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Let $g$ be continuous on $[a,b]$. Let $\{f_n\}_{n=1}^\infty$ be a sequence of continuous function, and $\{f_n\}_{n=1}^\infty$ converges to $f$ uniformly on $[a,b]$. Prove that $$\displaystyle \lim_{n\to\infty}\int_a^bf_ng = \int_a^bfg.$$

I'm having some trouble actually proving this result. I had a general outline of what I think needs to happen, but I'm having quite a difficult time filling in the details.

Because $f_n\rightrightarrows f$ (that is, $\{f_n\}_{n=1}^\infty$ converges uniformly to $f$ on $[a,b]$), we can say that given $\epsilon > 0$, there exists $N\in\mathbb N$ such that for all $n\in\mathbb N$ satisfying $n\ge N$, we have $\left|f_n-f\right|<\epsilon/\square.$ So,$$\begin{align}\left|\int_a^bf_ng-\int_a^bf\right| &= \left|\int_a^b(f_n-f)g\right|\\ & < \left|\int_a^b{\epsilon\over \square}|g|\right|\\&={\epsilon\over\square}\left|\int_a^b|g|\right|.\end{align}$$

What happens here though? Do I need to break this into cases? I can't quite make sense of what happens when $\displaystyle \left|\int_a^b|g|\right| =0$. If $\displaystyle \left|\int_a^b|g|\right| > 0$ then we're fine. That's all we need to put in the box. However, if $\displaystyle \left|\int_a^b|g|\right| = 0$, what happens exactly? Was this even the correct approach? I don't feel like I used all the assumptions.

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  • $\begingroup$ What does $C^1$ have to do with it? A better result with an easy proof would be "Suppose $f_n$ is a sequence of Riemann integrable functions that converge uniformly to $f.$ Let $g$ be Riemann integrable. Then $\int_a^b f_ng \to \int_a^b fg.$" $\endgroup$ – zhw. Mar 10 '17 at 18:41
  • $\begingroup$ @zhw., I have fixed the title. It should've just been $C[a,b]$. And that would certainly yield a much quicker result. However, that uses a result presented in the next section of my textbook. $\endgroup$ – Decaf-Math Mar 10 '17 at 18:45
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Stick closer to the estimates your textbook gives you. Compare these estimates with your estimates: $$ |\int_a^b f_n g - \int_a^b fg | \leq | \int_a^b (f_n -f)g|\le \int_a^b |f_n-f||g| \leq \int_a^b \epsilon|g| \leq C(a-b) \epsilon $$ Where $C \geq0$ is chosen such that $|g| \leq C$. This is well defined since we have a continuous function on a compact set.
Looking at your estimate, you had to shortly explain why $(f_n-f)g \leq \epsilon |g|$. Also, since your integrand is positive, there is no need for absolute value around the whole expression. More generally, there is the estimate: $$ |\int_a^b f |\leq\int_a^b|f| $$ To answer your other question: If $\int_a^b |g|=0$ and $g$ is continuous, there must be that $g=0$. You can proof this with the help of the $\epsilon -\delta$ defintion of continuity for $|g|$ and monoticity of integrals. Also, I dont see a problem with $g=0$. You dont need to assume $g \neq 0$ at any point. The calculation becomes trivial once you assume $g=0$.

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  • $\begingroup$ Is it better to say that since $g$ is continuous on $[a,b]$, we must have that $|g|$ is bounded, i.e. there exists $M>0$ such that $|g(x)| \le M$ for all $x\in[a,b]$. So we could say that $|f_n(x) - f(x)| < {\epsilon \over M(b-a)}$ to give us $$\begin{align}\left|\int_a^bf_ng-\int_a^bfg\right| &= \left|\int_a^b(f_n-f)g\right|\\&\le\int_a^b|f_n-f||g|\\& < \int{\epsilon\over M(b-a)}|g| \\ &\le {\epsilon\over M(b-a)}\int_a^bM \\ &= {\epsilon\over M(b-a)}M(b-a) \\ &= \epsilon.\end{align}$$ $\endgroup$ – Decaf-Math Mar 10 '17 at 13:24
  • $\begingroup$ It is more clean, but having something like $constant \cdot \epsilon $ does not really change your proof too much. Stick to your textbook and/or instructor if you are not sure how "sloppy" you can be. $\endgroup$ – F. Conrad Mar 10 '17 at 13:28

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