The standard knapsack problem imagines a thief trying to stick the most items in his knapsack as possible. It assumes that having, say, two Picasso paintings is twice as good as having one.
We might extend this problem by noting that people have diminishing returns, i.e. two Xs are not twice as good as one X. For example, maybe we think that the utility of having $n$ items is $\log n$ times as good as having one.
Is this a known problem? I didn't see it in Wikipedia's list of knapsack problems. It seems like it should be NP-Hard, but I'm having trouble proving a reduction.
EDIT: As some examples, we could consider $u(x)=0$ in which case there's a constant-time solution to maximize. We could also consider $u(S)=|\{x\in S: \textbf{x is the encoding of a TM that halts}\}|$ in which case maximizing the utility isn't even computable.
So changing the the function to maximize has a dramatic effect on the solubility of knapsack. Can we make a claim like "If $f$ is a non-constant, computable function, then maximizing $f$ in a knapsack is NP-Hard?"