# converge uniformly?

• December 18th 2012, 05:40 PM
converge uniformly?
On $[a,b]$, the function sequence $\{f_n\},\{g_n\}$ converge uniformly to $f,g$ respectively. Suppose there exists positive sequence $M_n$ such that $f_n(x)\leq M_n, g_n(x)\leq M_n,\ \forall\ x\in [a,b]$. Prove that $f_ng_n$ converge unformly to $fg$ on $[a,b]$

PS: If $M_n=M$, I know how to prove. But this?....Would you help me?
• January 2nd 2013, 11:51 PM
Re: converge uniformly?
To show that $f_ng_n$ converges uniformly to $fg$ on $[a,b]$, we have to show there exists an N such that for all x in the interval $[a,b]$ and $n \geq N$,

$|f_ng_n - fg|\leq \epsilon$

I'm sorry I'm a bit rusty, but perhaps it would help to note that you can consider an upper bound for your positive sequence $M_n$. If the least upper bound for $M_n$ is a real number, say M, then you can say $f_n \leq M$ and $g_n \leq M$ for all n. If there is no least upper bound for your positive sequence, I'm not sure how to continue.
• January 3rd 2013, 12:53 AM
hollywood
Re: converge uniformly?
So what you're saying is that $f_n$ and $g_n$ are sequences of bounded functions that converge uniformly to $f$ and $g$ respectively on $[a,b]$, and you want to prove that $f_ng_n$ converges uniformly to $fg$.

You said you know how to do it if $f_n$ and $g_n$ are uniformly bounded - that is, there are $M_f$ and $M_g$ such that $|f_n(x)| on $[a,b]$ and $|g_n(x)| on $[a,b]$. So prove that first:

Let $N$ be such that $|f_n(x)-f(x)|<\frac{1}{2}$ for all $x\in[a,b]$. Then for all $n>N$:

$|f_n(x)|\le |f_n(x)-f(x)|+|f(x)-f_N(x)|+|f_N(x)|\le \frac{1}{2}+\frac{1}{2}+M_N=M_N+1$

and I think you can probably fill in the rest of the proof.

- Hollywood
• January 3rd 2013, 06:29 PM