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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.

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

1 Answer 1

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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.

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