I have a problem in the proof of the following result: Given $f \in L^1(\mathbb{R}^n)$, we have that
$$|\hat{f}(\xi)| \rightarrow 0, \;\;\; as |\xi| \rightarrow \infty.$$
This result is known as the Riemann-Lebesgue Lemma (Proposition 2.2.17 from Grafakos's book Classical Fourier Analysis, third edition). In the proof of this proposition, one considers the function
$$g := \prod_{j=1}^n \chi_{[a_j,b_j]},$$
that I suppose is the characteristic function of the cube $\prod_{j=1}^n[a_j,b_j] \subset \mathbb{R}^n$, and whose Fourier transform is
$$\hat{g}(\xi) = \prod_{j=1}^n \frac{e^{-2\pi i \xi_ja_j} – e^{-2\pi i \xi_jb_j} }{2\pi i \xi_j},$$
in the meaning that if some $\xi_j = 0$, the correspondent factor is equal $b_j-a_j$. Now, if $\xi =(\xi_1, …, \xi_n) \neq 0$, choose $j_0$ such that $|\xi_{j_0}| \geq |\xi|/\sqrt{n}$. So that
$$\left| \prod_{j=1}^n \frac{e^{-2\pi i \xi_ja_j} – e^{-2\pi i \xi_jb_j} }{2\pi i \xi_j} \right| \leq
\frac{2\sqrt{n}}{2\pi|\xi|}\sup_{1\leq j_0\leq n}\prod_{j\neq j_0}(b_j-a_j).$$
This inequality is what I'm not being able to prove. Once it's proved, I have the desired result.
Best Answer
For the $j_0$ term in the product, we bound by $\frac{2\sqrt{n}}{2\pi |\xi|}$ via the triangle inequality on the complex exponentials. For the other terms, we are simply using the fact that each is bounded by $b_j-a_j$. To see this, note that $$|e^{-2\pi i\xi_ja_j}-e^{-2\pi \xi_jb_j}|=|e^{2\pi i\xi_j(b_j-a_j)}-1|=2\sin(\pi(b_j-a_j)|\xi_j|)$$
Thus, what you have to do is bound $\frac{\sin\pi(b_j-a_j) |\xi_j|}{\pi|\xi_j|}$ by $b_j-a_j$, which you may do via calculus.