(19.4)
Prove that if \(f\) is uniformly continuous on a bounded set \(S\), then \(f\) is a bounded function on \(S\). (Hint: Try proof by contradiction.)
Proof. Assume that \(f\) is uniformly continuous on \(S\) but suppose to the contrary that \(f\) is unbounded on \(S\). Then for all \(n \in \mathbb N\), there exists as number \(x_n\) so that \(f(x_n) > n\). Use these to define the sequence \((x_n)\).
\(S\) is bounded, so by the Bolzano-Weierstrauss theorem, there exists a convergent, hence Cauchy, subsequence \((x_{n_k})\). As \(f\) is assumed to be uniformly continuous, this means that \((f(x_{n_k}))\) is itself a Cauchy, hence convergent, sequence. Say it converges to the number \(y\).
But for all \(n_k > y\), we have that \(f(x_{n_k}) > n_k > y\). This implies that the sequence cannot possibly converge to \(y\), a contradiction. Hence \(f\) must be bounded on \(S\). \(\Box\)
Explain why (a) gives a proof that \(1/x^2\) is not uniformly continuous on \((0, 1)\).
\(1/x^2\) is unbounded on \((0,1)\), 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.
\(\displaystyle\sum \left(\dfrac xn\right)^n\)
The interval of convergence is \((-\infty, \infty)\).
\(\displaystyle\sum \left(\dfrac {n^3}{3^n}\right)x^n\)
The interval of convergence is \((-3, 3)\).
\(\displaystyle\sum \left( \dfrac{3^n}{n 4^n} \right)x^n\)
The interval of convergence is \([-4/3, 4/3)\).
\(\displaystyle\sum x^{n!}\)
Notice that the sequence of terms is \(a_n = 1\) if \(n = k!\) for some \(k \in \mathbb N\), or \(a_n = 0\) otherwise (if the series index starts at 0 instead of 1, then \(a_1 = 2\), but this does not change the tail of the sequence). Hence the limit superior of the roots of terms is going to be \(1\).
The interval of convergence is \((-1, 1)\).
(24.4) For \(x \in [0, \infty)\), let \(f_n(x) = \dfrac{x^n}{1 + x^n}\).
Find \(f(x) = \lim f_n(x)\), the pointwise limit of \((f_n)\).
The pointwise limit is the function
\[f = \left\{ \begin{array}{3} 0 &\;& 0 \le x < 1 \\ 1/2 &\;& x = 1 \\ 1 &\;& 1 < x \end{array}\right.\]To see this, take \(x\) to be in any of these cases and evaluate the limit \(\lim x^n/(1 + x^n)\).
Determine whether \(f_n \to f\) uniformly on \([0, 1]\).
It cannot, as the pointwise limit is not continuous on \([0, 1]\), while each of the functions \(f_n\) are.
Determine whether \(f_n \to f\) uniformly on \([0, \infty)\).
It cannot, as the pointwise limit is not continuous on \([0, \infty)\), while each of the functions \(f_n\) are.