 Mlejd5

2022-01-29

Evaluating $\underset{n\to \mathrm{\infty }}{lim}\frac{1}{n}{\int }_{0}^{\frac{\pi }{2}}\frac{{\mathrm{sin}}^{2}nx}{{\mathrm{sin}}^{2}x}f\left(x\right)dx$ for continuous f Emilie Booker

The limit is $\frac{\pi }{2}f\left(0\right)$ Note that $|{\int }_{0}^{\pi /2}\frac{{\mathrm{sin}}^{2}\left(nx\right)}{n{\mathrm{sin}}^{2}x}f\left(x\right)\phantom{\rule{0.167em}{0ex}}dx-\frac{\pi }{2}f\left(0\right)|$ (*) $⩽|{\int }_{0}^{\delta }\frac{{\mathrm{sin}}^{2}\left(nx\right)}{n{\mathrm{sin}}^{2}x}\left[f\left(x\right)-f\left(0\right)\right]\phantom{\rule{0.167em}{0ex}}dx|+|{\int }_{\delta }^{\pi /2}\frac{{\mathrm{sin}}^{2}\left(nx\right)}{n{\mathrm{sin}}^{2}x}\left[f\left(x\right)-f\left(0\right)\right],dx|$ Since f is continuous on $\left[0,\pi /2\right]$ and, hence, bounded we have $|f\left(x\right)|⩽M$ and $|f\left(x\right)-f\left(0\right)|⩽2M$ and an estimate for the second integral on the RHS of (*) is $|{\int }_{\delta }^{\pi /2}\frac{{\mathrm{sin}}^{2}\left(nx\right)}{n{\mathrm{sin}}^{2}x}\left[f\left(x\right)-f\left(0\right)\right]\phantom{\rule{0.167em}{0ex}}dx|⩽\frac{2M}{n}{\int }_{\delta }^{\pi /2}\frac{1}{{\mathrm{sin}}^{2}x}\phantom{\rule{0.167em}{0ex}}dx{\to }_{n\to \mathrm{\infty }}0$ Since the second integral converges to 0 for any $\delta$, it remains to prove that the first integral on the RHS of (*) can be made arbitrarily close to 0 with a suitable choice for $\delta$. For any $ϵ>0$ we can chose $\delta$ such that $|f\left(x\right)-f\left(0\right)|<ϵ$ for $0⩽x⩽\delta$ Using $\mathrm{sin}x>2x/\pi$ we get $|{\int }_{0}^{\delta }\frac{{\mathrm{sin}}^{2}\left(nx\right)}{n{\mathrm{sin}}^{2}x}\left[f\left(x\right)-f\left(0\right)\right]\phantom{\rule{0.167em}{0ex}}dx|⩽\frac{{\pi }^{2}ϵ}{4n}{\int }_{0}^{\delta }\frac{{\mathrm{sin}}^{2}\left(nx\right)}{{x}^{2}},dx$ $=\frac{{\pi }^{2}ϵ}{4}{\int }_{0}^{n\delta }\frac{{\mathrm{sin}}^{2}u}{{u}^{2}}du$ $⩽\frac{{\pi }^{2}ϵ}{4}{\int }_{0}^{\mathrm{\infty }}\frac{{\mathrm{sin}}^{2}u}{{u}^{2}}\phantom{\rule{0.167em}{0ex}}du=\frac{{\pi }^{3}ϵ}{8}$ Since, $ϵ$ can be arbitrarily close to 0 we are done.