Let \((X, \mathcal{M}, \mu)\) be a measure space and prove the following properties of \(L^ \infty (X, \mathcal{M}, \mu)\):
If \(f, g\) are measurable on \(X\) then \[ \norm{fg}_1 \leq \norm{f}_1 \norm{g}_{\infty } .\]
\(\norm{\wait}_{\infty }\) is a norm on \(L^{\infty }\) making it a Banach space.
\(\norm{f_n - f}_{\infty } \converges{n\to \infty }\to 0 \iff\) there exists an \(E\in \mathcal{M}\) such that \(\mu(X\sm E) = 0\) and \(f_n \to f\) uniformly on \(E\).
Simple functions are dense in \(L^{\infty }\).
Solution
Recall \(\|f\|_\infty = \inf\theset{M \geq 0 : |f| \le M \text{ a.e.}}\), and the infimum is attained: \(|f| \le \|f\|_\infty\) a.e., because \(\theset{|f| > \|f\|_\infty} = \bigcup_k \theset{|f| > \|f\|_\infty + 1/k}\) is a countable union of null sets.
1. \(\|fg\|_1 \le \|f\|_1\|g\|_\infty\) for measurable \(f, g\).
\(|fg| \le |f|\,\|g\|_\infty\) a.e., so \(\int|fg| \le \|g\|_\infty\int|f|\).
2. \(\|\cdot\|_\infty\) is a norm on \(L^\infty\), the space of a.e.-equivalence classes of essentially bounded functions, and \(L^\infty\) is complete.
2.1. \(\|\cdot\|_\infty\) is a norm.
\(\|f\|_\infty = 0\) if and only if \(f = 0\) a.e., and \(\|\alpha f\|_\infty = |\alpha|\,\|f\|_\infty\). Since \(|f + g| \le |f| + |g| \le \|f\|_\infty + \|g\|_\infty\) a.e., \(\|f + g\|_\infty \le \|f\|_\infty + \|g\|_\infty\).
2.2. Every Cauchy sequence \((f_n)\) in \(L^\infty\) converges in \(L^\infty\).
For \(n, m\) let \(Z_{n,m}\) be a null set off which \(|f_n - f_m| \le \|f_n - f_m\|_\infty\), and let \(Z = \bigcup_{n,m} Z_{n,m}\), a null set. On \(X \setminus Z\), \(\sup|f_n - f_m| \le \|f_n - f_m\|_\infty\), so \((f_n)\) is uniformly Cauchy there and converges uniformly to a bounded measurable \(f\); put \(f = 0\) on \(Z\). Given \(\eps > 0\) choose \(N\) with \(\|f_n - f_m\|_\infty < \eps\) for \(n, m \ge N\); letting \(m \to \infty\) gives \(|f_n - f| \le \eps\) on \(X \setminus Z\), so \(\|f_n - f\|_\infty \le \eps\) for \(n \geq N\).
2.3. Q.E.D.
Steps 1 and 2.
3. \(\|f_n - f\|_\infty \to 0\) if and only if there is \(E \in \mathcal M\) with \(\mu(X \setminus E) = 0\) and \(f_n \to f\) uniformly on \(E\).
If \(f_n \to f\) uniformly on such an \(E\), then \(\|f_n - f\|_\infty \le \sup_{E}|f_n - f| \to 0\). Conversely, let \(Z_n\) be a null set off which \(|f_n - f| \le \|f_n - f\|_\infty\), and put \(E = X \setminus \bigcup_n Z_n\). Then \(\sup_E|f_n - f| \le \|f_n - f\|_\infty \to 0\).
4. Simple functions are dense in \(L^\infty\).
Let \(f \in L^\infty\) be real-valued, \(M = \|f\|_\infty\), and \(Z = \theset{|f| > M}\), a null set. For \(k \ge 1\) let \[ s_k = \sum_{j} \frac{j}{2^k}\,\chi_{\theset{x \notin Z\,:\, j2^{-k} \le f(x) < (j+1)2^{-k}}}, \] the sum over the finitely many integers \(j\) with \(|j| \le M2^k + 1\). Then \(s_k\) is simple and \(|f - s_k| < 2^{-k}\) on \(X \setminus Z\), so \(\|f - s_k\|_\infty \le 2^{-k}\). For complex \(f\) apply this to the real and imaginary parts.