Let \(f:\mathbb R\to\mathbb R\) be monotonically increasing, perhaps discontinuous. Suppose \(0<f(0)\) and \(f(100)<100\). Prove that \(f(x)=x\) for some \(x\).
Solution
Set \[ A=\{x\in[0,100]:x\le f(x)\}. \]
1. The set \(A\) is nonempty and bounded above, so \(c=\sup A\) exists andlies in \([0,100]\).
Since \(0<f(0)\), one has \(0\in A\). By definition \(A\subseteq[0,100]\).
2. One has \(c\le f(c)\).
For every \(x\in A\), one has \(x\le c\). Because \(f\) is monotonically increasing, \[ f(x)\le f(c). \] But \(x\in A\) also gives \(x\le f(x)\). Thus \[ x\le f(c) \qquad\text{for every }x\in A, \] so \(f(c)\) is an upper bound for \(A\). Since \(c\) is the least upper bound, \(c\le f(c)\).
3. One has \(c<100\).
If \(c=100\), then step 2 would give \(100\le f(100)\), contrary to the hypothesis \(f(100)<100\).
4. One has \(f(c)\le c\).
Suppose for contradiction that \[ c<f(c). \] Monotonicity gives \[ f(c)\le f(f(c)). \] Since \(c<100\) by step 3, monotonicity also gives \[ f(c)\le f(100)<100, \] and \(f(c)>c\ge0\). Hence \(f(c)\in[0,100]\) and \(f(c)\le f(f(c))\), so \(f(c)\in A\). But \(f(c)>c=\sup A\), a contradiction.
5. \(f(c)=c\).
Steps 2 and 4 give \(c\le f(c)\le c\).
6. Q.E.D.
Step 5 gives the fixed point \(x=c\).