Show that \(f\in L^1(\RR^n) \implies \abs{f(x)} < \infty\) almost everywhere.
Show that if \(\ts{f_k} \subseteq L^1(\RR^n)\) with \(\sum \norm{f_k}_1 < \infty\) then \(\sum f_k\) converges almost everywhere and in \(L^1\).
Use the Dominated Convergence Theorem to evaluate \[ \lim_{t\to 0} \int_0^1 {e^{tx^2} - 1 \over t} \dx .\]
Solution
1. If \(f \in L^1(\RR^n)\), then \(|f(x)| < \infty\) for a.e. \(x\).
For every \(M > 0\), \(\theset{|f| = \infty} \subseteq \theset{|f| \ge M}\), and Markov’s inequality gives \(m\theset{|f| \ge M} \le \|f\|_1/M\). Letting \(M \to \infty\) shows \(m\theset{|f| = \infty} = 0\).
2. If \(f_k \in L^1(\RR^n)\) and \(\sum_k \|f_k\|_1 < \infty\), then \(\sum_k f_k\) converges a.e. and in \(L^1\).
2.1. \(g = \sum_k |f_k|\) is in \(L^1\), and \(g < \infty\) a.e.
By the monotone convergence theorem applied to the partial sums, \(\int g = \sum_k \|f_k\|_1 < \infty\). Step 1 applied to \(g\) gives \(g < \infty\) a.e.
2.2. Q.E.D.
Where \(g(x) < \infty\), the series \(\sum_k f_k(x)\) converges absolutely; call its sum \(F(x)\), and put \(F = 0\) elsewhere. Then \(|F| \le g\), so \(F \in L^1\), and \(\left\|F - \sum_{k=1}^N f_k\right\|_1 \le \int \sum_{k > N}|f_k| = \sum_{k>N}\|f_k\|_1 \to 0\).
3. \(\lim_{t \to 0}\int_0^1 \frac{e^{tx^2} - 1}{t}\,dx = \boxed{\tfrac{1}{3}}\).
3.1. For \(0 < |t| \le 1\) and \(x \in [0,1]\), \(\left|\frac{e^{tx^2} - 1}{t}\right| \le e\).
By the mean value theorem, \(e^u - 1 = e^\xi u\) for some \(\xi\) between \(0\) and \(u\), so \(|e^u - 1| \le e|u|\) for \(|u| \le 1\). Take \(u = tx^2\).
3.2. \(\frac{e^{tx^2} - 1}{t} \to x^2\) as \(t \to 0\) for each \(x\).
This is the derivative of \(t \mapsto e^{tx^2}\) at \(t = 0\).
3.3. Q.E.D.
By steps 1 and 2, dominated convergence with dominating function \(e\) on \([0,1]\) gives the limit \(\int_0^1 x^2\,dx = \frac13\) along every sequence \(t_k \to 0\).