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Define a convolution type operator $T_m$ by $$T_m(f) = p.v.\int_\mathbb{R}f(x-y)\frac{\log^m|y|}{y}dy.$$ Here $m\ge0$ is an integer. Consider $f \in H^s (s > 0)$ which is the usual Sobolev space. We know that if $m = 0$, $T$ is the Hilbert transform and is a bounded operator on $H^s$. What can we say about the case $m\ge 1$? Can we get the same conclusion?

Edit: a further question

From the answer of Christian we know that $T_m$ cannot be a bounded operator on the Sobolev space $H^s$. It seems there is only slight loss of regularity. Is it possible to show that $$|\mathcal{F}(\frac{\log^m|x|}{x})| \lesssim A + B |\log^m|\xi||?$$ If this holds, we can then show that if $f \in H^s $ for some $ s > -1/2$, then for any $\epsilon > 0$ and $m \ge 1$, $T_m(f) \in H^{s-\epsilon}$. The condition $s > -1/2$ is required to make sure $\hat{f}(\xi)$ is locally bounded around the origin.

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    $\begingroup$ This isn't defined (as an integral). For $m=0$, the usual procedure would be to regularize by taking the principal value. I assume for $m\ge 1$, you wanted to do something similar? Also, did you mean $\log |y|$ (or else, what is $\log y$ for $y<0$)? $\endgroup$
    – Christian Remling
    Commented May 9, 2021 at 18:38
  • $\begingroup$ @ChristianRemling Thanks for pointing this out! I agree that the integral should be regularized by taking the principal value. Then we need $\log |y|$ instead $\log y$. $\endgroup$
    – Jacob Lu
    Commented May 11, 2021 at 3:57

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No, this isn't true. I interpret $T$ as suggested in my comment, as $Tf = t*f$, with $t\in\mathcal S'$ being the tempered distribution $$ t =\textrm{PV}\frac{\log^m |x|}{x} . $$

Also, I'll focus on $m=1$. Now $xt = \log |x|$ has Fourier transform $a(\textrm{sgn}(\xi)\log|\xi|)' + b\delta$ (in distributional sense), see here. Thus $$ \widehat{t} = A\, \textrm{sgn}(\xi)\log|\xi| + B\chi_{(0,\infty)}(\xi) + C $$ is not bounded and since $(t*f)\widehat{ }=\widehat{t}\widehat{f}$, $T$ is unbounded on $H^s$ for every $s\ge 0$.

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  • $\begingroup$ Many thanks for the answer! This is very helpful! Since the Fourier transform of $t$ is just slightly away (just a $\log$ error) from a bounded function, it seems to be true that $T(f) \in H^{s-\epsilon}$ for every $m\ge 1$ and $\epsilon > 0$ assuming $f \in H^s$. Do you agree with this? $\endgroup$
    – Jacob Lu
    Commented May 11, 2021 at 4:13
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    $\begingroup$ @JacobLu: No, this doesn't quite follow (it would if $\widehat{t}$ were locally bounded, with $\log |\xi|$ increase only at large $\xi$). The problem is that $\widehat{t}$ is also unbounded near $\xi =0$. So if $\int_{|\xi|<1}|\widehat{f}(\xi)|^2\log^2|\xi| \, d\xi = \infty$ (which can happen for $f\in H^s$, no matter how large $s$ is), then $Tf\notin L^2=H^0$. $\endgroup$
    – Christian Remling
    Commented May 11, 2021 at 14:49
  • $\begingroup$ Oh yes, that's indeed an issue. A sufficient condition would be to make $\hat{f}(\xi)$ bounded near the origin, so $f \in L^1$ would suffice. By the Sobolev embedding, we should require $f \in H^s$ for $s > -1/2$. Then $Tf \in H^{s-\epsilon}$ for any $\epsilon$. Do you think we can prove that the Fourier transform of $\frac{\log^m|x|}{x}$ is bounded above by $\log^m|\xi|$? $\endgroup$
    – Jacob Lu
    Commented May 11, 2021 at 20:06
  • $\begingroup$ @JacobLu: I don't understand the first part of your comment: $f\in H^s$ will never imply that $f\in L^1$ (just focus on the Fourier transforms: the condition $(1+|\xi|)^s\widehat{f}\in L^2$ doesn't restrict local singularities of $\widehat{f}$ other than by the $L^2$ condition that we always have). The conjecture on the FT of $\textrm{PV}(\log^m |x|/x)$ sounds reasonable, and probably we can get this from the $m=1$ case by taking repeated convolutions with $(\log |x|)\widehat{}$. $\endgroup$
    – Christian Remling
    Commented May 11, 2021 at 20:29

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