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Prove that a particular function has a fixed point on [0,1]

by RoRi

Let f be a continuous, real function on the interval [0,1]. Assume

    \[ 0 \leq f(x) \leq 1 \qquad \text{for all } x \in [0,1]. \]

Prove that there exists a real number c \in [0,1] such that f(c) = c.

Proof. Let g(x) = f(x) - x. Then g is continuous on [0,1] since it is the difference of functions which are continuous on [0,1].

Then, g(0) = f(0) and g(1)= f(1) - 1. If f(0) = 0, then c = 0 and we are done. Similarly, if f(1) = 1, then c = 1 and we are done as well.

Assume then that 0 < f(0) \leq 1, and 0 \leq f(1) < 1. Then

    \begin{align*} f(0) > 0 \quad \implies \quad g(0) &>0 \\ f(1) < 1 \quad \implies \quad g(1) &= f(1) - 1 < 0. \end{align*}

Hence, applying Bolzano’s theorem to g, there is some c \in [0,1] such that g(c) = 0. This implies f(c) - c = 0, or f(c) = c. \qquad \blacksquare

One comment

  1. H says:

    First of all, thank you for the great solutions for this book, it helps a lot.

    you can prove it by the intermediate-value theorem instead of Bolzano’s theorem.

    since g is continuous on [0,1] ,by the intermediate-value theorem, it takes on every value on [0,1] (including 0).

    hence, we can find a number c such that g(c)=0 .
    then, f(c)-c=0 -> f(c)=c.

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