Linked Questions

1 vote
3 answers
824 views

Show for $f$ defined on $\mathbb Q \cap [0,1]$ uniformly continuous, exists extension $g$ defined on $[0,1]$ that is continuous [duplicate]

Suppose that $f$ is a uniformly continuous function defined on $\mathbb Q \cap [0,1]$. Show that there is a unique continuous function $g$ defined on $[0,1]$ so that $$g(x)=f(x) \text{ for every } x \...
user5826's user avatar
  • 12k
1 vote
0 answers
30 views

Does the extension of uniformly continuous functions uniformly continuous? [duplicate]

I already know that if $E\subset\mathbb{R}$, $E_{1}$ is a dense subset of $E$, and there is a uniformly continuous function $f_{1}(x)$ on $E_{1}$, then there is a unique function $f(x)$ such that $f(x)...
mio's user avatar
  • 1,038
27 votes
3 answers
18k views

Extension of a Uniformly Continuous Function between Metric Spaces

Let $(X,d_X)$ and $(Y,d_Y)$ be metric spaces with $(Y,d_Y)$ 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 $\bar{A}$ ...
Clayton's user avatar
  • 24.7k
4 votes
4 answers
4k views

If $f : [0,1) \rightarrow \mathbb{R}$ is uniformly continous, $f$ is bounded

Prove or Disprove - a) If $f:[0,1) \rightarrow \mathbb{R}$ is uniformly continuous in it's domain, $f$ is bounded. b) If $f:[0,1] \rightarrow \mathbb{R}$ is uniformly continuous in it's ...
Yariv Levy's user avatar
  • 1,145
6 votes
1 answer
3k views

Extending a uniformly continous function to the closure of its domain

Suppose $X$ is a normal space and $f:X \rightarrow X$ is continous on X, and also uniformly continuous on a subset $A \subseteq X$. In this setting, can one conclude that f is uniformly continuous on ...
Mike's user avatar
  • 996
3 votes
1 answer
330 views

Obvious but unprovable: If $f'$ is the zero function on $\mathbb Q$, then $f$ is constant [duplicate]

Let $f:\mathbb Q\to\mathbb R$ be a uniformly continuous function and assume that $f'(x)=0$ for all $x\in\mathbb Q$. That $f$ is constant is obvious...and, as far as I can tell, unprovable. Please tell ...
Casper's user avatar
  • 1,024
0 votes
2 answers
565 views

Is it always possible extend a uniformly continuous function?

Suppose we have some bounded Borel set $B\subset {\mathbb R}^n$. Let a function $f:B\to \mathbb R$ be uniformly continuous on B in the sense that $$w(r):=\sup_{x,y\in B\ :\ |x-y|\le r}|f(x)-f(y)|<\...
Viktor's user avatar
  • 221
1 vote
1 answer
1k views

Let $r \geq 0$ and $f(x)=x^r \sin \frac{1}{x}$, for $x\neq 0$ and $f(0)=0$. Which values of $r$, $f(x)$ is uniformly continuous or differentiable?

Let $r\geq 0$ and $f(x)=x^r \sin \frac{1}{x}$, for $x\neq 0$ and $f(0)=0$. a) For which values of $r$, $f(x)$ is uniformly continuous on $(0,\pi)$? b) For which values of $r$, $f$ is ...
Ross's user avatar
  • 403
0 votes
1 answer
932 views

A uniformly continuous function can be extended on the boundary

Suppose $X$ a metric space, $Y$ a complete metric space and $f: S \rightarrow Y$ a uniformly continuous function from $S \subseteq X$ to $Y$. Prove that $f$ can be extended to a uniformly continuous ...
Giovanni Barbarani's user avatar
2 votes
1 answer
182 views

Trouble understanding uniform continuity

I recognize intuitively what it means for a function to be continuous (i.e. no jumps or breaks in the function), but the concept of being uniformly continuous seems to be over my head. I'm looking at ...
the_new_guy's user avatar
0 votes
1 answer
94 views

A property of complete metric spaces makes them length (path or inner) metric spaces, Clarification of a proof

In the book "Metric Structures for Riemannian and Non-Riemannian Spaces", by Misha Gromov, I found a proof of the following statement (of Theorem 1.8. restated here more concentrated) Let $(...
Physor's user avatar
  • 4,586