Proof. To determine the Schur-concavity or Schur-convexity of the Sharma-Mittal entropy \(S_{\alpha,\beta}(\mathbf{p})\), we apply the Schur-Ostrowski criterion MO?: a function \(\phi:D\subset\mathbb{R}^n\to \mathbb{R}\) is Schur-concave (resp. Schur-convex) if and only if \(\phi\) is symmetric (that is, it is invariant under any permutation of its arguments), continuously differentiable, and for all \(i\neq j\) and \(z=(z_1,\dots,z_n)\in D\) it holds that \[(z_i - z_j)\left( \frac{\partial \phi}{\partial p_i} - \frac{\partial \phi}{\partial p_j} \right) \leq 0 \quad (\text{resp. } \geq 0).\] Thus, because \(S_{\alpha,\beta}(\mathbf{p})\) is symmetric and continuously differentiable, it follows that it is Schur-concave (resp. Schur-convex) if and only if for all \(i \neq j\): \[\label{eq:nec95suff95cond} (p_i - p_j)\left( \frac{\partial S_{\alpha,\beta}}{\partial p_i} - \frac{\partial S_{\alpha,\beta}}{\partial p_j} \right) \leq 0 \quad (\text{resp. } \geq 0).\tag{8}\] To this end, let us compute the partial derivative of \(S_{\alpha,\beta}(\mathbf{p})\) with respect to a generic component \(p_i\) of \(\mathbf{p}\). For the sake of notation, let \(A=\sum_{k=1}^n p_k^\alpha\). It follows: \[\frac{\partial S_{\alpha,\beta}}{\partial p_i} = \frac{1}{1-\beta} \left[ \frac{1-\beta}{1-\alpha} A^{\frac{1-\beta}{1-\alpha} - 1} \cdot \alpha p_i^{\alpha-1} \right] = \frac{\alpha}{1-\alpha} A^{\frac{\alpha-\beta}{1-\alpha}} p_i^{\alpha-1}.\] Thus, we need to study the behaviour of the following expression: \[\label{eq:diff} (p_i-p_j)\left(\frac{\partial S_{\alpha,\beta}}{\partial p_i} - \frac{\partial S_{\alpha,\beta}}{\partial p_j}\right) =(p_i-p_j) \frac{\alpha}{1-\alpha} A^{\frac{\alpha-\beta}{1-\alpha}} \left( p_i^{\alpha-1} - p_j^{\alpha-1} \right).\tag{9}\] Since the term \(A^{\frac{\alpha-\beta}{1-\alpha}}\) is always strictly positive, it does not affect the sign of the expression 9 . Thus, we just need to study the sign of the function \[g(\alpha)=(p_i-p_j) \frac{\alpha}{1-\alpha} \left( p_i^{\alpha-1} - p_j^{\alpha-1} \right).\] We now examine how \(g(\alpha)\) behaves with respect to \(\alpha\). We assume \(p_i\neq p_j\) to avoid trivialities.

Case: \(\alpha\geq 0\). We consider two subcases: \(0<\alpha<1\) and \(\alpha>1\). For the boundary cases \(\alpha=0\) and \(\alpha=1\), we observe that \(S_{\alpha,\beta}\) is Schur-concave. In fact, for \(\alpha=0\), it holds trivially since \(g(0)=0\), while for \(\alpha=1\), it holds because \(\lim_{\alpha\to 1} g(\alpha)=-(p_i-p_j)(\ln p_i-\ln p_j)\leq0\).

Let \(0<\alpha<1\). In this case, since the function \(x\to x^{\alpha-1}\) is strictly decreasing, the factors \[(p_i-p_j)\quad\text{and}\quad (p_i^{\alpha-1} - p_j^{\alpha-1})\] always have opposite signs. Hence, \[(p_i-p_j)(p_i^{\alpha-1} - p_j^{\alpha-1})< 0.\] Therefore, since the term \(\frac{\alpha}{1-\alpha}> 0\), it follows that \(g(\alpha)< 0\).

Let \(\alpha>1\). Since the function \(x\to x^{\alpha-1}\) is strictly increasing, the factors \[(p_i-p_j)\quad\text{and}\quad (p_i^{\alpha-1} - p_j^{\alpha-1})\] always have the same sign. Hence, \[(p_i-p_j)(p_i^{\alpha-1} - p_j^{\alpha-1})> 0.\] From this, since the term \(\frac{\alpha}{1-\alpha}<0\), it follows that \(g(\alpha)< 0\).

