Exercise 87Calculus › Single-variable calculus › Continuity › Intermediate Value TheoremHigh schoolAdded 8 Nov 2026

Fixed point on [0,1]

#fixed point#IVT#auxiliary function

Statement

Let $f:[0,1]\to[0,1]$ be continuous. Show that $f$ has a fixed point, i.e. there exists $x_0\in[0,1]$ with $f(x_0)=x_0$ .

Hint

Apply the IVT to $g(x)=f(x)-x$ .

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Solution

Define $g(x)=f(x)-x$ on $[0,1]$ . Since $f$ and $x\mapsto x$ are continuous, $g$ is continuous on $[0,1]$ .

At the endpoints: Since $f([0,1])\subseteq[0,1]$ , we have $0 \leq f \leq 1$ , hence $g(0)=f(0)\ge0$ and $g(1)=f(1)-1\le0$ .
The IVT provides $x_0\in[0,1]$ with $g(x_0)=0$ , i.e. $f(x_0)=x_0$ .

Fixed point on [0,1]

Statement

Solution

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