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Real Analysis Problem

  1. Sep 9, 2011 #1
    I Just started Analysis 1 this week and I've encountered some tricky problems in the Assignment

    1. The problem statement, all variables and given/known data
    Let f,g : [0,1] -> R be bounded functions.
    Prove that inf{ f(x) + g(1-x) : x (element of) [0,1]} >= inf{f(x) : x (element of) [0,1]} + inf{g(x) : x (element of) [0,1]}


    2. Relevant equations
    Perhaps the triangle inequality?


    3. The attempt at a solution
    I know that g(1-x) is still the same function over the same domain as g(x) however it runs in the opposite direction of the original function, thus making f(x) + g(1-x) a different function that f(x) + g(x). I do not know however, how that function is different.
     
  2. jcsd
  3. Sep 9, 2011 #2

    micromass

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    The g(1-x) is just a read herring, since

    [itex]inf_{x\in [0,1]}{g(1-x)}=inf_{x\in [0,1]}{g(x)}[/itex]

    Convince yourself of that.

    Anyway. To prove that inequality of yours. Take an [itex]x\in [0,1][/itex] arbitrary. Can you prove that

    [tex]f(x)+g(1-x)\geq inf_{x\in [0,1]}{f(x)}+inf_{x\in [0,1]}{g(x)}[/tex]
    ??
     
  4. Sep 9, 2011 #3
    I'm having a bit of trouble with it.

    I know that I can use the triangle inequality to say that |f(x) + g(1-x)| <= |f(x)| + |g(1-x)|
    I am unsure as to how to proceed from there, I have been thinking perhaps using the Completeness Property to say that |f(x)| + |g(1-x)| >= -[f(x) + g(1-x)] but that doesnt seem right.


    P.s. thank you for the quick reply
     
  5. Sep 9, 2011 #4

    micromass

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    Why use the triangle inequality anyway?? I see no absolute value signs anywhere...
     
  6. Sep 9, 2011 #5
    Perhaps I am misunderstanding the way to solve the problem here.
    In my mind I would
    a) show that Inf{ f(x) + g(1-x) } >= inf{f(x)} + inf {g(1-x)}
    b) show that g(1-x) = g(x) for the domain [0,1] and hence that Inf{ f(x) + g(1-x) } >= inf{f(x)} + inf {g(x)}

    I dont know how to show (a). nothing in my textbook is leading me to the conclusion that the infimum of the sum of two functions is greater than or equal to the sum of the infima of the two functions
     
  7. Sep 9, 2011 #6

    micromass

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    Well, what is the definition of an inf??? It's the greatest lower bound right??
    Why not show that [itex]inf(g(x))+inf(f(x))[/itex] is a lower bound. Then you have immediately your inequality since [itex]inf(f(x)+g(1-x))[/itex] is the greatest lower bound.

    (b) is false, by the way. It's not true in general that g(1-x)=g(x). It is true that

    [itex]inf(g(x))=inf(g(1-x))[/itex]

    though...
     
  8. Sep 9, 2011 #7
    sorry thats what i meant for b
     
  9. Sep 9, 2011 #8

    micromass

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    Can you prove that for t in [0,1] that

    [tex]f(t)\geq inf_{x\in [0,1]}f(x)[/tex]

    and

    [tex]g(1-t)\geq inf_{x\in [0,1]}g(x)[/tex]

    ??
     
  10. Sep 9, 2011 #9
    ahhh got it, thank you so much
     
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