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I have a constrained maximization problem (maximizing a functional), with number of constraints being uncountable infinite.

It looks something like this. I want to maximize the convex functional $C(f)$ over $f \in S$ with the constraint that look like $G(f,\phi) = 0 \forall \phi \in C^{\infty}(\Omega)$. Clearly these constraints are infinite in number (infact uncountably infinite).

I don't know how to use Lagrange multiplier technique in this context. Hence I request for a reference for a theory in this regard. I am not aware of any such concepts and no idea on what terminolgy I should search on google.

Edit : The example cited may not represent a typical case, neverthless I want a reference to the concerned generic theory.

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You are looking for optimization in vector spaces (preferably normed spaces). There is a rich theory available. The choice of spaces and norms is not always clear and often the key to the solution. In general, if you have a constraint $g(x)=0$ and $g$ maps into a normed space $X$, then the Lagrange multiplier $\lambda$ for this constraint is an element in the dual space $X^*$. You would add $\langle \lambda,g(x)\rangle$ to the objective to form the Lagrangian. A classic reference for the practitioner is "Optimization by Vector Space Methods" by Luenberger.

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  • $\begingroup$ To be more specific, the functional is $C(f,\alpha)$ where $f$ is a function and $\alpha$ a positive real scalar. It needs to be maximized with respect to both $f$ and $\alpha$. The constraint are of the form $G(f,\alpha,\phi) = 0, \forall \phi \in C^{\infty}(\Omega)$. I hope reference is still applicable as $f$ lies in a vector space. Please confirm. Also the case of Lagrange multipliers in infinite dimensions does not apply here? $\endgroup$ – Rajesh Dachiraju May 11 at 15:05
  • $\begingroup$ Yes, $f$ in a vector space works. $\endgroup$ – Dirk May 11 at 15:46
  • $\begingroup$ Thanks for the answer @Dirk $\endgroup$ – Rajesh Dachiraju May 11 at 16:06
  • $\begingroup$ In my question the constraints are $G(f,\phi) = 0 \forall \phi$, while in your answer, you are suggesting to add $\langle \lambda,g(x) \rangle$ to form the Lagrangian. Do you mean $\phi$ itself plays the role of $\lambda$, that is the Lagrange multiplier? (..I want to get clarified before I buy this book). $\endgroup$ – Rajesh Dachiraju May 14 at 10:46
  • $\begingroup$ The constraint $G(f,\phi)=0$ for all $\phi\in C^\infty$ is a problem in this form. I suspect that the constraint can be modeled in a different way, e.g. as a constraint on $f$ (without using $\phi$), but this is just a guess since I do not know any details. $\endgroup$ – Dirk May 14 at 12:49

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