Case: \(\alpha<0\). Since the function \(x\to x^{\alpha-1}\) is strictly decreasing, the factors \[(p_i-p_j)\quad\text{and}\quad (p_i^{\alpha-1} - p_j^{\alpha-1})\] always have opposite signs. Thus, since \[(p_i-p_j)(p_i^{\alpha-1} - p_j^{\alpha-1})< 0,\] and the term \(\frac{\alpha}{1-\alpha}< 0\), it follows that \(g(\alpha)>0\). ◻

Proof. We first observe that as \(\beta\to 1\) and as \((\alpha,\beta)\to(1,1)\), the Sharma-Mittal entropy reduces to the Rényi and Shannon entropies, respectively, for which the equality holds. Thus, we can restrict our attention to \(\alpha,\beta\in\mathbb{R}\setminus\{1\}\). We observe that from 4 , we have that \[1+(1-\beta)S_{\alpha,\beta}(\mathbf{p})=\left(\sum_{i=1}^n p_i^\alpha\right)^{\frac{1-\beta}{1-\alpha}}.\] Thus, it follows that \[\begin{align} 1+(1-\beta)S_{\alpha,\beta}(\mathbf{p}\otimes\mathbf{q})&=\left(\sum_{i,j} (p_iq_j)^\alpha\right)^{\frac{1-\beta}{1-\alpha}}\nonumber\\ &=\left(\sum_{i,j} p_i^\alpha q_j^\alpha\right)^{\frac{1-\beta}{1-\alpha}}\nonumber\\ &=\left(\left(\sum_{i} p_i^\alpha\right) \left(\sum_{j}q_j^\alpha\right)\right)^{\frac{1-\beta}{1-\alpha}}\nonumber\\ &=\left(\sum_{i} p_i^\alpha\right)^{\frac{1-\beta}{1-\alpha}}\left(\sum_{j}q_j^\alpha\right)^{\frac{1-\beta}{1-\alpha}}\nonumber\\ &=\left(1+(1-\beta)S_{\alpha,\beta}(\mathbf{p})\right)\left(1+(1-\beta)S_{\alpha,\beta}(\mathbf{q})\right)\nonumber\\ &=1+(1-\beta)S_{\alpha,\beta}(\mathbf{p})+(1-\beta)S_{\alpha,\beta}(\mathbf{q})+(1-\beta)^2S_{\alpha,\beta}(\mathbf{p})S_{\alpha,\beta}(\mathbf{q}).\label{eq:last95step} \end{align}\tag{10}\] Rearranging 10 to isolate \(S_{\alpha,\beta}(\mathbf{p}\otimes\mathbf{q})\) and dividing by \((1-\beta)\), we obtain \[S_{\alpha,\beta}(\mathbf{p}\otimes\mathbf{q})=S_{\alpha,\beta}(\mathbf{p})+S_{\alpha,\beta}(\mathbf{q})+(1-\beta)S_{\alpha,\beta}(\mathbf{p})S_{\alpha,\beta}(\mathbf{q}),\] which concludes the proof. ◻

Proof. First, we observe that \(\mathbf{p}\) and \(\mathbf{q}\) are aggregations of the tensor product \(\mathbf{p}\otimes\mathbf{q}\), that is, \(p_i=\sum_{j} p_iq_j\) for all \(i\) and \(q_j=\sum_i p_iq_j\) for all \(j\). In the context of majorization, an aggregation of a distribution always majorizes the original one. Consequently, we have \[\mathbf{p}\otimes\mathbf{q}\preceq \mathbf{p}\quad \text{and} \quad\mathbf{p}\otimes\mathbf{q}\preceq\mathbf{q}.\] Moreover, by definition of the greatest lower bound \(\mathbf{p}\wedge\mathbf{q}\), it immediately follows that \[\label{eq:prod95maj} \mathbf{p}\otimes\mathbf{q}\preceq\mathbf{p}\wedge\mathbf{q}.\tag{11}\] Now, from 11 and Lemma 1, we have that for \(\alpha\geq 0\), the Sharma-Mittal entropy \(S_{\alpha,\beta}\) is Schur-concave, yielding \[\label{eq:conc} S_{\alpha,\beta}(\mathbf{p}\wedge\mathbf{q})\leq S_{\alpha,\beta}(\mathbf{p}\otimes\mathbf{q}).\tag{12}\] Conversely, for \(\alpha<0\), it is Schur-convex, yielding \[\label{eq:convex} S_{\alpha,\beta}(\mathbf{p}\wedge\mathbf{q})\geq S_{\alpha,\beta}(\mathbf{p}\otimes\mathbf{q}).\tag{13}\] Finally, applying Lemma 2 to expand \(S_{\alpha,\beta}(\mathbf{p}\otimes\mathbf{q})\) in 12 and 13 yields the desired inequalities, concluding the proof. ◻

