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$f\colon(-1,1)\rightarrow \mathbb{R}$ is bounded and continuous does it mean that $f$ is uniformly continuous?

Well, $f(x)=x\sin(1/x)$ does the job for counterexample? Please help!

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For continuity to lead to uniform continuity, domain has to be compact, and as you can see the domain is not compact here. Also, rightly $f(x)=\sin(\frac{1}{x+1}) $ serves as a counterexample or even $ \sin(e^x)$ for that matter.

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    $\begingroup$ $\sin(e^x)$ is uniformly continuous on $(-1,1)$. $\endgroup$ – commenter Oct 25 '12 at 11:17
  • $\begingroup$ Yes it is. Thanks. Don't know what I was thinking. $\endgroup$ – Vishesh Oct 25 '12 at 11:18
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You're close: $$\sin\frac{1}{x+1}$$ is a counterexample to the statement.

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$\sin(x^2)$ is also a nice example and it's happening because it's not periodic.

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    $\begingroup$ This function is continuous on $[-1,1]$, so it is uniformly continuous there. A fortiori on $(-1,1)$. $\endgroup$ – Julien Apr 25 '13 at 16:37
  • $\begingroup$ Over the real line, this is, not on any bounded interval. $\endgroup$ – Pedro Tamaroff Apr 25 '13 at 16:39
  • $\begingroup$ This answer is wrong. Although its a nice counterexample to show that a bounded function on a closed, but unbounded interval is not necessarily uniformly continuous. $\endgroup$ – jodag Feb 28 '18 at 0:13
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$F(x)=\tan(x)$ with domain: $(\frac{\pi}{2},\frac{\pi}{2})$ .

For example, take $F(x)=\tan(\frac{\pi x}{2}):\text{ where, dom(F)}= (-1, 1)$ .

A 'necessary condition' (not sufficient) condition for continuous $f$ on $(-1,1)$ to be non- uniformly continuous over $(-1, 1)\,\text{ is}\, $(A)$:

$(A)$: the extension of $f: F$, where: $$\text{dom(f)}= (-1,1):\, \text{dom(F)}=[-1,1];\text{where}\,F=G\text{ on}\, (-1,1)\text{but}, \,F \text{is non-continuous at one the end points of the extended domain}.$$

But $F$ is non-continuous at one of the end points of the extended domain: $$[-1, 1]: -1\,\text{, or}\, 1$$.

Otherwise,if $F$ were continuous at the end points, $F$ would be continuous over $[-1,1]$, given $F=G$ on $(-1,1)$ .

And by hypothesis of the question, $f$ is continuous on $(-1,1)$ . Thus, the extension of $f$, $F$ is continuous on $(-1,1)$ ( say this tentatively)and would also be continuous at ${-1}\land \text{at}\, {1}$ .

Thus, $F$ would be continuous on $[-1,1]$.

As $[-1,1]$ on a closed compact subset of the reals. Then $F$, however, would be uniformly continuous, over $[-1,1]$, by the Heine-Borel theorem.

And thus , $F$ is uniformly continuous on $(-1,1)$.

Presumably $f\,\text{on}\, (-1,1)$ would be, as a result of this , uniformly continuous as well. As: $F=f\text{on}\,(-1,1)$. However, I stress that non-continuity of $F$ or the extension of $f$ to the closed domain,$[-1,1]$ at the end points $-1, 1$ :

At least I think so. I say I think so, relates again the query about uniform continuity below under different designations :

$(1)$ As,complete uniform continuity of the restricted function $f$ on the entirety of f's domain: $(-1,1)$ on the one hand.

And $(2)$: The uniform continuity of the extension /unrestricted function, $F$, on part of its domain: $(-1,1)$.

$$F: \text{F@}(-1,1),\,(-1,1)\subset[-1,1]= \text{Dom(F)}$$

As ,uniform continuity is a a global property of the function and its domain. There may be difference due to a technicality in the designation of the name 'evaluating the uniform continuity" of a restriction $f$:

$$(1)\, \text{where uniform continuity of the function on} \,(-1,1)\, text{as}\, f\text{ rather than} \,F\,\text{ where dom(F)}=(-1,1)\neq =[-1,1]\,\text{dom(F), onn}\, (-1,1)\subsetneq\[-1,1]=\text{dom(F)}$$.

$$ \text{where this is evaluated as the uniform continuity of} \,f\text{on f's entire entire domain}\land f\text{is undefined}@ 1,\land -1$$ .

$$\text{Where: dom(f)}=(-1,1)\subset[-1,1]=\text{dom(F) }$$.

on the one hand, and $(2)$:

$$(2):\text{the uniform continuity of F on a sub part of F's domain}:(-1,1)\, \subsetneq [-1,1]=\text{dom(F)}.$$

$$\text{despite} \,F=f \, \text{on:} (-1,1)$$? .

