*4.1. Proof of Theorem 1*

We prove Item (a), and then readily prove Items (b)–(d). Define the auxiliary sequence

$$f\_k^{(n)} \stackrel{\triangle}{=} \frac{1}{\binom{n}{k}} \sum\_{\substack{\mathcal{T} \subseteq \Omega \colon |\mathcal{T}| = k}} f(\mathcal{T}), \qquad k \in [0:n], \tag{55}$$

averaging *f* over all *k*-element subsets of the *n*-element set Ω - {*ω*1, ... , *ωn*}. Let the permutation *π*: [*n*] → [*n*] be arbitrary. For *k* ∈ [*n* − 1], let

$$\mathcal{S}\_1 \stackrel{\triangle}{=} \{ \omega\_{\pi(1)}, \dots, \omega\_{\pi(k-1)}, \omega\_{\pi(k)} \}\_{\prime} \tag{56a}$$

$$\mathcal{S}\_2 \stackrel{\triangle}{=} \{ \omega\_{\pi(1)}, \dots, \omega\_{\pi(k-1)}, \omega\_{\pi(k+1)} \},\tag{56b}$$

which are *k*-element subsets of Ω with *k* − 1 elements in common. Then,

$$f(\mathcal{S}\_1) + f(\mathcal{S}\_2) \ge f(\mathcal{S}\_1 \cup \mathcal{S}\_2) + f(\mathcal{S}\_1 \cap \mathcal{S}\_2),\tag{57}$$

which holds by the submodularity of *f* (by assumption), i.e.,

$$f\left(\{\omega\_{\pi(1)},\ldots,\omega\_{\pi(k)}\}\right) + f\left(\{\omega\_{\pi(1)},\ldots,\omega\_{\pi(k-1)},\omega\_{\pi(k+1)}\}\right)$$

$$\geq f\left(\{\omega\_{\pi(1)},\ldots,\omega\_{\pi(k+1)}\}\right) + f\left(\{\omega\_{\pi(1)},\ldots,\omega\_{\pi(k-1)}\}\right).\tag{58}$$

Averaging the terms on both sides of (58) over all the *n*! permutations *π* of [*n*] gives

$$\frac{1}{n!} \sum\_{\pi} f\left(\{\omega\_{\pi(1)}, \dots, \omega\_{\pi(k)}\}\right) = \frac{k! \left(n - k\right)!}{n!} \sum\_{\substack{\mathcal{T} \subseteq \Omega \colon |\mathcal{T}| = k}} f(\mathcal{T}) \tag{59a}$$

$$\mathbf{I} = \frac{1}{\binom{n}{k}} \sum\_{\substack{\mathcal{T} \subseteq \Omega \colon |\mathcal{T}| = k}} f(\mathcal{T}) \tag{59b}$$

= *f* (*n*) *<sup>k</sup>* , (59c)

and similarly

$$\frac{1}{n!} \sum\_{\pi} f\left(\{\omega\_{\pi(1)'}, \dots, \omega\_{\pi(k-1)'}\omega\_{\pi(k+1)}\}\right) = f\_k^{(n)},\tag{60a}$$

$$\frac{1}{n!} \sum\_{\pi} f\left(\{\omega\_{\pi(1)\prime}, \dots, \omega\_{\pi(k+1)}\}\right) = f\_{k+1\prime}^{(n)}\tag{60b}$$

$$\frac{1}{n!} \sum\_{\pi} f\left(\{\omega\_{\pi(1)}, \dots, \omega\_{\pi(k-1)}\}\right) = f\_{k-1'}^{(n)}\tag{60c}$$

with *f* (*n*) <sup>0</sup> <sup>=</sup> 0 since by assumption *<sup>f</sup>*(∅) = 0. Combining (58)–(60) gives

$$2f\_k^{(n)} \ge f\_{k+1}^{(n)} + f\_{k-1'}^{(n)} \quad k \in [n-1],\tag{61}$$

which is rewritten as

$$f\_k^{(n)} - f\_{k-1}^{(n)} \ge f\_{k+1}^{(n)} - f\_k^{(n)}, \quad k \in [n-1]. \tag{62}$$

