Sobolev space is spanned by distributions supported on half-lines?

Question

I asked this question on Mathematics Stack Exchange previously. This seems to be a very basic property of Sobolev spaces, but I wasn't able to find a proof for it.

For any $s \leq 1/2$, $$H^s(\mathbb{R}) = H^s_-(\mathbb{R}) + H^s_+(\mathbb{R})$$ where $H^s_-(\mathbb{R}) = \{ u \in H^s(\mathbb{R}) \mid \operatorname{supp} u \subset (-\infty, 0] \}$ and $H^s_+(\mathbb{R}) = \{ u \in H^s(\mathbb{R}) \mid \operatorname{supp} u \subset [0, \infty) \}$.

I am only really interested in the case $s < 1/2$, but I am led to believe the result also holds for $s = 1/2$.

Context

This question stems from my attempt to prove subjectivity of the natural map $$H^s_+(\mathbb{R}) \rightarrow H^s([0, \infty)) = H^s(\mathbb{R}) / H^s_-(\mathbb{R})$$ and its higher dimensional analogues $H^s_+(\mathbb{R} \times \mathbb{R}^d) \rightarrow H^s([0, \infty) \times \mathbb{R}^d)$. This property is claimed in Melrose's notes, Proposition 3.5.1. The proof involves taking the dual of the injection $H^{-s}_+(\mathbb{R}) \rightarrow H^{-s}([0,\infty))$. But this requires the natural topology of $H^{-s}_+(\mathbb{R})$ to coincide with the subspace topology from $H^{-s}([0,\infty))$, which is not justified in the notes. I believe the latter property is equivalent to $H^{-s}_-(\mathbb{R}) + H^{-s}_+(\mathbb{R})$ being closed in $H^{-s}(\mathbb{R})$.

My thoughts

In the case $s < 1/2$, I can show that $H^s_-(\mathbb{R}) + H^s_+(\mathbb{R})$ contains the Schwartz functions $\mathcal{S}(\mathbb{R})$, but this does not imply the desired result unless $H^s_-(\mathbb{R}) + H^s_+(\mathbb{R})$ is closed.

The inclusion $\mathcal{S}(\mathbb{R}) \subset H^s_-(\mathbb{R}) + H^s_+(\mathbb{R})$ is obtained by multiplying $u \in \mathcal{S}(\mathbb{R})$ by sequences of smooth functions approximating the indicators on $(-\infty, 0]$ and $[0, \infty)$, and showing that the resulting sequences converge in $H^s$. For example, fixing a smooth $\phi$ supported on $[0, \infty)$ and eventually equal to 1, then consider its scalings $\phi_{\epsilon}(x) = \phi(x /\epsilon)$. One can show that $\lim_{\epsilon \rightarrow 0^+} u \phi_{\epsilon}$ converges in $H^s$, $s \in [0, 1/2)$ using the inequality $\|v\|_{H^s} \lesssim \|v\|_{H^0}^{1-s} \|v\|_{H^1}^s$ and the crude estimates $$\|u(\phi_\epsilon - \phi_\delta)\|_{L^2} \leq \|u\|_{L^{\infty}} \|\phi_\epsilon - \phi_\delta\|_{L^\infty} \|\phi_\epsilon - \phi_\delta\|_{L^0}^{1/2} \lesssim \epsilon^{1/2}$$ $$\|u \cdot \partial(\phi_\epsilon - \phi_\delta)\|_{L^2} \leq \|u\|_{L^\infty} \|\partial(\phi_\epsilon - \phi_\delta)\|_{L^\infty} \|\partial(\phi_\epsilon - \phi_\delta)\|_{L^0}^{1/2} \lesssim \epsilon^{-1} \epsilon^{1/2}$$ I am unable to generalize this to the case $u \in H^s$ or $s = 1/2$. Any suggestion or reference is greatly appreciated.

Answer 1

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.