Continuous, bounded, and not uniform?

[icantadd]

Homework Statement

Give an example of a function f : R -> R where f is continuous and bounded but not uniformly continuous.

Homework Equations

A function f : D -> R and R contains D, with Xo in D, and | X - Xo | < delta (X in D), implies | f(X) - f(Xo) | < epsilon. Then f is continuous at Xo.

f is continuous if it is continuous at all Xo.

f is uniformly continuous is close to continuity:
" " x and y in D, | x - y | < delta then | f(x) - f(y) | < epsilon.

The Attempt at a Solution

The problem I am having is creating a bounded function over the entirety of R, and then showing that the function is uniformly continuous. First I tried,
f(x) = 0 when x = 0
x*sin(1/x) else
Thinking that the craziness around 0 would make it not u.c.
But this doesn't work because f is uniformly continuous, just not pretty.

What I am thinking is that if a function that is continuous, then it is uniformly continuous on any closed interval for its domain. So, if I could find a function that is continuous but contains an open interval it would not be uniformly continuous. But as I say those words, I don't understand them. Any help that is great would be greatly appreciated.

One more thing, the text I am using has not yet given a definition of a bounded function, but something very close, that if a function has a limit (or is continuous) at a point, then there is some bounded interval around that point.

 

[Dick]

You want a function that is bounded and continuous but has an unbounded derivative. How about sin(x^2)? Why would that work?

 

[icantadd]

You want a function that is bounded and continuous but has an unbounded derivative. How about sin(x^2)? Why would that work?

Yes, thank you. At a calculus two level, taking the derivative yields a function whose rate of change is not bounded. i.e. the 2*x part. On a conceptual level the graph oscilates more differently as x gets big making it not uniformly continuous, meaning I can't pick to values in the domain, and know that there is well behaved delta. Is that correct?

 

[Dick]

I think you have the right picture. No matter how small an epsilon you pick, for x large enough f(x) will range between 2 and -2 for values in the interval (x-e,x+e).

 

[Mark44]

I think you have the right picture. No matter how small an epsilon you pick, for x large enough f(x) will range between 2 and -2 for values in the interval (x-e,x+e).

If we're talking about f(x) = sin(x^2), f(x) can never be larger than 1 or less than -1.

 

[Dick]

If we're talking about f(x) = sin(x^2), f(x) can never be larger than 1 or less than -1.

Right, sure. Thanks.