Proof. It follows from Theorem 1 since the Sharma-Mittal entropy is non-negative and for \(\beta\geq1\) the term \((1-\beta)S_{\alpha,\beta}(\mathbf{p})S_{\alpha,\beta}(\mathbf{q})\) is smaller or equal to \(0\). ◻

Proof. It follows from Theorem 1 since the Sharma-Mittal entropy is non-negative and for \(\beta\leq1\) the term \((1-\beta)S_{\alpha,\beta}(\mathbf{p})S_{\alpha,\beta}(\mathbf{q})\) is greater or equal to \(0\). ◻

Proof. Let \[\mathbf{w}=\mathbf{p}\wedge \mathbf{q}, \qquad \mathbf{v}=\mathbf{p}\vee \mathbf{q},\] and define the function \(g_\alpha(\mathbf{p}):{\cal P}_n\to \mathbb{R}\) as \[g_\alpha(\mathbf{p})=\sum_{i=1}^n p_i^\alpha.\] Recalling the definition of the Tsallis entropy of order \(\alpha\in(0,\infty)\) \[T_\alpha(\mathbf{p})=\frac{1-\sum_{i=1}^np_i^\alpha}{\alpha-1},\] it follows that \[g_\alpha(\mathbf{p})=(1-\alpha)T_\alpha(\mathbf{p})+1.\] Moreover, define the function \(h_{\alpha,\beta}:\mathbb{R^+}\to \mathbb{R}\) as \[\label{unzidvgp} h_{\alpha,\beta}(x)=\frac{1}{1-\beta}\left(x^{\frac{1-\beta}{1-\alpha}}-1\right).\tag{14}\] We can express the Sharma-Mittal entropy as a function of \(g_\alpha(\mathbf{p})\) as follows: \[\label{rqzcxygw} S_{\alpha,\beta}(\mathbf{p})=h_{\alpha,\beta}(g_\alpha(\mathbf{p})).\tag{15}\]

We now proceed by considering three distinct cases depending on the value of \(\alpha\).

Case: \(0<\alpha<1\). In this case, since \(1-\alpha> 0\), and since the Tsallis entropy is Schur-concave and subadditive on the majorization lattice [29], and supermodular for \(\alpha\in[0,\infty)\) [34], we have that the function \(g_\alpha(\mathbf{p})\) is also Schur-concave, subadditive and supermodular.

Moreover, since \(\alpha\in(0,1)\) and \(\beta\leq \alpha\), the exponent \(\frac{1-\beta}{1-\alpha}\) in ([eq:h95def]) is greater or equal to \(1\), and both the first derivative \(h'_{\alpha,\beta}(x)=\frac{1}{1-\alpha}x^{\frac{1-\beta}{1-\alpha}-1}\) and the second derivative \(h''_{\alpha,\beta}(x)=\frac{\frac{1-\beta}{1-\alpha}-1}{1-\alpha}x^{\frac{1-\beta}{1-\alpha}-2}\) are non negative. Thus, both the function \(h_{\alpha,\beta}(x)\) and its derivative \(h'_{\alpha,\beta}(x)\) are increasing in \(x\).