Where one evaluates uniform continuity : $$\text{of}\, F\,\text{on the open interval:} \,(-1,1) \subset\,\text{dom(F)}?$$ .

That is, must one keep in mind: $$\text{that in (2) unlike\, (1) that we are considering F and :}\,(-1,1)\subsetneq[-1,1] \text{dom(F)}$$.

Where ,$F$ ,is the function, under consideration, not $f$ .when evaluating global properties on a subset of its domain, like uniform continuity @ $(-1,1)\text{ of F}.$?

$$\text{Where in} (2)\,\text{the entire domain of F:dom(F)}=[-1,1]\, ;\text{where} \,[-1,1]\neq( -1,1)=\text{dom(f) unlike in (1)}.$$?

This being ,in contrast to $(1)$. Where we consider uniform continuity of $f@(-1,1)$, under the aspect of the restriction $f$:

Namely, uniform continuity of $f$ over $f$'s entire domain, as $f$'s entire domain is $(-1,1)$?.

Is there a difference, between $(1)$ and $(2)$ between the unfiorom contunity of an entirey function as a restriction and between uniform continuity of the unrestricted function on a subpart of its domain, where the bounded interval of interest is the same )-1,1) for example, in both cases and $f=F \text{on} (-,1)$ Or is this philosophical pedantantry.

That is without restricting the domain of $F: =[-1,1]$ to $(-1,1)$, call the restriction $f$ and consider whether $f$ is completely uniformly continuous on its entire domain, where the restriction is undefined at the end points.

$$F @ (-1,1)\text{, where}\, (-1,1)\subsetneq[-1,1]=\text{dom(F)}.$$

Whilst Bearing in mind that, $F$ is defined at the end points, unlike $f$. Or, rather whilst bearing in mind that,$(-1,1)$ is only a sub-part of $\text{dom(F)}=[-1,1]$.

As I believe that uniform continuity (at a point,if it makes sense) depends not merely on the point of interest, but is evaluated globally on the properties and depend of the entire function over its entire domain?

Would be a necessary condition to finding a candidate, of:

'non uniformly continuous, continuous function $f\text{where} \text{dom}=(-1,1)$.' I do not think that it is sufficient conditions. One merely needs to consider:

$G(x)=x\text{on} (-1,1); G(-1)=-10\land G(10)=10$ ,constituting a failure of continuity.

That is, some kind of end-point, dis-continuity .Whether this can be really considered a jump discontinuity or a removable discontinuity or something else, is hard to say, given that each end-point only has a single one side limit, whether continuous or not.

In any case, the function function value of $\text{ at}\, 1, -1\neq\text{ the appropriate one-sided limits of f from the right at}-1\,\land \text{the left at} \, 1$.

That is, at the end points of the domain: $$ [1,1]\,G(1)=10\neq \lim_{x\to 1_{-1}}=1$$.

Whilst, the restriction, $F$, of the extension, $G$ ,where $F:F=G(-1,1)\land \text{ F is defined only on:} (-1,1):

$$F:\forall(x\in(-1,1)):F(x)=x$$.

$$ \text{where}\,, F\text{ is the identity function }\land \text{dmo(F)}=(-1,1).$$

And $F$ , the identity function, is clearly uniformly continuous unlike $G$ . I presume, however that $G:[-1,-1]\text{to, Im(G)}$ ,**may not technically qualify.

I am not sure as uniformly continuous on $(-1,1)$** despite being identical to $F$ on this open interval.

As uniform continuity is a global property unlike point-wise continuity, and the domain of $G$ so defined is: $[-1,1]$.

I am unclear about this (it might sound like pedantry).

That is despite appearances, I am not sure if one can completely ignore the title given to the function, $G$ or $F$, when considering whether the function so denoted, is uniformly continuous on $(-1,1)$ .

That is will there be a difference, due to some technicality in terms, depending on whether we are considering uniform continuity of this function on the open interval $(-1,1)$ as $G$ rather than $F$?

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