Show that $\lim_{n\to \infty} \int_a^bf(x)g(nx)dx=0$ Let $f$ be differentiable on the interval $[a,b]$, so that the derivative is also continuous on the interval $[a,b]$. And let $g$ be continuous on $[0,\infty)$ such that for every $0\le a \le x$ :
$$\big| \int_a^x g(t) dt\big| \le C$$ Where $C$ is some constant.
Show that $\lim_{n\to \infty} \int_a^bf(x)g(nx)dx=0$
I know this has to do with integration by parts, I tried getting to a point where I can use triangle inequalities but I cannot get anywhere. Thoughts?
 A: Suggestion:
Let $G(x)=\int^x_ag$ and Change variable $u=nx$ to get
$$
I_n=\frac{1}{n}\int^{nb}_{na}f(x/n)dG(x)
$$
integrate by parts.
\begin{aligned}
I_n &=\frac{1}{n}(G(nb)f(b)-G(na)f(a)-\frac{1}{n^2}\int^{nb}_{nb}G(x)f'(x/n)\,dx
\end{aligned}
The first term on the right-hand-side is bounded by $\frac{2C}{n}(|f(b)|+|f(a)|)$; the second term on the right hand side is bounded by
$\frac{C}{n}\int^b_a|f'(x)|\,dx$.
A: Define $G:[0, \infty) \to \mathbb{R}$ as
$$G(x) = \int_a^xg(t)dt.$$
Since $g$ is continuous, we have that $G$ is differnetiable with $G' = g$.
Moreover, we have that $|G(x)| \le C$ for all $x \ge 0$.  
Now, for a fixed $n$, let us calculate the desired integral as follows:
$$\begin{align}
\left|\int_a^bf(x)g(nx)dx\right| &= \left|\frac{1}{n}\int_a^bf(x)\left[\dfrac{d}{dx}G(nx)\right]dx\right|\\~\\
&= \dfrac{1}{n}\left|\left[f(x)G(nx)\big|_a^b - \int_a^bf'(x)G(nx)dx\right]\right|\\~\\
&\le\dfrac{1}{n}\left[|f(b)G(nb)| + |f(a)G(na)| + \left|\int_a^bf'(x)G(nx)dx\right| \right] \\~\\
&\le\dfrac{1}{n}\left[C\left(|f(b)| + |f(a)| + \int_a^b|f'(x)|dx\right) \right] \\~\\
&= \dfrac{C}{n}\left(|f(b)| + |f(a)| + \int_a^b|f'(x)|dx\right) 
\end{align}$$
The result now follows as we let $n \to \infty$. (Everything within the brackets is independent of $n$.)
A: Because $f(x)$ is continuously differentiable on $[a,b]$, $f'(x) $ is bounded on $[a,b]$.  S owe can find $B$ such that $| f(x) | \leqslant B$.
Now define 
$$
G(x) = \int_a^x g(t) dt.
$$
Because $g$ is continuous, the function $G$ is differentiable with $G'(x) = g(x)$ and we are told also that $G(x)$ is bounded by $C$.  
Then
$$
\begin{align}
\int_a^b f(x) g(nx) dx &= \int_a^b f(x) G'(nx) dx \\
    &= \left[ \frac{1}{n} G(nx) f(x) dx\right]_a^b - \int_a^b \frac{1}{n} G(nx) f'(x) dx \\
&= \frac{1}{n}\Bigg( G(nb)f(b) - G(na)f(a) - \int_a^b G(nx) f'(x) dx\Bigg)
\end{align}
$$
We can now apply the bounds.  $|G(nb)f(b)| \leqslant Cf(b)$, $|G(na)f(a)|\leqslant Cf(a)$ and $|G(nx)f'(x)| \leqslant CB$.  Using the fact that the absolute value of an integral is always less than the integral of the absolute value,
$$
\left |\int_a^b f(x)g(nx) dx \right |\leqslant \frac{1}{n} \Bigg( Cf(b) + Cf(a) + (b-a) CB \Bigg).
$$
As $n \to\infty$ the terms on the right hand side in parentheses are fixed and the right hand side has limit zero.  The result you want then follows.
