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Thread: Injective

  1. #1
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    Injective

    Happy Thanksgiving to those of you in the States.

    I have a problem:


    Suppose that a function f [a,b] to R is continuous such that f attains its minimum or maximum on [a,b] at a point c such that
    a<c<b.
    Prove that f cannot be injective.

    How would you work through this proof?
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  2. #2
    Grand Panjandrum
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    Can you tell us why you need three accounts?

    RonL
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  3. #3
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    Quote Originally Posted by OntarioStud View Post
    Happy Thanksgiving to those of you in the States.

    I have a problem:


    Suppose that a function f [a,b] to R is continuous such that f attains its minimum or maximum on [a,b] at a point c such that
    a<c<b.
    Prove that f cannot be injective.

    How would you work through this proof?
    I am going to make this proof as if it was a maximum point the minimum point is anagolous.

    Assume,
    $\displaystyle f:[a,b]\to \mathbb{R}$ is an injective map.
    Also that the function has a maximum $\displaystyle c\in (a,b)$.
    Subdivide your interval into,
    $\displaystyle [a,c],[c,b]$
    Since the function is continous on the full interval it is continous on these.
    Without lose of generality assume $\displaystyle f(a)\geq f(b)$. Pick any number $\displaystyle d$ such that $\displaystyle f(b) \leq d< f(c)$
    Then that number (by our conditions) also exists on,
    $\displaystyle f(a)\leq d< f(c)$
    By the intermediate value theorem there is a point $\displaystyle m_{1,2}$ such that $\displaystyle f(m_1)=d$ on $\displaystyle [a,c]$ and $\displaystyle f(m_2)=d$ on $\displaystyle [c,b]$. Thus, $\displaystyle f(m_1)=f(m_2)$, and if $\displaystyle m_1\not = m_2$ the function fails to be injective. The only way it can is when $\displaystyle m_1=m_2=c$ but that is not possible.

    I hope you can understand it, mathematicians are by nature lazy.
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