\) in \(BGV^{\varphi,2}_\alpha(\Omega)\). ◻

Proof. Let us define \(X_2 := C^2_0(\Omega; \mathbb{R}^{n \times n}_{\rm sym})\), \(X_1 := C^1_0(\Omega; \mathbb{R}^n)\) and \(\Lambda := \mathrm{div}\in \mathcal{L}(X_2,X_1)\). We stress that \[\label{chiusure} X_2 = \overline{C^2_c(\Omega;\mathbb{R}^{n \times n}_{\rm sym})}^{\|\cdot\|_{C^2}} \qquad \textrm{ and } \qquad X_1 = \overline{C^1_c(\Omega;\mathbb{R}^n) }^{\|\cdot\|_{C^1}}.\tag{20}\] Chosen \(u \in L^1(\Omega)\), we define \[F_1(M) := I_{\{ \|\cdot\|_{\varphi*} \le \alpha_2\}}(M), \quad \textrm{ for any } M \in X_2\] and \[F_2^u(v) := I_{\{\| \cdot\|_{\varphi^*} \le \alpha_1\}}(v) - \int_\Omega u\,\mathrm{div}v \, dx, \quad \textrm{ for any } v \in X_1.\] Since \(\varphi\) satisfies (A0), then by Lemma 1(iv) also \(\varphi^*\) satisfies (A0) and, by Lemma 2, we have that \(F_1\) and \(F^u_2\) are continuous in the \(C^2\)- and \(C^1\)-topologies, respectively. Then, thanks to a density argument and to 20 , we have that \[TGV^{\varphi, 2}_\alpha(u) = -\inf_{M \in X_2} \left[ F_1(M) + F_2^u(\Lambda M) \right].\] Let us remark that we can write the space \(X_1\) equivalently as \[\label{unione95domini} X_1 = \bigcup_{\lambda \ge 0} \lambda \left(\mathrm{dom}F_2^u - \Lambda \, \mathrm{dom}F_1\right).\tag{21}\] Indeed, if \(z \in \mathrm{dom}F_2^u\) then, trivially, \(z \in X_1\) implies \(z \in \bigcup_{\lambda \ge 0}\lambda\left(\mathrm{dom}F_2^u - \Lambda \, \mathrm{dom}F_1\right)\) and for some \(\lambda \ge 0\) it is always verified that \(\tfrac{z}{\lambda} \in \mathrm{dom}F^u_2\). This implies that \(X_1 \subset \bigcup_{\lambda \ge 0}\lambda\left(\mathrm{dom}F_2^u - \Lambda \, \mathrm{dom}F_1\right)\). On the other hand, the converse is true since \(X_1\) is a closed vector space (and \(\mathrm{dom}F^u_2\) and \(\mathrm{dom}F_1\) are subsets of \(X_1\)).
Now, since \(X_1\) is a closed set, it means that also the right-hand side of 21 is closed, which allows us to use [42], a Fenchel-Rockafellar duality result, getting \[TGV^{\varphi, 2}_\alpha(u) = -\inf_{M \in X_2} \left[F_1(M) + F_2^u(\Lambda M) \right] = \min_{w \in X_1^*} \left[ (F_2^u)^*(w) + F_1^*(-\Lambda^* w) \right].\] Now, exploiting the definition of Fenchel conjugate, we get \[(F_2^u)^*(w) = \alpha_2 \sup_{v \in X_1} \left\{ \langle w , v \rangle_Y + \int_\Omega u \, \mathrm{div}v \, dx \ : \ \|v\|_{\varphi^*} \le 1 \right\},\] for every \(w\in {\mathcal{D}}'(\Omega;\mathbb{R}^n)\), and \[F_1^*(-\Lambda^* w) = \alpha_1 \sup_{M \in X_2} \left\{ \langle -\Lambda^* w, M \rangle_X \ : \ \|M\|_{\varphi^*} \le 1 \right\},\] for every \(w\in {\mathcal{D}}'(\Omega;\mathbb{R}^n)\) with \(-\Lambda^* w \in {\mathcal{D}}'(\Omega;\mathbb{R}^{n \times n})\), where \(\langle\cdot,\cdot\rangle_{X_1}\) and \(\langle\cdot,\cdot\rangle_{X_2}\) denote, respectively, the duality products between \(X_1\) and \(X_1^*\) and between \(X_2\) and \(X_2^*\). Now, we observe that if \(F_1^*(-\Lambda^* w) = +\infty\), then by