- Let \(f _ { n }\) be a sequence of continuous functions on R satisfying \(0 \leq f _ { n } \leq f _ { n + 1 } \leq 1\) for all \(x \in \mathbb { R }\) and \(n \in \mathbb { Z } ^ { + }\) . Let \(f ( x ) = \operatorname* { l i m } _ { n \to \infty } f _ { n } ( x )\) . Show that if \(f\) is continuous at x, then for any \(\epsilon > 0\) there exist δ and N such that \(| f _ { n } ( y ) - f _ { n } ( x ) | < \epsilon\) whenever \(| y - x | < \delta\) and \(n > N\)
Solution
1. Choose a sufficiently late index at the base point.
Fix \(x\in\mathbb R\) at which \(f\) is continuous, and let \(\varepsilon>0\). Since \[ f_n(x)\uparrow f(x), \] choose \(N\) so large that \[ 0\le f(x)-f_N(x)<\frac\varepsilon3. \]
2. Choose a common neighborhood using continuity of \(f\) and \(f_N\).
Because both \(f\) and \(f_N\) are continuous at \(x\), there exists \(\delta>0\) such that whenever \(|y-x|<\delta\), \[ |f(y)-f(x)|<\frac\varepsilon3 \] and \[ |f_N(y)-f_N(x)|<\frac\varepsilon3. \]
3. Bound both one-sided differences uniformly for all later indices.
Let \(n\ge N\) and \(|y-x|<\delta\). Since \[ f_N\le f_n\le f, \] we have \[ \begin{aligned} f_n(y)-f_n(x) &\le f(y)-f_N(x)\\ &=[f(y)-f(x)]+[f(x)-f_N(x)]\\ &<\frac{2\varepsilon}{3}, \end{aligned} \] while \[ \begin{aligned} f_n(x)-f_n(y) &\le f(x)-f_N(y)\\ &=[f(x)-f_N(x)]+[f_N(x)-f_N(y)]\\ &<\frac{2\varepsilon}{3}. \end{aligned} \] Hence \[ |f_n(y)-f_n(x)|<\varepsilon \] for every \(n\ge N\) and every \(y\) with \(|y-x|<\delta\).
4. Q.E.D.
Steps <1>1–<1>3 produce \(N\) and \(\delta\) with the required estimate for every \(n\ge N\).