Is $\sqrt{x}$ uniformly continuous on $(0,1)$?

Question

And can anybody give me a function that is uniformly continuous on some interval but $f^2$ is not?

@abiessu OP likely means $f(x) \equiv \sqrt{2}$ and that would be defined for all real $x$, including $x \in (0,1)$... — gt6989b, Nov 12 '15 at 22:19
These questions are not related enough; they should probably be asked in two separate posts. — Brian Tung, Nov 12 '15 at 22:20
Over $\mathbb{R}$, $f(x)=x$ is uniformly continuous but $f^2(x)=x^2$ is not. — MrMazgari, Nov 12 '15 at 22:22
Oops, yes, only works over $\mathbb{R}$. Over any finite interval, if $f$ is uniformly continuous, then $f^2$ is uniformly continuous. — Brian Tung, Nov 12 '15 at 22:50

score 1 · Answer 1 · answered Nov 12 '15 at 22:19

1

$\sqrt{2}$ is a constant, not a function. If you mean $f(x) = \sqrt{2}$ for all values of $x$, it would be uniformly continuous.

answered Nov 12 '15 at 22:19

gt6989b

54,422

sorry should ve been square root of x – jessie Nov 12 '15 at 22:23

Idris Addou · Answer 2 · 2015-11-13T00:19:47.560

1

Weierstrass theorem. Every continuous function is uniformly continuous over any compact set. So constant functions are continuous then they are uniformly over any compact , particularly over 0,1

EDIT1: Particularly over [0,1] and therefore over any subset of [0,1], for example (0,1).

EDIT2: I am sorry to credit this theorem to Weierstrass.

edited Nov 13 '15 at 00:19

answered Nov 12 '15 at 22:32

Idris Addou

4,193

I have heard this called Heine-Cantor, and I have heard it unnamed (usually mentioned shortly after proving Heine-Borel, since there is a very elegant proof using topological compactness). I have never heard it attributed to Weierstrass. Also, a function can be continuous on $(0,1)$ but not extensible to $[0,1]$, in which case Heine-Cantor is not applicable. For instance $f(x)=\tan(\pi x - \pi/2)$. – Ian Nov 12 '15 at 22:37

score 0 · Answer 3 · edited Apr 13 '17 at 12:21

JVV has given an example that fails on $\Bbb R$.

But on $(0,1)$ (or any bounded interval), which is totally bounded, a uniformly continuous function $f$ can be extended to a continuous function $\tilde f$ on $[0,1]$ (see this question). In particular, $\tilde f$ is continuous, and so is $\tilde f^2$. And a continuous function on a compact is uniformly continuous (Heine-Cantor theorem), hence $\tilde f^2$ is uniformy continuous, and so is $f^2$, on $(0,1)$.

Is $\sqrt{x}$ uniformly continuous on $(0,1)$?

3 Answers3