Let \[a=g_\alpha(\mathbf{p}), \qquad b=g_\alpha(\mathbf{q}), \qquad c=g_\alpha(\mathbf{w}), \qquad d=g_\alpha(\mathbf{v}).\] Because \(g_\alpha\) is Schur-concave and \[\mathbf{w}\preceq \mathbf{p},\; \mathbf{q}\preceq \mathbf{v},\] it follows \[c\ge a, \qquad c\ge b, \qquad a\ge d, \qquad b\ge d.\] Moreover, without loss of generality, assuming \(a\geq b\), yields \[\label{tykuxwgl} c\ge a\ge b\ge d.\tag{16}\] Now, since \(g_\alpha\) is supermodular on the majorization lattice, we have that \[\label{yrazebph} a+b\leq c+d.\tag{17}\] Let us define \[\delta=(c-a)\ge0.\] From ([eq:g95supermod]), it holds that \[d-b\ge -\delta.\] Hence, there exists \(\varepsilon\ge0\) such that \[\label{kfsrtpua} d=b-\delta+\varepsilon.\tag{18}\] Thus, we have that \(c=a+\delta\) and \(d=b-\delta+\varepsilon\).

Define the function \(G(t)\) as \[G(t)=h_{\alpha,\beta}(a+t)+h_{\alpha,\beta}(b-t), \qquad 0\le t\le \delta,\] where \(h_{\alpha,\beta}\) is defined in ([eq:h95def]). First, we observe that the argument \(b-t\) is always non negative for \(t\in[0,\delta]\). This holds because \(g_\alpha\) is subadditive and, therefore, since \(g_\alpha(\mathbf{p}\wedge\mathbf{q})=c\leq a+b=g_\alpha(\mathbf{p})+g_\alpha(\mathbf{q})\), we have \(b-t\geq b-\delta=a+b-c\geq0\) for \(t\in[0,\delta]\).

Let \[G'(t)=h'_{\alpha,\beta}(a+t)-h'_{\alpha,\beta}(b-t)\] be the derivative of \(G(t)\). Since, from the chain of inequalities in ([eq:order]), we have that \[a+t\ge b-t,\] and since \(h'_{\alpha,\beta}\) is increasing, it follows that \[G'(t)\ge0.\] Thus, the function \(G\) is increasing and, evaluating \(G(t)\) at \(t=\delta\) and \(t=0\) we obtain

\[\label{xkyaedgr} G(\delta)=h_{\alpha,\beta}(a+\delta)+h_{\alpha,\beta}(b-\delta)\geq h_{\alpha,\beta}(a)+h_{\alpha,\beta}(b)=G(0).\tag{19}\] Moreover, since \(h_{\alpha,\beta}\) is increasing and \(\varepsilon\geq 0\), we have that \[\label{bmxhlrcd} h_{\alpha,\beta}(b-\delta+\varepsilon)\geq h_{\alpha,\beta}(b-\delta).\tag{20}\] Combining [eq:b43eps] with [eq:G], we finally get \[h_{\alpha,\beta}(c)+h_{\alpha,\beta}(d)=h_{\alpha,\beta}(a+\delta)+h_{\alpha,\beta}(b-\delta+\varepsilon)\geq h_{\alpha,\beta}(a)+h_{\alpha,\beta}(b).\] Recalling the definitions of \(a,b,c\) and \(d\), and [eq:sharma95m95as95tsa] we obtain \[S_{\alpha,\beta}(\mathbf{w})+S_{\alpha,\beta}(\mathbf{v}) \ge S_{\alpha,\beta}(\mathbf{p})+S_{\alpha,\beta}(\mathbf{q}),\] which demonstrates that \(S_{\alpha,\beta}\) is supermodular in the range \(0<\alpha < 1\) and \(\beta\le \alpha\).

