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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 ?

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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

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    @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.2011-09-20
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    request you to kindly provide some reference or hint/clue to the proof of your statement.2011-09-20
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    @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 $(f^{(n)})_h=(f_h)^{(n)}$?2011-09-20
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    \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 ?2011-09-21
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    @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 $\mathbb{R}$.2011-09-21
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    @Rajesh: The Mean Value Theorem says $g(x)=g(d)+g'(\xi)(x-d)$ for some $\xi\in(x,d)$. Thus, $g(t-x)=D+O(t-d-x)$, where $|O(x)|$x$ near $0$ (see [Big-O Notation](http://en.wikipedia.org/wiki/Big_O_notation)). $$\begin{align}&\frac{1}{\pi}\int_{t-d}^{t-d+1}\frac{D+O(t-d-x)}{x}\;\mathrm{d}x\\ &=\frac{1}{\pi}\int_{t-d}^{t-d+1}\frac{D}{x}\;\mathrm{d}x+\frac{1}{\pi}\int_{t-d}^{t-d+1}\frac{O(t-d-x)}{x}\;\mathrm{d}x\\&=\frac{D}{\pi}(\log|t-d+1|-\log|t-d|)+O\left(\frac{1}{\pi}\int_0^1\frac{-x}{x+t-d}\;\mathrm{d}x\right)\tag{1}\end{align}$$ – 2011-09-21
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    @Rajesh: Since we are looking at $t>d$ near $d$, say $0$0\le\log|t-d+1|\le\log(2)$ and $0\le\frac{x}{x+t-d}\le1$ on $[0,1]$. Thus,$$(1)=-\frac{D}{\pi}\log|t-d|+O(1)$$ Does that help? – 2011-09-21
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    Thank you very much, i have understood it. Please consider adding these comments to the answer at the bottom. And just a small question at the end, what do you think of the comment by Andrew about amending the conjecture, I think it is not possible as the way stated by him.(it still won't be finite), please comment as he is not available to clarify. Thanks again for the explanation.2011-09-21
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    Would we really need $f+g$ to be smooth, to apply mean value theorem we need it only to be continuously differentiable. Please clarify.2011-09-21
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    @Rajesh: by "smooth" I meant "as smooth as needed" (so that $f_h$ is continuous). If we are looking at $f^{(k)}$, then we need more smoothness, but if we are simply looking at $f$, then $C^1$ suffices.2011-09-21
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    thank you for the comment.2011-09-21
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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.

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    Thank you very much for the clearly written answer.2011-09-20
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    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 $1/\pi$).2011-09-20
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    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.2011-09-21
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    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.2011-09-21
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    @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.2011-09-21
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    As for the integral, $$ \int_{-\infty}^\infty \frac{e^{-|x-y|}}y\,dy= \int_{-\infty}^x\frac{e^{y-x}}y\,dy+\int_x^{\infty}\frac{e^{x-y}}y\,dy= $$ $$ e^{-x}\int_{-\infty}^x\frac{e^{y}}y\,dy+e^x\int_x^{\infty}\frac{e^{-y}}y\,dy= $$ $$ e^{-x}\int_{-\infty}^x\frac{e^{y}}y\,dy+e^x\int_{-\infty}^{-x}\frac{e^{y}}y\,dy= e^{-x} \text{Ei}(x)-e^x \text{Ei}(-x). $$ The series for $\text{Ei}(x)$ is on the page cited in my answer.2011-09-21