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First of all, as explained in my comment, this is the same as asking if the Hilbert transform is bounded on $L^2(\mathbb R, w\, dx)$, with $w(x)=(1+|x|)^{2s}$. Or, to state this one more time, this reformulation follows because restricting $u$ to a half line is essentially the same as applying the Hilbert transform to $\widehat{u}$. Hunt, Muckenhoupt, Wheeden proved that this is equivalent to $w\in A_2$ (see Theorem 9 there).

It is easy to check, using the defining condition $$ \int_I w \int_I w^{-1}\lesssim |I|^2 $$ of $A_2$, that indeed $w\in A_2$ for our weight $w(x)=(1+|x|)^{2s}$ as long as $-1/2<s<1/2$.

The argument cannot handle $s=1/2$ (or $s\le -1/2$), which one would perhaps also expect to be more difficult since it is the borderline case. For $s>1/2$, $H^s$ functions are continuous, so the result clearly no longer holds in this case.

First of all, as explained in my comment, this is the same as asking if the Hilbert transform is bounded on $L^2(\mathbb R, w\, dx)$, with $w(x)=(1+|x|)^{2s}$. Or, to state this one more time, this reformulation follows because restricting $u$ to a half line is essentially the same as applying the Hilbert transform to $\widehat{u}$. Hunt, Muckenhoupt, Wheeden proved that this is equivalent to $w\in A_2$ (see Theorem 9 there).

It is easy to check, using the defining condition $$ \int_I w \int_I w^{-1}\lesssim |I|^2 $$ of $A_2$, that indeed $w\in A_2$ for our weight $w(x)=(1+|x|)^{2s}$ as long as $-1/2<s<1/2$.

If $s=1/2$, then in fact $w(x)=1+|x|\notin A_2$, so the argument shows that $u\in H^{1/2}$ does not in general admit a $u=u_++u_-$ decomposition. Thank you to Giorgio for pointing this out, and I hope no one read the rather nonsensical comment I made here originally.

First of all, as explained in my comment, this is the same as asking if the Hilbert transform is bounded on $L^2(\mathbb R, w\, dx)$, with $w(x)=(1+|x|)^{2s}$. Or, to state this one more time, this reformulation follows because restricting $u$ to a half line is essentially the same as applying the Hilbert transform to $\widehat{u}$. Hunt, Muckenhoupt, Wheeden proved that this is equivalent to $w\in A_2$.

It is easy to check, using the defining condition $$ \int_I w \int_I w^{-1}\lesssim |I|^2 $$ of $A_2$, that indeed $w\in A_2$ for our weight $w(x)=(1+|x|)^{2s}$ as long as $-1/2<s<1/2$.

The argument cannot handle $s=1/2$ (or $s\le -1/2$), which one would perhaps also expect to be more difficult since it is the borderline case. For $s>1/2$, $H^s$ functions are continuous, so the result clearly no longer holds in this case.

First of all, as explained in my comment, this is the same as asking if the Hilbert transform is bounded on $L^2(\mathbb R, w\, dx)$, with $w(x)=(1+|x|)^{2s}$. Or, to state this one more time, this reformulation follows because restricting $u$ to a half line is essentially the same as applying the Hilbert transform to $\widehat{u}$. Hunt, Muckenhoupt, Wheeden proved that this is equivalent to $w\in A_2$ (see Theorem 9 there).

It is easy to check, using the defining condition $$ \int_I w \int_I w^{-1}\lesssim |I|^2 $$ of $A_2$, that indeed $w\in A_2$ for our weight $w(x)=(1+|x|)^{2s}$ as long as $-1/2<s<1/2$.

The argument cannot handle $s=1/2$ (or $s\le -1/2$), which one would perhaps also expect to be more difficult since it is the borderline case. For $s>1/2$, $H^s$ functions are continuous, so the result clearly no longer holds in this case.

First of all, as explained in my comment, this is the same as asking if the Hilbert transform is bounded on $L^2(\mathbb R, w\, dx)$, with $w(x)=(1+|x|)^{1/2}$. Or, to state this one more time, this reformulation follows because restricting $u$ to a half line is essentially the same as applying the Hilbert transform to $\widehat{u}$. Hunt, Muckenhoupt, Wheeden proved that this is equivalent to $w\in A_2$.

It is easy to check, using the defining condition $$ \int_I w \int_I w^{-1}\lesssim |I|^2 $$ of $A_2$, that indeed $w\in A_2$ for our weight $w(x)=(1+|x|)^{1/2}$.

First of all, as explained in my comment, this is the same as asking if the Hilbert transform is bounded on $L^2(\mathbb R, w\, dx)$, with $w(x)=(1+|x|)^{2s}$. Or, to state this one more time, this reformulation follows because restricting $u$ to a half line is essentially the same as applying the Hilbert transform to $\widehat{u}$. Hunt, Muckenhoupt, Wheeden proved that this is equivalent to $w\in A_2$.

It is easy to check, using the defining condition $$ \int_I w \int_I w^{-1}\lesssim |I|^2 $$ of $A_2$, that indeed $w\in A_2$ for our weight $w(x)=(1+|x|)^{2s}$ as long as $-1/2<s<1/2$.

The argument cannot handle $s=1/2$ (or $s\le -1/2$), which one would perhaps also expect to be more difficult since it is the borderline case. For $s>1/2$, $H^s$ functions are continuous, so the result clearly no longer holds in this case.