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    If I have a function  f: [0,2] \rightarrow [0,2] , does this mean it's a contraction because it maps the interval back to itself? If so how would I show that this function is a contraction? Would it be sufficient to show that the absolute value of the gradient on the function is strictly less than 1? Thanks.
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    Now here was me thinking this was all about the later stages of pregnancy when I read the title...
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    (Original post by JBKProductions)
    If I have a function  f: [0,2] \rightarrow [0,2] , does this mean it's a contraction because it maps the interval back to itself? If so how would I show that this function is a contraction? Would it be sufficient to show that the absolute value of the gradient on the function is strictly less than 1? Thanks.
    It's not necessarily true: take f on rational to be 1 and f on irrational numbers to be 0.
    Also with just the information you've provided it doesn't follow that f has a gradient.
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    (Original post by IrrationalNumber)
    It's not necessarily true: take f on rational to be 42 and f on irrational numbers to be 0.
    Also with just the information you've provided it doesn't follow that f has a gradient.
    Sorry I forgot to add that the function is  f(x)= \frac{x^2}{32}+cos(\frac{x\pi}{4  })
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    my mind went to pregnancy when I saw this title..
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    (Original post by JBKProductions)
    Sorry I forgot to add that the function is  f(x)= \frac{x^2}{32}+cos(\frac{x\pi}{4  })
    A gradient less than 1 will suffice in this case. If f is C^1 and |f'(x)|<1 then |f'(x)|<d for some d<1 (why?) and then you can use the mean value theorem.
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    (Original post by IrrationalNumber)
    A gradient less than 1 will suffice in this case. If f is C^1 and |f'(x)|<1 then |f'(x)|<d for some d<1 (why?) and then you can use the mean value theorem.
    I'm not sure why, can you explain it a bit further? Also I'm not sure why I'm supposed to bring the mean value theorem into this. Thanks.
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    (Original post by JBKProductions)
    I'm not sure why, can you explain it a bit further? Also I'm not sure why I'm supposed to bring the mean value theorem into this. Thanks.
    The MVT tells you that if f is differentiable on (0,2) and continuous on [0,2] then for all x,y in [0,2] we can find a c in (x,y) such that
     f'(c)=\frac{f(x)-f(y)}{x-y}
    In particular then
     |f'(c)| =\frac{|f(x)-f(y)|}{|x-y|} Can you see why that might be useful now?

    Now if f belongs to C^1, the derivative is continuous. The modulus of any continuous function is still continuous. What can we say about the maximum value of a continuous function on a compact set(you may know this result as the maximum value of a continuous function on a closed interval)?
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    (Original post by IrrationalNumber)
    The MVT tells you that if f is differentiable on (0,2) and continuous on [0,2] then for all x,y in [0,2] we can find a c in (x,y) such that
     f'(c)=\frac{f(x)-f(y)}{x-y}
    In particular then
     |f'(c)| =\frac{|f(x)-f(y)|}{|x-y|} Can you see why that might be useful now?

    Now if f belongs to C^1, the derivative is continuous. The modulus of any continuous function is still continuous. What can we say about the maximum value of a continuous function on a compact set(you may know this result as the maximum value of a continuous function on a closed interval)?
    So the maximum value of the continuous function on the closed interval is when  |f'(c)| is at the maximum?
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    (Original post by JBKProductions)
    So the maximum value of the continuous function on the closed interval is when  |f'(c)| is at the maximum?
    Right, in particular there exists a d such that  |f'(c)|\leq |f'(d)|
    So if you can show that |f'|<1 then you get that |f'|<C<1 for some C. Now suppose that there are x and y in [0,2] such that  |f(x)-f(y)|&gt;C|x-y| and argue by contradiction.
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    I'm not sure why we're worrying about compactness or where f'(x) attains a maximum here. f'(x) = x/16 - (pi/4) sin( x pi/4). |sin x| <=1 and |x/16| <=2/16. Since pi/4+2/16 < 1, you're done. (Still need to use MVT afterwards, of course).
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    (Original post by DFranklin)
    I'm not sure why we're worrying about compactness or where f'(x) attains a maximum here. f'(x) = x/16 - (pi/4) sin( x pi/4). |sin x| <=1 and |x/16| <=2/16. Since pi/4+2/16 < 1, you're done. (Still need to use MVT afterwards, of course).
    Is it possible to construct a function f for which |f'|<1 everywhere but f is not a contraction? I haven't thought about it too hard but I think it should be possible by getting the derivative to approach 1. It certainly won't be possible to do that on a compact space.
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    (Original post by IrrationalNumber)
    Right, in particular there exists a d such that  |f'(c)|\leq |f'(d)|
    So if you can show that |f'|<1 then you get that |f'|<C<1 for some C. Now suppose that there are x and y in [0,2] such that  |f(x)-f(y)|&gt;C|x-y| and argue by contradiction.
    Thanks!
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    (Original post by DFranklin)
    I'm not sure why we're worrying about compactness or where f'(x) attains a maximum here. f'(x) = x/16 - (pi/4) sin( x pi/4). |sin x| &lt;=1 and |x/16| &lt;=2/16. Since pi/4+2/16 &lt; 1, you're done. (Still need to use MVT afterwards, of course).

    (Original post by IrrationalNumber)
    Is it possible to construct a function f for which |f'|&lt;1 everywhere but f is not a contraction? I haven't thought about it too hard but I think it should be possible by getting the derivative to approach 1. It certainly won't be possible to do that on a compact space.
    Yes, if we consider f(x)=x^2 on (0,1/2) and consider  x_n = \frac{1}{2} - \frac{1}{n} and  y_n = \frac{1}{2}-\frac{2}{n} then  |f(x_n)-f(y_n)| =|x_n-y_n||1-\frac{3}{n}| which cannot be always less than or equal to  K |x-y| for some fixed K<1.
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    (Original post by IrrationalNumber)
    Is it possible to construct a function f for which |f'|<1 everywhere but f is not a contraction?
    Yes.

    E.g. g(x) = x - exp(-x) on [0,\infty).

    But in the original case it's easy to show |f'| < k < 1 directly, so no need to do anything more.
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    (Original post by DFranklin)
    Yes.

    E.g. g(x) = x - exp(-x) on [0,\infty).

    But in the original case it's easy to show |f'| < k < 1 directly, so no need to do anything more.
    Ah, you're quite right.
 
 
 
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