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Math Help - Continuous functions with fixed points

  1. #1
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    Continuous functions with fixed points

    1. Let f : [0,1] \rightarrow [0,1] be continuous. Prove that f has a fixed point.

    2. let f: (0,1] \rightarrow (0,1] be continuous. Give a counterexample to show that f need not have a fixed point.

    For both the questions, I supposed a sequence {x_n} = \frac{1}{n}, which converges to 0, but this does not look correct. Can anyone tell how this is proved?
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    MHF Contributor Drexel28's Avatar
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    Quote Originally Posted by serious331 View Post
    1. Let f : [0,1] \rightarrow [0,1] be continuous. Prove that f has a fixed point.
    Hint:

    Spoiler:


    The mapping \varphi:[0,1]\to[0,1]:x\mapsto f(x)-x is continuous. Think IVT


    2. let f: (0,1] \rightarrow (0,1] be continuous. Give a counterexample to show that f need not have a fixed point.
    What do you think?
    For both the questions, I supposed a sequence {x_n} = \frac{1}{n}, which converges to 0, but this does not look correct. Can anyone tell how this is proved?
    What does that sequence help with?
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    Quote Originally Posted by Drexel28 View Post
    Hint:

    Spoiler:


    The mapping \varphi:[0,1]\to[0,1]:x\mapsto f(x)-x is continuous. Think IVT




    What do you think?


    What does that sequence help with?
    I am sorry I am not good at writing proofs. but how can these questions be shown from the IVT?
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  4. #4
    MHF Contributor Drexel28's Avatar
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    Quote Originally Posted by serious331 View Post
    I am sorry I am not good at writing proofs. but how can these questions be shown from the IVT?
    I'll give you one more hint and let you take one more stab.

    Consider the \varphi(x) above.

    Now, \varphi being the difference of two continuous real-valued functions is continuous. Also, \varphi(0)=f(0)-0=f(0). Now, if \varphi(0)=0 we're done, right? So what can you assume?

    Similarly \varphi(1)=f(1)-1. If it's zero we're done again, so we can assume not.

    So now you should have two conditions which should immediately make you jump for joy at knowing the IVT, what are they?
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    Quote Originally Posted by Drexel28 View Post
    I'll give you one more hint and let you take one more stab.

    Consider the \varphi(x) above.

    Now, \varphi being the difference of two continuous real-valued functions is continuous. Also, \varphi(0)=f(0)-0=f(0). Now, if \varphi(0)=0 we're done, right? So what can you assume?

    Similarly \varphi(1)=f(1)-1. If it's zero we're done again, so we can assume not.

    So now you should have two conditions which should immediately make you jump for joy at knowing the IVT, what are they?
    So, do we have \varphi(0) \geq 0 \geq \varphi(1), and from IVT there is a fixed point x in [0,1] such that \varphi(x) = 0?
    Is that right?
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  6. #6
    MHF Contributor Drexel28's Avatar
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    Quote Originally Posted by serious331 View Post
    So, do we have \varphi(0) \geq 0 \geq \varphi(1), and from IVT there is a fixed point x in [0,1] such that \varphi(x) = 0?
    Is that right?
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    Quote Originally Posted by Drexel28 View Post
    Thank you so much Drexel28. Can you give me a hint on what counterexample I can present on the second question?
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  8. #8
    MHF Contributor Drexel28's Avatar
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    Quote Originally Posted by serious331 View Post
    Thank you so much Drexel28. Can you give me a hint on what counterexample I can present on the second question?
    Honestly, it's not that bad to come up with one. It may sound stupid, but just draw a picture.
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