By assuming the linear approximation is good enough, you beg the question. Consider that if that is to be assumed, you could just take U, A and B to be linear functions to begin with...
It’s not so much an assumption, as an intermediate conclusion that follows from the dependence of U on A and B being smooth enough (across the kind of change that dM is capable of making) and nothing else. My only claim is that under this assumption, the non-diversification conclusion follows. U(A,B) being linear is a hugely more restrictive assumption.
that follows from the dependence of U on A and B being smooth enough (across the kind of change that dM is capable of making) and nothing else.
This is incorrect. Being smooth is not a license to freely replace function values with linear approximation values; that’s not what smoothness means. You have to analyze the error term, most easily presented as the higher-order remainder in the Taylor approximation. Such an analysis is what I’d tried to supply in the post I linked to.
It’s not so much an assumption, as an intermediate conclusion that follows from the dependence of U on A and B being smooth enough (across the kind of change that dM is capable of making) and nothing else. My only claim is that under this assumption, the non-diversification conclusion follows. U(A,B) being linear is a hugely more restrictive assumption.
This is incorrect. Being smooth is not a license to freely replace function values with linear approximation values; that’s not what smoothness means. You have to analyze the error term, most easily presented as the higher-order remainder in the Taylor approximation. Such an analysis is what I’d tried to supply in the post I linked to.