Lp Convergence of Fourier Integrals Using Hilbert Transform

Question

I am reading about Hilbert transform and its application on Fourier analyisis, and I am triying to prove a statement given by Terence Tao in his notes on Fourier Analysis. He says that if $\varphi\in\mathcal{S}(\mathbb{R})$ (Schwartz space), then $$\int_{-R}^R\hat{\varphi}(\xi)e^{2\pi ix\xi}\, d\xi\to\varphi\text{ in }L^p(\mathbb{R}) \text{ with } 1<p<\infty, \text{ as } R\to\infty.$$ The hint given in the proof suggests to use the dominated convergence theorem. Obviously, from the inversion formula follows the pointwise convergence because $\varphi$ is a Schwartz function, but I don't know how to find a function $f\in L^p(\mathbb{R})$ such that for a.e point $x\in\mathbb{R}$, $$\left|{\int_{-R}^R\hat{\varphi}(\xi)e^{2\pi ix\xi}\, d\xi}\right|\leq |f(x)|.$$ The link where notes can be found is https://www.math.ucla.edu/~tao/247a.1.06f/notes4.pdf , and the statement is in page 9. Thanks four your help.

Answer 1

Pick a constant $M>0$. For $|x|\leq M$, the left-hand side is dominated by $\|\hat \varphi\|_1$.

For $|x|>M$, integrate by parts: $$\begin{split} \left|{\int_{-R}^R\hat{\varphi}(\xi)e^{2\pi ix\xi}\, d\xi}\right| &= \left| \frac{e^{2i\pi xR}\hat \varphi(R) - e^{-2i\pi xR}\hat \varphi(-R)}{2i\pi x}- \frac 1 {2i\pi x}\int_{-R}^R\hat{\varphi}^\prime(\xi)e^{2\pi ix\xi}\, d\xi\right|\\ &\leq \frac{2\|\hat\varphi\|_\infty+\|\hat\varphi^\prime\|_1}{2\pi |x|} \end{split}$$ and $x\mapsto \mathbb 1_{|x|>M}\cdot \frac 1 x$ is in $L^p$ for $p>1$. This is a form of the Riemman-Lebesgue Lemma.