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Thread: 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], $\displaystyle C[0,1]$. Define $\displaystyle p(f,g) = \sup _{x \in [0,1] } \mid f(x)-g(x) \mid $.

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

    Proof so far.

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

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

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

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

    So now, I have $\displaystyle 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], $\displaystyle C[0,1]$. Define $\displaystyle p(f,g) = \sup _{x \in [0,1] } \mid f(x)-g(x) \mid $.

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

    Proof so far.

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

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

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





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

    So now, I have $\displaystyle 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 :

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

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

    BUT:

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

    Thanks!
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