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Proof about continuous function related to balls and sets

  1. Aug 25, 2012 #1
    1. The problem statement, all variables and given/known data
    Let [tex] \displaystyle f:{{\mathbb{R}}^{n}}\to \mathbb{R}[/tex] a continuous function. Proove that:

    If [tex] \displaystyle f\left( p \right)>0[/tex] then there's a ball [tex] \displaystyle {{B}_{p}}[/tex] centered at p such that [tex] \displaystyle \forall x\in {{B}_{p}}[/tex] we have [tex] \displaystyle f\left( x \right)>0[/tex].



    2. Relevant equations



    3. The attempt at a solution

    f is continuous, that is:

    [tex] \displaystyle \forall \varepsilon >0[/tex] [tex] \displaystyle \exists \delta >0[/tex] such that [tex] \displaystyle f\left( B\left( p,\delta \right) \right)\subset B\left( f\left( p \right),\varepsilon \right)[/tex].

    Let f(p)>0, then I cannot conclude anthing with the above hypothesis. What's missing?

    Thank you!
     
  2. jcsd
  3. Aug 25, 2012 #2

    LCKurtz

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    Let's talk for a moment about a continuous f(x) on the real line. If f(p) > 0 do you see how to find an interval about p where f(x) > 0? What happens if ##\epsilon = \frac {f(p)}{2}##?
     
  4. Aug 25, 2012 #3
    I don't understand why you ask that and why you say that at the end. Is there seomthing wrong with my attempt?

    Thank you!
     
  5. Aug 25, 2012 #4

    HallsofIvy

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    He is making a specific choice for [itex]\epsilon[/itex]. While there are many possible, f(p)/2 is one choice that will help here. Do you see why?
     
  6. Aug 25, 2012 #5

    LCKurtz

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    Because I am trying to help you see how to work your problem by working a simpler one. How about addressing my questions? You might learn something in the process.
     
  7. Aug 25, 2012 #6
    Hmm.. alright. Answering your question I honestly don't remember exactly how to find an interval about p where f(x) > 0. Could you please remind me?

    Thank you both!
     
  8. Aug 25, 2012 #7

    LCKurtz

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    If ##\epsilon = \frac {f(p)}{2}##, write out carefully what the ##\delta,\epsilon## definition of continuity of ##f## at ##p## tells you.
     
  9. Aug 25, 2012 #8
    Yes. It would be:

    [tex] \displaystyle \forall \varepsilon >0[/tex] [tex] \displaystyle \exists \delta >0[/tex] such that if: [tex] \displaystyle |x-p|<\delta [/tex] then [tex] \displaystyle |f\left( x \right)-f\left( p \right)|<\frac{f\left( p \right)}{2}[/tex].

    But what is this for? Just setting [tex] \displaystyle \varepsilon =\frac{f\left( p \right)}{2}[/tex] it's not enough to find an interval about p such that f(x)>0. It also depends on delta... doesn't it?

    Thank you
     
  10. Aug 25, 2012 #9
    Ahhh I see it now!. If you choose that epsilon you get forced to choose a delta such that the interval about p of radious delta has positive image. Is it right?
     
  11. Aug 25, 2012 #10

    LCKurtz

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    In particular for ##\epsilon = \frac{f(p)}{2}##

    So you have a ##\delta## that works for this value of ##\epsilon##. And doesn't ##|x-p|<\delta## describe an interval about ##p## that goes from ##p-\delta## to ##p+\delta##?
    What does that last inequality tell you about ##f(x)##? Unscramble it without the absolute value signs.
     
  12. Aug 25, 2012 #11
    Yeah, now I get it. It implies that:

    [tex] \displaystyle \frac{f\left( p \right)}{2}<f\left( x \right)<\frac{3}{2}f\left( p \right)[/tex]

    That is: f(x)>0 . So the interval about p exists and it's the one of radious delta for that epsilon. Now it's almost the same but for balls, isn't it?

    Thank you very much!!

    Edit. I think the definition of continuity I used in my attempt is not useful in this demonstration. I think I should use the equivalent definition with distances, shouln't I?
     
  13. Aug 25, 2012 #12

    LCKurtz

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    Yes, it's almost the same proof. It is frequently a good idea, when trying to prove something for ##R^n## to make sure you understand it first for ##R^1## or ##R^2##.

    You're welcome.
     
  14. Aug 25, 2012 #13
    Yeah, I think the problem was that we were given three equivalent definitions and in my attempt I used the definition of continuityn (for several variables) involving balls but I think in this case I should have used the definition involving distances or norm, those are closer to the one for one variable.
     
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