Let \(S\) be a set and let \((f_n)\) and \((g_n)\) be sequences of functions \(S\to\mathbb R\).
Show that if \(f_n\to f\) and \(g_n\to g\) uniformly and \(f,g\) are bounded, then \(f_ng_n\to fg\) uniformly.
Give a counterexample showing the conclusion can fail when \(f\) is unbounded.
Solution
1. Suppose the hypotheses of part (a) hold. Choose constants\(B,C\ge0\) such that\[ |f(s)|\le B, \qquad |g(s)|\le C \]for every \(s\in S\).
Such constants exist because \(f\) and \(g\) are bounded by hypothesis.
2. There is \(N_0\) such that\[ |f_n(s)|\le B+1 \]for every \(n\ge N_0\) and every \(s\in S\).
Since \(f_n\to f\) uniformly, there is \(N_0\) such that \[ |f_n(s)-f(s)|<1 \] for all \(n\ge N_0\) and all \(s\in S\). Then step 1 and the triangle inequality give \[ |f_n(s)| \le |f(s)|+|f_n(s)-f(s)| < B+1. \]
3. The sequence \((f_ng_n)\) converges uniformly to \(fg\) on \(S\).
Let \(\varepsilon>0\). Uniform convergence supplies integers \(N_1,N_2\) such that, for every \(s\in S\), \[ |g_n(s)-g(s)| < \frac{\varepsilon}{2(B+1)} \] when \(n\ge N_1\), and \[ |f_n(s)-f(s)| < \frac{\varepsilon}{2(C+1)} \] when \(n\ge N_2\).
Take \[ N\coloneqq\max\{N_0,N_1,N_2\}. \] For \(n\ge N\) and every \(s\in S\), steps <1>1–<1>2 give \[ \begin{aligned} |f_n(s)g_n(s)-f(s)g(s)| &= |f_n(s)(g_n(s)-g(s))+(f_n(s)-f(s))g(s)|\\ &\le |f_n(s)|\,|g_n(s)-g(s)| + |f_n(s)-f(s)|\,|g(s)|\\ &< (B+1)\frac{\varepsilon}{2(B+1)} + C\frac{\varepsilon}{2(C+1)}\\ &< \varepsilon. \end{aligned} \] The same \(N\) works for every \(s\in S\), so the convergence is uniform. This proves part (a).
4. For part (b), take\[ S=\RR, \qquad f_n(x)=x, \qquad g_n(x)=1+\frac1n. \]Then\[ f_n\to f, \qquad f(x)=x, \]and\[ g_n\to g, \qquad g(x)=1, \]uniformly on \(\RR\), while \(f\) is unbounded.
For every \(n\), \[ \sup_{x\in\RR}|f_n(x)-f(x)|=0. \] Also \[ \sup_{x\in\RR}|g_n(x)-g(x)| = \frac1n \longrightarrow 0. \] Thus both convergences are uniform, and \(f(x)=x\) is unbounded.
5. The products in step 4 do not converge uniformly to \(fg\).
For every \(x\in\RR\), \[ f_n(x)g_n(x)-f(x)g(x) = \frac{x}{n}. \] Hence for every fixed \(n\), \[ \sup_{x\in\RR} |f_n(x)g_n(x)-f(x)g(x)| = \sup_{x\in\RR}\frac{|x|}{n} = \infty. \] In particular, these suprema do not tend to \(0\), so convergence is not uniform. This proves part (b).
6. Q.E.D.
Step 3 proves part (a), and steps <1>4–<1>5 give the required counterexample for part (b).