$\int_a^b f(x)g(x)dx = \sum \int_a^b f_n(x)g(x)dx.$

Question

Let $\sum f_n(x) $ be uniformly convergent to $f(x)$ on $[a,b]$ where each $f_n$ is continuous on $[a,b]$. If $g: [a,b] \to \mathbb R$ be integrable on $[a,b]$, then $$\int_a^b f(x)g(x)dx = \sum \int_a^b f_n(x)g(x)dx.$$

we know that if $\sum f_n(x) $ be uniformly convergent implies $$\int_a^b f(x)dx = \sum \int_a^b f_n(x)dx.$$

If we can show that $\sum f_n(x)g(x)$ is uniformly convergent to $f(x)g(x)$. But how to proceed? will integration by parts help out here??

Answer 1

Since $g$ is integrable on $[a,b]$, then it is bounded. Thus, there exists a number $B>0$ such that $|g(x)|\le B$ for all $x\in[a,b]$. Now, we have

$$\begin{align} \left|\int_a^b \left(g(x)\sum_{n}^N\,f_n(x)-g(x)f(x)\right)dx\right|&\le\int_a^b \left|g(x)\sum_{n}^N\,f_n(x)-g(x)f(x)\right|dx\\\\ &=\int_a^b \,|g(x)|\,\left|\sum_{n}^N\,f_n(x)-f(x)\right|dx\\\\ &\le B\int_a^b \left|\sum_{n}^N\,f_n(x)-f(x)\right|dx \end{align}$$

Since the series $\sum_{n}f_n(x)$ converges uniformly to $f$, then for any given $\epsilon>0$ we can find a number $N'$ such that $\int_a^b \left|\sum_{n}^N\,f_n(x)-f(x)\right|dx <\frac{\epsilon}{B}$ whenever $N>N'$ .

And we are done!

Answer 2

A complement about uniform convergence, too long to be in comment :

Notice that $\sum f_n(x) g(x)$ is not necessarily uniformly convergent if $g$ is only $L^1$.

Indeed, take ( on $[0,1]$ for simplicity) a serie of function that converge uniformly to $f(x) = \sqrt{x}$ and that verify $\forall n, \ \forall x \in [0,\frac{1}{2^n} ],\ f_n( x ) = 0$.

It's indeed possible to construct such a sequence by reccurence, defining :

$f_0(x) = \sqrt{ x}$ on $[\frac{1}{2},1]$ and linear on $[\frac{1}{2^2} ,\frac{1}{2}]$ such that its continuous

$f_n(x) = \sqrt{x} - \sum_{k=2}^{n-1} f_k(x)$ on $[\frac{1}{2^n} , 1]$ and linear on $[\frac{1}{2^{n+1}} ,\frac{1}{2^n}]$ such that its continuous

And you can verify that $\sum_n f_n$ converge uniformly to $\sqrt{x}$

Now take $g(x) = \frac{1}{\sqrt{x}}$ (and 0 if x=0). It's in $L^1$, and $f(x) g(x) = \mathbb{1}_{]0,1]}$

But as $f_n = 0$ on $[0,\frac{1}{2^n}]$ for any n, you have that

$$\| f(x)g(x) - \sum_{k=1}^n f_n(x)g(x) \|_{\infty} \geq \sup_{x\in [0,\frac{1}{2^n}]} | f(x)g(x) - \sum_{k=1}^n f_n(x)g(x) | \geq 1$$

So there is no uniform convergence