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Implicit Function Thm proof question.

  1. Jul 5, 2012 #1
    Assume the function f(x, y) is continuous on R2 and that for point (a, b) in R2, we have f(a, b) > 0.
    Prove that there exists a real number r > 0 such that for all x, y, |x−a| < r,&|y−b| < r ==> f(x, y) > 0.
    (this is relevant to the beginning of the proof of IFT.)

    Attempt:

    So we know f(x,y) is continuous on R2 and at (a,b), f(a,b) >0

    ==> there exists an r1 > 0 s.t |x-a|< r1 and |y-b|< r1 ==> f(a,b) > 0.

    So from what I got to now which was essentially just a restatement of my assumptions, I have to somehow show that this is the case not for one point, but for all (x,y). My issue is that it does not say if the function is C1 so I'm kind of stuck when it comes to trying to apply the implicit function theorem which is what the objective of this question is.
     
  2. jcsd
  3. Jul 5, 2012 #2

    micromass

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    What is your definition of continuous??
     
  4. Jul 5, 2012 #3
    The regular epsilon-delta definition:

    there exists a delta > 0 such that if |x-a|< (delta) ==> |f(x)-f(a)|< (epsilon).

    This is separate, but I was under the impression that since I've learned about uniform continuity that I would be using that definition a lot more, but it hasn't really come up.

    Oh btw, I talked to my prof about the question and he forgot to add that the derivative of f ' (x,y) > 0, so it's strictly increasing. I guess now I might be able to apply the intermediate value thm.
     
  5. Jul 6, 2012 #4

    HallsofIvy

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    What do you ean "show that this is the case not for one point but for all (x,y)"? Show what is the case? If you mean "prove f(x,y)> 0 for all (x,y)", you can't. You can only prove that f(x,y)> 0 for some neighborhood of (a, b)- which is all that is asked.
     
  6. Jul 6, 2012 #5
    did he not give you a specific treatment for continuity in higher dimension?
    and thinking back to the one dimensional case... your neighborhood around a point a (|x-a|<e) was an interval... how would you define points around a single point in 2 dimensions? maybe even consider all the points in the neighborhood at a specific distance away from your single point
     
  7. Jul 8, 2012 #6

    Ok I think I might have worked something out, but it doesn't seem like much.

    Since the function is continuous on R2 and at (a,b), f(a,b)>0

    then this means there exists a neighbourhood around the point (a,b) at which for all ε > 0 there exists a [itex]\delta[/itex] > 0 s.t |(x,y) - (a,b)|<[itex]\delta[/itex]
    ==> |f(x,y) - f(a,b)|< ε. So let [itex]\delta[/itex] = r

    ∴ |x-a| < r and |y-b| < r
     
  8. Jul 8, 2012 #7
    alright, but what does |(x,y)-(a,b)|<δ look like?
    literally draw the neighborhood on paper
    |x-a|<r and |y-b|<r ... would look like a square right? think about how you would formulate your neighborhood
     
  9. Jul 8, 2012 #8
    I don't know if it's the same thing as what you mean by formulating my neighborhood, but I was working on another solution before I saw this reply that might be what your talking about.

    at a = (a,b) there exists a set S = B(r, a). Let x=(x,y) be in B(r, a.

    ==> |x-a|< r

    Since f is continuous

    ==> |x-a| < r < [itex]\delta[/itex]
    ==> |f(x) - f(a) < ε

    ∴ r exists s.t. f(x,y) > 0.


    My only beef with this is that for some reason I feel that I should be using the triangle inequality to show that r < [itex]\delta[/itex]
     
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