$$f_n(x)=x(1+1/n) \text{ if } x \in \mathbb{R}$$
$$g_n(x) = \begin{cases} (1/n)& x = 0, \text{ or } x \in \mathbb{I} \\
b+1/n& \text{if } x \in \mathbb{Q} \text{ with } x=a/b
\end{cases}$$
I need to show that $h_n(x) = f_n(x)g_n(x)$ does not converge uniformly on any bounded interval. I already showed the uniform convergence of $f_n$ and $g_n$ on any bounded interval but I am having a hard time with showing $h_n(x)$ does not converge unifly. Thanks!
Best Answer
For $h_n$ to be well defined you need a unique representation $x = a/b$ for $x \in \mathbb{Q}$. Presumably you mean that $a/b$ is in lowest terms where $(a,b) = 1$, that is $a$ and $b$ are coprime.
In this case, we have for a bounded interval $[\alpha, \beta]$ that $f_n(x)$ converges uniformly to $f(x) = x$ and $g_n(x)$ converges uniformly to
$$g(x) = \begin{cases}0, &x = 0 \text{ or }x \in \mathbb{I}& \\ b,& x \in \mathbb{Q} \text{ with } x = a/b, \,(a,b) =1 \end{cases}$$
The convergence $g_n \to g$ is uniform since $|g_n(x) - g(x) | = 1/n$.
Consider the convergence $f_ng_n \to fg$ on the bounded interval $[0,1]$. We have
$$\tag{*}|f_n(x)g_n(x) - f(x)g(x)| = \begin{cases}x(1/n + 1/n^2), & x= 0 \text{ or }x \in \mathbb{I}\cap [\alpha,\beta]& \\ x(1/n + 1/n^2) + (xb)/n,& x \in \mathbb{Q}\cap [\alpha,\beta] \text{ with } x = a/b, \,(a,b) =1 \end{cases}$$
where $(xb)/n = a/n$.
For any $m \in \mathbb{N}$ the rational $a/b = m/(m+1)$ is in $[0,1]$. Thus $a$ is unbounded and the supremum of (*) is unbounded and does not converge to $0$ as $n \to \infty$. Therefore, convergence is not uniform. We can make a similar argument for other bounded intervals.