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Let \(f : [0, 1] \to [0, 1]\) be a continuous function. Then which of the following is true ?

  • A.For all \(x, y \in [0, 1]\), \(x \le y \Rightarrow f(x) \le f(y)\)
  • B.\(\exists\, x \in [0, 1]\) such that \(f(x) = x\)
  • C.If for some \(x, y \in [0, 1]\) with \(x \le y\), \(f(x) \le f(y)\) then for all \(x \le y\), \(f(x) \le f(y)\)
  • D.If there is \(x_{0} \in [0, 1]\), \(f(x_{0}) = x_{0}\) then \(f(x) = x\) for all \(x \in X\)

Answer: B

The true statement is there exists \(x \in [0, 1]\) with \(f(x) = x\).

Let \(g(x) = f(x) - x\). Since \(f\) maps into [0, 1], \(g(0) = f(0) \ge 0\) and \(g(1) = f(1) - 1 \le 0\). By the intermediate value theorem \(g\) has a zero, which is a fixed point of \(f\). This is the one-dimensional Brouwer fixed point theorem.

The other statements fail for \(f(x) = 1 - x\). It is not increasing, which refutes A. C fails because its hypothesis holds trivially for \(x = y = 0\) while \(f\) is decreasing. And its only fixed point is 1/2, so having one fixed point does not make \(f\) the identity (D).

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