convergence dominate of Lebesgue-Application

Question

my question is: let $f(x)= \dfrac{\sin (nx)}{x}.\varphi(x)$, where $\varphi \in C^\infty_c(\mathbb{R})$. We can found an function $g \in L^1(\mathbb{R})$ such as $|f(x)| \leq g$? Please. I think that, supposing that $Supp(\varphi) \subset [-a,a]$ with $a>0$. We have $$ |\displaystyle\int_{\mathbb{R}} \dfrac{\sin (nx)}{x}.\varphi(x)|\leq C \displaystyle\int_{\mathbb{R}} |\dfrac{\sin(nx)}{x}|dx $$ where $C=\sup_x|\varphi(x)|$. then since $\displaystyle\int_{\mathbb{R}} |\dfrac{\sin(nx)}{x}|dx=\pi$, there exist $g(x)= \dfrac{\sin(nx)}{x} dx \in L^1(\mathbb{R})$. Is it correct?

Answer 1

You are wrong, $\sin(x)/x\not \in L^1(\mathbb{R})$. We have that $$\int_{\mathbb{R}} \dfrac{\sin(x)}{x}dx=\pi\quad\text{and}\quad \int_{\mathbb{R}} \left|\dfrac{\sin(x)}{x}\right|dx=+\infty.$$ See Evaluating the integral $\int_0^\infty \frac{\sin x} x \ dx = \frac \pi 2$? and Improper integral $sin(x)/x $ converges absolutely, conditionaly or diverges? .

On the other hand, since $|\sin(x)|\leq |x|$ for all $x\in \mathbb{R}$, it follows that $$\left|\dfrac{\sin (nx)}{x}\cdot \varphi(x)\right|\leq n|\varphi(x)|$$ and $g=n|\varphi|\in L^1(\mathbb{R})$.