$f\colon(-1,1)\rightarrow \mathbb{R}$ 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!
$f\colon(-1,1)\rightarrow \mathbb{R}$ 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!
$F(x)=\tan(x)$ 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 $F$?
You're close: $\sin\frac{1}{x+1}$ is a counterexample to the statement.
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.
$\sin(x^2)$ is also a nice example and it's happening because it's not periodic.