Let $f:\mathbb{R}\to \mathbb{R}$ and $f \in BV(0,1)$ with a jump discontinuity at $x = a\in(0,1)$. Let $f_h$ be its Hilbert transform and let $$f_A(x) = f(x) + i f_h(x)$$ Is it true that the function $$\phi(x) = \angle{f_A(x)}$$ also has a jump discontinuity at $x = a$?
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$\begingroup$ reason for down vote? $\endgroup$– Rajesh DDec 25, 2014 at 3:59
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$\begingroup$ what is $\angle$? is it $\arg$? $\endgroup$– robjohn ♦Dec 25, 2014 at 4:12
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$\begingroup$ @robjohn : yes, it is same as argument of a complex number. $\endgroup$– Rajesh DDec 25, 2014 at 4:17
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$\begingroup$ @robjohn : On these lines : math.stackexchange.com/a/66095/2987 $\endgroup$– Rajesh DDec 25, 2014 at 4:19
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1 Answer
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I think the easiest way to tackle this problem is to write $f = g+h$, where $h$ is a piecewise linear function such that $h(0) = h(1) = 0$, chosen so that $g = f-h$ doesn't have a jump discontinuity at $x = a$. Now presumably you have some results that ensure that $g_A$ doesn't have a jump discontinuity at $x = a$. You should be able to compute $h_A$ explicitly, and at this point the answer should be easy to determine.
(If memory serves me correctly, $h_h(x)$ will probably converge to $\pm\infty$ at $x = a$, with the $\pm$ being different on either side. If this is the case, then the jump discontinuity of $\text{arg}(f_A)$ will be $\pm \pi$.)
Added later: looking at the comments below, it seems like memory did not serve me correctly, and the blow up at $x=a$ occurs with the same sign around both sides. In that case $\text{arg}(f_A)$ will be either $\pi/2$ or $-\pi/2$, with no jump at all, at $x=a$.
answered Dec 25, 2014 at 5:40
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$\begingroup$ Oops - I think I just filled in the gaps for Rajesh D's proposed answer. Well if my explanation helped, you can upvote me. Otherwise thank Rajesh D for giving you a good hint. $\endgroup$ Dec 25, 2014 at 5:43
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$\begingroup$ (+1) the Hilbert transform may blow up logarithmically at a jump discontinuity, but the argument of $f_A$ cannot jump more than $\pi$. $\endgroup$– robjohn ♦Dec 25, 2014 at 6:21
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$\begingroup$ @robjohn : I doubt if $f_h$ blows up with opposite sign on both sides of jump, my guess is blows up with same sign on both sides. So my question is whether $arg(f_A)$ even jumps? $\endgroup$– Rajesh DDec 25, 2014 at 6:24
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$\begingroup$ @RajeshD: You are correct. $\mathrm{PV}\int_{-1}^x\frac1t\,\mathrm{d}t=-\log(|x|)$ and there is no change of sign. $\endgroup$– robjohn ♦Dec 25, 2014 at 7:43