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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. Can we show that if $f \in H^s$, then for any $\epsilon > 0$ and $m \ge 1$, $T_m(f) \in H^{s-\epsilon}$?

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.

Define a convolution type operator $T$ by $$T(f) = p.v.\int_\mathbb{R}f(x-y)\frac{\log^m|y|}{y}dy.$$ 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?

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. Can we show that if $f \in H^s$, then for any $\epsilon > 0$ and $m \ge 1$, $T_m(f) \in H^{s-\epsilon}$?

Define a convolution type operator $T$ by $$T(f) = \int_\mathbb{R}f(x-y)\frac{\log^m(y)}{y}dy.$$ 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?

Define a convolution type operator $T$ by $$T(f) = p.v.\int_\mathbb{R}f(x-y)\frac{\log^m|y|}{y}dy.$$ 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?