Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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 $f\cdot g$。

share|cite|improve this question
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.

share|cite|improve this answer
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

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.