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Thread: Small Challenge # 2

  1. #1
    Senior Member Dinkydoe's Avatar
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    Small Challenge # 2

    This is a challenge question:

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


    Moderator edit: Approved Challenge question.
    Last edited by mr fantastic; Aug 1st 2010 at 12:34 PM. Reason: Approved.
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  2. #2
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    Lemma 1:
    Let $\displaystyle \epsilon>0$ and $\displaystyle x \in [0,1]$.
    There can be no sequence $\displaystyle x_{n}\uparrow x$ with $\displaystyle f(x)+\epsilon\leq f(x_{n})$ for all n
    and
    no sequence $\displaystyle x_{n}\downarrow x $with $\displaystyle f(x)-\epsilon\geq f(x_{n})$ for all n

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

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

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

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

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

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

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

    Therefor
    $\displaystyle f(y)=0$.
    qed

    Very good !
    Last edited by Dinkydoe; Aug 3rd 2010 at 06:11 AM.
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  4. #4
    Senior Member Dinkydoe's Avatar
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    You guys seem to have no trouble solving my Small challenges.... hehe

    Here's what I did:

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

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

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

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

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

    Conclusion: $\displaystyle h(u)=g(u)$
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