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Let $(X,d)$ and $(Y,d')$ be metric spaces with $(Y,d')$ complete. Let $A\subseteq X$. I need to show that if $f:A\to Y$ is uniformly continuous, then $f$ can be uniquely extended to $\bar{A}$ maintaining the uniform continuity.

My attempt at this has involved taking each point $a\in \bar{A}-A$ and forming a Cauchy sequence to it by considering open balls $B_{\frac{1}{n}}(a)-B_{\frac{1}{n+1}}(a)$ beginning with $n$ large enough so there is such a sequence, and defining $g(a)$ to be the limit in $Y$. The uniqueness seems to be obvious just by thinking about the uniqueness of limits (referring to the sequence in $Y$), but I have to admit I don't know how to rigorously show it. The uniform continuity seems natural, but I don't know how to show it, either.

This seems to be correct, but I'm not entirely sure... Any help would be very appreciated!

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3 Answers 3

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You may want to see the answers for this question, which answer yours, Extending a function by continuity from a dense subset of a space.

I built the proof myself based on Srivatsan's answer for that question. If anybody still needs it, here it goes:

Theorem

If $X$ and $Y$ are metric spaces and $f:S \to Y$ is uniformly continuous with $S$ dense in $X$, and $Y$ is complete, then there exists a unique continuous extension of $f$ in $\overline{S}$ which by the way is uniformly continuous.

Proof

Let $d$ and $D$ be the metrics of $X$ and $Y$ respectively.

Let $g:\overline{S} \to Y$ be given by $g(a) = \lim f(x_n)$, where $(x_n)$ is any sequence of points in $S$, with $x_n \to a$.

$g$ is well defined:

  • $\lim f(x_n)$ exists:

    Let $\varepsilon > 0$. Because of the uniform continuity of $f$, there exists $\delta>0$ such that for every $a,b \in S$, if $d(a,b) < \delta$, then $D(f(a),f(b)) < \varepsilon$.

    Since $x_n \to a$, $(x_n)$ is Cauchy, there exists $N \in \mathbb{Z}^{+}$ such that if $n,m \geq N$, $d(x_n,x_m)<\delta$.

    Hence, if $n,m \geq N$, $D(f(x_n),f(x_m))<\varepsilon$. Then $(f(x_n))$ is Cauchy, and since $Y$ is complete, $\lim f(x_n)$ exists.

  • If $x_n \to a$ and $y_n \to a$ then $\lim f(x_n) = \lim f(y_n)$:

    Let $(z_n) = (x_1,y_1,x_2,y_2,...)$. If $\varepsilon>0$, there exists $N \in \mathbb{Z}^{+}$ with $d(x_n,a) < \varepsilon$ and $d(y_n,a) < \varepsilon$ for each $n \geq N$.

    Consequently, if $n \geq 2N$, then $n/2,(n+1)/2 \geq N$ and so, if $n$ is even, $d(z_n,a) = d(y_{n/2},a) < \varepsilon$, and if $n$ is odd, $d(z_n,a) = d(y_{(n+1)/2},a) < \varepsilon$. Therefore $z_n \to a$.

    So, $\lim f(z_n)$ exists and since $(f(x_n))$ and $(f(y_n))$ are subsequences of $(f(z_n))$, $\lim f(x_n) = \lim f(z_n) = \lim f(y_n)$.

$g$ is an extension of $f$:

  • If $a \in S$, $a \to a$, therefore $g(a) = \lim f(a) = f(a)$.

$g$ is uniformly continuous:

  • Let $\varepsilon > 0$. Since $f$ is uniformly continuous, there exists $\delta > 0$ such that $D(f(a),f(b))<\varepsilon/3$ for every $a,b \in S$ with $d(a,b)<\delta$.

    Let $a,b \in \overline{S}$ with $d(a,b)<\delta/3$.

    There exist sequences in $S$, $(x_n)$ and $(y_n)$ with $x_n \to a$ and $y_n \to b$. Since $x_n \to a$ and $y_n \to b$, there exists $N_1 \in \mathbb{Z}^{+}$ with $d(x_n,a)<\delta/3$ and $d(y_n,b)<\delta/3$ for every $n\geq N_1$.

    If $n \geq N_1$, $d(x_n,y_n) \leq d(x_n,a) + d(a,b) + d(b,y_n) < \delta$ and so, $D(f(x_n),f(y_n)) < \varepsilon/3$.

    Also, since $f(x_n) \to g(a)$ and $f(y_n) \to g(b)$, there exists $N_2 \in \mathbb{Z}^{+}$ with $D(f(x_n),g(a))<\varepsilon/3$ and $D(f(y_n),g(b))<\varepsilon/3$ for every $n\geq N_2$.

