1. ## converge uniformly?

On $\displaystyle [a,b]$, the function sequence $\displaystyle \{f_n\},\{g_n\}$ converge uniformly to $\displaystyle f,g$ respectively. Suppose there exists positive sequence $\displaystyle M_n$ such that $\displaystyle f_n(x)\leq M_n, g_n(x)\leq M_n,\ \forall\ x\in [a,b]$. Prove that $\displaystyle f_ng_n$ converge unformly to $\displaystyle fg$ on $\displaystyle [a,b]$

PS: If $\displaystyle M_n=M$, I know how to prove. But this?....Would you help me?

2. ## Re: converge uniformly?

To show that $\displaystyle f_ng_n$ converges uniformly to $\displaystyle fg$ on $\displaystyle [a,b]$, we have to show there exists an N such that for all x in the interval $\displaystyle [a,b]$ and $\displaystyle n \geq N$,

$\displaystyle |f_ng_n - fg|\leq \epsilon$

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 $\displaystyle M_n$. If the least upper bound for $\displaystyle M_n$ is a real number, say M, then you can say $\displaystyle f_n \leq M$ and $\displaystyle g_n \leq M$ for all n. If there is no least upper bound for your positive sequence, I'm not sure how to continue.

3. ## Re: converge uniformly?

So what you're saying is that $\displaystyle f_n$ and $\displaystyle g_n$ are sequences of bounded functions that converge uniformly to $\displaystyle f$ and $\displaystyle g$ respectively on $\displaystyle [a,b]$, and you want to prove that $\displaystyle f_ng_n$ converges uniformly to $\displaystyle fg$.

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

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

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

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

- Hollywood

4. ## Re: converge uniformly?

Thank you very much indeed.