# Reference request : How to use Lagrange multiplier technique with infinite (infact uncountably) number of constraints?

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.

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.
• 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? – Rajesh Dachiraju May 11 at 15:05
• Yes, $f$ in a vector space works. – Dirk May 11 at 15:46
• 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). – Rajesh Dachiraju May 14 at 10:46
• 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. – Dirk May 14 at 12:49