# A convolution type singular integral operator with log

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.

• 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 $y<0$)? – Christian Remling May 9 at 18:38
• @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 $\log y$. – Jacob Lu May 11 at 3:57

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 $$s\ge 0$$.
• 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? – Jacob Lu May 11 at 4:13
• @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 $Tf\notin L^2=H^0$. – Christian Remling May 11 at 14:49
• 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 $\log^m|\xi|$? – Jacob Lu May 11 at 20:06
• @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 $(\log |x|)\widehat{}$. – Christian Remling May 11 at 20:29