converge uniformly?

**xinglongdada** · December 18th 2012, 04:40 PM

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?

**MacstersUndead** · January 2nd 2013, 10:51 PM

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.

**hollywood** · January 2nd 2013, 11:53 PM

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 $|f_n(x)|<M_f$ on $[a,b]$ and $|g_n(x)|<M_g$ 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

**xinglongdada** · January 3rd 2013, 05:29 PM

Thank you very much indeed.

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