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Math Help - Real Analysis

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    Real Analysis

    Show that if f is differentiable on R and |f ' (x)| < 1 for all x in R, then f has at most one fixed point.
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    Quote Originally Posted by dhhnerd View Post
    Show that if f is differentiable on R and |f ' (x)| < 1 for all x in R, then f has at most one fixed point.
    Assume that there are two fixed points x<y. Then f on [a,b] satisfies the continous of the Mean Value Theorem and therefore f(y) - f(x) = f'(c)(y-x) where x<c<y. But that gives us (y-x)=f'(c)(y-x)\implies f'(c) = 1. This is a contradiction because |f'| < 1.
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