Exercise 8.5.6

Question

Show that $\left|\int_a^b h(x) \sin(nx) dx\right| < \epsilon/n$, and use this fact to complete the proof.

Ulisse Mini · Accepted Answer

Slight change to the premise - we'll require that $\left|h(x) - h(y)\right| < \epsilon / (2\pi)$ when $\left|x-y\right| < \delta$.

Define $\Delta h(x) $ satisfying $h(x) = h(\frac{a+b}{2}) + \Delta h(x) $ and note that by uniform continuity, $\left|\Delta h(x)\right| < \epsilon / {2\pi}$. Note also that $\int_a^b \sin(nx) dx= 0$.
$$
    \begin{aligned}
    \left|\int_a^b h(x) \sin(nx) dx\right| &= \left|h\left(\frac{a+b}{2}\right) \int_a^b \sin(nx) dx + \int_a^b \Delta h(x) dx\right| \leq \int_a^b \left|\Delta h(x)\right| dx \
    &< \frac{\epsilon}{2\pi} \frac{2\pi}{n} = \epsilon/n
    \end{aligned}
$$
Now, subdivide $[-\pi, \pi]$ into $n$ subintervals, each of size $2\pi/n$, and apply the above result to each interval; this shows that $$\left|\int_{-\pi}^\pi h(x) \sin(nx)\right| < \epsilon$$. The process can be repeated with $\cos(nx)$ replacing $\sin(nx)$ to get $$\left|\int_{-\pi}^\pi h(x) \cos(nx)\right| < \epsilon$$

Exercise 8.5.6

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Exercise 8.5.6

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