Given \(n\) and \(\epsilon\), define a continuous function \(f: I \to \mathbb{R}\) such that \(f \in U_n\) and \(\abs{f(x)} \leq \epsilon\) for all \(x\).
Solution
Here \(U_n = \{f \in \mathcal{C}(I, \mathbb{R}) : \forall x \in I, \exists t \in I, 0 < |x - t| < \frac{1}{n} \text{ and } |\frac{f(x) - f(t)}{x - t}| > n\}\).
1. Parameter selection:Given \(n \in \mathbb{Z}_+\) and \(\varepsilon > 0\), choose an integer \(k \in \mathbb{Z}_+\) sufficiently large such that:\[k > 2n \quad \text{and} \quad k > \frac{2n}{\varepsilon}.\]Let \(h = \frac{1}{2k}\). Then \(h < \frac{1}{4n} < \frac{1}{n}\) and \(\frac{\varepsilon}{h} = 2k\varepsilon > 4n > n\).
2. Construction of the sawtooth function \(f\):Partition \([0, 1]\) into \(2k\) subintervals of equal length \(h = \frac{1}{2k}\) by the division points \(t_j = jh\) for \(j = 0, 1, \dots, 2k\).Define \(f: [0, 1] \to \mathbb{R}\) to be the continuous piecewise linear function with values at the vertices:\[f(t_j) = \begin{cases} \varepsilon & \text{if } j \text{ is odd}, \\ 0 & \text{if } j \text{ is even}, \end{cases}\]and linearly interpolated on each subinterval \([t_{j-1}, t_j]\).
3. Uniform bound on \(f\):For all \(x \in [0, 1]\), \(0 \le f(x) \le \varepsilon\), so \(|f(x)| \le \varepsilon\).Proof: On each subinterval \([t_{j-1}, t_j]\), \(f\) is linear with endpoint values in \(\{0, \varepsilon\}\). By convexity, \(0 \le f(x) \le \varepsilon\).
4. Secant slope property (\(f \in U_n\)):For every \(x \in [0, 1]\), there exists \(t \in [0, 1]\) such that \(0 < |x - t| < \frac{1}{n}\) and \(\left|\frac{f(x) - f(t)}{x - t}\right| > n\).Proof:
4.1. For any \(x \in [0, 1]\), \(x\) belongs to some subinterval \([t_{j-1}, t_j]\) of length \(h\).
4.2. The function \(f\) is strictly linear on \([t_{j-1}, t_j]\) with constant slope magnitude:\[\left|\frac{f(t_j) - f(t_{j-1})}{t_j - t_{j-1}}\right| = \frac{\varepsilon}{h} = 2k\varepsilon > 4n > n.\]
4.3. If \(x \neq t_j\), choose \(t = t_j\). Then \(0 < |t - x| \le h < \frac{1}{n}\), and by linearity on \([t_{j-1}, t_j]\), \(\left|\frac{f(x) - f(t)}{x - t}\right| = \frac{\varepsilon}{h} > n\).
4.4. If \(x = t_j\), choose \(t = t_{j-1}\) (or \(t_{j+1}\) if \(j = 0\)). Then \(0 < |t - x| = h < \frac{1}{n}\), and the secant slope is \(\frac{\varepsilon}{h} > n\).
4.5. Hence \(f \in U_n\).
5. Q.E.D.By steps 3 and 4, the continuous function \(f\) satisfies \(f \in U_n\) and \(|f(x)| \le \varepsilon\) for all \(x \in [0, 1]\).