Case: \(\alpha>1\). In this case, since \(1-\alpha< 0\), we have that the function \(g_\alpha(\mathbf{p})\) is Schur-convex and submodular on the majorization lattice. Moreover, since \(\alpha>1\) and \(\beta\leq \alpha\), we have that the first derivative \(h'_{\alpha,\beta}(x)=\frac{1}{1-\alpha}x^{\frac{1-\beta}{1-\alpha}-1}\) is negative and the second derivative \(h''_{\alpha,\beta}(x)=\frac{\frac{1-\beta}{1-\alpha}-1}{1-\alpha}x^{\frac{1-\beta}{1-\alpha}-2}\) is positive. Thus, the function \(h(x)\) is convex and decreasing in \(x\), and its derivative \(h'(x)\) is increasing in \(x\). Let \[a=g_\alpha(\mathbf{p}), \qquad b=g_\alpha(\mathbf{q}), \qquad c=g_\alpha(\mathbf{w}), \qquad d=g_\alpha(\mathbf{v}).\] Because \(g_\alpha\) is Schur-convex and \[\mathbf{w}\preceq \mathbf{p},\; \mathbf{q}\preceq \mathbf{v},\] it follows \[c\leq a, \qquad c\leq b, \qquad a\leq d, \qquad b\leq d.\] Moreover, without loss of generality, assuming \(a\geq b\), yields \[\label{eq:order95convex} c\leq b\leq a\leq d.\tag{21}\] Since \(g_\alpha\) is submodular on the majorization lattice, we have that \[\label{eq:g95submodular} a+b\geq c+d.\tag{22}\] Let us define \[\delta=(a-c)\ge0.\] From (22 ), it holds that \[d-b\leq \delta.\] Hence, there exists \(\varepsilon\ge0\) such that \[\label{eq:d95convex} d=b+\delta-\varepsilon.\tag{23}\] Thus, we have that \(c=a-\delta\) and \(d=b+\delta-\varepsilon\). Define the function \(G(t)\) as \[G(t)=h_{\alpha,\beta}(a-t)+h_{\alpha,\beta}(b+t), \qquad 0\le t\le \delta,\] where \(h_{\alpha,\beta}\) is defined in ([eq:h95def]). First, we observe that the argument \(a-t\geq a-\delta=c\) is always non negative for \(t\in[0,\delta]\). Let \[G'(t)=-h'_{\alpha,\beta}(a-t)+h'_{\alpha,\beta}(b+t)\] be the derivative of \(G(t)\). We observe that \(G'(t)\geq 0\) for all \(t\) which satisfy \[h'_{\alpha,\beta}(b+t)\geq h'_{\alpha,\beta}(a-t).\] But since \(h'\) is increasing, this implies that \(G'(t)\geq 0\) for all \(t\) such that \[t\geq \frac{a-b}{2}.\] Thus, the function \(G\) is increasing for \(t\in[\frac{a-b}{2},\delta]\). Moreover, evaluating \(G(t)\) at \(t=a-b\leq a-c=\delta\), we have that \[\label{eq:g40a-b41} G(a-b)=h_{\alpha,\beta}(a-(a-b))+h_{\alpha,\beta}(b+(a-b))=h_{\alpha,\beta}(b)+h_{\alpha,\beta}(a)=G(0).\tag{24}\] Therefore, from 24 , we have that \(G(0)=G(a-b)\), and since the function \(G\) is increasing for \(t\in[\frac{a-b}{2},\delta]\), it follows that \[\label{eq:G95convex} G(\delta)=h_{\alpha,\beta}(a-\delta)+h_{\alpha,\beta}(b+\delta)\geq h_{\alpha,\beta}(a)+h_{\alpha,\beta}(b)=G(0).\tag{25}\] Moreover, since \(h_{\alpha,\beta}\) is decreasing and \(\varepsilon\geq 0\), we have that \[\label{eq:b-eps} h_{\alpha,\beta}(b+\delta-\varepsilon)\geq h_{\alpha,\beta}(b+\delta).\tag{26}\] Combining 26 with 25 , we finally get \[h_{\alpha,\beta}(c)+h_{\alpha,\beta}(d)=h_{\alpha,\beta}(a-\delta)+h_{\alpha,\beta}(b+\delta-\varepsilon)\geq h_{\alpha,\beta}(a)+h_{\alpha,\beta}(b).\] Recalling the definitions of \(a,b,c\) and \(d\), and [eq:sharma95m95as95tsa] we obtain \[S_{\alpha,\beta}(\mathbf{w})+S_{\alpha,\beta}(\mathbf{v}) \ge S_{\alpha,\beta}(\mathbf{p})+S_{\alpha,\beta}(\mathbf{q}),\] which demonstrates that \(S_{\alpha,\beta}\) is supermodular in the range \(\alpha>1\) and \(\beta\le \alpha\).

