Supremum of an integral in distributions?

Question

I am reading a book on distributions and they are trying to show that $u = \sum_{n=-\infty}^{\infty} a_n e^{inx}$ converges to a distribution. I understand most of the argument the book is showing but at some point when analyzing the behavior of each of the terms of this integral, they obtain the integral $\int_{-\infty}^{\infty} e^{ikx}\partial^{n+2}\phi(x)dx$ and when taking the absolute value of this, they claim that $ |\int_{-\infty}^{\infty} e^{ikx}\partial^{n+2}\phi(x)dx|\leq \sup|\partial^{n+2}\phi(x)|$, how can argue that this is true? I just don't see how the integral is bounded by the supremum of the partial derivative. Any guide, help or references to understand this is highly appreciated. Thanks!

Answer 1

If $\phi \in C_c^\infty(\mathbb R)$ then there is $R>0$ such that $$ \left| \int_{-\infty}^{\infty} e^{ikx} \partial^{n+2}\phi(x) \, dx \right| =\left| \int_{-R}^{R} e^{ikx} \partial^{n+2}\phi(x) \, dx \right| \\ \leq \int_{-R}^{R} \left| e^{ikx} \partial^{n+2}\phi(x) \right| \, dx = \int_{-R}^{R} \left| \partial^{n+2}\phi(x) \right| \, dx \\ \leq \int_{-R}^{R} \sup \left| \partial^{n+2}\phi(x) \right| \, dx = \int_{-R}^{R} \sup \left| \partial^{n+2}\phi(x) \right| \, dx \\ = 2R \sup \left| \partial^{n+2}\phi(x) \right| $$