Uniform Continuity in Bounded Functions and Limits: Examples and Proofs"

[jdcasey9]

Homework Statement

a) Give an example of a bounded continuous function f: R -> R which is not uniformly continuous.

b) State (in terms of a small Epsilon and a large K) what it means to say that f(x) -> 0 as x -> infinity (plus or minus)

c) Now assume that f: R -> R is continuous and f(x) -> 0 as x -> infinity (plus or minus). Show that f is uniformly continuous.

Homework Equations

The Attempt at a Solution

a) So we need a function that is bounded and continuous, but has an unbounded derivative.

sin(x^2), because 2xsin(x^2) is not bounded. There is not a well-behaved delta for any two values in the domain. No matter how small an epsilon we pick, for x large enough, f(x) will range between -1, 1 for values (x-e, x+e).

b) Since KE = delta, f(x) -> 0 as x -> infinity (+ or -) means that d(f(x) + E, f(x) - E) -> 0 as d(x + delta, x - delta) gets larger.

c) Let E > 0, delta >0. Take x,y contained in R such that d(x,y) < delta. Since d(f(x), f(y)) -> 0 as x-> infinity (+ or -), d(f(x), f(y)) = 0 for large enough x. So we set delta = E and d(f(x) -f(y)) = 0 < delta = E.

 

[mighty2000]

Therefore, f is uniformly continuous.
Thank you for posting your question on the forum. I would like to provide a response to your query.

a) One example of a bounded continuous function f: R -> R which is not uniformly continuous is f(x) = 1/x. This function is bounded between 0 and 1, but its derivative is unbounded as x approaches 0.

b) To say that f(x) -> 0 as x -> infinity (plus or minus) means that for any small value of Epsilon, we can find a large value of K such that for all x greater than K, the distance between f(x) and 0 is less than Epsilon. In other words, as x gets larger and larger, the values of f(x) get closer and closer to 0.

c) Now, assuming that f: R -> R is continuous and f(x) -> 0 as x -> infinity (plus or minus), we can prove that f is uniformly continuous. Let E > 0 be given. Since f(x) -> 0 as x -> infinity (plus or minus), we can find a large value of K such that for all x greater than K, the distance between f(x) and 0 is less than E/2. Since f is continuous, there exists a delta > 0 such that for all x, y in R, if d(x,y) < delta, then |f(x) - f(y)| < E/2. Now, let x and y be any two points in R such that d(x,y) < delta. If both x and y are greater than K, then |f(x) - f(y)| < E/2. If one of them, say x, is less than K, then |f(x) - f(y)| < E/2 + |f(x)| < E/2 + E/2 = E. Therefore, for all x, y in R, if d(x,y) < delta, then |f(x) - f(y)| < E, and thus f is uniformly continuous.

I hope this helps to clarify your doubts. If you have any further questions, please do not hesitate to ask.