## Winter 1993

Problem Statement: Suppose that $\{f_n\}$ converges to $f$ uniformly and $\{g_n\}$ converges to $g$ uniformly on $[a,b]$ where each $f_n,g_n$ is continuous on $[a,b]$. Prove that $f_ng_n$ converges uniformly to $fg$ on $[a,b]$.

Proof: Since $[a,b]$ is a compact set and $f_n,g_n$ are each continuous on $[a,b]$ it follows that each $f_n,g_n$ is bounded on $[a,b]$. So, there exist $M,P\in\mathbb{R}$ such that $|f_n(x)|\leq M$ and $|g_n(x)|\leq P$ for every $x\in [a,b]$ and for every $n\in\mathbb{N}$.

Let $\varepsilon>0$. Then there exists $N_1\in\mathbb{N}$ such that $|f_n(x)-f(x)|<\dfrac{\varepsilon}{2P}$ for every $n\geq N_1$ and for every $x\in [a,b]$ since $f_n\rightarrow f$ uniformly. Similarly, there exists $N_2\in\mathbb{N}$ such that $|g_n(x)-g(x)|<\dfrac{\varepsilon}{2M}$ for every $n\geq N_2$ and for every $x\in [a,b]$.

Let $N=max\{N_1,N_2\}$. Then for $n\geq N$ and $x\in [a,b]$ it follows that

$|f_ng_n(x)-fg(x)|=|(f_n(x)-f(x))g_n(x)+f(x)(g_n(x)-g(x))|$

$\leq |f_n(x)-f(x)|P+M|g_n(x)-g(x)|$ for every $x\in [a,b]$

$<\dfrac{\varepsilon}{2P}P+\dfrac{\varepsilon}{2M}M$

$=\dfrac{\varepsilon}{2}+\dfrac{\varepsilon}{2}=\varepsilon$

Thus, $f_ng_n$ converges to $fg$ uniformly on $[a,b]$.

$\Box$

Reflection: This one really followed straight from the definitions. The key being in that we knew each of the functions was bounded. Also, we used a little bit of “math magic” to make the inside of the absolute values look like things we knew about. The thing to keep in mind when doing all of these problem is what you know information about. You can never make a conclusion about some inequality if you don’t have any information about it!