It’s a problem when I read the Fourier Series in one calculus textbook.
Suppose that $0<a<\pi$,$f\in C^0[0,a]$,show that $$ \underset{\lambda \rightarrow \infty}{\lim}\frac{1}{\lambda}\int_0^a{f\left( x \right) \frac{\sin ^2\lambda x}{\sin ^2x}}dx=\frac{\pi}{2}f\left( 0 \right) . $$
I have made two attempts to solve this problem. One is the approximation of $f$ by trigonometric polynomials, and the other is the fourier expansion for $sin^2x$.However,I have failed.
Hints: First verify that this is true when $f$ is a constant by making the change of variable $t=\lambda x$. You will need the fact that $\int_0^{\infty} \frac {\sin ^{2}t} {t^{2}} dt=\frac {\pi} 2$.
Next reduce the proof to the case $f(0)=0$ by considering the function $f(x)-f(0)$.
Choose $r\in (0,a)$ such that $|\int_0^{r} f(x) \frac {\sin ^{2}(\lambda x)} {\sin^{2}x} dx| <\epsilon$. Now observe that $|\int_r^{a} f(x) \frac {\sin ^{2}(\lambda x)} {\sin^{2}x} dx| \leq M |\int_r^{a} f(x) \frac {\sin ^{2}(\lambda x)} {\sin^{2}x} dx|$ where $M=sup |f(x)|$. Now use the inequlity $\sin t \geq \frac 2 {\pi}t$ for $t \in (0,\pi)$. Also $\int_r^{a} f(x)\sin^{2}(\lambda x) dx=\frac 1 2 \int_r^{a} f(x)[1-\cos (2\lambda x)d$. Use Riemann Lebesgue Lemma to show that $\int_r^{a} f(x) \frac {\sin ^{2}(\lambda x)} {\sin^{2}x} dx $ is bounded.