No. For the purposes of absolute integrability, one is only integrating non-negative functions (the magnitude of the integrand) and so there is no possibility of cancellation.

]]>Are you assuming that is nonnegative outside the interval ? Is it possible that one has some cancellation?

]]>If for instance for then the integral that would formally be used to define will not be absolutely convergent regardless of what is doing outside of the interval .

]]>Up to irrelevant constants, yes.

]]>In this post, instead of is used in the definition of the Hilbert transform. Are the two versions the same?

]]>By writing and using the triangle inequality one can easily verify the asymptotic limits for and . If is non-zero, then is bounded from below by for some and all sufficiently large , which implies that is not absolutely integrable.

]]>Also, why does (*) imply that is not bounded on ?

]]>for . You mentioned this in the very beginning of notes 4. Can one prove this without using the result that maps the Schwartz class to ? It seems that (*) is equivalent to

for . ]]>