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Showing sup{f(x)} - inf{f(x)} >= sup{|f(x)|} - inf{|f(x)|}

  1. Feb 6, 2012 #1
    Hello, below I have typed the problem and drew a diagram to help me think about the problem.

    MMMMM.png

    It seems intuitive to me that [itex]M \leq M'[/itex] since M' might be the absolute value of the most extreme negative f value that is greater in magnitude than the most extreme positive f value. Also, [itex]m \leq m' [/itex] since [itex]m' \geq 0[/itex] and m could be negative.

    Is this type of problem best handled with cases such as Case 1: [itex]f(x) \leq 0[/itex]?
     
  2. jcsd
  3. Feb 7, 2012 #2

    Bacle2

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    Science Advisor

    I think working out the individual cases may do it. It seems to come down to showing
    that:

    A-B ≥ |A|-|B|

    So that you get equality. Only real case is when A,B have different signs. Maybe
    you can show it geometrically, using |X| asthe distance from X to a fixed value
    (thinking of 0 in the real line).
     
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