Convergence of integral with compact support

Question

Let $a,b \in \mathbb{R}$ with $a<b$ and let $f:[a,b] \longrightarrow \mathbb{R}$ be continuous. Show that $$\int_a^bf(x)\text{cos}(nx)\:dx \longrightarrow 0 \quad as \quad n \longrightarrow \infty.$$ My attempt

Let $\epsilon >0.$ By uniform continuity of $f$, there exist $\delta>0$ such that $$|s-t|< \delta \Longrightarrow |f(s)-f(t)| < \frac{\epsilon}{2(b-a)}.$$Pick $a=x_0<x_1<...<x_m$ such that $|x_{k-1}-x_k|< \delta.$ Then we may write $$\int_a^bf(x)cos(nx)\:dx= A_n+B_n$$where $$A_n=\sum_{k=1}^m \int_{x_{k-1}}^{x_k}(f(x)-f(x_k))\text{cos}(nx)\;dx$$ and $$B_n=\sum_{k=1}^mf(x_k) \int_{x_{k-1}}^{x_k}\text{cos}(nx)\;dx.\text{(from here i dont know what to do).}$$ Next use the preceding to show that $$\int_a^b\rho(x)\text{cos}(nx)\:dx\longrightarrow 0 \quad as \quad n \longrightarrow \infty.$$ where $\rho \in L^1(\mathbb{R})$. I know that i should the denseness of $C_c(\mathbb{R}) \in L^1(\mathbb{R})$.

Any help,suggestions, outlines, or hints for this one. I've been really messing it up. Thanks.

Answer 1

Hint: If $f$ is continuous on $[a,b]$, that implies that it is bounded on that set, and therefore $$\left|\int_{a}^{b}f(x)\cos(nx) ~\mathrm{d} x\right| \leq \|f\|_{\infty}\int_{a}^{b}|\cos(nx)| ~\mathrm{d} x$$ Now use the substitution $u = nx$.