How Can You Prove Uniform Continuity on a Closed Interval?

[steven187]

hello all

i have been working on this problem, see i can see how it could be true but i don't know how to prove it,would anybody have any ideas on how to prove this ?

let [a,b] be a closed interval in R and f:[a,b]->R be a continuous function. prove that f is uniformly continuous.

please help

thanxs

 

[fourier jr]

tell us the definitions of uniform continuity & continuity

 

[matt grime]

take the negation of "is uniformly continuous" what can you do with that?

 

[steven187]

well the definition of continuity is

let X be a subset of R and f:X->R and xo is an element of X then f is said to be continuous if given epsilon >o there exists a delta>0 such that
x an element of X, |x-xo|<delta =>|f(x)-f(xo)|<epsilon

and for uniform continuity

let X be a subset of R and f:X->R and x,y is an element of X then f is said to be uniformly continuous if given epsilon >o there exists a delta>0 such that
x,y an element of X, |y-x|<delta =>|f(y)-f(x)|<epsilon

they are the definitions that i understand but i don't know how to apply them for this question, what do i do?

 

[steven187]

take the negation of "is uniformly continuous" what can you do with that?

hello Matt

well when you say negation do you mean proving that
|y-x|<delta =>|f(y)-f(x)|>epsilon

 

[matt grime]

no, i mean the negation of the statement f is uniformly continuous if...

the negation of A => B is not A=>not(B), is it? it is A and not B

 

[fourier jr]

edit: never mind i got confused

 

[steven187]

well i think your saying that since uniform continuity implies continuity
then the negation is that uniform continuity implies that it is not continuous,
i still can't see the link?

 

[matt grime]

eh? write down the full statement of "f is uniformly continuous if..." now take its negation, thus if f is not uniformly continuous what can you deduce? how about there is an e>0 such that for all n there are elements x(n) and y(n) such that

|x(n)-y(n)| <1/n and |f(x(n)) - f(y(n))| >e now, passing to a subsequence we may assume that the x(n) and y(n) are all distinct, since [a,b] is closed and bounded they must converge to x and y, but |x-y| < 1/n for all n hence x=y, but then f(x(n)) and f(y(n)) should converge to the same thing (f is continuous), but they can't since they are always at least e apart.


this seems long and unwieldy , so what was the "proper" proof?

 

[HallsofIvy]

The difference between "continuity" and "uniform continuity" is that, for f continuous at a point c in some set, the &delta; required to get "|f(x)-f(c)|<&epsilon; whenever |x-c|< &delta;" may depend on c. Uniform continuity, on a set (and uniform continuity is not defined at individual points), the &delta; for a given &epsilon; cannot depend on c.

Okay, if f is continuous on the compact set [a,b], then there exist a &delta; for each point in [a,b]. Show that that set of &deltas; is also compact and so contains a smallest &delta;. What happens if you choose that delta at each point?

 

[matt grime]

is there a nice proof that doesn't directly invoke compactness (which is obviously the 'proper' proof)?

 

[steven187]

hello all

well Matt this the way I ended up doing it and I did use compactness, hopefully it should be correct

choose r>0 such that for every delta>0 there exists x,y an element of[a,B] such that
|x-y|<delta and |f(x)-f(y)|>=r (my assumption) then
for each n an element of N choose xn,yn an element of [a,b] such that
|xn-yn|<1/n, |f(xn)-f(yn)|>=r,
since [a,b] is compact {xn} has a convergent subsequence {xnk}, we let xnk->x since xnk-ynk is a subsequence of
{xn-yn} and xn-yn->0 so xnk-ynk->0
it follows that
ynk=xnk-(xnk-ynk)->x-0=x,
but f is continuous at x. so
f(xnk)->f(x) and f(ynk)->f(x) thus f(xnk)-f(ynk)->0,
but for every k an element of N
|f(xnk)-f(ynk)|>=r>0

and so this contradicts my assumption and so the opposite must be true, would this be correct? it took me ages to get this result, and no I couldn't think of any nice way of directly proving this, but i would definitely love to see one? muchly appreciated if some one can help us there thanxs

Steven

 

[matt grime]

you didn't use compactness per se, you used the bolzano weierstrass property, which is weaker (aka sequential compactness). however they are equivalent for metric spaces.

if you know about compactness then a simple direct proof is available and doesn't require this messiness.

given e>0 for each x in [a,b] pick d(x) such that |x-y|< d(x) => |f(x)-f(y)| <e, the balls B(x,d(x)) are an open cover - pick a finite subcover and take d to be the min of the d(x)'s of the finite subcover.

 

[steven187]

hello Matt

I have to admit that didnt look so messy, but the thing is it seems like its way beyond my level,I have only researched into all of one dimensional theory of analysis, would there be any other direct proof based on my level? also I wanted to ask, see with what I already know I can use my imagination to analyse it, would that be possible with multiple dimensional theory of analysis? as I will be looking forward to reseach into it, and any advice on where to start would be great?

thanxs

 

[mathwonk]

there is an approach via the lebesgue covering number. i.e. given any open cover of a closed setk, the glb of the distances if any point from the outside of the open seys of the cover is a continuous function, hence has a minimum on a closed bounded interval.

But since the cioollection of open sets coivers the minimum must be positive, so there is some number d such that every point is further than d inside some open set of the cover.

thus if f is continuous on [a.b] and we let e be given, and choose for each point x of [a,b] an open set on which the variation of f is less than e, then there is a d>0 such that every interval of lenbgth d is fully inside one fo these open intervals. Thus if we considert any two, points of [a,b] which are within d of each other, then their f values are hopefully within e of each other. is this right? i am perhaps not totally with it at the moment.