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

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.

• Since everything is a function, the only way to obtain such a decomposition $u=u_++u_-$ is the obvious one, by restricting to the half lines. But this is the same as applying (something like) $1\pm H$ to $\widehat{u}$, $H$ denoting the Hilbert transform. So your question is closely related to this one: mathoverflow.net/questions/378388/… Commented Aug 27, 2022 at 19:38
• A related question is whether $H^s_- + H^s_+$ is closed in $H^s$. This is equivalent to these closed subspaces (quotient by intersect) having a positive gap. In the case $s > -1/2$, another equivalent statement is $H^s(\mathbb{R})$ and $H^s([0,\infty))$ induce the same topology on $H^s_+$. If they do, then $H^s_+$ can be taken as an alternative definition of $H^s_0([0,\infty))$. In view of Prof. Remling's answer and $H^{1/2}_0([0,\infty)) = H^{1/2}([0,\infty))$, I suspect that $H^{1/2}_+ \neq H^{1/2}_0([0,\infty))$. Commented Aug 28, 2022 at 21:49
• It just occurred to me writing the previous comment that this is related to the question of zero-extension for members of $H^{s}_0$. Commented Aug 28, 2022 at 21:51

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 $$-1/2.
If $$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.
• Thank you Prof. Remling. This is a useful characterization. However, I am guessing from symmetry that the Hilbert transform is only bounded for $w$ between $\langle x \rangle^{-1/2}$ and $\langle x \rangle^{1/2}$. So this method would only cover the case $s \in [-1/2,1/2]$. Do you have any idea about $s < -1/2$? In that case, the decomposition is no longer unique. Also, I am wondering whether there is simple proof that cover both cases, possibly ignoring $s=1/2$. It seems a pity that such an essential and elementary fact requires so much analysis to justify. Commented Aug 28, 2022 at 8:46
• @TianXia: Actually, the argument only covers $-1/2<s<1/2$, I made a trivial (but rather confusing to me, for a while, until I finally cleared it up) mistake in the bookkeeping, now corrected. Commented Aug 28, 2022 at 17:35
• @ChristianRemling If $s=1/2$ the weight $(1+|x|)^{2s}$ is not in $A_2$, so your argument should give that the result does not hold. Am I overlooking something? Commented Aug 28, 2022 at 17:50
• @GiorgioMetafune: In fact, it's perfectly obvious. If $u=1$ near $x=0$ (and $u\in C_0^{\infty}$, let's say), then the half line restrictions are not in $H^{1/2}$ (their FTs behave like those of $\chi_I$). Commented Aug 28, 2022 at 18:41
• @ChristianRemling True but another thing is maybe worth noticing: $H^s_-+H^s_+$ is dense in $H_s$ also for $s=1/2$. Take $u$ in the Schwartz class and $u_\epsilon=u\phi(x/\epsilon)$ where $\phi$ is even and equal to 0/1 for small/large $|x|$. Then, as in the OP text, $\|u_\epsilon -u\|_{H^{1/2}} \leq C$ and, by weak compactness,$u_\epsilon \to u$ weakly. This does not work on one side only, I mean if $\phi$ is the same for positive $x$ and zero for negative beacuse we need $u_\epsilon-u$ small. Commented Aug 28, 2022 at 21:51