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Closedness and Lower Semicontinuity

  1. Jan 11, 2006 #1
    Given a real-valued function [tex]f:X \mapsto \mathbb{R}[/tex] where [tex]X[/tex] is a subset of [tex]\mathbb{R}^n[/tex]. Consider a closed extended real-valued function [tex]\tilde{f}:\mathbb{R}^n \mapsto (-\infty, \infty][/tex] defined as

    \tilde{f}(x) = \left \{
    f(x) & \quad , \textup{ if } x \in X \\
    \infty & \quad , \textup{ otherwise}

    I want to show that if [tex]X[/tex] is not closed, then for a sequence [tex]\{x_k\} \subset X[/tex] converging to a point [tex]\bar{x}[/tex] on the boundary of [tex]X[/tex] that does not belong to [tex]X[/tex] necessarily results in [tex]\liminf_{k \to \infty} f(x_k) = \infty[/tex]. I use the following definition for closed function (definition used by my text book): A function is said to be closed if its epigraph is closed.

    The following is my proof, but I am a bit unsure of its correctness. Hope someone can give me some feedback.

    Suppose otherwise that [tex]\liminf_{k \to \infty} f(x_k) = M[/tex] for some [tex]M < \infty[/tex]. Since [tex]\tilde{f}[/tex] is closed, it is lower semicontinuous. Hence,

    \tilde{f}(\bar{x}) \leq \liminf_{k \to \infty} \tilde{f}(x_k) = \liminf_{k \to \infty} f(x_k) = M < \infty.

    But [tex]\tilde{f}(\bar{x}) = \infty[/tex], thus a contradiction.
    Last edited: Jan 11, 2006
  2. jcsd
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