Prove that uniform convergence implies pointwise convergence implies a.e. convergence, but none of the implications may be reversed.
Show that \(\sum {x^n \over n!}\) converges uniformly on any compact subset of \(\RR\).
Solution
1. If \(f_n \to f\) uniformly, then \(f_n \to f\) pointwise.
\(|f_n(x) - f(x)| \leq \sup_y|f_n(y) - f(y)| \to 0\) for each fixed \(x\).
2. If \(f_n \to f\) pointwise, then \(f_n \to f\) a.e.
The set where \(f_n(x) \not\to f(x)\) is empty, hence null.
3. \(f_n(x) = x^n\) on \([0,1]\) converges pointwise but not uniformly.
\(x^n \to 0\) for \(0 \le x < 1\) and \(1^n = 1\), so \(f_n \to f\) pointwise with \(f = 0\) on \([0,1)\) and \(f(1) = 1\). At \(x = 1 - 1/n\), \(|f_n(x) - f(x)| = (1 - 1/n)^n \to e^{-1}\), so \(\sup_{[0,1]}|f_n - f| \not\to 0\).
4. \(f_n = \chi_{[0, 1/n)}\) on \([0,1]\) converges to \(0\) a.e. but not pointwise to \(0\).
For every \(x > 0\), \(f_n(x) = 0\) once \(1/n \le x\), so \(f_n \to 0\) off the null set \(\theset{0}\). But \(f_n(0) = 1\) for every \(n\).
5. \(\sum_{n=0}^\infty \frac{x^n}{n!}\) converges uniformly on every compact \(K \subseteq \RR\).
Choose \(M\) with \(K \subseteq [-M, M]\). On \(K\), \(\left|\frac{x^n}{n!}\right| \le \frac{M^n}{n!}\), and \(\sum_n \frac{M^n}{n!}\) converges by the ratio test, since \(\frac{M^{n+1}/(n+1)!}{M^n/n!} = \frac{M}{n+1} \to 0\). The Weierstrass M-test gives uniform convergence on \(K\).
6. Q.E.D.
Steps 1 and 2 are the implications, steps 3 and 4 show that neither reverses, and step 5 is the second part.