A simple theoretical optimization problem

Question

Let $C$ be a finite-dimensional compact smooth manifold (in my application it's just a projective space $\mathbb{C} P^n$ but it doesn't matter), and consider the map $\lambda \in C \mapsto g(\lambda) \in L^2(\mathbb{R}^d,\mathbb{R})$, where $\int_{\mathbb{R}^d} g(\lambda) = 0$. Consider the problem \begin{align*} \sup_{f \in L^2(\mathbb{R}^d,\mathbb{R})} \min_{\lambda \in C} \int g(\lambda) f. \end{align*} Can we say that the sup is attained ? What can we say about it ?

Answer 1

The supremum is always $\infty$ or $0$. For any function $f\in L_2(\Bbb R^d, \Bbb R)$ and any real number $a \ge 0, af\in L_2(\Bbb R^d, \Bbb R)$, and if $\int g_\lambda f > 0$, $$\min_{\lambda \in C} \int g_\lambda (af) = a \min_{\lambda \in C} \int g_\lambda f$$

If the map $\lambda \mapsto g_\lambda$ is such that the minimum for all $f$ is $\le 0$, then the supremum will occur for $f \equiv 0$. For example, this will happen if for every $\lambda$, there is a $\lambda'$ with $g_{\lambda'} = -g_\lambda$.

On the other hand, if there is an $f$ for which $\min_{\lambda \in C} \int g_\lambda f > 0$, then by increasing $a$, the value of $\min_{\lambda \in C} \int g_\lambda (af)$ can be made as large as desired. Since there is no $f$ with $\int g_\lambda f = \infty$, this supremum is never obtained.

If you restrict $f$ to range over a compact set, and also require $\lambda \mapsto g_\lambda$ to be continuous, then the supremum will be obtained, as this is always true of continuous maps over compact sets. But if I recall correctly (it has been a long time, so I may not), requiring $\|f\|_2 = 1$ is not sufficient to get a compact set.