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I don't understand it. The term g(x)(x-y) cannot be positive for all y close to x, since suppose $y_1>x$, then it requires g(x)<0, but then for $y_2<x$ close to x, it gives $g(x)(x-y_2)<0$.
Well, the optimality condition gives you HJB, where the Hamiltonian term (see notes I linked to) is maximized over all possible controls, i.e., to do the sup, you have to allow control to have arbitary dependence on $v,x,\sigma$.. So you cannot prescribe the optimal control to be constant: it comes out of the maximization (sup ) process. Now depending upon the system, it may depend just on $\nabla v$, .e.g., in linear quadratic case (example given in the paper you mention), and yes, then you can omit other terms
Agree it is horrible notation. f is still the same f, but now implicitly depends on $\sigma$, $v$,and its gradient because the $a$ is taken to be $a(x,v,\nabla v,\Delta v,\sigma)$. See these notes e.g.: columbia.edu/~dl3133/MFGSpring2018.pdf pages 63-70. In other words, $f=f(x,a(x,v,\nabla v,\Delta v,\sigma))$
Optimal control equations result in BVP to be solved on 1d (time) domain. So essentially, you are solving for intersection of solution of two Cauchy (IVP) problems with half of boundary values known on one end, and half on other end. By guesssing the rest, one integration is started from t=0, and one is started back from t=T (final time). That makes shooting a natural candidate,