$f\colon(-1,1)\rightarrow \mathbb{R}$ 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!


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

You're close: $$\sin\frac{1}{x+1}$$ is a counterexample to the statement.


$\sin(x^2)$ 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

$F(x)=\tan(x)$ 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):


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

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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