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Convolution inequality

  1. Nov 29, 2006 #1


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    Given f,g in L^1(R), I'm given the inequality:

    [tex] ||f * g||_1 \leq ||f||_1 ||g||_1[/tex]

    And now I'm supposed to derive the conditions for equality to hold. I keep going around in circles. So far, I can show equality holds iff there is some real function [itex]\phi(x)[/itex] such that [itex]e^{i \phi(x+y)} f(x) g(y) \geq 0[/itex] holds almost everywhere, but I can't easily translate this into conditions on f and g. If f and g were both almost everywhere non-zero, this would just say they need to have linear phases with the same slope, but this doesn't help when they are zero on sets of positive measure.

    For the case when f is real and g is non-negative, equality holds iff the support of [itex]f^+*g[/itex] and [itex]f^-*g[/itex] don't intersect, where [itex]f^\pm[/itex] are the positive and negative parts of f (I use the fact that convolutions of L^1 functions are continuous). The best I can do with this is say it is equivalent to there being some x such that the support of g(y) intersects both the support of [itex]f^+(x-y)[/itex] and that of [itex]f^-(x-y)[/itex] in sets of positive measure. Is this the best I can do? And how would I extend it to the general case? Like I said, I'm not getting anywhere.
    Last edited: Nov 29, 2006
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  3. Nov 30, 2006 #2


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    I'm doing other problems now and I keep coming back to this same related question: when does |f*g(x)|=|f|*|g|(x) hold almost everywhere. This is all I need to know, but I can't figure it out. It seems like I'm overlooking something really simple. Can anyone help me out?
    Last edited: Nov 30, 2006
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