Let \((X, d_X)\) and \((Y, d_Y)\) be metric spaces; let \(Y\) be complete. Let \(A \subset X\). Show that if \(f: A \to Y\) is uniformly continuous, then \(f\) can be uniquely extended to a continuous function \(g: \overline{A} \to Y\), and \(g\) is uniformly continuous.
Solution
1. Definition and well-definedness of the extension \(g\):Proof:
1.1. Let \(x \in \overline{A}\). There exists a sequence \((x_n)_{n=1}^\infty\) in \(A\) such that \(\lim_{n \to \infty} x_n = x\).
1.2. Since \((x_n)\) converges in \(X\), it is a Cauchy sequence in \(X\).
1.3. Because \(f\) is uniformly continuous, it preserves Cauchy sequences: for any \(\varepsilon > 0\), there exists \(\delta > 0\) such that \(d_X(a, b) < \delta \implies d_Y(f(a), f(b)) < \varepsilon\).
1.4. For large \(n, m\), \(d_X(x_n, x_m) < \delta\), so \(d_Y(f(x_n), f(x_m)) < \varepsilon\). Thus \((f(x_n))_{n=1}^\infty\) is Cauchy in \(Y\).
1.5. Completeness of \((Y, d_Y)\) ensures that \((f(x_n))\) converges to a unique limit \(L \in Y\). Define \(g(x) = \lim_{n \to \infty} f(x_n)\).
1.6. If \((x_n')\) is another sequence in \(A\) converging to \(x\), then \(d_X(x_n, x_n') \to 0\), so \(d_Y(f(x_n), f(x_n')) \to 0\) by uniform continuity. Hence \(\lim f(x_n) = \lim f(x_n')\), making \(g(x)\) well-defined.
1.7. For \(x \in A\), taking \(x_n = x\) gives \(g(x) = f(x)\), so \(g|_A = f\).
2. Uniform continuity of the extension \(g\):Proof:
2.1. Given \(\varepsilon > 0\), choose \(\delta > 0\) such that for all \(a, b \in A\), \(d_X(a, b) < \delta \implies d_Y(f(a), f(b)) < \varepsilon/3\).
2.2. Let \(x, y \in \overline{A}\) satisfy \(d_X(x, y) < \delta/3\).
2.3. Choose sequences \((x_n), (y_n)\) in \(A\) converging to \(x, y\) respectively.
2.4. For all sufficiently large \(n\), we have \(d_X(x_n, x) < \delta/3\), \(d_X(y_n, y) < \delta/3\), \(d_Y(f(x_n), g(x)) < \varepsilon/3\), and \(d_Y(f(y_n), g(y)) < \varepsilon/3\).
2.5. By the triangle inequality:\[d_X(x_n, y_n) \le d_X(x_n, x) + d_X(x, y) + d_X(y, y_n) < \frac{\delta}{3} + \frac{\delta}{3} + \frac{\delta}{3} = \delta.\]
2.6. Therefore \(d_Y(f(x_n), f(y_n)) < \varepsilon/3\).
2.7. Applying the triangle inequality in \(Y\):\[d_Y(g(x), g(y)) \le d_Y(g(x), f(x_n)) + d_Y(f(x_n), f(y_n)) + d_Y(f(y_n), g(y)) < \frac{\varepsilon}{3} + \frac{\varepsilon}{3} + \frac{\varepsilon}{3} = \varepsilon.\]
2.8. Thus \(g\) is uniformly continuous on \(\overline{A}\) (which also implies \(g\) is continuous).
3. Uniqueness of the continuous extension:Proof:
3.1. \(A\) is dense in \(\overline{A}\), and the metric codomain \(Y\) is Hausdorff.
3.2. Any two continuous functions on \(\overline{A}\) that agree on the dense subset \(A\) are identically equal on all of \(\overline{A}\).
4. Q.E.D.Steps 1, 2, and 3 give existence, uniform continuity, and uniqueness of the extension \(g\).