Evaluating $\lim_{n\to\infty}\frac1{n}\int_{0}^{\pi /2}{\frac{\sin^2nx}{\sin^2 x}f(x)\;dx}$, for continuous $f$

Question

Let $f:\left[ 0,{\pi }/{2}\; \right]\to \mathbb{R}$ be a continuous function. What is the value of the following limit?

$$\underset{n\to \infty }{\mathop{\lim }}\,\frac{1}{n}\int_{0}^{{\pi }/{2}\;}{\frac{{{\sin }^{2}}\left( nx \right)}{{{\sin }^{2}}\left( x \right)}f\left( x \right)dx}$$

I strongly believe that the solution depends on this integral

$$\int_{0}^{{\pi }/{2}\;}{\frac{{{\sin }^{2}}\left( nx \right)}{{{\sin }^{2}}\left( x \right)}dx}=\frac{n\pi }{2}$$ and using Weierstrass Approximation Theorem.

I didn't exactly work out but I think Stolz cesaro theorem might help — Rohan Shinde, Jul 31 '18 at 18:17
Are you sure no other information about f is given like if $f$ is an even function or odd function — Rohan Shinde, Jul 31 '18 at 18:26
As much as I could work out (If I have correctly applied Stolz Cesaro and used Dirichlet kernel appropriately) I reached up-to $$\frac 12\int_0^{\pi} f(t)dt +\sum_{k=1}^{\infty}\left(\int_0^{\pi} \left(\cos (kt) f(t)\right) dt\right)$$ — Rohan Shinde, Jul 31 '18 at 18:31
This converges to $\pi f(0)$ (possibly with a constant I've lost) by Fejér's theorem. — Chappers, Jul 31 '18 at 18:48

RRL · Accepted Answer · 2018-08-01T02:44:29.050

The limit is $\displaystyle \frac{\pi}{2}f(0)$.

Note that

$$\tag{*} \left|\int_0^{\pi/2} \frac{\sin^2(nx)}{n\sin^2 x}f(x) \, dx - \frac{\pi}{2}f(0)\right| \\\leqslant \left|\int_0^{\delta} \frac{\sin^2(nx)}{n\sin^2 x}[f(x)-f(0)] \, dx\right|+ \left|\int_\delta^{\pi/2} \frac{\sin^2(nx)}{n\sin^2 x}[f(x)-f(0)] \, dx\right|$$

Since $f$ is continuous on $[0,\pi/2]$ and, hence, bounded we have $|f(x)| \leqslant M$ and $|f(x) - f(0)| \leqslant 2M$ and an estimate for the second integral on the RHS of (*) is

$$\left|\int_\delta^{\pi/2} \frac{\sin^2(nx)}{n\sin^2 x}[f(x)-f(0)] \, dx\right| \leqslant \frac{2M}{n}\int_\delta^{\pi/2} \frac{1}{\sin^2 x} \, dx \to_{n \to \infty} 0$$

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 $\epsilon >0$ we can chose $\delta$ such that $|f(x) - f(0)| < \epsilon$ for $0 \leqslant x \leqslant \delta$.

Using $\sin x > 2x/\pi$ we get

$$\left|\int_0^{\delta} \frac{\sin^2(nx)}{n\sin^2 x}[f(x)-f(0)] \, dx\right| \leqslant \frac{\pi^2 \epsilon}{4n} \int_0^\delta \frac{\sin^2 (nx)}{x^2} \, dx \\ = \frac{\pi^2 \epsilon}{4} \int_0^{n\delta} \frac{\sin^2 u}{u^2} \, du \\ \leqslant \frac{\pi^2 \epsilon}{4} \int_0^{\infty} \frac{\sin^2 u}{u^2} \, du = \frac{\pi^3 \epsilon}{8}$$

Since, $\epsilon$ can be arbitrarily close to $0$ we are done.

@Manthanein: With $u= nx$ we have $\frac{\sin^2(nx)}{nx^2} , dx = \frac{\sin^2(u)}{n(u/n)^2} , d(u/n) = \frac{\sin^2(u)}{u^2} , du$ — RRL, Aug 01 '18 at 02:28
@Manthanein: $f$ is continuous on the compact interval $[0,\pi/2]$. Please read the very first sentence of OP. — RRL, Aug 01 '18 at 02:29
@Manthanein: A continuous function on a closed, bounded interval is bounded. — RRL, Aug 01 '18 at 02:37

Evaluating $\lim_{n\to\infty}\frac1{n}\int_{0}^{\pi /2}{\frac{\sin^2nx}{\sin^2 x}f(x)\;dx}$, for continuous $f$

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