Wisniewool

2022-07-04

The purpose of this problem is that I want to prove that for any $\lambda$ integrable function f on a bounded closed interval [a,b] holds
$\underset{n\to \mathrm{\infty }}{lim}{\int }_{\left[a,b\right]}f\left(x\right)\mathrm{sin}\left(nx\right)d\lambda =0.$
I have submitted a proof below.

treccinair

Expert

The smooth functions with compact suppport ${\mathcal{C}}_{0}^{\mathrm{\infty }}\left(\left[a,b\right]\right)$ functions are dense over the Lebesgue integral functions ${L}^{1}\left[a,b\right]$. Then for every f there is a sequence $\left({f}_{k}{\right)}_{k}\subset {\mathcal{C}}_{0}^{\mathrm{\infty }}\left(\left[a,b\right]\right)$ such that $‖f-{f}_{k}{‖}_{1}\stackrel{k\to \mathrm{\infty }}{\to }0$. We observe that
$\begin{array}{r}|{\int }_{a}^{b}{f}_{k}\left(x\right)\mathrm{sin}\left(nx\right)dx|=|-\frac{1}{n}{\int }_{a}^{b}{f}_{k}^{\prime }\left(x\right)\mathrm{cos}\left(nx\right)dx|=\frac{1}{{n}^{2}}|{\int }_{a}^{b}{f}_{k}^{″}\left(x\right)\mathrm{sin}\left(nx\right)dx|\le \frac{{M}_{k}}{{n}^{2}}\stackrel{n\to \mathrm{\infty }}{\to }0\end{array}$
Where ${M}_{k}=\underset{x\in \left[a,b\right]}{sup}|{f}_{k}^{″}\left(x\right)|\left(b-a\right)<\mathrm{\infty }$. Now
$\begin{array}{rl}|{\int }_{a}^{b}f\left(x\right)\mathrm{sin}\left(nx\right)dx|& \le |{\int }_{a}^{b}{f}_{k}\left(x\right)\mathrm{sin}\left(nx\right)dx|+|{\int }_{a}^{b}\left(f\left(x\right)-{f}_{k}\left(x\right)\right)\mathrm{sin}\left(nx\right)dx|\\ & \le \frac{{M}_{k}}{{n}^{2}}+‖{f}_{k}-f{‖}_{1}\stackrel{n\to \mathrm{\infty }}{\to }‖{f}_{k}-f{‖}_{1}\stackrel{k\to \mathrm{\infty }}{\to }0.\end{array}$

Do you have a similar question?