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Let $\{f_n\},\{g_n\}$ be functions on $[a,b]$.$f_n$ and $g_n$ are uniformly convergent to $f$ and $g$. Supose there exits sequence $M_n>0$, such that $|f_n|<M_n$ and $|g_n|<M_n$. How to prove $f_n\cdot g_n$ are uniformly convergent to $f\cdot g$。

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1 Answer 1

up vote 3 down vote accepted

The basic idea when studying convergence of a product is the following inequality: $$|f(x)g(x)-f_n(x)g_n(x)|=|f(x)(g(x)-g_n(x))+g_n(x)(f(x)-f_n(x))|\leq $$ $$\leq |f(x)||g(x)-g_n(x)|+|g_n(x)||f_n(x)-f(x)|$$

If the sequence $M_n$ is bounded, then you can finalize the proof immediatley.


Take $\varepsilon >0$ and $n \geq n_0$ where $n_0$ is given by $|f(x)-f_n(x)|< \varepsilon, \forall x \in [a,b],\forall n \geq n_0$. Then it follows easily that $|f(x)| \leq M_n+\varepsilon$. Therefore $f$ is bounded by a constant $M$.

We do the argument the other way around (change $f$ and $f_n$) and we obtain that $|f_n(x)|\leq M+\varepsilon,\forall x \in [a,b],\ \forall n \geq n_0$. So we can choose $M_n$ to be bounded.

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This is the solution I had - but as you say it relies on $M_n$ being bounded. –  Stephen Harris Feb 2 '12 at 12:07
    
how about $M_n$ is not bounded? –  Leitingok Feb 2 '12 at 12:08

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