Does a function and its Hilbert transform have the same behaviour? Let $f:\mathbb{R} \to \mathbb{R}$ be a function of bounded variation and let $f_h:\mathbb{R} \to \mathbb{R}$ be its Hilbert transform. If the $k^{th}$ derivative of $f$, $f^{(k)}(x)$ has a jump discontinuity at a point $x_o$ with $f^{(k)}(x_o)$ being not defined and $$|f^{(k)}(x_o^+)-f^{(k)}(x_o^-)| = L$$ I'd like to know if a similar thing can be said about $f_h^{(k)}(x)$ ? I mean whether the statement $$|f_h^{(k)}(x_o^+)-f_h^{(k)}(x_o^-)| = L$$ is true ?
 A: No. Consider the function $f(x)=e^{-|x|}\;$. Its derivative has a jump discontinuity at $x=0$: $|f'(0^+)-f'(0^-)| = 2\ $. But
$$
f_h(x)=e^{-x} \text{Ei}(x)-e^x \text{Ei}(-x)=-2 x (\log |x|+\gamma -1)+O\left(x^2\right),
\quad x\to0,
$$
there $\text{Ei}(x)$ is the exponential integral function. So 
the first derivative $f'_h(x)=-2 \log |x|+O(1)\ $ has a logarithmic singularity at the origin.
A: Since the Hilbert Transform is defined as
$$
f_h(t)=P.V.\frac{1}{\pi}\int_{-\infty}^\infty\frac{f(t-x)}{x}\;\mathrm{d}x\tag{1}
$$
at any jump discontinuity, $d$, of $f$, $f_h$ blows up like $\frac{D}{\pi}\log|t-d|$, where $D$ is the size of the jump discontinuity at $d$.  Furthermore, $(f^{(n)})_h=(f_h)^{(n)}$, so the same can be said for any derivatives of $f$ and $f_h$.
Explanation: Suppose that $f$ is nice (integrable and smooth) away from $d$ and that $\lim\limits_{x\to d^+}f(x)-\lim\limits_{x\to d^-}f(x)=D$. Assume that $t>d$. We can find a $g$ that is nice away from $d$, supported on $[d-1,d]$, and so that $f+g$ is smooth, thus, $f_h+g_h$ is smooth. For such a $g$, $\lim\limits_{x\to d^-}g(x)=D$.
$$
\begin{align}
g_h(t)&=P.V.\frac{1}{\pi}\int_{-\infty}^\infty\frac{g(t-x)}{x}\;\mathrm{d}x\\
&=\frac{1}{\pi}\int_{t-d}^{t-d+1}\frac{g(t-x)}{x}\;\mathrm{d}x\\
&=\frac{1}{\pi}\int_{t-d}^{t-d+1}\frac{D+O(t-d-x)}{x}\;\mathrm{d}x\\
&=\frac{1}{\pi}\int_{t-d}^{t-d+1}\frac{D}{x}\;\mathrm{d}x+O(1)\\
&=-\frac{D}{\pi}\log|t-d|+O(1)\\
\end{align}
$$
Thus, $f_h(t)=\frac{D}{\pi}\log|t-d|+O(1)$. We can do the same for $t<d$ by choosing a $g$ supported on $[d,d+1]$.
