# *Unique* extension of a uniformly continuous function to the closure of its domain

Let $$(S,d_S)$$ and $$(T,d_T)$$ be two metric spaces with $$T$$ complete and suppose that a mapping $$f:A\subseteq S\to T$$ is uniformly continuous on its domain. I want to show that $$f$$ can be uniquely extended to the closure $$\overline{A}$$ with the extension uniformly continuous on $$\overline{A}$$.

Actually, this is a homework exercise from my introductory-analysis class, and before this exercise, several subsidiary questions had been deployed so I kind of cracked this question. For example, with every $$x\in\overline{A}$$, there comes a sequence $$\{x_n\}_{n=1}^\infty$$ in $$A$$ s.t. $$x_n\to x$$. This sequence is mapped to a Cauchy sequence $$\{f(x_n)\}_{n=1}^\infty$$ since $$f$$ is uniformly continuous on $$A$$. Then, thanks to completeness of $$T$$, we are assured of convergence of this Cauchy sequence, and hence able to designate the limit to be the value of a newborn $$g:\overline{A}\to T$$ at $$x$$. This $$g$$ will be the desired extension once we justify the following claims.

First, we have to show that $$g$$ is well-defined. By this, I mean another sequence $$\{x_n\}_{n=1}^\infty\subseteq A$$ converging to $$x$$ does not yield a different limit of $$\{f(x_n)\}_{n=1}^\infty$$. After establishing this claim, we need to show that $$g$$ is uniformly continuous on its domain, which requires employing uniform continuity of $$f$$. Last but not least, we have to show that $$g$$ is the restriction of $$f$$ to $$A$$. This can be done by recalling that $$f$$ is a continuous mapping.

So far, I've been able to resolve all the doubts about the claims mentioned above, but as indicated in the first paragraph, I fail to explain why a uniformly continuous extension of $$f$$ to $$\overline{A}$$ is unique. Does anyone have an idea? Thank you.

• Assume we have such an extension function $f$. If $a$ is a limit point of $A$ then choose any sequence $\{a_i\}_{i=1}^{\infty}$ of points in $A$ that converges to $a$. Then $f(a)$ must have the value $\lim_{n\rightarrow\infty} f(a_i)$ (and the limit must exist) else $f$ is not continuous. It is not possible to choose $f(a)$ to have any other value. There were no choices about $f(a_i)$ since the function was already defined at those points. Jul 15 at 0:41
• At most one continuous extension to the closure: requires only that $f$ be continuous. Exists at least one continuous extension: this follows from uniform continuity. Jul 15 at 1:27

Uniqueness (Alternative)

Suppose $$f_1, f_2$$ be two uniform extension of $$f$$ on $$\overline{A}$$

Then $$f_1=f=f_2$$ on $$A$$

Let $$D=\{x\in \overline{A} : f_1(x) =f_2(x) \}$$

Then clearly $$D$$ is a closed set i.e $$\overline{D}=D$$ (the set of all points where two continuous map agrees is a closed set, provided the target space is Hausdorff.)

$$A\subset D\subset \overline{A}$$

Then $$\overline{A}\subset\overline{ D}\subset \overline{A}$$

Implies $$\overline{A}\subset{ D}\subset \overline{A}$$

Hence $$D=\overline{A}$$ , implies $$f_1=f_2$$ on $$\overline{A}$$.

• Thank you. Since $\overline{A}$ is the smallest closed set that contains $A$, that $\overline{A}\subseteq S$ should be obvious. Jul 15 at 2:46
• Exactly. You are correct. Jul 15 at 2:51
• Your argument that $\mathcal{S}$ (to be distinguished from the metric space $(S,d_S)$) is closed in $S$ is new to me. Does that take a large amount of knowledge of topology? Jul 15 at 3:18
• I used two facts : 1) set of all points where two continuous functions (between two metric spaces) agree is closed set . 2) If $D \subset S$ is closed then $D\cap \overline{A}$ is closed in $(\overline{A}, d_{\overline{A}})$ Jul 15 at 3:55
• Since $\overline{A}$ is closed, for any $D\subset \overline{A}$ is closed in the subspace iff colsed in the whole space. Hence no need to worry about closed (doesn't matter where) set $D$ Jul 15 at 4:02

A brief note in the contra style.

Let $$(x_n)_n$$ and $$(y_n)_n$$ be any sequences in $$A,$$ both converging to $$a\in\overline A.$$ Let $$c_{2n}=x_n$$ and $$c_{2n+1}=y_n.$$ Then $$(c_n)_n$$ converges to $$a.$$

Suppose by contradiction that $$(f(c_n))_n$$ is not a Cauchy sequence in $$T.$$ Then for some $$r>0$$ we have $$\lim_{m\to\infty}\sup_{mr.$$ But $$\lim_{m\to\infty}\sup_{m So for every $$e>0$$ there exist $$n,n'$$ with $$d(c_n,c_{n'}) and $$d_T(f(c_n),f(c_{n'}))>r,$$ contrary to the uniform continuity of $$f$$.

Therefore $$(f(c_n))_n$$ is Cauchy in $$T.$$ By completeness of $$d_T,$$ therefore $$(f(c_n))_n$$ converges to some $$b\in T$$. For $$f$$ to be continuous at $$a$$ it is necessary that $$b=\lim_n f(c_n)=f(\lim_n c_n)=f(a).$$ And $$(f(x_n))_n ,\, (f(y_n))_n$$ are subsequences of $$(f(c_n))_n$$ so they both converge to $$b.$$

Owing to the suggestions of @Michael and @GEdgar, some thoughts came into my mind. I hope I didn't misunderstand them. Suppose there is another such extension $$h$$. We need to show that $$\forall x\in\overline{A}$$, $$g(x)=h(x)$$. Grab a sequence $$\{x_n\}_{n=1}^\infty$$ in $$A$$ that converges to $$x$$. Because uniform continuity implies continuity, we have $$\text{g(x_n)\to g(x) and h(x_n)\to h(x)}.$$ Now that both $$g$$ and $$h$$ restrict to $$f$$ on $$A$$, each pair of $$g(x_n)$$ and $$h(x_n)$$ agree. In fact, they are all equal to $$f(x_n)$$, which makes those two limits $$g(x)$$ and $$h(x)$$ identical. The work is done!