Results 1 to 4 of 4

Math Help - Small Challenge # 2

  1. #1
    Senior Member Dinkydoe's Avatar
    Joined
    Dec 2009
    Posts
    411

    Small Challenge # 2

    This is a challenge question:

    Let f:[0,1]\to \mathbb{R} with f(0)>0 and  f(1)<0. If there exists a continuous function g:[0,1]\to  \mathbb{R} such that f+g is non-decreasing, show there exists a x_0\in (0,1) such that f(x_0)=0.


    Moderator edit: Approved Challenge question.
    Last edited by mr fantastic; August 1st 2010 at 01:34 PM. Reason: Approved.
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Junior Member
    Joined
    Aug 2010
    Posts
    44
    Lemma 1:
    Let \epsilon>0 and x \in [0,1].
    There can be no sequence x_{n}\uparrow x with f(x)+\epsilon\leq f(x_{n}) for all n
    and
    no sequence x_{n}\downarrow x with f(x)-\epsilon\geq f(x_{n}) for all n

    Proof:
    Towards a contradiction suppose there is such a sequence with x_{n} \uparrow x .
    Since f+g is non-decreasing and x_{n}\leq x , we have
     f(x)+\epsilon+g(x_{n}) \leq f(x_{n})+g(x_{n})\leq f(x)+g(x)
    Since g is continous, \lim g(x_{n})=g(x), so that taking limits on both sides leads to the contradiction
    f(x)+\epsilon+g(x) \leq f(x)+g(x).
    Analogous in the second case.
    qed

    Now consider y:=\sup \{x\in [0,1] : f(x)\geq 0\}. y \in [0,1] , because [0,1] is closed.
    Assume f(y) \neq 0.

    case 1:
    f(y)>\epsilon>0 for some \epsilon.
    Then y<1 and f(x)<0 for all x>y according to the definition of y.
    So (y,1] is nonempty such that there is a sequence x_{n} \in (y,1] with x_{n}\downarrow y , which violates lemma 1.
    case 2:
    f(y)<-\epsilon<0 for some \epsilon>0
    Then y>0.
    So [0,y) is nonempty such that there is a sequence x_{n} \in [0,y) with x_{n}\uparrow y and f(x_n)\geq 0 which violates lemma 1.

    Therefor
     f(y)=0.
    qed
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Senior Member Dinkydoe's Avatar
    Joined
    Dec 2009
    Posts
    411
    Quote Originally Posted by Iondor View Post
    Lemma 1:
    Let \epsilon>0 and x \in [0,1].
    There can be no sequence x_{n}\uparrow x with f(x)+\epsilon\leq f(x_{n}) for all n
    and
    no sequence x_{n}\downarrow x with f(x)-\epsilon\geq f(x_{n}) for all n

    Proof:
    Towards a contradiction suppose there is such a sequence with x_{n} \uparrow x .
    Since f+g is non-decreasing and x_{n}\leq x , we have
     f(x)+\epsilon+g(x_{n}) \leq f(x_{n})+g(x_{n})\leq f(x)+g(x)
    Since g is continous, \lim g(x_{n})=g(x), so that taking limits on both sides leads to the contradiction
    f(x)+\epsilon+g(x) \leq f(x)+g(x).
    Analogous in the second case.
    qed

    Now consider y:=\sup \{x\in [0,1] : f(x)\geq 0\}. y \in [0,1] , because [0,1] is closed.
    Assume f(y) \neq 0.

    case 1:
    f(y)>\epsilon>0 for some \epsilon.
    Then y<1 and f(x)<0 for all x>y according to the definition of y.
    So (y,1] is nonempty such that there is a sequence x_{n} \in (y,1] with x_{n}\downarrow y , which violates lemma 1.
    case 2:
    f(y)<-\epsilon<0 for some \epsilon>0
    Then y>0.
    So [0,y) is nonempty such that there is a sequence x_{n} \in [0,y) with x_{n}\uparrow y and f(x_n)\geq 0 which violates lemma 1.

    Therefor
     f(y)=0.
    qed

    Very good !
    Last edited by Dinkydoe; August 3rd 2010 at 07:11 AM.
    Follow Math Help Forum on Facebook and Google+

  4. #4
    Senior Member Dinkydoe's Avatar
    Joined
    Dec 2009
    Posts
    411
    You guys seem to have no trouble solving my Small challenges.... hehe

    Here's what I did:

    Let h= f+g. Then h(0)>g(0) and h(1)<g(1). We need to show the existence of a x_0\in (0,1) such that h(x_0)=g(x_0).

    Let S = \left\{x: h(x) \geq g(x)\right\} and u=\sup(S).

    1.Suppose h(u)>g(u).
    Then (u,1) is nonempty and by definition of u we have h(x)<g(x) for all x> u.

    Thus we must have \lim_{x\to u^{+}}h(x)< \lim_{x\to u^{+}}g(x)=g(u) by continuity of g . But then h(u)>\lim_{x\to u^{+}}h(x). This contradicts the fact that h is non-decreasing.

    2. If h(u)<g(u) we get the same contradiction: \lim_{x\to u^{-}}h(x)>\lim_{x\to u^{-}}g(x)=g(u). Hence \lim_{x\to u^{-}}h(x)>h(u)

    Conclusion: h(u)=g(u)
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Small Challenge
    Posted in the Math Challenge Problems Forum
    Replies: 17
    Last Post: July 23rd 2010, 02:27 AM
  2. 5a(SMALL 3) =625
    Posted in the Algebra Forum
    Replies: 2
    Last Post: May 28th 2010, 07:08 PM
  3. How small?
    Posted in the Calculus Forum
    Replies: 1
    Last Post: March 25th 2010, 06:10 PM
  4. small change from small changes
    Posted in the Calculus Forum
    Replies: 2
    Last Post: June 30th 2009, 09:30 AM
  5. Is A small?
    Posted in the Differential Equations Forum
    Replies: 3
    Last Post: May 28th 2009, 03:39 AM

Search Tags


/mathhelpforum @mathhelpforum