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Math Help - Sup norm is metric proof.

  1. #1
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    Sup norm is metric proof.

    Consider the set of all continuous functions on [0,1], C[0,1]. Define p(f,g) = \sup _{x \in [0,1] } \mid f(x)-g(x) \mid .

    Prove that  p(f,g)+p(g,h) \geq p(f,h)

    Proof so far.

    Let a= \sup _{x \in [0,1] } \mid f(x)-g(x) \mid and b= \sup _{x \in [0,1] } \mid g(x)-h(x) \mid

    As well as c= \sup _{x \in [0,1] } \mid f(x)-h(x) \mid

    Then  \forall \epsilon > 0 , there exists x-1 , x_2 , x_3 \in [0,1] such that:

    a- \epsilon < \mid f(x_1)-g(x_1) \mid
    b- \epsilon < \mid g(x_2)-h(x_2) \mid
    c- \epsilon < \mid f(x_3)-h(x_3) \mid

    So now, I have a+b = a - \epsilon + b - \epsilon + 2 \epsilon ... I'm stuck here...

    How should I go on from here? Thanks!
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  2. #2
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    Quote Originally Posted by tttcomrader View Post
    Consider the set of all continuous functions on [0,1], C[0,1]. Define p(f,g) = \sup _{x \in [0,1] } \mid f(x)-g(x) \mid .

    Prove that  p(f,g)+p(g,h) \geq p(f,h)

    Proof so far.

    Let a= \sup _{x \in [0,1] } \mid f(x)-g(x) \mid and b= \sup _{x \in [0,1] } \mid g(x)-h(x) \mid

    As well as c= \sup _{x \in [0,1] } \mid f(x)-h(x) \mid

    Then  \forall \epsilon > 0 , there exists x-1 , x_2 , x_3 \in [0,1] such that:





    a- \epsilon < \mid f(x_1)-g(x_1) \mid
    b- \epsilon < \mid g(x_2)-h(x_2) \mid
    c- \epsilon < \mid f(x_3)-h(x_3) \mid

    So now, I have a+b = a - \epsilon + b - \epsilon + 2 \epsilon ... I'm stuck here...

    How should I go on from here? Thanks!
    if a+b = a -ε +b -ε +2ε then 0=0



    For all xε[0,1] we have :

    |f(x)-g(x)|\leq sup{|f(x)-g(x)|:xε[0,1]} and

    |g(x)-h(x)|\leq sup{|g(x)-h(x)|: xε[0,1]}

    BUT:

    |f(x)-h(x)|\leq |f(x)-g(x)| + |g(x)-h(x)|\leq sup{|f(x)-g(x)|:xε[0,1]} + sup{|g(x)-h(x)|:xε[0,1]} = p(f,g) + p(g,h).

    Hence sup{|f(x)-h(x)|:xε[0,1]} \leq p(f,g) + p(g,h) ,
    thus p(f,h) \leq p(f,g) + p(g,h)
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  3. #3
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    But does sup \mid f(x)-h(x) \mid necessarily takes on one of the value of  \mid f(x_0) - h(x_0) \mid for some x_0 \in [0,1] ? If the sup is outside of this set, then the inequality doesn't hold.

    Thanks!
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