Lemma 2 we find \(\|-\Lambda^* w\|_\mathcal{M}= +\infty\). Moreover, since \(-\Lambda^* = E\), we infer that for every \(w\in {\mathcal{D}}'(\Omega;\mathbb{R}^n)\) with \(-\Lambda^* w \in {\mathcal{D}}'(\Omega;\mathbb{R}^{n \times n})\) we have \[F_1^*(-\Lambda^* w) = \begin{dcases} \alpha_1 \widetilde{V}_\varphi^{n \times n}(E w) & \textrm{if } E w\in \mathcal{M}(\Omega;\mathbb{R}^{n \times n}) \\ +\infty & \textrm{if } E w \in {\mathcal{D}}'(\Omega;\mathbb{R}^{n\times n}) \setminus \mathcal{M}(\Omega;\mathbb{R}^{n \times n}). \end{dcases}\] Analogously, by observing that, since \(u\in L^1(\Omega)\), \[(F_2^u)^*(w) = \alpha_2 \sup_{v \in X_1} \left\{ \langle Du-w , v\rangle_{X_1} \ : \ \|v\|_{\varphi^*} \le 1 \right\},\] for every \(w\in {\mathcal{D}}'(\Omega;\mathbb{R}^n)\), the same argument as for \(F_1^*\) yields \[(F_2^u)^*(w) = \begin{dcases} \alpha_2 \widetilde{V}_\varphi^n (Du-w) &\textrm{if } Du - w \in \mathcal{M}(\Omega;\mathbb{R}^n) \\ +\infty & \textrm{if } Du-w \in {\mathcal{D}}'(\Omega;\mathbb{R}^n) \setminus \mathcal{M}(\Omega;\mathbb{R}^n), \end{dcases}\] again for every \(w\in {\mathcal{D}}'(\Omega;\mathbb{R}^n)\). Thus, to summarize, by classical properties of \(BD\) and \(BV\), see [31] and [30], we have that the following characterization holds: \[\label{eq95min95inf} TGV^{\varphi, 2}_\alpha(u) = \begin{dcases} \min_{w \in BD(\Omega)} \left\{ \alpha_2 \widetilde{V}_\varphi^n (Du-w) + \alpha_1 \widetilde{V}_\varphi^{n \times n}(E w) \right\} & \textrm{if } u\in BV(\Omega;\mathbb{R}^n), \\ +\infty & \textrm{otherwise in } L^1(\Omega;\mathbb{R}^n). \end{dcases}\tag{22}\] To conclude the proof we observe that if \(u \in BV^\varphi(\Omega)\), then by [11] the above infimum problem can be considered on the subspace of \(BD(\Omega)\) where additionally \(\|w\|_{\varphi^{**}} < +\infty\), namely \(BD^\varphi(\Omega)\), see Remark 8. ◻

Proof. By Remark 4, choosing \(w = 0\) as a test function in the characterization of \(TGV^{\varphi,2}_\alpha(u)\), namely ?? , we directly infer \[TGV^{\varphi,2}_\alpha(u) \le \alpha_2 \widetilde{V}_\varphi^n(Du) = \alpha_2 V_\varphi(u),\] (see 2 ), which, in turn, yields \[\|u\|_\varphi+ TGV^{\varphi,2}_\alpha(u) \le \max\{1,\alpha_2 \} \|u\|_{BV^\varphi(\Omega)}.\] Conversely, since \(\widetilde{V}^n_\varphi\) satisfies the triangle inequality, letting \(w_u\) be the minimum in ?? , we deduce \[\alpha_2 V_{\varphi}(u) = \alpha_2 \widetilde{V}_\varphi^n(Du) \le \alpha_2 \widetilde{V}_\varphi^n(Du - w_u) + \alpha_2 \widetilde{V}_\varphi^n(w_u) \le TGV^{\varphi,2}_\alpha(u) + \alpha_2 \widetilde{V}_\varphi^n(w_u).\] ◻

