Correct on nos. 1 and 2. For 3, split the interval into two subintervals, [0,1] and [1,∞], and ask why the function should be uniformly continuous on both of them.
Discussing the uniform continuity of the following:
1.
Claim: f is uniformly continuous on (-1,1)
Proof.
Given , pick , then for each with , we would have
Q.E.D.
2.
Claim: f is uniformly continuous.
Proof.
Given , pick , then for each with , we would have
Q.E.D.
3.
Claim: f is not uniformly continuous.
So far, I know that , which is not below above as x approaches to 0.
But is it possible to prove this by defintion?
Proof so far.
Pick , then with , I need to show that But I'm stuck in here, any hints?
Thanks!!!
For the first interval of : f is continuous on [0,1], a compact set, meaning f is also uniformly continuous.
For the second interval of : The derivative is bounded by 0 below and 1/3 above, therefore it is uniformly continuous.
Is this right? So if a function is uniformly continuous on two subinterval, then it is also uniformly continuous on the whole thing?
I was rewriting my solutions, and I found a problem in my answer to the function of
Is ? I'm thinking now it is not, since y can be something really big, that would bring the distance out.
Should I consider the series and prove this series is uniformly convergence?
Then I use the M test, since , and converges by the ratio test, so f is uniformly convergence, is that right? Thanks.
You seem to be confusing uniform continuity with uniform convergence. If you want to show that the series is uniformly convergent on the interval [0,∞) then you use the M-test. But the question asks you to show that the function is uniformly continuous on the interval [0,∞). The easiest way to do that is to observe that the derivative is bounded on that interval.