# 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 ?

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 $[d,d+1]$.

• @all : While the answer by Andrew is complete with a good example (for which i thank him), I've selected this answer (by robjohn) as I can select only one and it provides useful information in a generic sense. Please not that none of these answers contain any proof and hence i had to select the answer based on information they provide. – Rajesh D Sep 20 '11 at 17:53
• request you to kindly provide some reference or hint/clue to the proof of your statement. – Rajesh D Sep 20 '11 at 18:03
• @Rajesh: I have ammended my answer to include an explanation of the estimate for $f_h$. It was unclear which statement you wanted explained. Did you also need a proof that $(f^{(n)})_h=(f_h)^{(n)}$? – robjohn Sep 20 '11 at 21:13
• \begin{align}&=\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\\ \end{align} I do not understand how you have got the second step from the first step ? I do not understand anything about $D+O(t-d-x)$, like what is meant by $O(t-d-x)$ ? Also \begin{align} &=\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}, how is this integral evaluated, how did the limits substituted after integration ? – Rajesh D Sep 21 '11 at 6:14
• @Rajesh: Since you posited that $|f^{(k)}(x_0^+)-f^{(k)}(x_0^-)| = L$, I use that $f$ could be extended to a smooth function on $[-\infty,d]$ and to a smooth function on $[d,\infty]$, in which case, we can find a $g$ so that $g(d)=D$ and $g(x)=0$ for $x\not\in[d-1,d]$ and $f+g$ is smooth on $\mathbb{R}$ and an $h$ so that $h(d)=-D$ and $h(x)=0$ for $x\not\in[d,d+1]$ and $f+h$ is smooth on $\mathbb{R}$. – robjohn Sep 21 '11 at 12:42

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.

• Thank you very much for the clearly written answer. – Rajesh D Sep 20 '11 at 17:54
• May be the conjecture still could be amended if to consider the symmetric difference, so to say the principal value: $$\lim_{t\to+0}|f_h^{(k)}(x_o+t)-f_h^{(k)}(x_o-t)| = C L,$$ where $C$ depends upon chosen constant before after V.P., say $C=2/\pi$ in the form of the Hilbert transform written in the answer of robjohn or $C=2$ for my answer (since I omitted the multiplier $1/\pi$). – Andrew Sep 20 '11 at 21:46
• How can you say that $e^{-x} \text{Ei}(x)-e^x \text{Ei}(-x)=-2 x (\log |x|+\gamma -1)+O\left(x^2\right), \quad x\to0,$, I do not understand how to derive this, also i do not know well about the $O(x^2)$ business. Kindly let me know how to derive this. – Rajesh D Sep 21 '11 at 10:43
• regarding your comment on amending the conjecture, as we make the $t$ small, the thing blows up, proportional to $L$ (amount of jump) lograthamically. Hence the difference also blows up logarthamically, only thing can be said is that the proportionality is $CL$ and not equal to $CL$. Please correct me if i have misunderstood. – Rajesh D Sep 21 '11 at 10:49
• @Rajesh I was mistaken about amendment. Though the logarithmic singularity is even and for the function in my answer $\lim_{t\to0^+}|f'_h(x_o+t)-f'_h(x_o-t)| = 0\$, so the value $C=0$ is not excluded. – Andrew Sep 21 '11 at 17:55