Results 1 to 4 of 4

Math Help - extended real valued function bounded below and attains its minimal value

  1. #1
    Newbie
    Joined
    Sep 2010
    Posts
    10

    extended real valued function bounded below and attains its minimal value

    Let f(x) be a real-valued function (an extended real valued function) on a metric space (M; d). Then
    \lim\sup_{x\rightarrow y}f(x)=\lim_{\epsilon\rightarrow0}\sup_{0<d(x,y)<\  epsilon}f(x)
    and
    \lim\inf_{x\rightarrow y}f(x)=\lim_{\epsilon\rightarrow0}\inf_{0<d(x,y)<\  epsilon}f(x)

    A function is called lower (upper) semicontinuous at a point y if f(y)\neq\pm\infty
    and
    f(y)=\leq\liminf_{x\rightarrow y}f(x) and
    f(y)=\geq\limsup_{x\rightarrow y}f(x)


    Let a metric space M be complete, and let K be a compact subset in M. Prove that a lower semicontinuous function on K is bounded from below and attains its minimal value and an upper semicontinuous function on K is bounded from above and attains its maximal value.

    Any idea of how to approach this??
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Super Member
    Joined
    Apr 2009
    From
    México
    Posts
    721
    Dealing with the first one (lower semicontinuity), assume f is not bounded below, then there exists a sequence (x_n) \subset K such that f(x_n) \rightarrow -\infty, but K is compact, so it's sequentially compact, so without loss of generality we get x_n\rightarrow x. What can you say about f(x).
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Newbie
    Joined
    Sep 2010
    Posts
    10
    Hum... nice explanation,

    so f(x)\subset K which is a contradiction, right?

    so that would do it
    Follow Math Help Forum on Facebook and Google+

  4. #4
    Super Member
    Joined
    Apr 2009
    From
    México
    Posts
    721
    Quote Originally Posted by morito14 View Post
    Hum... nice explanation,

    so f(x)\subset K which is a contradiction, right?

    so that would do it
    Not really, since f:K\rightarrow \mathbb{R} so f(x) \in \mathbb{R}, the point is that f(x) \leq \liminf_{x_n} f(x_n) = -\infty (and obviously x\in K since K is compact (maybe this is what you meant?)).
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. an extended real-valued function
    Posted in the Differential Geometry Forum
    Replies: 2
    Last Post: March 21st 2010, 09:28 PM
  2. Real Valued Function
    Posted in the Discrete Math Forum
    Replies: 1
    Last Post: November 22nd 2008, 03:01 AM
  3. Real-valued function
    Posted in the Calculus Forum
    Replies: 5
    Last Post: November 14th 2008, 08:10 AM
  4. Real valued function on [0,1]
    Posted in the Discrete Math Forum
    Replies: 1
    Last Post: October 15th 2008, 09:34 AM
  5. real valued function
    Posted in the Calculus Forum
    Replies: 0
    Last Post: February 26th 2008, 04:05 AM

Search Tags


/mathhelpforum @mathhelpforum