Proof. Because \(\varphi\) satisfies (A0), by Remark 2 we have \(BV^\varphi(\Omega) \hookrightarrow BV(\Omega)\) and so \[|Du - w|(\Omega) \le \widetilde{V}_\varphi^n(Du - w).\] Analogously, thanks to Remark 7 we get \(BD^\varphi(\Omega) \hookrightarrow BD(\Omega)\), entailing \[|Ew|(\Omega) \le \widetilde{V}_\varphi^{n\times n}(Ew),\] for every \(w \in BD(\Omega)\). In view of [5], we infer the existence of a constant \(c > 0\) such that \[\begin{align} \|u\|_{L^p(\Omega)} &\le c \, TGV^{2}_\alpha(u) \le c \left[ \alpha_2 |Du - w|(\Omega) + \alpha_1 |Ew|(\Omega) \right] \\ &\le c \left[ \alpha_2 \widetilde{V}_\varphi^n(Du - w) + \alpha_1 \widetilde{V}_\varphi^{n \times n}(Ew) \right], \end{align}\] for every \(w \in BD(\Omega)\). The thesis follows then by Theorem 19. ◻

Proof. Arguing exactly as in the proof of [5], in view of Propositions 17 and 22, we find that any minimizing sequence for ?? is uniformly bounded in \(L^p(\Omega)\) for every \(1\leq p \le \tfrac{n}{n-1}\). The thesis follows then by Lemma 9. ◻

Proof. The limsup inequality is immediately verified by considering for each \(u\in L^p(\Omega)\) the constant recovery sequence \(\bar{u}_k = u\) for every \(k \in \mathbb{N}\). To show the liminf inequality, let \(u \in L^p(\Omega)\) and let \((u_k)_k \subset L^p(\Omega)\) be such that \(u_k \rightharpoonup u\) weakly in \(L^p(\Omega)\). The fact that \[\mathcal{F}_0(u) \le \liminf_{k \to +\infty} \mathcal{F}_k(u_k)\] is a direct consequence of Lemma 9. ◻

Proof. The result follows from the Fundamental Theorem of Gamma-convergence, cf. [43], by combining Theorem 24 and Proposition 22. ◻

Proof. By the definitions of total variation of a measure and of weak derivative, together with the fact that \(Eu = \mathcal{E}u + E^s u\), we see that \[\begin{align} |Eu|(\Omega) &= \sup_{\psi \in C^1_c(\Omega; \mathbb{R}^{n \times n}_{\rm sym}), |\psi| \le 1} \left( \int_\Omega \psi : d\mathcal{E}u + \int_\Omega \psi : dE^su \right) \\ &\le \sup_{\psi \in L^\infty(\Omega; \mathbb{R}^{n \times n}_{\rm sym}), |\psi| \le 1} \int_\Omega \psi : d \mathcal{E}u + \sup_{\psi \in L^\infty(\Omega; \mathbb{R}^{n \times n}_{\rm sym}), |\psi| \le 1} \int_\Omega \psi : dE^su \\ &= |\mathcal{E}u|(\Omega) + |E^su|(\Omega). \end{align}\] Since \(|\mathcal{E}u|(\Omega) + |E^su|(\Omega) = |Eu|(\Omega)\) as \(\mathcal{E}\) and \(E^s\) are mutually singular, each inequality has to be an equality, and so the claim follows. ◻

Proof of Proposition 11. By the definition of the total variation of a measure, Lemma 7 and the definition of \(T^\varphi\), see ?? , we have \[\sup_{\psi \in T^\varphi} \int_\Omega \psi : dE^su \le \sup_{\psi \in T^\varphi} \int_\Omega |\psi| \, d|E^su| \le \int_\Omega \varphi'_\infty \, d|E^su|.\] In order to tackle the opposite inequality, let us define \(h_k : \Omega \to [0,+\infty]\) by \[h_k(x) := \lim_{r \to 0^+} \inf_{y \in B(x,r)} \frac{\varphi(y, k)}{k} \qquad \textrm{ for any } x \in \Omega.\] Notice that \(h_k\) is lower semicontinuous with \[\label{leqhk} h_k \le \tfrac{\varphi(\cdot, k)}{k} \le \varphi^\infty.\tag{23}\] From the first inequality it follows that \(\varphi^*(\cdot, h_k) \le \varphi(\cdot, k)\), implying that \(\varrho_{\varphi^*}(h_k) \le \varrho_\varphi(k) < +\infty\) since \(\varphi\) satisfies (A0) and (aDec), and \(\Omega\) is bounded.