Consequently, it follows that

$$\frac{f\_k^{(n)}}{k} - \frac{f\_{k+1}^{(n)}}{k+1} = \frac{1}{k} \sum\_{j=1}^k \left( f\_j^{(n)} - f\_{j-1}^{(n)} \right) - \frac{1}{k+1} \sum\_{j=1}^{k+1} \left( f\_j^{(n)} - f\_{j-1}^{(n)} \right) \tag{63a}$$

$$= \left(\frac{1}{k} - \frac{1}{k+1}\right) \sum\_{j=1}^{k} \left(f\_j^{(n)} - f\_{j-1}^{(n)}\right) - \frac{1}{k+1} \left(f\_{k+1}^{(n)} - f\_k^{(n)}\right) \tag{63b}$$

$$=\frac{1}{k(k+1)}\sum\_{j=1}^{k}\left\{ \left(f\_{j}^{(n)} - f\_{j-1}^{(n)}\right) - \left(f\_{k+1}^{(n)} - f\_{k}^{(n)}\right) \right\} \tag{63c}$$

$$\geq 0,\tag{63d}$$

where equality (63a) holds since *f* (*n*) <sup>0</sup> = 0, and inequality (63d) holds by (62). The sequence *f* (*n*) *k k* !*n k*=1 is therefore monotonically decreasing, and in particular

$$f\_k^{(n)} \ge \frac{k f\_n^{(n)}}{n} = \frac{k}{n}.\tag{64}$$

We next prove (25) when *α* = 1, and then proceed to prove Theorem 1. By (64)

$$\frac{f\_n^{(n)}}{n} \le \frac{f\_{n-1}^{(n)}}{n-1},\tag{65}$$

where, by (55),

$$f\_n^{(n)} = f(\Omega), \qquad f\_{n-1}^{(n)} = \frac{1}{n} \sum\_{\substack{\mathcal{T} \subseteq \Omega \colon |\mathcal{T}| = n-1}} f(\mathcal{T}).\tag{66}$$

Combining (65) and (66) gives

$$\left(\left(n-1\right)f\left(\Omega\right)\leq\sum\_{\substack{\mathcal{T}\subseteq\Omega:\ |\mathcal{T}|=n-1}}f(\mathcal{T}).\tag{67}$$

Since there are *n* subsets T ⊆ Ω with |T | = *n* − 1, rearranging terms in (67) gives (25) for *α* = 1; it is should be noted that, for *α* = 1, the set function *f* does not need to be nonnegative for the satisfiability of (25) (however, this will be required for *α* > 1).

We next prove Item (a). By (20), for *k* ∈ [*n*],

$$t\_k^{(n)} = \frac{1}{\binom{n}{k}} \sum\_{\substack{\mathcal{T} \subseteq \Omega \colon |\mathcal{T}| = k}} g\left(\frac{f(\mathcal{T})}{k}\right) \tag{68a}$$

$$\mathcal{I} = \frac{1}{\binom{n}{k}} \sum\_{\mathcal{T} = \{t\_1, \dots, t\_k\} \subseteq \Omega} \mathcal{S}\left(\frac{f\left(\{t\_1, \dots, t\_k\}\right)}{k}\right). \tag{68b}$$

Fix Ω*<sup>k</sup>* - {*t*1, ... , *tk*} ⊆ <sup>Ω</sup>, and let ˜ *<sup>f</sup>* : <sup>2</sup>Ω*<sup>k</sup>* <sup>→</sup> <sup>R</sup> be the restriction of the function *<sup>f</sup>* to the subsets of Ω*k*. Then, ˜ *f* is a submodular set function with ˜ *f*(∅) = 0; similarly to (55), (65) and (66) with *f* replaced by ˜ *<sup>f</sup>* , and *<sup>n</sup>* replaced by *<sup>k</sup>*, the sequence ˜ *f* (*k*) *j j* !*k j*=1 is monotonically decreasing. Hence, for *k* ∈ [2 : *n*],

$$\frac{\tilde{f}\_k^{(k)}}{k} \le \frac{\tilde{f}\_{k-1}^{(k)}}{k-1},\tag{69}$$

where

$$f\_k^{(k)} = f(\Omega\_k) = f(\{t\_1, \dots, t\_k\}),\tag{70a}$$

$$\tilde{f}\_{k-1}^{(k)} = \frac{1}{k} \sum\_{\substack{\mathcal{T} \subseteq \Omega\_k \colon |\mathcal{T}| = k-1}} \tilde{f}(\mathcal{T}) \tag{70b}$$

$$=\frac{1}{k}\sum\_{\substack{\mathcal{T}\subseteq\Omega\_{k}:|\mathcal{T}|=k-1}}f(\mathcal{T})\tag{70c}$$

$$=\frac{1}{k}\sum\_{i=1}^{k}f\left(\{t\_{1\prime},\ldots,t\_{i-1\prime},t\_{i+1\prime},\ldots,t\_k\}\right).\tag{70d}$$

