Suppose \((f_n)\) is a sequence of bounded functions from \(\mathbb R\) to \(\mathbb R\) that converges uniformly to a function \(f\). Prove that there is an \(M>0\) such that \[ |f_n(x)|\le M \] for all natural numbers \(n\) and all real numbers \(x\).
Solution
- Suppose \(f_n \rightrightarrows g\) with each \(f_n\) bounded; we want to show that all of the \(f_n\) are uniformly bounded by some \(M\), i.e. \[ \exists M \suchthat \forall x\in \RR, \forall n\in \NN, \quad \abs{f_n(x)} \leq M. \]
- Since each \(f_n\) is bounded, we can produce some \(M_n\) such that \(\abs{f_n(x)} \leq M < \infty\).
- Since \(f_n \rightrightarrows g\), we can give ourselves an epsilon of room and get an \(N\) such that \(n\geq N \implies \abs{f_n(x) - g(x)} < \varepsilon\). We then write \[ f_n(x) = f_n(x) - g(x) + g(x) - f_N(x) + f_N(x) \\ \implies \abs{f_n(x)} \leq \abs{f_n(x) - g(x)} + \abs{g(x) - f_N(x)} + \abs{f_N(x)} \\ \leq \varepsilon + \varepsilon + M_N \]
by the above two statements. But \(N<\infty\), so we can choose \(M = \max\theset{M_1, M_2, \cdots M_{N-1}, 2\varepsilon + M_N}\) as a uniform bound. Then just take \(\varepsilon \to 0\). (Maybe not necessary?)
Solution
Goal: Let \(E\) be a set, and let \((f_n)_{n=1}^\infty\) be a sequence of functions \(f_n: E \to \mathbb{R}\) such that each \(f_n\) is bounded on \(E\). Suppose \(f_n \to g\) uniformly on \(E\). Prove that the sequence \((f_n)_{n=1}^\infty\) is uniformly bounded on \(E\), i.e., there exists a constant \(M > 0\) such that \(|f_n(x)| \le M\) for all \(n \in \mathbb{N}\) and all \(x \in E\).
1. Definition of uniform convergence: There exists an integer \(N \ge 1\) such that for all \(n \ge N\) and all \(x \in E\), \(|f_n(x) - g(x)| < 1\).
By the definition of uniform convergence with \(\varepsilon = 1 > 0\).
2. For all \(n \ge N\) and all \(x \in E\), \(|f_n(x)| \le |f_N(x)| + 2\).Proof:
2.1. By the triangle inequality:\[|f_n(x) - f_N(x)| \le |f_n(x) - g(x)| + |g(x) - f_N(x)|.\]
2.2. For any \(n \ge N\), applying 1 gives \(|f_n(x) - g(x)| < 1\) and \(|f_N(x) - g(x)| < 1\).Thus \(|f_n(x) - f_N(x)| < 1 + 1 = 2\).
2.3. Again by the triangle inequality:\[|f_n(x)| \le |f_N(x)| + |f_n(x) - f_N(x)| < |f_N(x)| + 2.\]
3. For each \(k \in \{1, 2, \dots, N\}\), there exists a constant \(M_k < \infty\) such that \(|f_k(x)| \le M_k\) for all \(x \in E\).
By hypothesis, each function \(f_k\) is bounded on \(E\).
4. Define \(M = \max\{M_1, M_2, \dots, M_{N-1}, M_N + 2\}\). Then \(|f_n(x)| \le M\) for all \(n \in \mathbb{N}\) and all \(x \in E\).Proof:
4.1. If \(1 \le n < N\), then for all \(x \in E\), \(|f_n(x)| \le M_n \le M\) by 3 and definition of \(M\).
4.2. If \(n \ge N\), then by 2 and 3:\[|f_n(x)| < |f_N(x)| + 2 \le M_N + 2 \le M.\]
4.3. In all cases, \(|f_n(x)| \le M\) for all \(n \in \mathbb{N}\) and all \(x \in E\).
5. Conclusion: \((f_n)_{n=1}^\infty\) is uniformly bounded on \(E\).
4. Q.E.D.