Let us now remark that \(h_k \to \varphi^\infty\) as \(k \to +\infty\). Indeed, if \(\varphi^\infty(x) = +\infty\), then we can use (A1) in all sufficiently small balls, since \(\varphi^+(k) < +\infty\), to conclude that \[\label{geqhk} h_k(x) = \lim_{r \to 0^+} \inf_{y \in B(x,r)} \frac{\varphi(y, k)}{k} \ge \frac{\varphi(x, \beta k) - 1}{k} \to \beta \varphi^\infty(x) = +\infty\tag{24}\] as \(k \to +\infty\). If instead \(\varphi^\infty(x)< +\infty\), then we use the same inequality but now with \(\beta := \tfrac{1}{1 + \omega(r)}\), where \(\omega\) is the modulus of continuity coming from the restricted (VA1) condition, and we obtain the desired convergence as \(\beta \to 1^-\).
Note that \(h_k\) is increasing in \(k\) since \(\varphi\) is convex. Then, by monotone convergence, \[\int_\Omega \varphi^\infty \, d|E^su| = \lim_{k \to \infty} \int_\Omega h_k \, d|E^su|.\] Let us fix \(\varepsilon> 0\) and assume that \(\int_\Omega \varphi^\infty \, d|E^su| < +\infty\). Then, we can find \(h = h_k\) and \(K > 0\) such that \[\begin{align} \int_\Omega \varphi^\infty \, d|E^su| &\le \int_\Omega h \, d|E^su| + \varepsilon \le \sum_{j = 1}^{K^2} \int_\Omega \frac{1}{K} \, \chi_{\{h > \tfrac{j}{K}\}} \, d|E^su| + 2\varepsilon\\ &= \frac{1}{K} \sum_{j = 1}^{K^2} |E^su|\left(\left\{h > \frac{j}{K}\right\}\right) + 2\varepsilon. \end{align}\] Since \(h\) is lower semicontinuous, then \(\{h > \tfrac{j}{K}\}\) is open, and hence by Lemma 10 we can choose \(\psi_j \in C^1_c(\Omega; \mathbb{R}^{n \times n}_{\rm sym})\) with \(|\psi_j| \le 1\) such that \[\int_\Omega \varphi^\infty \, d|E^su| \le \frac{1}{K} \sum_{j = 1}^{K^2} \int_{\{h > \tfrac{j}{K}\}} \psi_j : dE^su + 3\varepsilon = \int_\Omega \left(\sum_{j = 1}^{K^2} \frac{1}{K} \psi_j \right) : dE^su + 3\varepsilon.\] Defining \(\psi_\varepsilon:= \sum_{j = 1}^{K^2} \tfrac{1}{K} \psi_j\), we can notice that \(\psi_\varepsilon\in C^1_c(\Omega; \mathbb{R}^{n \times n}_{\rm sym})\) and \[|\psi_\varepsilon| \le \frac{1}{K} \sum_{j = 1}^{K^2} |\psi_j| \le \frac{1}{K} \sum_{j = 1}^{K^2} \chi_{\{h > \frac{j}{K}\}} \le h,\] giving \(\varrho_{\varphi^*}(|\psi_\varepsilon|) < +\infty\). Therefore \(\psi_\varepsilon\in T^\varphi\) and \[\int_\Omega \varphi^\infty \, d|E^su| \le \int_\Omega \psi_\varepsilon: dE^su + 3\varepsilon \le \sup_{\psi \in T^\varphi} \int_\Omega \psi : dE^su +3\varepsilon.\] The upper bound then follows letting \(\varepsilon\to 0^+\). If instead \(\int_\Omega \varphi^\infty \, d|E^su| = +\infty\), then a similar argument gives \(\tfrac{1}{3\varepsilon} \le \sup_{\psi \in T^\varphi} \int_\Omega \psi : dE^su\) and the claim again follows. ◻