Combining (69) and (70) gives

$$f\left(\{t\_1,\ldots,t\_k\}\right) \le \frac{1}{k-1} \sum\_{i=1}^k f\left(\{t\_{1'},\ldots,t\_{i-1},t\_{i+1'},\ldots,t\_k\}\right),\tag{71}$$

and, since by assumption *g* is monotonically increasing,

$$\mathbb{S}\left(\frac{f\left(\{t\_1,\ldots,t\_k\}\right)}{k}\right) \le \mathbb{S}\left(\frac{1}{k}\sum\_{i=1}^k \frac{f\left(\{t\_1,\ldots,t\_{i-1},t\_{i+1},\ldots,t\_k\}\right)}{k-1}\right).\tag{72}$$

From (68) and (72), for all *k* ∈ [2 : *n*],

$$\mathbf{t}\_k^{(n)} \le \frac{1}{\binom{n}{k}} \sum\_{\mathcal{T} = \{t\_1, \dots, t\_k\} \subseteq \Omega} \mathbf{g}\left(\frac{1}{k} \sum\_{i=1}^k \frac{f\left(\{t\_1, \dots, t\_{i-1}, t\_{i+1}, \dots, t\_k\}\right)}{k-1}\right),\tag{73}$$

and

$$\mathbf{t}\_k^{(n)} \le \frac{1}{k \binom{n}{k}} \sum\_{i=1}^k \sum\_{\substack{\mathcal{T} = \{t\_1, \dots, t\_k\} \subseteq \Omega}} \mathbf{g} \left( \frac{f\left(\{t\_1, \dots, t\_{i-1}, t\_{i+1}, \dots, t\_k\}\right)}{k-1} \right) \tag{74a}$$

$$\mathbf{y} = \frac{n - k + 1}{k \binom{n}{k}} \sum\_{i=1}^{k} \left\{ \sum\_{\{t\_1, \ldots, t\_{i-1}, t\_{i+1}, \ldots, t\_k\} \subseteq \Omega} g\left(\frac{f\left(\{t\_1, \ldots, t\_{i-1}, t\_{i+1}, \ldots, t\_k\}\right)}{k - 1}\right) \right\} \tag{74b}$$

$$=\frac{k!\left(n-k\right)!\left(n-k+1\right)}{n!\ k}\sum\_{\substack{S\subseteq\Omega\ \left|S\right|=k-1}}g\left(\frac{f(\mathcal{S})}{k-1}\right)\tag{74c}$$

$$=\frac{(k-1)!\left(n-k+1\right)!}{n!} \sum\_{\substack{\mathcal{S}\subseteq\Omega:\ |\mathcal{S}|=k-1}} \mathcal{g}\left(\frac{f(\mathcal{S})}{k-1}\right) \tag{74d}$$

$$\mathcal{S} = \frac{1}{\binom{n}{k-1}} \sum\_{\substack{\mathcal{S} \subseteq \Omega \colon |\mathcal{S}| = k-1}} \mathcal{g}\left(\frac{f(\mathcal{S})}{k-1}\right) \tag{74e}$$

$$t = t\_{k-1'}^{(n)} \tag{74f}$$

where (74a) holds by invoking Jensen's inequality to the convex function *g*; (74b) holds since the term of the inner summation in the right-hand side of (74a) does not depend on *ti*, so for every (*<sup>k</sup>* − 1)-element subset S = {*t*1, ... , *ti*−1, *ti*+1, ... , *tk*} ⊆ <sup>Ω</sup>, there are *<sup>n</sup>* − *<sup>k</sup>* + 1 possibilities to extend it by a single element (*ti*) into a *k*-element subset T = {*t*1, ... , *tk*} ⊆ Ω; (74e) is straightforward, and (74f) holds by the definition in (20). This proves Item (a).

Item (b) follows from Item (a), and similarly Item (d) follows from Item (c), by replacing *<sup>g</sup>* with <sup>−</sup>*g*. Item (c) is next verified. If *<sup>f</sup>* is a supermodular set function with *<sup>f</sup>*(∅) = 0, then (57) and (58), and (61)–(63) hold with flipped inequality signs. Hence, if *g* is monotonically decreasing, then inequalities (72) and (73) are reversed; finally, if *g* is also concave, then (by Jensen's inequality) (74) holds with a flipped inequality sign, which proves Item (c).