    Then, if $N=max\{N_1,N_2\}$, $D(g(a),g(b)) \leq D(g(a),f(x_N)) + D(f(x_N),f(y_N)) + D(f(y_N),g(b)) < \varepsilon.$

$g$ is unique:

  • If $h$ is a continuous extension of $f$ in $\overline{S}$ and $a\in \overline{S}$, there exists a sequence $(x_n)$ in $S$ with $x_n \to a$. Since $h$ is continuous, $h(x_n) \to h(a)$. But $(h(x_n)) = (f(x_n))$ and $f(x_n) \to g(a)$, then $h(a) = g(a)$ must hold.
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  • $\begingroup$ When proving that $\lim f(x_n)$ exists, you need a sequence $(x_n)$ converging to $a$. AFAIK, constructing it requires $a\in\overline{S}$ and $d$. $\endgroup$
    – beroal
    May 13, 2016 at 14:53
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If $ a \in \overline{A} $ then $ a = \lim_n a_n $ where $ a _n \in A $. Then, $ a_n $ is Cauchy and as $ f $ is uniformily continuous n $ A $, $ f(a_n) $ is Cauchy and as $ (Y,d´) $ is complete you can define $ f(a) : = \lim_n f(a_n) $. For example, if $ b \in A $ you have that \begin{eqnarray} d(f(a),f(b)) &\le& d(f(a_n),f(b)) + d(f(a),f(a_n)) \end{eqnarray} and by the definition of uniform continuity it is clear that the extension of $f$ is uniformily continuous. Analogously, if $ b \in \overline{A}, b =\lim_n b_n $ where $ b_n \in \overline{A} $ and \begin{eqnarray} d(f(a),f(b)) &\le& d(f(a_n),f(a)) + d(f(b_n),f(a_n)) + d(f(b_n), f(b)) \end{eqnarray}

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[This is, except for a few details, same as Carlos Pinzon's answer above]

Let ${ X,Y }$ be metric spaces and ${ A (\subseteq X ) \overset{f}{\to} Y }$ a continuous map. Both ${ d _X, d _Y }$ will be denoted by ${ d }$ for brevity.

The following question arises. Can we impose a "general enough" constraint such that : There is a unique continuous map ${ \overline{A} \overset{\overline{f}}{\to} Y }$ with the property ${ \overline{f} \vert _{A} = f }$ ?

Turns out imposing that ${ f }$ is uniformly continuous and ${ Y }$ is complete will do. Further, in this case ${ \overline{f} }$ is uniformly continuous.


${ \underline{ \textbf{Defining} \text{ } \overline{f} } }$

For every ${ p \in \overline{A}, }$ there exists a seq ${ (x _n) \subseteq A }$ with ${ d(x _n, p) \to 0 }.$

Let ${ p \in \overline{A} }.$ For every ${ n \in \mathbb{Z} _{\gt 0} },$ pick an ${ x _n \in A }$ with ${ d(x _n, p) \lt \frac{1}{n} }.$

So if we ensure that

${ {\color{purple}{(1)}} }$ If ${ (x _n) \subseteq A }$ and ${ d(x _n, p) \to 0 }$ for some ${ p \in X }$ then ${ \lim _{n \to \infty} f(x _n) }$ exists
${ {\color{purple}{(2)} } }$ If ${ \lbrace (x _n) \subseteq A; d(x _n, p) \to 0 \rbrace }$ and ${ \lbrace (y _n) \subseteq A; d(y _n, p) \to 0 \rbrace }$ for some ${ p \in X }$ then ${ \lim _{n \to \infty} f(x _n) = \lim _{n \to \infty} f(y _n) }$

then we can define a map ${ \overline{A} \overset{\overline{f}}{\to} Y }$ naturally by : Let ${ p \in \overline{A} }.$ Pick an ${ (x _n) \subseteq A }$ with ${ d(x _n, p) \to 0 },$ and set ${ \overline{f}(p) := \lim _{n \to \infty} f(x _n) }.$

Ensuring ${ {\color{purple}{(1)} } }$: Let ${ (x _n) \subseteq A }$ and ${ d(x _n, p) \to 0 }$ for some ${ p \in X }.$ So ${ (x _n) }$ is Cauchy. So imposing that ${ f }$ is uniformly continuous ensures ${ (f(x _n)) }$ is Cauchy.

Let ${ \epsilon \gt 0 }.$ Pick ${ \delta \gt 0 }$ such that ${ x,y \in A },$ ${ d(x,y) \lt \delta }$ implies ${ d(f(x), f(y)) \lt \epsilon }.$ Pick ${ N }$ such that ${ d(x _m, x _n) \lt \delta }$ whenever ${ m, n \geq N }.$ Now ${ d(f(x _m), f(x _n)) \lt \epsilon }$ whenever ${ m, n \geq N }.$

Further imposing that ${ Y }$ is complete ensures ${ (f (x _n)) }$ is convergent.

