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I have asked the same question on MathSE. I was thinking about the following problem.

Problem. Given $\alpha>0$, find all values of $\beta\geq 0$ such that the following estimate is true for all $\varphi\in \mathscr D(\mathbb R)$: $$ \|\langle x\rangle^{-\beta}\;|\partial_x|^{1/2}\varphi\|_{L^2 (\mathbb R)}\lesssim \|\partial_x \varphi\|_{L^2 (\mathbb R)}+\|\langle x\rangle^{-\alpha}\;\varphi\|_{L^2 (\mathbb R)}.\qquad(1) $$

Context and notation. The “half derivative” is defined as a Fourier multiplier $$ \mathscr F(|\partial_x|^s\varphi)(\xi)=|\xi|^s\hat\varphi(\xi), $$ where $\mathscr F$ and $\widehat\cdot$ denote the Fourier transform, and I define the japanese bracket as $\langle x\rangle:=(1+x^2)^{1/2}$. One could consider more general weights of course and different derivation orders. I considered a fractional derivative because it is less immediate to try and prove the estimate from scratch (and also because I actually need to prove one of them).

A few preliminary observations.

  • Taking $\alpha$ larger and larger eventually does not change the right hand side up to equivalent norms. In fact, a function such that $\partial_x\varphi\in L^2$ satisfies $$ |\varphi(x)-\varphi(y)|\lesssim |x-y|^{1/2}, $$ and in particular the quantity $\|\langle x\rangle^{-\alpha}\varphi\|_{L^2}$ is automatically finite if $\alpha>1$. With this (being a bit careful since $\partial_x\varphi$ determines $\varphi$ only up to a constant) it is feasible to prove that, if we take $\alpha>1$, the right hand side of $(1)$ is equivalent to $$ \|\partial_x \varphi\|_{L^2}+|\varphi(0)|. $$ So the problem is split in two cases: the case $0<\alpha\leq 1$ and the case $\alpha>1$, the latter consisting in fact of one single norm on the right hand side.

  • It is possible to show through the Fourier transform that $|\partial_x|^{1/2}\varphi(x)=[\operatorname{sgn}(\cdot)|\cdot|^{-1/2}*\partial_x\varphi](x)$ and since $\partial_x\varphi\in L^2$, then $|\partial_x|^{1/2}\varphi\in BMO(\mathbb R)$. This implies that $|\partial_x|^{1/2}\varphi$ is, for instance, locally in $L^2$, so I am confident that (unless my eyes are cheated by some spell) the following estimate should hold $$ \|\mathbb \chi_{[-1,1]}(\cdot) |\partial_x|^{1/2}\varphi \|_{L^2} \lesssim \|\partial_x \varphi\|_{L^2}+ |\varphi(0)| $$ and by a summation trick (i.e., taking infinitely many shifted copies of the above estimate to cover all the real line), using the Hölder-continuity of $\varphi$ which I stated above, one should obtain that estimate $(1)$ holds whenever $\alpha\geq 0$ and $\beta>1$.

My questions:

  1. Do you have any clue on how one would prove the estimate for $\alpha<1$, besides doing it from scratch? Simply put, does this kind of estimates fall into some more general framework for which there are strong techniques to prove the estimates whenever I need to use one in a paper without reinventing the wheel? Would abstract interpolation theory help in this context? Any references you think could be useful in understanding how to work out these estimates, or research papers dealing with this kind of estimates?
  2. (Forgive me if this question is not suitable for MO) Does what I have said above in the second observation make sense to you? I did a bit of hand waving after saying that $|\partial_x|^{1/2}\varphi\in BMO(\mathbb R)$ and I am not sure how to make the next part of the proof rigorous.
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  • $\begingroup$ Can you define |D_x|^{1/2}? $\endgroup$ Commented Dec 12, 2022 at 11:39
  • $\begingroup$ Sure. I edited my question, thank you for the comment $\endgroup$ Commented Dec 12, 2022 at 12:36
  • $\begingroup$ Assume that $\beta=0$ and let $\phi_a(x)=\phi(x-a)$. The LHS in (1) is independent of $a$ as well as the first term on the RHS. But the second one tends to 0 as $a \to \infty$ if $\alpha>0$. Therefore, (1) holds without the second term on the RHS, but this is not true, by taking the Fourier transform. $\endgroup$ Commented Dec 12, 2022 at 22:19
  • $\begingroup$ Sure. This says that for any given $0<\alpha$, the estimate is not true for $\beta=0$. And from what I wrote, the estimate is always true for $\beta>1$. I would expect the optimal $\beta$ is exactly equal to $\alpha$ in the range $0\leq\alpha\leq 1$ (it is not clear to me whether the endpoint case would be achieved). $\endgroup$ Commented Dec 13, 2022 at 7:29

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