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Showing a function in R2 is unbounded (no least upper bound)

  1. Nov 5, 2014 #1
    1. The problem statement, all variables and given/known data
    Show that this function has no absolute max by showing that it is unbounded

    2. Relevant equations
    f(x,y) = (x-1)^2 + (y+2)^2 -4

    3. The attempt at a solution
    my initial idea is to construct a sequence of points {(xk, yk)} so that the sequence {f(xk, yk)} becomes unbounded.

    to show that : Let M=f(x,y)
    ∨M>0 ∃xk, yk s.t xk,yk∉ B(M,(1,-2)). This issue i have is determining an adequate sequence of values to use.
     
    Last edited: Nov 5, 2014
  2. jcsd
  3. Nov 5, 2014 #2

    pasmith

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    Observe that [itex]f(x,y) \geq (x - 1)^2 -4[/itex]. Is that bounded above?
     
  4. Nov 5, 2014 #3
    I'm assuming by f(x,y) you're referring to the original equation and no it isn't bounded above. I'm sorry, i have a feeling i'm supposed to make some intellectual leap with that example,but im still somewhat lost.
     
  5. Nov 5, 2014 #4
    Actually, i noticed that you reduced it to a single variable? To what end?
     
  6. Nov 5, 2014 #5

    pasmith

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    If [itex](x - 1)^2 - 4[/itex] is not bounded above, and [itex]f(x,y) \geq (x-1)^2 -4[/itex], can it be the case that [itex]f(x,y)[/itex] is bounded above?
     
  7. Nov 5, 2014 #6

    PeroK

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    Perhaps this will help, but I'm not sure:

    Can you think of any function (even of a single variable) that is not bounded above?
     
  8. Nov 5, 2014 #7
    f(x,y) = X^2 for instance, For the sake of a concrete proof. Would it be sufficient to say that if f(x) is unbounded and f(x,y) >f(x) then f(x,y) is also unbounded. Seems incomplete, or have we just skimmed the surface in terms of reasoning?
     
  9. Nov 5, 2014 #8

    PeroK

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    Ok, so you know that ##f(x) = x^2## is not bounded above. What about ##f(x) = (x-1)^2##? Not bounded above?
     
  10. Nov 6, 2014 #9

    HallsofIvy

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    The sequence or points (1, 1), (2, 2), ...., (n, n) for any integer n leaps out at you.
     
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