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I ask myself what $$ {\mathscr F}^{-1}( e^{it\xi} ({\mathscr F} \phi)'(\xi) )(s) $$ is. If it was just about $$ {\mathscr F}^{-1}( e^{it\xi} ({\mathscr F} \phi)(\xi) )(s) $$ it would be clear (a shift by $t$), the same is with $$ {\mathscr F}^{-1}( ({\mathscr F} \phi)'(\xi) )(s), $$ which gives a multiplication by $-is$.

But what is about the combination of exponential and derivative? Any hints? Thanks, Eric

Austin Mohr
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Eric
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1 Answers1

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Why don't you just compute it? $$ {\mathscr F}^{-1}( e^{it\xi} ({\mathscr F} \phi)'(\xi) )(s)={\mathscr F}^{-1}( ({\mathscr F} \phi)'(\xi) )(s+t)=-i(s+t)\phi(s+t) $$

Thomas Klimpel
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