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First, I will clarify some of the notation: $$ x_i \in S_i,\; i\in \{1,2,\dots, n\} \quad x\in S, \quad S\equiv \prod_{i=1}^nS_i \text{ (direct product set)} $$ So basically, we have $x \in S$ which is, if $x_i \in \mathbb{R}$, $x= (x_1,x_2,\dots , x_n)$.

Next, we have the optimization problem: $$\max_{x\in \prod_{i=1}^n S_i} \sum_{i=1}^n\left [H_i(x_i) - \sum_{k=1}^m \lambda^k C_i^k(x_i)\right ]$$ where $C_i^k$ is some real valued function.

Since $S\equiv \prod_{i=1}^n S_i$, the choices $x_i$ may be made independently. Why is this?

I think it is because the $x_i$ don't depend on each other (intuitively, they obviously don't. I don't know how to say this rigorously), and I think that maximizing over a sum of a function of different independent variables -- where the independent variables are indexed by what we are summing over -- is the same as summing the max of the function of each independent variable (I am trying to figure out why this is exactly, but I intuitively it seems true).

Thank you.

Edit: If necessary I can try to be more specific, but I think there is enough here/I have included everything relevant?

Also, example is taken from "Generalized Lagrange Multiplier Method" by Hugh Everett III

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  • $\begingroup$ like $S_1 \times ... \times S_n$? $\endgroup$ – BCLC Jan 31 '16 at 5:24
  • $\begingroup$ It is cartesian product if $x_i \in \mathbb{R}$, but if $x_i \not \in \mathbb{R}$ I don't think it will necessarily be cartesian product. My understanding isn't much on the subject. $\endgroup$ – majmun Jan 31 '16 at 5:26
  • $\begingroup$ What do you mean by "independent"? $\endgroup$ – user228113 Jan 31 '16 at 6:15
  • $\begingroup$ @G.Sassatelli I meant to say independent variables. But I guess that doesn't make much sense either, so I rolled the question back to a previous form. Sorry $\endgroup$ – majmun Jan 31 '16 at 6:34
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I'm not sure, but if I were to guess I would say that it is because $x\in S \equiv \prod_{i=1}^n S_i$ implies that $x= (f(1),f(2),f(3),\dots ,f(n))$ which in turn implies $x_1 = f(1), x_2 = f(2) \dots x_n = f(n)$ so we have the sum of functions (or the composition of functions) of different independent variables.

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