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This question already has an answer here:

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

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

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marked as duplicate by Simon S, Clement C., Leucippus, hardmath, ncmathsadist Nov 13 '15 at 2:20

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    $\begingroup$ Huh... I always thought $\sqrt 2\notin (0,1)$... $\endgroup$ – abiessu Nov 12 '15 at 22:17
  • $\begingroup$ @abiessu OP likely means $f(x) \equiv \sqrt{2}$ and that would be defined for all real $x$, including $x \in (0,1)$... $\endgroup$ – gt6989b Nov 12 '15 at 22:19
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    $\begingroup$ But $f(x) = x$ should work. $\endgroup$ – Brian Tung Nov 12 '15 at 22:21
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    $\begingroup$ Over $\mathbb{R}$, $f(x)=x$ is uniformly continuous but $f^2(x)=x^2$ is not. $\endgroup$ – JVV Nov 12 '15 at 22:22
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    $\begingroup$ Oops, yes, only works over $\mathbb{R}$. Over any finite interval, if $f$ is uniformly continuous, then $f^2$ is uniformly continuous. $\endgroup$ – Brian Tung Nov 12 '15 at 22:50
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$\sqrt{2}$ is a constant, not a function. If you mean $f(x) = \sqrt{2}$ for all values of $x$, it would be uniformly continuous.

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  • $\begingroup$ sorry should ve been square root of x $\endgroup$ – jessie Nov 12 '15 at 22:23
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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.

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  • $\begingroup$ 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 $f(x)=\tan(\pi x - \pi/2)$. $\endgroup$ – Ian Nov 12 '15 at 22:37
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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 $(0,1)$.

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