Proof of Proposition 12. The upper bound follows directly from Young’s inequality, indeed \[\sup_{\psi \in T^\varphi} \int_\Omega [\mathcal{E}u : \psi - \varphi^*(x, |\psi|)] \, dx \le \int_\Omega \varphi(x, |\mathcal{E}u|) \, dx = \varrho_\varphi(|\mathcal{E}u|).\] In order to prove the lower bound let us choose \(g_i \in C(\Omega; \mathbb{R}^{n \times n}_{\rm sym}) \cap L^\varphi(\Omega; \mathbb{R}^{n \times n}_{\rm sym})\) such that \(g_i \to \mathcal{E}u\) pointwise a.e and in \(L^1(\Omega; \mathbb{R}^{n \times n}_{\rm sym})\). Then Fatou’s Lemma and \(L^1\)-convergence yield \[\int_\Omega \varphi(x, |\mathcal{E}u|) \, dx \le \liminf_{i \to +\infty} \int_\Omega \varphi(x, |g_i|) \, dx \quad \textrm{ and } \quad \lim_{i \to +\infty} \int_\Omega g_i : \psi \, dx = \int_\Omega \mathcal{E}u : \psi \, dx\] for any \(\psi \in T^\varphi\). Thus, it suffices to show that \[\int_\Omega \varphi(x, |g|) \, dx \le \sup_{\psi \in T^\varphi} \int_\Omega [g : \psi - \varphi^*(x, |\psi|)] \, dx\] for any \(g \in C(\Omega; \mathbb{R}^{n \times n}_{\rm sym}) \cap L^\varphi(\Omega; \mathbb{R}^{n \times n}_{\rm sym})\). Moreover, replacing \(\psi\) by \(\frac{g}{\varepsilon+ |g|} |\psi|\) and letting \(\varepsilon\to 0^+\), we see that \(g : \frac{g}{\varepsilon+ |g|} |\psi| \to |g| \, |w|\) pointwise. Thus, by monotone convergence, the vector-values of \(g\) and \(\psi\) are irrelevant and we only need to show that \[\label{eq:thesis952} \int_\Omega \varphi(x, |g|) \, dx \le \sup_{\psi \in C^1_c(\Omega)} \int_\Omega [|g \psi| - \varphi^*(x, |\psi|)] \, dx\tag{25}\] for any \(g\in C(\Omega) \cap L^\varphi(\Omega)\). We can also exclude test functions such that \(\varrho_{\varphi^*}(\psi) = +\infty\), see Remark 9.
Now let us denote with \(\varphi'\) the left-derivative of \(\varphi\) with respect to the second variable. Then \(\varphi'\) is left-continuous and \(\varphi(x,s) \ge \varphi(x,s_0) + \varphi'(x,s_0)(s-s_0)\) by convexity. For any \(\varepsilon> 0\), let us define \[\label{final95choice} \psi_\varepsilon(x,t) := \int_{-\infty}^{+\infty} \varphi(x, \max\{\tau,0\}) \, \zeta_\varepsilon(t-\tau) \, d\tau = (\varphi*_t \zeta_\varepsilon)(x, t),\tag{26}\] where \(\zeta_\varepsilon\) is a mollifier in \(\mathbb{R}\) with support in \([0,\varepsilon]\) and where \(*_t\) denotes the convolution only in the second variable. Since \(\varphi\) and \(\varphi'\) are increasing in the second variable and left-continuous, we have that \(\psi_\varepsilon\nearrow \varphi\) and \(\psi_\varepsilon' \nearrow \varphi'\) as \(\varepsilon\to 0^+\). Moreover, \(\psi_\varepsilon' = \varphi*_t \zeta_\varepsilon'\) is continuous in both \(x\), since \(\varphi\) is, and \(t\), as a convolution with a smooth function.
