Proof of Riemann_Lebesgue Lemma(Spivak Calculus 15-26)

Question

The following question is from Spivak's Calculus Chap 15 Problem 26.

Note that Problem 13-26 proves the existence of step functions arbitrary close to any integrable function.

I want someone to confirm that by answer to part (c) is legit. My answer (abbreviated)

By problem 13-26 we know that there exists $s_1$ and $s_2$ such that $s_1(x) \leq f(x) \leq s_2(x) $. Then $ \int s_1\sin(\lambda x) \leq \int f(x)\sin(\lambda x) \leq \int s_2(x)\sin(\lambda x)$. We apply limit as $\lambda \to \infty$ and using part (b), $$\lim_{\lambda \to \infty} \int f(x)\sin(\lambda x)=0$$

I am not sure if you can just times the inequality with $\sin(\lambda x)$. Can anyone discuss if my approach is okay?

Answer 1

This is incorrect because the inequality $$s_1(x) \le f(x) \le s_2(x)$$ does not imply that $$s_1(x)\sin(\lambda x) \le f(x)\sin(\lambda x) \le s_2(x)\sin(\lambda x)$$ for all $\lambda$. Moreover, if your method was correct, then simply bounding $f$ by $-M \le f(x) \le M$ for some $M \in \mathbf{R}$ would "imply" $ \lim_{\lambda \to \infty} \int_a^b f(x)\sin(\lambda x) \,dx = 0$ without having to do anything more sophisticated. This would be too easy and should be your first indication that something is off.

Instead you need to find a step function that is close to $f$ in some way that implies

$$ \left\lvert \int_a^b s(x) \sin(\lambda x)\,dx - \int_a^b f(x) \sin(\lambda(x)) \,dx \right\rvert < \varepsilon $$

for all $\lambda$. Then take $\lambda \to \infty$ to obtain

$$ \left\lvert \lim_{\lambda \to \infty} \int_a^b f(x) \sin(\lambda(x)) \,dx \right\rvert < \varepsilon $$

and finally take $\varepsilon \to 0$.

Answer 2

The inequality is not valid since $\sin(\lambda x)$ changes sign (many times in the limit!). Explicitly, consider the interval $[a,b]=[0,2\pi]$ and function $$f(x)=\begin{cases}0 & \text{if }x < \pi \\ \frac{x}{\pi}-1 & \text{if }x>\pi\end{cases}$$ Let $s_1(x)=0$ and $s_2(x)=\chi_{[\pi,2\pi]}$. Note that $s_1\leq f\leq s_2$. Let $\lambda=2$ and we see that for any $x\in [\pi,2\pi]$ that $s_1(x)\sin(\lambda x)=0$, but $s_2(x)\sin(\lambda x)=\sin(\lambda x)$ takes on both positive and negative values on the interval $[\pi,2\pi]$.