Let \(f, g: \mathbb{R} \to \mathbb{R}\) be continuous for all \(x\). Using an \(\varepsilon\)-\(\delta\) argument, show that \(f(x)g(x)\) is continuous for all \(x\).
Solution
Fix \(a\in\mathbb R\) and let \(\varepsilon>0\). Since \(f\) is continuous at \(a\), choose \(\delta_0>0\) such that \[ |x-a|<\delta_0\Longrightarrow |f(x)-f(a)|<1. \] Then, for such \(x\), \[ |f(x)|\le |f(a)|+1=:M. \] By continuity of \(g\) at \(a\), choose \(\delta_1>0\) such that \[ |x-a|<\delta_1\Longrightarrow |g(x)-g(a)|<\frac{\varepsilon}{2M}. \] By continuity of \(f\) at \(a\), choose \(\delta_2>0\) such that \[ |x-a|<\delta_2\Longrightarrow |f(x)-f(a)|<\frac{\varepsilon}{2(|g(a)|+1)}. \] Set \[ \delta=\min\{\delta_0,\delta_1,\delta_2\}. \] If \(|x-a|<\delta\), then \[ \begin{aligned} |f(x)g(x)-f(a)g(a)| &\le |f(x)|\,|g(x)-g(a)|+|g(a)|\,|f(x)-f(a)|\\ &< M\frac{\varepsilon}{2M} +|g(a)|\frac{\varepsilon}{2(|g(a)|+1)}\\ &<\varepsilon. \end{aligned} \] Thus \(fg\) is continuous at \(a\). Since \(a\) was arbitrary, \(fg\) is continuous on \(\mathbb R\).