Let \(v_i \in C_c(\Omega)\) with \(0 \le v_i \le 1\) and \(v_i \nearrow 1\) as \(i \to +\infty\). By uniform continuity in \({\rm supp}\, v_i\), we can choose \(\delta = \delta_{\varepsilon, i} > 0\) such that, for any \(\varepsilon> 0\) and \(i \in \mathbb{N}\), \[\psi_\varepsilon'(x, |g(x)| \, v_i(x)) - \varepsilon\le \psi_\varepsilon'(y, |g(y)| \, v_i(y))\] for all \(x \in B(y,\delta)\) and \(y \in \Omega\). Then \[\psi_{\varepsilon, i} := \max\{\psi_\varepsilon'(\,\cdot\,, |g| \, v_i) - \varepsilon, 0\} *_x \eta_\delta \le \psi_\varepsilon'(\cdot, |g|) \le \varphi'(\cdot, |g|),\] where \(\eta_\delta\) is a mollifier in \(\mathbb{R}\) with support in \([0,\delta]\) and where \(*_x\) denotes the convolution only in the first variable. We have that \(\psi_{\varepsilon,i} \to \varphi'(\cdot, |g|)\), so we conclude by Fatou’s Lemma that \[\int_\Omega |g| \, \varphi'(x, |g|) \, dx \le \liminf_{i \to +\infty, \, \varepsilon\to 0} \int_\Omega |g \, \psi_{\varepsilon, i}| \, dx.\] Since \(\varphi\) satisfies (A0) and (aDec), we see that \[\varphi^*(x, |\psi_{\varepsilon,i}|) \le \varphi^*(x, \varphi'(x, |g|)) \le |g| \, \varphi'(x, |g|) \le c \, \varphi(x, |g|),\] where \(c > 0\) is a constant. Since \(g \in L^\varphi(\Omega)\) and \(\varphi\) satisfies (aDec), then \(\varrho_\varphi(g) < +\infty\). Thus, dominated convergence with upper bound \(c \, \varphi(\cdot, g)\) yields \[\int_\Omega \varphi^*(x, \varphi'(x, |g|)) \, dx = \lim_{i \to +\infty, \, \varepsilon\to 0} \int_\Omega \varphi^*(x, |\psi_{\varepsilon,i}|) \, dx.\] Since \(\psi_\varepsilon\) is a valid test function and \(\varphi'(\cdot, |g|) < +\infty\) a.e. in \(\Omega\), the last equality, together with “Young’s equality”, see [44], implies that \[\begin{align} \sup_{\psi \in C^1_c(\Omega), \, \varrho_{\varphi^*}(\psi) < +\infty} &\int_\Omega [|g \psi| - \varphi^*(x, |\psi|)] \, dx \ge \liminf_{i \to +\infty, \, \varepsilon\to 0} \int_\Omega [|g| \, |\psi_{\varepsilon,i}| - \varphi^*(x, |\psi_{\varepsilon,i}|)]\, dx \\ \ge &\int_\Omega [|g| \, \varphi'(x, |g|) - \varphi^*(x, \varphi'(x, |g|))] \, dx = \int_\Omega \varphi(x, |g|) \, dx, \end{align}\] proving 25 and completing the proof of the lower bound. ◻

Proof. By definition \[f^*(x^*) = \sup_{x \in X} [\langle x^*, x \rangle - f(x)] = \sup_{x \in X} [\langle x^*, x \rangle - \phi(\|x\|_{X})],\] which can be equivalently written as \[\begin{align} \label{24613Amar} f^*(x^*) &= \sup_{t \ge 0} \sup_{\|x\|_X = t} [\langle x^*, x \rangle - \phi(t)] \\ &= \sup_{t \ge 0} \left\{ \left[ \sup_{\|x\|_X = t} \langle x^*, x \rangle \right] - \phi(t) \right\} = \sup_{t \ge 0} [t \|x^*\|_{X^*} - \phi(t)]. \end{align}\tag{27}\] On the other hand, by definition \[\label{24614Amar} \phi^*(\|x^*\|_{X^*}) = \sup_{t \in \mathbb{R}} [t \|x^*\|_{X^*} - \phi(t)],\tag{28}\] hence it suffices to prove that 27 coincides with 28 , i.e. that it suffices to take the supremum on nonnegative real numbers. Using the fact that \(\phi\) is even, for any \(t > 0\) we have \[-t \|x^*\|_{X^*} - \phi(-t) = -t \|x^*\|_{X^*} - \phi(t) \le t \|x^*\|_{X^*} - \phi(t),\] hence \[\sup_{t \in \mathbb{R}} [t\|x^*\|_{X^*} - \phi(t)] = \sup_{t \ge 0} [t\|x^*\|_{X^*} - \phi(t)],\] which concludes the proof. ◻