The answers to Q3, Q4 and Q6 are all no. I will give a sketchy argument here.
Consider the one dimensional case, where the lotteries are represented by real numbers in the interval L=[0,1], and consider the function u:L→[0,1] given by u(x)=12−(x−13)3(x−23). Let ⪰ be the preference order given by x⪰y if and only if u(x)≥u(y).
u is continuous and quasi-concave, which means ⪰ is going to satisfy A1, A2, A3, A4, and B2. Further, since u is monotonically increasing up to the unique argmax, and then monotonically decreasing, ⪰ is going to satisfy A5.
u is not concave, but we need to show there is not another concave function giving the same preference relation as u. The only way to keep the same preference relation is to compose u with a strictly monotonic function f, so v(x)=f(u(x)).
If f is smooth, we have a problem, since v′(13)=f′(u(13))u′(13)=f′(12)0=0. However, since, v′ must be on some x>13, but concavity would require v′ to be decreasing.
In order to remove the inflection point at x=13, we need to flatten it out with some f that has infinite slope at 12. For example, we could take f(z)=3√z−12. However, any f that removes the inflection point at x=13, will end up adding an inflection point at x=23, which will have a infinite negate slope. This newly created inflection point will cause a problem for similar reasons.
I am skeptical that it will be possible to salvage any nice VNM-like theorem here that makes it all the way to concavity. It seems like the jump necessary to fix this counterexample will be hard to express in terms of only a preference relation.
The answers to Q3, Q4 and Q6 are all no. I will give a sketchy argument here.
Consider the one dimensional case, where the lotteries are represented by real numbers in the interval L=[0,1], and consider the function u:L→[0,1] given by u(x)=12−(x−13)3(x−23). Let ⪰ be the preference order given by x⪰y if and only if u(x)≥u(y).
u is continuous and quasi-concave, which means ⪰ is going to satisfy A1, A2, A3, A4, and B2. Further, since u is monotonically increasing up to the unique argmax, and then monotonically decreasing, ⪰ is going to satisfy A5.
u is not concave, but we need to show there is not another concave function giving the same preference relation as u. The only way to keep the same preference relation is to compose u with a strictly monotonic function f, so v(x)=f(u(x)).
If f is smooth, we have a problem, since v′(13)=f′(u(13))u′(13)=f′(12)0=0. However, since, v′ must be on some x>13, but concavity would require v′ to be decreasing.
In order to remove the inflection point at x=13, we need to flatten it out with some f that has infinite slope at 12. For example, we could take f(z)=3√z−12. However, any f that removes the inflection point at x=13, will end up adding an inflection point at x=23, which will have a infinite negate slope. This newly created inflection point will cause a problem for similar reasons.
I am skeptical that it will be possible to salvage any nice VNM-like theorem here that makes it all the way to concavity. It seems like the jump necessary to fix this counterexample will be hard to express in terms of only a preference relation.