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I am trying to solve an optimization problem with an objective function $$f(x)+h(y)$$ with constraints $$g(x)+y=0$$ Using the Lagrangian method the problem I cast the original constrained problem into an unconstrained problem as, $$\mathcal{L} (x,y,\lambda) = f(x)+h(y)-\lambda (g(x)+y)$$ But I would like to cast an optimization problem including a delayed update of Lagrangian multiplier in the objective function. Something like this, $$\mathcal{L} (x,y,\lambda) = f(x)+h(y)+j(\lambda_{i-1})-\lambda_{i} (g(x)+y)$$
where i-1 and i refers to variables from earlier and present steps of optimization. Can the Lagrangian variable be included in the objective function? I am confused since the Lagrange multiplier is the dual for the primal problem. Any suggestions appreciated.

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  • $\begingroup$ Why would you even want to do that? What is the benefit? What guarantees convergence to a KKT point? Usually using the $t-1$ data is used to accelerate algorithms (momentum) and when doing so we require a very detailed analysis of the algorithm + additional assumptions on the problem structure. It's not just "something to do". $\endgroup$ – iarbel84 Aug 1 '20 at 11:23
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Perhaps a different approach may be more useful, although it is not what you are asking for.

Your constraint implies $y=-g(x)$ so now you are seeking to optimize $f(x) + h(-g(x))$, a regular one-variable optimization, which is usually qualitatively easier to deal with than the 2-variable problem.

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  • $\begingroup$ Thanks. But my question is more concerned about if I can use the Lagrange multiplier in the objective function to optimise? Because in the case where we have a primal-dual problem, when we try to minimise the primal implies maximisation of the dual. $\endgroup$ – math123 Jun 12 '20 at 22:11

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