# $\bigg|\int_{0}^{\infty}f(s)e^{-ias}ds\bigg| \leq C$ [closed]

I would like to know for which conditions on the function $$f \in C^{1}([0,\infty]) \cap L^{1}([0,\infty])$$ there is a constant $$C > 0$$ such that $$\bigg|\int_{0}^{\infty}f(s)e^{-ias}ds\bigg| \leq C$$ with $$a \in \mathbb{R}$$

I have no idea how to do this problem.

• What have you tried? Commented Mar 6, 2023 at 18:02
• @KamalSaleh I have no idea how to do this problem. Commented Mar 6, 2023 at 18:04
• Okay, so where did you get this problem from? This would make it easier for us to help you. Commented Mar 6, 2023 at 18:05
• It's on a list of exercises my teacher gave me. Commented Mar 6, 2023 at 18:06
• Have you heard of the Laplace transform? you could re-write the problem that way. Commented Mar 6, 2023 at 18:10

From the triangle inequality for complex integrals you have $$\bigg|\int_{0}^{\infty}f(s)e^{-ias}ds\bigg| \leq \int_{0}^{\infty}|f(s)e^{-ias}|ds=\int_{0}^{\infty}|f(s)|ds$$
So you have to require $$\int_{0}^{\infty}|f(s)|ds<\infty$$ but this is the definition of $$f$$ belonging to $$L^1$$, and you are done.