Prove that a $\mathcal{C}^1$ function in $[a,b]$ with $f(a) =f(b)$ has this limit

Question

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Prove that if $f \in \mathcal{C}^1([a,b])$ with $f(a) = f(b)$ then $$ \lim_{n \to \infty} \int_a^b f(x) \sin nx \, \mathrm{d} x = 0 $$

Thank you.

Answer 1

You can prove the Riemann-Lebesgue Lemma under much weaker conditions, but just assume $f \in C([a,b])$. Choose a partition $a=x_0 < x_1 < \dots, x_m=b.$

Then

$$\int_{a}^{b}f(x)\sin nx \, dx=\sum_{j=1}^{m}\int_{x_{j-1}}^{x_j}[f(x)-f(x_j)]\sin nx\,dx + \sum_{j=1}^{m}\int_{x_{j-1}}^{x_j}f(x_j)\sin nx\,dx$$

and using the triangle inequality

$$\Big|\int_{a}^{b}f(x)\sin nx \, dx\Big|\leq \\ \sum_{j=1}^{m}\int_{x_{j-1}}^{x_j}|f(x)-f(x_j)||\sin nx|\,dx + \sum_{j=1}^{m}|f(x_j)|\frac{|\cos nx_j-\cos nx_{j-1}|}{n} \\ \leq \sum_{j=1}^{m}\int_{x_{j-1}}^{x_j}|f(x)-f(x_j)|\,dx + \sum_{j=1}^{m}|f(x_j)|\frac{2}{n}$$

Since $f$ is continuous on $[a,b]$ it is bounded and uniformly continuous. Hence there exist $M > 0$ such that for all $x \in [a,b],$

$$|f(x)| \leq M,$$

and for every $\epsilon > 0$ there exists $\delta >0$ such that for all $|x-y| < \delta,$

$$|f(x)-f(y)| < \frac{\epsilon}{2(b-a)}.$$

Let the norm of the partition be less than $\delta$.

Then

$$\Big|\int_{a}^{b}f(x)\sin nx \, dx\Big|\leq \frac{\epsilon}{2} + \frac{2Mm}{n}.$$

For all $n > 4Mm/\epsilon$, we have

$$\Big|\int_{a}^{b}f(x)\sin nx \, dx\Big|\leq \frac{\epsilon}{2} + \frac{\epsilon}{2}= \epsilon,$$

and

$$\lim_{n \rightarrow \infty}\int_{a}^{b}f(x)\sin nx \, dx=0$$

Answer 2

Hints: $$ \begin{array}{cc} u = f(x) & v = ? \\ \mathrm{d} u = ? & \mathrm{d} v = \sin nx \, \mathrm{d} x \end{array} $$ $$ -1 \le \cos u \le 1 \, \forall u \in \mathbb{R} $$