Establish if $g_n (\alpha)=\int_a^b \ \alpha(x) \ \sin (nx) \ \cos(nx) $ converges uniformly

Question

$$X=\{ \alpha:[a,b] \rightarrow \mathbb{R} \}$$ $\alpha''$ exists and it is continuous $$\exists \ K>0 \ : \forall \ x \in [a,b], \forall \alpha \in X: \\ \ \\ \rvert \alpha(x) \rvert, \rvert \alpha''(x) \rvert \le K$$ $n=1,2,... \qquad (g_n): X \rightarrow \mathbb{R}$ $$g_n (\alpha)=\int_a^b \alpha(x) \ \sin (nx) \ \cos(nx) $$
Establish if $g_n (\alpha) $ converges uniformly

I'm thinking about taking limits under the integral sign.

But, first I have to study the convergence of: $ f_n(x)=\alpha(x) \ \sin (nx) \ \cos(nx)$

$lim_{n \rightarrow \infty} \ f_n(x) $ does not exist for these terms: $\sin (nx) \ \cos(nx)$ .

But, I know that $\rvert f_n(x) \rvert \le K $

How can I continue the exercise?

Thanks!

Answer 1

Observe first that $$ \int_a^b \alpha(x)\,\sin(nx)\,\cos(nx)\,dx=\frac{1}{2}\int_a^b \alpha(x)\,\sin(2nx)\,dx. $$ Then, using integration by parts, we obtain that $$ \frac{1}{2}\int_a^b \alpha(x)\,\sin(2nx)\,dx=\frac{1}{2}\alpha(x)\left(-\frac{\cos(2nx)}{2n}\right)\,\bigg|_a^b+\frac{1}{2n}\int_a^b \alpha'(x)\,\cos(2nx)\,dx $$ Clearly $$ \left|\frac{1}{2}\alpha(x)\left(-\frac{\cos(2nx)}{2n}\right)\,\bigg|_a^b\right|\le \frac{1}{4n}\big(\left|\alpha(a)\cos(2na)\right|+\left|\alpha(b)\cos(2nb)\right|\big)\le \frac{2K}{4n}=\frac{K}{2n} $$ and $$ \left|\frac{1}{2n}\int_a^b \alpha'(x)\,\cos(2nx)\,dx\,\right|\le \frac{K(b-a)}{2n}. $$ Hence $$ \left|\int_a^b \alpha(x)\,\sin(nx)\,\cos(nx)\,dx\,\right|\le\frac{1}{n}\left(\frac{K}{2}+\frac{K(b-a)}{2}\right) $$

Answer 2

Hints: 1. $\sin (nx)\cos (nx) = (1/2)\sin (2nx).$

2. There exists $K'$ such that $|\alpha'(x)|\le K'$ for all $\alpha \in X.$

3. Integration by parts.