# Lagrange Multipliers for linear functionals

Say I have a Banach-space $X$ and linear (!) functionals $f,g$, and I'm trying to solve the constrained optimization problem

$$\max~f(x)\quad \mbox{s.t.}~g(x)= 0,~\Vert x\Vert\le 1.$$

Suppose I can show that a unique solution $\tilde{x}$ exists. Does this imply the existence of a Lagrange Multiplier $\lambda$ for which $\tilde{x}$ is a solution to

$$\max~\left(f(x)+\lambda g(x)\right) \quad \mbox{s.t.}~\Vert x\Vert\le 1$$

(without the constraint $g(x)=0$)? If yes, does the same hold for multiple constraints $g_1,\ldots,g_n$?

It appears that this is somewhat different from the "classical" use of Lagrange Multipliers, due to the additional constraint $\Vert x\Vert\le 1$, which restricts $x$ to the closed unit ball. Also note that the functionals are assumed to be linear, so this might allow for some simplifications. Any help is appreciated.

• Have you solved your problem in the finite dimensional case? That might give you some ideas of what might happen. The condition $g(x)=0$ restricts you to a subspace (a closed one if $g$ is continuous) so you are just trying to minimize $f(x)$ in the closed unit ball of the Banach space $\ker g\subset X$. The same happens with any (finite) number of $g$s. A maximizer $x$ will necessarily satisfy $\|x\|=1$ so you get Lagrange coefficients for this condition. – Joonas Ilmavirta Jan 14 '15 at 20:35
• For the finite dimensional case I'd decompose X according to the respective kernels of $g_1,...,g_n$ and use orthogonality relations afterwards. However, in the context of a Banach Space, there is no "orthogonality" to use. – Yoan Jan 14 '15 at 22:50