Ensuring ${ {\color{purple}{(2)} } }$: Say ${ f }$ is uniformly continuous and ${ Y }$ is complete, which ensures ${ {\color{purple}{(1)} } }.$ Let ${ (x _n), (y _n) \subseteq A }$ with ${ d(x _n, p) \to 0 },$ ${ d(y _n, p) \to 0 }$ for some ${ p \in X }.$ So ${ \lim _{n \to \infty} f(x _n) = \ell _1 }$ and ${ \lim _{n \to \infty} f(y _n) = \ell _2 }$ exist. We'll show ${ \ell _1 = \ell _2 }.$
As ${ d(\ell _1, \ell _2) }$ ${ \leq \underbrace{ d(\ell _1, f(x _n)) } _{\to 0} }$ ${ + d(f (x _n), f(y _n)) }$ ${ + \underbrace{ d(f(y _n), \ell _2) } _{\to 0} },$ it suffices to show ${ d(f(x _n), f(y _n)) \to 0 }.$

Let ${ \epsilon \gt 0 }.$ Pick ${ \delta \gt 0 }$ such that ${ x,y \in A, }$ ${ d(x,y) \lt \delta }$ implies ${ d(f(x), f(y)) \lt \epsilon }.$
From ${ d(x _n, y _n) }$ ${ \leq d(x _n, p) + d(p, y _n) \to 0 },$ we have ${ d(x _n, y _n) \to 0 }.$ So pick ${ N }$ such that ${ d(x _n, y _n ) \lt \delta }$ for all ${ n \geq N }.$ Now ${ d(f(x _n), f(y _n)) \lt \epsilon }$ for all ${ n \geq N },$ as needed.


${ \underline{ \textbf{Properties of} \text{ } \overline{f} } }$

From now, ${ A \overset{f}{\to} Y }$ is uniformly continuous and ${ Y }$ is complete, and ${ \overline{A} \overset{\overline{f}}{\to} Y }$ is as defined above.

Note ${ \overline{f} \vert _{A} = f }.$ We'll show ${ \overline{f} }$ is uniformly continuous.
Let ${ \epsilon \gt 0 },$ and ${ p, q \in \overline{A} }.$ We want a ${ \delta \gt 0 }$ independent of ${ p,q },$ such that ${ d(p,q) \lt \delta }$ implies ${ d(\overline{f}(p), \overline{f}(q) ) \lt \epsilon }.$

Pick ${ \eta \gt 0 }$ such that ${ x,y \in A, d(x,y) \lt \eta }$ implies ${ d(f(x), f(y)) \lt \frac{\epsilon}{10} }.$
We'll show ${ \delta := \frac{1}{10} \min\lbrace \epsilon, \eta \rbrace }$ will work.

Suppose ${ {\color{green}{d(p,q) \lt \delta}} }.$
Pick seqs ${ (x _n), (y _n) \subseteq A }$ with ${ d(x _n, p) \to 0 },$ ${ d(y _n, q) \to 0 }.$ Now ${ d(f(x _n), \overline{f}(p)) \to 0 }$ and ${ d(f(y _n), \overline{f}(q)) \to 0 }$ too.
So pick an ${ m }$ such that ${ {\color{green}{d(x _m, p)}} },$ ${ {\color{green}{d(y _m, q)}} },$ ${ {\color{green}{d(f(x _m), \overline{f}(p))}} }$ and ${ {\color{green}{d(f(y _m), \overline{f}(q))}} }$ are all ${ {\color{green}{\lt \delta}} }.$
Now ${ d(\overline{f}(p), \overline{f}(q)) }$ ${ \leq d(\overline{f}(p), f(x _m)) }$ ${ + d(f(x _m), f(y _m)) }$ ${ + d(f (y _m), \overline{f}(q)) }$ ${ \leq 2\delta + {\color{red}{d(f(x _m), f(y _m))}} . }$
But as ${ d(x _m, y _m) }$ ${ \leq d(x _m, p) }$ ${ + d(p,q) }$ ${ + d(q, y _m) }$ ${ \lt 3 \delta \lt \eta },$ we have ${ {\color{red}{d(f (x _m), f(y _m)) \lt \frac{\epsilon}{10}}} }.$ This gives ${ d(\overline{f}(p), \overline{f}(q)) }$ ${ \leq 2\delta + \frac{\epsilon}{10} }$ ${ \lt \epsilon }.$
Finally ${ d(\overline{f}(p), \overline{f}(q)) \lt \epsilon },$ as needed.

[This shows we could've set ${ \delta := \frac{1}{3} \min \lbrace \epsilon, \eta \rbrace }$ to begin with. But ${ \frac{1}{10} }$ was a "safer factor" to carry out the estimates]

[Uniqueness] Suppose ${ \overline{A} \overset{g}{\to} Y }$ is continuous and ${ g \vert _{A} = f }.$ We see ${ g = \overline{f} }$ :
Let ${ p \in \overline{A}. }$ Pick a seq ${ (x _n) \subseteq A }$ with ${ d(x _n, p) \to 0 }.$ As ${ g }$ is continuous, ${ g(p) = \lim _{n \to \infty} g(x _n) }.$ But ${ g(x _n) = f(x _n) }$ and ${ \overline{f}(p) = \lim _{n \to \infty} f(x _n). }$ So ${ \overline{f}(p) = g(p) },$ as needed.

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