(19.4)
Prove that if is uniformly continuous on a bounded set , then is a bounded function on . (Hint: Try proof by contradiction.)
Proof. Assume that is uniformly continuous on but suppose to the contrary that is unbounded on . Then for all , there exists as number so that . Use these to define the sequence .
is bounded, so by the Bolzano-Weierstrauss theorem, there exists a convergent, hence Cauchy, subsequence . As is assumed to be uniformly continuous, this means that is itself a Cauchy, hence convergent, sequence. Say it converges to the number .
But for all , we have that . This implies that the sequence cannot possibly converge to , a contradiction. Hence must be bounded on .
Explain why (a) gives a proof that is not uniformly continuous on .
is unbounded on , so it cannot be uniformly continuous.
(23.1, 23.2) For each of the following power series, find the radius of convergence and determine the exact interval of convergence.
The interval of convergence is .
The interval of convergence is .
The interval of convergence is .
Notice that the sequence of terms is if for some , or otherwise (if the series index starts at 0 instead of 1, then , but this does not change the tail of the sequence). Hence the limit superior of the roots of terms is going to be .
The interval of convergence is .
(24.4) For , let .
Find , the pointwise limit of .
The pointwise limit is the function
To see this, take to be in any of these cases and evaluate the limit .
Determine whether uniformly on .
It cannot, as the pointwise limit is not continuous on , while each of the functions are.
Determine whether uniformly on .
It cannot, as the pointwise limit is not continuous on , while each of the functions are.