如何证明满足一致连续条件的映射f可延拓为X闭包上的连续函数？

First, let's clarify a key necessary assumption for the statement to hold: Y must be a complete metric space. Without completeness, the result fails (we’ll give a concrete example later). Assuming Y is complete, here’s a step-by-step proof:

Step 1: Define the extension $g$ on $\overline{X}$ #

Take any point $a \in \overline{X}$. By definition of closure, there exists a sequence ${x_n} \subseteq X$ such that $x_n \to a$ in $\mathbb{R}^2$.

Since convergent sequences in $\mathbb{R}^2$ are bounded, the set $B = {x_n \mid n \in \mathbb{N}}$ is a bounded subset of $X$. By the problem’s condition, $f$ is uniformly continuous on $B$. Uniform continuity preserves Cauchy sequences: since ${x_n}$ is Cauchy (it converges), ${f(x_n)}$ must be Cauchy in $Y$.

Because Y is complete, Cauchy sequences converge. Let $g(a) = \lim_{n \to \infty} f(x_n)$.

Step 2: Show $g$ is well-defined #

We need to confirm $g(a)$ doesn’t depend on the choice of sequence ${x_n}$. Suppose ${y_n} \subseteq X$ is another sequence with $y_n \to a$.

The union ${x_n} \cup {y_n}$ is bounded (both sequences are bounded), so $f$ is uniformly continuous on this set. For any $\varepsilon > 0$, there exists $\delta > 0$ such that $d(x,y) < \delta$ implies $d(f(x),f(y)) < \varepsilon$. Since $x_n \to a$ and $y_n \to a$, for large enough $n$, $d(x_n, y_n) < \delta$, so $d(f(x_n), f(y_n)) < \varepsilon$. Taking the limit as $n \to \infty$, the limits of ${f(x_n)}$ and ${f(y_n)}$ must be identical. Thus $g(a)$ is well-defined.

Step 3: Verify $g$ extends $f$ #

For any $a \in X$, take the constant sequence $x_n = a$ for all $n$. Then $g(a) = \lim_{n \to \infty} f(a) = f(a)$, so $g|_X = f$.

Step 4: Prove $g$ is continuous on $\overline{X}$ #

Let $a \in \overline{X}$ and $\varepsilon > 0$. We need to find $\delta > 0$ such that for all $b \in \overline{X}$ with $d(b,a) < \delta$, $d(g(b), g(a)) < \varepsilon$.

1. Choose a sequence ${x_n} \subseteq X$ with $x_n \to a$. The set $B = {x_n}$ is bounded, so $f$ is uniformly continuous on $B$. Let $\delta_1 > 0$ be the constant from uniform continuity for $\varepsilon/2$: if $x,y \in B$ and $d(x,y) < \delta_1$, then $d(f(x),f(y)) < \varepsilon/2$.
2. Pick $\delta = \delta_1/2$. For any $b \in \overline{X}$ with $d(b,a) < \delta$, take a sequence ${y_n} \subseteq X$ with $y_n \to b$. For large enough $n$, $d(x_n, a) < \delta/2$ and $d(y_n, b) < \delta/2$, so:
   $$d(x_n, y_n) \leq d(x_n,a) + d(a,b) + d(b,y_n) < \delta/2 + \delta + \delta/2 = 2\delta = \delta_1$$
   Thus $d(f(x_n), f(y_n)) < \varepsilon/2$. Taking the limit as $n \to \infty$, we get $d(g(a), g(b)) \leq \varepsilon/2 < \varepsilon$.

This proves $g$ is continuous at $a$, and since $a$ was arbitrary, $g$ is continuous on $\overline{X}$.

Addressing Your Proposed Approach #

Your idea of extending locally via bounded balls around points in $X$ is intuitive, but it has practical challenges:

• Patching together local extensions across overlapping balls requires ensuring consistency (i.e., the extension at a point in the intersection of two balls must match from both extensions). This is possible but more cumbersome than the sequence-based method.
• The sequence approach directly leverages the definition of closure and uniform continuity on bounded sets to avoid dealing with local patches, making the proof cleaner and more straightforward.

Note on Completeness of $Y$ #

Here’s an example where the statement fails if $Y$ is not complete:

• Let $X = (0,1) \subseteq \mathbb{R}^2$, $Y = (0,1)$ (with the usual Euclidean metric), and $f(x) = x$.
• $f$ is uniformly continuous on every bounded subset of $X$ (in fact, uniformly continuous on all of $X$).
• The closure of $X$ is $[0,1]$, but there’s no continuous map $g: [0,1] \to Y$ extending $f$, since $g(0)$ would have to be $0$ (not in $Y$) and $g(1)$ would have to be $1$ (also not in $Y$).

内容的提问来源于stack exchange，提问作者Eric