Let $H$ be a Hilbert space. For $1<p<\infty$, define the atomic space $U^{p}(\mathbb{R};H)$ as follows. We say that $a(t)$ is a $U^{p}$ atom if $\{t_{k}\}$ is a partition of $\mathbb{R}$, $a_{k}\in H$ with $\sum_{k}\|a_{k}\|_{H}^{p}=1$, and

$$a(t)=\sum_{k}\mathrm{1}_{[t_{k},t_{k+1})}(t)a_{k}$$

Let $u=\sum_{\lambda}c_{\lambda}u_{\lambda}$ be an atomic decomposition of a function $u: \mathbb{R}\rightarrow H$, where $u_{\lambda}$ are $U^{p}$-atoms and $c_{\lambda}\in\mathbb{C}$. We define

$$\|u\|_{U^{p}}:=\inf_{\{c_{\lambda}\}} \sum_{\lambda}|c_{\lambda}|,$$

where the infimum is taken over all atomic decompositions of $u$.

Now let $S(t)$ denote the propagator associated to the linear Schrodinger equation $iu_{t}+\Delta u =0$, and let $H=L^{2}(\mathbb{R}^{d})$. We define the space

$$U_{\Delta}^{p}:=\{u:\mathbb{R}\rightarrow L^{2}(\mathbb{R}^{d}) : S(-t)u\in U^{p}\}$$

with norm $\|S(-t)u\|_{U^{p}}$. One can verify that functions in $U_{\Delta}^{p}$ have the form

$$u=\sum_{\lambda}c_{\lambda}u_{\lambda},$$

where $u_{\lambda}=\sum_{k}1_{[t_{k},t_{k+1})}e^{it\Delta}u_{k}$, with $\{t_{k}\}$ as above and $\sum_{k}\|u_{k}\|_{L_{x}^{2}(\mathbb{R}^{d})}^{p}=1$ (i.e. $u_{\lambda}$ is a $U_{\Delta}^{p}$ atom). For more details on these spaces, see here.

Let $\mathcal{F}_{t,x}$ denote the space-time Fourier transform on $\mathbb{R}\times\mathbb{R}^{d}$, and $\mathcal{F}_{x}$ just denote the spatial Fourier transform. If we write define $v = \mathcal{F}_{t,x}^{-1}[Re(\mathcal{F}_{t,x}[u])^{+}]$ or $v = \mathcal{F}_{x}^{-1}[Re(\mathcal{F}_{x}[u(t)])^{+}]$ (i.e., so that either the space-time or space Fourier transform of $u$ is real and positive), is there an estimate of the form

$$\|v\|_{U_{\Delta}^{p}}\lesssim \|u\|_{U_{\Delta}^{p}}$$

where the implied constant is, of course, independent of $u$?

The motivation for my question is that the space $U_{\Delta}^{2}$ and the related space $V_{\Delta}^{2}$ have been used successfully as replacements for the Bourgain space $X^{0,\frac{1}{2}}(\mathbb{R}\times\mathbb{R}^{d})$. In applications, for instance see pg. 9 of this well-known paper by CKSTT on the I-method, I have seen it written that one can always assume that the spatial Fourier transform of functions to be positive when $X^{s,b}$ spaces are involved. It's not clear to me why this assumption is valid, although I do see from the definition of the $X^{s,b}$ norm that we may assume the space-time Fourier transform is positive. Perhaps, someone can clarify this point.

In any case, I would like to know if something similar can be done for $U_{\Delta}^{p}$ spaces, but it's not clear to me how to produce an atomic decomposition for $v$ above, which would yield the desired result.

Let $H$ be a Hilbert space. For $1<p<\infty$, define the atomic space $U^{p}(\mathbb{R};H)$ as follows. We say that $a(t)$ is a $U^{p}$ atom if $\{t_{k}\}$ is a partition of $\mathbb{R}$, $a_{k}\in H$ with $\sum_{k}\|a_{k}\|_{H}^{p}=1$, and

$$a(t)=\sum_{k}\mathrm{1}_{[t_{k},t_{k+1})}(t)a_{k}$$

Let $u=\sum_{\lambda}c_{\lambda}u_{\lambda}$ be an atomic decomposition of a function $u: \mathbb{R}\rightarrow H$, where $u_{\lambda}$ are $U^{p}$-atoms and $c_{\lambda}\in\mathbb{C}$. We define

$$\|u\|_{U^{p}}:=\inf_{\{c_{\lambda}\}} \sum_{\lambda}|c_{\lambda}|,$$

where the infimum is taken over all atomic decompositions of $u$.

Now let $S(t)$ denote the propagator associated to the linear Schrodinger equation $iu_{t}+\Delta u =0$, and let $H=L^{2}(\mathbb{R}^{d})$. We define the space

$$U_{\Delta}^{p}:=\{u:\mathbb{R}\rightarrow L^{2}(\mathbb{R}^{d}) : S(-t)u\in U^{p}\}$$

with norm $\|S(-t)u\|_{U^{p}}$. One can verify that functions in $U_{\Delta}^{p}$ have the form

$$u=\sum_{\lambda}c_{\lambda}u_{\lambda},$$

where $u_{\lambda}=\sum_{k}1_{[t_{k},t_{k+1})}e^{it\Delta}u_{k}$, with $\{t_{k}\}$ as above and $\sum_{k}\|u_{k}\|_{L_{x}^{2}(\mathbb{R}^{d})}^{p}=1$ (i.e. $u_{\lambda}$ is a $U_{\Delta}^{p}$ atom). For more details on these spaces, see here.

Question 1: Let $\mathcal{F}_{t,x}$ denote the space-time Fourier transform on $\mathbb{R}\times\mathbb{R}^{d}$, and $\mathcal{F}_{x}$ just denote the spatial Fourier transform. If we write define $v = \mathcal{F}_{t,x}^{-1}[Re(\mathcal{F}_{t,x}[u])^{+}]$ or $v = \mathcal{F}_{x}^{-1}[Re(\mathcal{F}_{x}[u(t)])^{+}]$ (i.e., so that either the space-time or space Fourier transform of $u$ is real and positive), is there an estimate of the form

$$\|v\|_{U_{\Delta}^{p}}\lesssim \|u\|_{U_{\Delta}^{p}}$$

where the implied constant is, of course, independent of $u$?

The motivation for my question is that the space $U_{\Delta}^{2}$ and the related space $V_{\Delta}^{2}$ have been used successfully as replacements for the Bourgain space $X^{0,\frac{1}{2}}(\mathbb{R}\times\mathbb{R}^{d})$. In applications, for instance see pg. 9 of this well-known paper by CKSTT on the I-method, I have seen it written that one can always assume that the spatial Fourier transform of functions to be positive when $X^{s,b}$ spaces are involved. It's not clear to me why this assumption is valid, although I do see from the definition of the $X^{s,b}$ norm that we may assume the space-time Fourier transform is positive. Perhaps, someone can clarify this point.

Question 2: Why when working with $X^{s,b}$ norms can one assume that the spatial Fourier transforms of functions are positive?

In any case, I would like to know if something similar can be done for $U_{\Delta}^{p}$ spaces, but it's not clear to me how to produce an atomic decomposition for $v$ above, which would yield the desired result.