Example [lcc-001I]

Consider the problem "Optimize f(Ax + b)", where A: X \to Y is a linear map between two finite-dimensional real vector spaces. This is clearly equivalent (in an informal sense) to "Optimize f(y) subject to y = Ax + b".

The first of these problems corresponds to L_1 = (X,*,L_1(x,*) = f(Ax + b))

The second corresponds to L_2 = (X \times Y, Y^*, L_2(x,y,\alpha ) = f(y) + \alpha (Ax + b---y)) (Where by Y^* I just mean the vector space dual of Y)

There is a morphism L_1 \to L_2, defined by \phi ^+(x) = (x, Ax+b). The fact that this is a morphism gives the inequality (L_1)_*(x) \geq (L_2)_*(\phi ^+(x)), which is one part of the "equivalence".

In the other direction, we have the map (x,y) \mapsto x, the first projection. Now, it is true that (L_2)_*(x,y) \geq (L_1)_*(x)---the former is \infty unless y = Ax + b, in which case they're both equal to the infimal value of f(Ax + b). But there is no map * \to Y^* picking out an \alpha which witnesses this---even if we pass to some sort of extended reals, the signs we want to choose depend on the sign of Ax + b---y, which could go either way. This is one sense in which the problems are "equivalent but not isomorphic".