A sufficient condition for a function to attain maximum/minimum

Let ${(\Omega, \mathcal{F}, \mathbb{P})}$ be a probability space. Consider a measurable function ${f: D\rightarrow \mathbb{R}}$ . We will find a sufficient condition under which ${f}$ attains maximum (minimum) in ${D}$ .

Proposition 1 Let ${D}$ is convex, bounded, convex in ${L^1}$ , and closed under ${\mathbb{P}}$ -a.e. convergence.

If ${f: D\rightarrow \mathbb{R}}$ is concave, and upper semicontinuous in ${L^1(\mathbb{P})}$ , then ${f}$ attains maximum in ${D}$ .
If ${f: D\rightarrow \mathbb{R}}$ is convex, and lower semicontinuous in ${L^1(\mathbb{P})}$ , then ${f}$ attains minimum in ${D}$ .

Proof:

Let ${V = \sup_D f(X)}$ , and ${\{X_n\}\subset D}$ be a maximizing sequence, i.e. ${f(X_n) \uparrow V}$ . By [Kom67], since ${D}$ is bounded in ${L^1(\mathbb{P})}$ , there exists ${\hat X\in L^1(\mathbb{P})}$ and relabelled subsequence ${\{X'_n\}}$ such that
$\displaystyle Y_n := \frac 1 n \sum_{j=1}^n X'_j \rightarrow \hat X, \ \mathbb{P}-a.e.$

By convexity, ${Y_n \in D}$ . Moreover, by closedness of ${D}$ under ${\mathbb{P}}$ -a.e. convergence, we have ${\hat X \in D}$ . Thus, ${f(\hat X) \le V}$ . On the other hand, we can see ${f(\hat X) = V}$ by following inequalities:

$\displaystyle \begin{array}{ll} f(\hat X) & = f(\lim_n Y_n) \\ & = \lim\sup_n f(Y_n), \ \hbox{ by upper semicontinuity}\\ & = \lim\sup_n f(\frac 1 n \sum_{j=1}^n X'_j), \\ & \ge \lim\sup_n \frac 1 n \sum_{j=1}^n f(X'_j), \ \hbox{ by concavity of } f \\ & = V. \end{array}$

This implies ${V = f(\hat X) <\infty}$ for some ${\hat X\in D}$ .
Second claim follows, if we apply first result to ${-f}$ .