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Math Help - A metric space of a collection of bounded functions

  1. #1
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    A metric space of a collection of bounded functions

    Let B[a,b] denotes the collection of all bounded functions from a closed interval [a,b] into R. Let \rho be a metric for B[a,b] such that \rho(f,g) = lub \{ |f(x) - g(x) | : x \in [a,b]\}.

    Let C[a,b] denotes the collection of all continuous real-valued functions on [a,b] such that C[a,b] is a closed subspace of B[a,b].
    Let \rho' be a metric for C[a,b] such that \rho'(f,g) = lub \{|f(x) -g(x) |: x \in [a,b] \}.

    (a) Show that C[a,b] is not a compact in B[a,b].
    (b) Show that C[a,b] is no where dense in B[a,b].

    (c) A sequence \{f_{n}\}_{n=1}^{\infty} in B[a,b] converges to a member f \in B[a,b] with respect to the metric \rho if and only if \{f_{n}\}_{n=1}^{\infty} converges to f uniformly.
    Last edited by aliceinwonderland; February 1st 2009 at 01:37 AM.
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  2. #2
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    Quote Originally Posted by aliceinwonderland View Post
    Let B[a,b] denotes the collection of all bounded functions from a closed interval [a,b] into R. Let \rho be a metric for B[a,b] such that \rho(f,g) = lub \{ |f(x) - g(x) | : x \in [a,b]\}.

    Let C[a,b] denotes the collection of all continuous real-valued functions on [a,b] such that C[a,b] is a closed subspace of B[a,b].
    Let \rho' be a metric for C[a,b] such that \rho'(f,g) = lub \{|f(x) -g(x) |: x \in [a,b] \}.

    (a) Show that C[a,b] is not a compact in B[a,b].
    (b) Show that C[a,b] is no where dense in B[a,b].

    (c) A sequence \{f_{n}\}_{n=1}^{\infty} in B[a,b] converges to a member f \in B[a,b] with respect to the metric \rho if and only if \{f_{n}\}_{n=1}^{\infty} converges to f uniformly.
    Question (c) shows an interesting case as the following proof indicates.

    For the forward implication we want to prove f_{n} converges uniformly to f.

    Hence let ε>0.Since  f_{n} converges to f w.r.t ρ in B[a,b],there exists a kεN (k belonging to natural Nos) such that:

    .......if  n\geq k ,then lub{| f_{n}(x)-f(x)|:xε[a,b]}<ε..........for all n................................................. ........1

    Let now  n\geq k and xε[a,b]...............

    Then |f_{n}(x)-f(x)|\leq lub \{|f_{n}(x)-f(x)|:x\in [a,b]\}=\rho(f_{n},f)<\epsilon.

    Since |f_{n}(x)-f(x)| is a member of the set { |f_{n}(x)-f(x)|:x \in [a,b]}.

    Thus f_{n} converges uniformly to f.

    Now for the converse they are two ways to do the proof:

    1) To prove that uniform convergence implies,that f_{n}(x) converges to f w.r.t ρ metric (or sup metric) ,and then prove that fεB[a,b],e.i f is bounded.


    2)To prove that uniform convergence implies that f is bounded and then prove that f_{n} converges uniformly to f w.r.t sup metric.


    I will follow the 2nd way and it is open for discussion in the forum to decide which procedure is the right one.

    Hence ,uniform convergence implies that for all ε>0 there exists a k integer such that:

    ............if  n\geq k and xε[a,b],then | f_{n}(x)- f(x)|<ε........for all n and x................................................. ...2


    Let now xε[a,b],put ε=1,then there exists an integer k such that :

    ....if  n\geq k and xε[a,b],then | f_{n}(x)-f(x)|<1 for all n ,x................................................ ........................................3

    Put now n=k+1,

    Then |f(x)|= |f(x)-f_{k+1}(x)+f_{k+1}(x)|\leq |f(x)-f_{k+1}(x)| + |f_{k+1}(x)|\leq 1+|f_{k+1}(x)|.

    Since f_{k+1} is bounded ,there exists M>0 such that:

    .............if xε[a,b] ,then |f_{k+1}(x)|\leq M..................

    So there exists a c for all xε[a,b] such that :

    ....... |f(x)|\leq c= M + 1|,implying that f is bounded.

    And now to prove that f_{n} converges to f w.r.t supmetric ρ.

    So let ε>0 .Again due to uniform convergence there exists an integer k such that:

    .........If  n\geq k and xε[a,b],then |f_{n}(x)-f(x)|<ε/2.....for all n , x................................................. .............

    Also let n\geq k.

    Since ε/2 is an upper bound for the set { |f_{n}(x)-f(x)|:x\in[a,b]},then lub{ |f_{n}(x)-f(x)|:x\in[a,b]}<ε/2<ε,

    and thus f_{n} converges w.r.t sup metric ρ.
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