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Three Problems on continuity.

  1. Oct 13, 2012 #1
    1. The problem statement, all variables and given/known data

    1- Let f be a continuous function for all real numbers such that :

    [tex]\lim_{x\rightarrow+\infty}f(x)=L[/tex] and [tex]\lim_{x\rightarrow-\infty}f(x)=L'[/tex]

    and that LL'≤0. Prove that f equals 0 at some point C in ℝ.

    2- Let f be a continuous function on [a,b] such that for every (x,x') in ([a,b])^2 and x≠x:

    [tex]|f(x)-f(x')|<k|x-x'|[/tex] .Prove that the equation f(x)=x has only one solution on [a,b].

    3-Let f and g be continuous functions on [0,1] such that for every x in [0,1]: f(x)<g(x).

    Prove that there exists a number m>0 such that for every x in [0,1]: f(x)+m<g(x).


    3. The attempt at a solution

    1- I know that since LL'<0 so that means that L>L' or L<L' . So the Intermediate value theorem states that there exists a number c such that f(c)=0, but i know how I'm going to show that.

    2-I think I am supposed to use the definition of a limit to solve it but i don't know where to start.

    3- I have no idea how to start this one. Any help would be very appreciated.
     
  2. jcsd
  3. Oct 13, 2012 #2

    HallsofIvy

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    That is only saying "L is not equal to L'" which says nothing about their relation to 0. What you meant to say, I believe, was "either L> 0> L' or L'> 0> L".

    IF the intermediate theorem states that, then that would "show" it. But it doesn't. The intermediate value theorem says that if f(a)> 0 and f(b)< 0 there exist x between a and b such that f(x)= 0. What you are given are limits, not values of the function at specific points.

    What you can say is that, because [itex]\lim_{x\to\infty} f(x)= L[/itex], if L> 0, there exist [itex]x_0[/itex] such that if [itex]x> x_0[/itex] then f(x)> L-1. Do the same with L' and x going to negative infinity.

    Take a guess and give it a try. See what you learn from trying.
     
  4. Oct 13, 2012 #3
    Ok for the negative infinity i can say that if L<0 then there exists x1 such that x>x1 then
    f(x)>L-1. Right?
     
  5. Oct 14, 2012 #4
    Okay for number 3 I did a proof by contradiction and I got:

    Let h(x)=f(x)-g(x) and we know that since g(x)>f(x) then h(x)=f(x)-g(x)<0. Now we have to prove that h(x)<0 for all x in [0.1]. Suppose that there is a point C in [0,1] such that f(c)>g(c) implying that h(c)=f(c)-g(c)>0 and that's a contradiction because h(x)<0 for every x in [0,1]. But does this imply that f(x)+m<g(x) for every x in [0,1].
     
    Last edited: Oct 14, 2012
  6. Oct 14, 2012 #5

    HallsofIvy

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    For 3 use the fact that every continuous function attains both maximum and minimum values on a closed and bounded interval.
     
  7. Oct 14, 2012 #6
    For number 2 i said that let g(x)=f(x)-x be a continuous function on the interval I=[a,b] as a difference of 2 continuous equations. and for every x in I f(x) is also in I.


    And f(I)=[m,M] so therefore f(a) is in I and f(b) is in I.

    That implies that m≤f(a)≤M and m≤f(b)≤M and that implies that f(a)-m≥0 and f(b)-M≤0 and that implies that f(a)*f(b)≤0.

    Hence the IVT states that there exists an x in I such that f(x)=x . But i think that i have to show that f is monotone increasing or monotone decreasing for this proof to hold. Am I correct??
     
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