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Function inequality

  1. Dec 10, 2006 #1
    1. The problem statement, all variables and given/known data

    Given a continuous function f(x):R->R with

    lim(f(x)/x^2)=0, x-->+-infinity

    Show that then an element t exist such that:

    x^2+f(x)>=t^2+f(t) for every x in R.

    2. Relevant equations

    -> The mathematical definition of continuous and limes
    (but I really don't know if these are needed)

    3. The attempt at a solution

    I really thought hours on that problem but didn't find a solution. I've no really good attempt. Well, I know that f(x) has to increase less rapidly than x^2 accordint to lim(f(x)/x^2)=0, x-->+-infinity.
     
  2. jcsd
  3. Dec 10, 2006 #2

    StatusX

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    Homework Helper

    You just need to show that the function x^2+f(x) has a minimum. Since f(x) is continuous, it is bounded on any finite interval, and the first condition shows f(x) goes to zero as x->infinity, so you can show that f(x) is bounded for all x. Combine this with the fact that x^2 is bounded below.
     
    Last edited: Dec 10, 2006
  4. Dec 10, 2006 #3
    In theory f(x) could be a function like that one in my picture. I don't really understand what you mean with "bounded". I mean in what case are they bounded?
     
  5. Dec 10, 2006 #4

    HallsofIvy

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    "Bounded" just means that it has upper and lower bounds: that there exist numbers m and M such that m<= f(x)<= M for all x. If lim f(x)/x^2= 0 for x-> +infinity, then There exist X such that if x> X, |f(x)/x^2|< 1: i.e. -1< f(x)/x^2. Similarly, since lim f(x)/x^2= 0 for x-> -infinity there exist X' so that the same is true for x< X'. Let m be the smallest value of f(x)/x^2 for X'<= X<= X and you know f(x)/x^2<= m (if m> -1, take m= -1) for all x.
     
  6. Dec 10, 2006 #5
    But consider f(x)=x. Then f(x)+x^2 has no real minimum and lim f(x)/x^2=x/x^2=1/x=0. So I must show that |f(x)+x^2| has a minimum.

    By the way:
    HallsofIvy: I understand your explanation but how would you integrate this in a proove of x^2+f(x)>=t^2+f(t) for every x in R?
     
  7. Dec 10, 2006 #6

    StatusX

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    f(x)+x^2=x^2+x, which has a minimum at x=-1/2.
     
  8. Dec 10, 2006 #7
    uffff, you're right, man, what's up with me :) . If I tell that x^2+x has no minimum, how can I ever solve this :) (seems I'm disturbed).
     
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