Integral limit of $\sin(x/n)f(x)$

Question

For any $f\in L^1[0,\pi]$, evaluate

$n\to \infty \int^\pi_0 n$sin$(x/n)f(x)dx$

My idea is, $n$sin$(x/n)f(x)\to xf(x)$ and it seems that it is increasing sequence. I am not able to show it is increasing. Next thing if it increasing how could we apply monotone convergence theorem unless f is positive.

Next idea, i tried substitution, taking $x/n$ as $t$ but i get $f(tn)$ after substitution. Then stopped there. Do you have any specific idea for this..

Answer 1

Since $$\sin\left(\frac x n\right)\sim_\infty \frac xn$$ then we have

$$ \forall x\in[0,\pi],\quad n \sin\left(\frac x n\right)f(x)\xrightarrow{n\to\infty} xf(x)$$ moreover using that $|\sin x|\le |x|,\;\forall x$ we have

$$\left|n\sin\left(\frac x n\right)f(x)\right|\le |xf(x)|\le \pi|f(x)|\in L^1[0,\pi]$$ so by the dominated convergence theorem we have $$\lim_{n\to\infty}\int_0^\pi n\sin\left(\frac x n\right)f(x)=\int_0^\pi xf(x)dx$$

Answer 2

As you write correctly, $n\sin(x/n) \to x$, pointwise. Now noting that $\sin x \le x$ on $[0,\pi]$, we have $n\sin(x/n) \le n \cdot x/n = x$. As $\pi|f|$ is integrable and majorizes $f(x)\sin(x/n)n$ on $[0,\pi]$, we have $$ \int_0^\pi n\sin(n^{-1}x)f(x) \,dx \to \int_0^\pi xf(x)\, dx $$ by dominated convergence.