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Proving two metrics are not equivalent

  1. Mar 25, 2012 #1
    1. The problem statement, all variables and given/known data
    The vector space of continuous functions on [0,1] is given the following metrics

    [itex] d_{∞} (f,g) = sup_{x\in [0,1]} |f(x)-g(x)| [/itex]

    [itex] d_{2} (f,g) = (∫^{1}_{0} |f(x)-g(x)|^{2} dx)^{1/2}[/itex]

    Are these two metrics equivalent?

    2. Relevant equations
    If [itex] d_{∞} [/itex] and [itex] d_{2} [/itex] are equivalent, then for every [itex]f \in C[0,1][/itex] and every [itex]\epsilon>0[/itex], there exist a [itex]\delta>0[/itex] such that [itex]B^{d_{∞}}_{\delta}(f)\subset B^{d_{2}}_{\epsilon}(f)[/itex] and [itex]B^{d_{2}}_{\delta}(f)\subset B^{d_{∞}}_{\epsilon}(f)[/itex]

    3. The attempt at a solution

    I know that these two metrics are not equivalent, but i can't find a function at which i can find an [itex]\epsilon>0[/itex] for every [itex]\delta>0[/itex] where they are not the subset of each other.

    I know that [itex] d_{∞} [/itex] is like the highest point of the g(x) if I let f(x)=0 and g(x) be some other function. [itex] d_{2} [/itex] is like the area under the graph between [0,1].
     
  2. jcsd
  3. Mar 25, 2012 #2

    tiny-tim

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    Hi Lily@pie! :smile:
    Yes, you're practically there …

    you need a sequence gn(x) with constant maximum but decreasing area :wink:
     
  4. Mar 25, 2012 #3
    I'm clueless :cry:

    any hints???? :bugeye:
     
  5. Mar 25, 2012 #4

    tiny-tim

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    how about a curve which starts at (0,1) and curves gracefully down to (1,0) ?

    then reduce the area :wink:
     
  6. Mar 25, 2012 #5
    Would this be a function g(x)=(x-1)2 where 0≤x≤1??

    And f(x) = 0 for all x

    For all δ>0 such that [itex]d_{2} (f,g)[/itex] will be undefined!! oh my god... what did i do...
     
  7. Mar 26, 2012 #6

    tiny-tim

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    i was thinking of somthing that gets flatter a lot quicker,

    like e-x (but adjusted to fit into [0,1]) :smile:
     
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