Jump discontinuity of a function and its analytic phase

Question

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 $x = a$?

Answer 1

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 $x=a$.