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.

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

Proof:

  1. 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}.

  2. Second claim follows, if we apply first result to {-f}.

\Box

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