It is the case that unbounded but continuous functions (such as X ==> 1/X or X ==> X[sup]2[/sup]) are not uniformly continuous.
On the other hand, is it true that all continuous bounded functions are uniformly continuous? .
In other words, what is the difference, if any, between a continuous bounded function and a uniformly continuous function?
Thanks!
F(x)=x is unbounded and uniformly continuous. I think any unbounded, continuous function with a bounded derivative will be uniformly continuous.
I believe the answer to your question is no if I recall the definitions correctly. The function sin(1/x) is not uniformly continuous since it takes on all values between -1 and 1 infinitely often as x -> 0. But it’s certainly bounded.
However, it’s not continuous at zero. It is uniformly continuous on any finite length interval which does not contain the origin by the Heine-Cantor theorem, which states that any function which is continuous on a finite length interval is uniformly continuous on that interval. If you want a continuous function to be uniformly continuous on an infinite length interval, you need something else. Boundedness is a sufficient condition, but it’s not necessary, as Lance Turbo notes.
Note that the concept of uniform continuity is essentially about comparing the ‘degree of continuity’ in various points. Changing the points under consideration will make a big difference. In other words, when giving an example, you should specify both the formula and the domain of the function.
f(x) = x[sup]2[/sup] is, as you note, not uniformly continuous as a function from ℝ to ℝ but, as a function from, say, the closed interval [2,4] to ℝ, it is uniformly continuous. Your other example can also easily been made into a uniformly continuous function by specifying a proper domain.
This has no direct bearing on your question though. By restricting f(x) = x[sup]2[/sup] to the domain [2,4] it has also become bounded and hence an example of a bounded continuous function that is uniformly continuous. Still, it’s good to understand the importance of the domain.
I realize you’re trying to avoid the technical term compact here, but it’s not working out. Indeed, boundedness is not a sufficient condition for a continuous function on a infinite length interval to be uniformly continuous, but neither is it a sufficient condition on a finite length interval.
OldGuy’s example f(x) = sin (1/x) is both continuous and bounded on the finite-length half-open interval (0,1], but having the troublesome spot x=0 at the boundary of this domain is still enough to cause trouble for the uniform continuity of sin (1/x) on (0,1].
Coming back to KarlGauss’s question, the following two statements are true:
[ul]
[li] A function that is continuous on a bounded and closed interval is uniformly continuous. (this follows from the Heine-Cantor theorem)[/li][li] A function that is continuous on a bounded and closed interval is bounded.[/li][/ul]
So while it’s very reasonable to suppose a connection between the two concepts, to find the difference between continous bounded function and uniformly contiuous function, one must consider functions on an interval that either isn’t closed, or isn’t bounded (or neither). OldGuy has provided an example of the first and it should be instructive to try and construct an example of the latter yourself.
That is, follow Lance Turbo’s lead and define a function f: ℝ -> ℝ and see if you can get the derivative to become unbounded, while keeping the function itself bounded and continuous).
Thank you all for your answers.
After reviewing 4.66’s very thoughtful and very helpful post and as a result of Lance Turbo’s neat example of F(x) = x as an unbounded yet uniformly continuous function (when compared with, say, F(x) = x[sup]2[/sup]) and OldGuy’s sin(1/x) counter-example, I now realize I had totally misunderstood the concept of uniform continuity.
So, before I even try to come up with an answer to 4.66’s ‘exercise’ and “see if (I) can get the derivative to become unbounded, while keeping the function itself bounded and continuous”, I’m afraid I need to make sure that I (finally, may) understand the definition of UC.