Past exam of the computer science course of the University of Oxford / 2023 / Combinatorial optimisation / 1 / c

Yes: this is a monotone set function. Assume $V\ne \varnothing$, as required by its definition at the empty set. If $\varnothing \ne A\subseteq B$, every weight considered in the maximum over $A$ is considered in the maximum over $B$, hence $f(A)\le f(B)$. If $A=\varnothing$ and $B\ne \varnothing$, the minimum over $V$ is at most every weight in $B$, hence again $f(A)\le f(B)$. No sign condition on the weights is needed.

For every $S\subseteq V$ and $x\notin S$, including $S=\varnothing$, the marginal increment isFor the empty set, this uses $w(x)\ge f(\varnothing )={\min }_{v\in V}w(v)$. The monotone set function property from (i) gives $f(A)\le f(B)$ when $A\subseteq B$. ConsequentlyThus diminishing returns holds, so $f$ is a submodular set function by (b).