Case: \(\alpha=1\). We recall that the Sharma-Mittal entropy \(S_{\alpha,\beta}(\mathbf{p})\) can be expressed as a function of the Rényi entropy \(H_\alpha(\mathbf{p})\) as shown in 7 . Thus, as \(\alpha\to1\), the Rényi entropy converges to the Shannon entropy \(H(p)\), yielding the following representation: \[\label{eq:S951} S_{1,\beta}(\mathbf{p}) =\phi_\beta(H(\mathbf{p})),\tag{27}\] where the function \(\phi_\beta(x)\) is defined as: \[\label{eq:phi} \phi_\beta(x)=\begin{cases} \frac{2^{(1-\beta)x}-1}{1-\beta}, &for \beta\neq 1,\\ \ln(2)x, & for\beta=1. \end{cases}\tag{28}\] Given that the Shannon entropy \(H(\mathbf{p})\) is Schur-concave, subadditive and supermodular on the majorization lattice [4], and that, for \(\beta\leq\alpha=1\), the first derivative \(\phi'_\beta(x)\) and second derivative \(\phi''_\beta\) of the function \(\phi_\beta(x)\) are non negative, we can proceed following the same idea of the Case \(\alpha\in(0,1)\). Let \[a=H(\mathbf{p}), \qquad b=H(\mathbf{q}), \qquad c=H(\mathbf{w}), \qquad d=H(\mathbf{v}).\] Because Shannon entropy \(H(\mathbf{p})\) is Schur-concave and \[\mathbf{w}\preceq \mathbf{p},\; \mathbf{q}\preceq \mathbf{v},\] it follows \[c\ge a, \qquad c\ge b, \qquad a\ge d, \qquad b\ge d.\] Moreover, without loss of generality, assuming \(a\geq b\), yields \[\label{exjvpbag} c\ge a\ge b\ge d.\tag{29}\] Since the Shannon entropy satisfies the supermodularity property, we have \[\label{uteqzobv} c+d\ge a+b.\tag{30}\] Let us define \[\delta=(c-a)\ge0.\] From ([eq:2]), it holds that \[d-b\ge -\delta.\] Hence there exists \(\varepsilon\ge0\) such that \[\label{ctqoriev} d=b-\delta+\varepsilon.\tag{31}\] Therefore \[(c,d) = (a+\delta,\;b-\delta+\varepsilon).\] Define the function \(F(t)\) as \[F(t)=\phi_\beta(a+t)+\phi_\beta(b-t), \qquad 0\le t\le \delta.\] where \(\phi_\beta\) is defined in (28 ). We first check that the argument \(b-t\geq b-\delta\) remains non negative, therefore it stays within the domain of \(\phi_\beta.\) Since the Shannon entropy is subadditive on the majorization lattice, we get that \(c=H(\mathbf{p}\wedge \mathbf{q})\leq H(\mathbf{p})+H(\mathbf{q})=a+b.\) Therefore \(c\leq a+b\) ensures \(b-t\geq b-\delta=a+b-c\geq 0\) for \(t\in[0,\delta].\)

Because \(\beta\le\alpha=1\), we have that \(\phi_\beta''(x)=(\ln 2)^2(1-\beta)2^{(1-\beta)x}\ge0\). Therefore, the function \(\phi_\beta\) is convex, and its derivative \(\phi_\beta'\) is increasing. Differentiating \(F(t)\) yields \[F'(t) = \phi_\beta'(a+t)-\phi_\beta'(b-t).\] From the chain of inequalities in ([eq:1]), it follows that \[a+t\ge b-t,\] and since \(\phi_\beta'\) is increasing, we obtain that \[F'(t)\ge0.\] Thus, the function \(F(t)\) is increasing in \(t\). Evaluating \(F(t)\) at \(t=\delta\) and \(t=0\) gives \[\label{iprsvbyw} F(\delta)=\phi_\beta(a+\delta)+\phi_\beta(b-\delta) \ge \phi_\beta(a)+\phi_\beta(b)=F(0).\tag{32}\] Finally, since \(\phi_\beta\) is increasing and \(\varepsilon\ge0\), we obtain the inequality \[\label{xfnpyqlg} \phi_\beta(b-\delta+\varepsilon) \ge \phi_\beta(b-\delta).\tag{33}\] Combining ([ineq]) with ([eq:4]),we get \[\phi_\beta(c)+\phi_\beta(d) = \phi_\beta(a+\delta)+\phi_\beta(b-\delta+\varepsilon) \ge \phi_\beta(a)+\phi_\beta(b).\] Recalling the definitions of \(a,b,c, d\), and (27 ) we obtain \[S_{1,\beta}(\mathbf{w})+S_{1,\beta}(\mathbf{v}) \ge S_{1,\beta}(\mathbf{p})+S_{1,\beta}(\mathbf{q}),\] which demonstrates that \(S_{\alpha,\beta}\) is also supermodular for \(\alpha=1\) and \(\beta\le \alpha\). ◻