2.22

This exercise is asking us to prove the Riemann-Lebesgue Lemma. Exactly as the hint prescribes, first observe that $\xi \cdot \xi' = \frac{1}{2}$, and then by the translation invariance of the Lebesgue integral: $$\hat{f}(\xi) = \int_{\mathbb{R}^d} f(x - \xi')e^{-2\pi i (x - \xi') \cdot \xi} \hspace{0.1cm}dx = \ldots$$ $$= \int_{\mathbb{R}^d} f(x - \xi')e^{-2\pi i x \cdot \xi} e^{-2\pi i \xi \cdot \xi'} \hspace{0.1cm}dx = -\int_{\mathbb{R}^d} f(x - \xi')e^{-2\pi i x \cdot \xi}\hspace{0.1cm}dx$$ So we can certainly rewrite $\hat{f}(\xi)$ as: $$\hat{f}(\xi) = \frac{1}{2} \int_{\mathbb{R}^d}\big(f(x)- f(x - \xi')\big)e^{-2\pi i x \cdot \xi}\hspace{0.1cm}dx$$ Now, observe that: $$\lim_{|\xi| \to \infty} |\hat{f}(\xi)| = \lim_{|\xi| \to \infty} \Bigg|\frac{1}{2} \int_{\mathbb{R}^d}\big(f(x)- f(x - \xi')\big)e^{-2\pi i x \cdot \xi}\hspace{0.1cm}dx\Bigg| = \dagger$$ ...and thus, by the triangle inequality, and since $\xi' \to 0$ if $|\xi| \to \infty$, it's clear that: $$\dagger \leq \lim_{\xi' \to 0} \frac{1}{2} \int_{\mathbb{R}^d}\big|f(x)- f(x - \xi')\big| \hspace{0.1cm}dx = 0$$ ...from Proposition 2.5 (p. 74).