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The Bourgain space is $X^{s,b} := X^{s,b}(\mathbb R \times \mathbb{T}^3)$ is the completion of $C^\infty (\mathbb R; H^s(\mathbb{T}^3))$ under the norm

$$\| u\|_{X^{s,b}}:= \|e^{- i t \triangle} u(t,x)\|_{H_t^b (\mathbb{R}; H_x^s(\mathbb{T}^3))}\\ =\left(\sum_{\xi \in \mathbb{Z}^3} \int_\mathbb{R} \langle\tau+\sum_{j=1}^3 \theta_j \xi^2_j\rangle^{2b} \langle\xi\rangle^{2s} |\hat{u}(\tau,\xi)|^2 d\tau \right)^{\frac{1}{2}}.$$

Then we have:

1- For $s_1 \leq s_2$ and $b_1 \leq b_2$, $X^{s_2,b_2} \hookrightarrow X^{s_1,b_2}$.

2- For $b > 1/2$, $X^{0,b} \hookrightarrow C_t L_x^2$.

3- $X^{0,1/4} \hookrightarrow L_t^4 L_x^2$

First, I would like some explanation for the meaning of how the Sobolev mixed norm of the semigroup for $u$, namely $\|e^{- i t \triangle} u(t,x)\|_{H_t^b (\mathbb{R}; H_x^s(\mathbb{T}^3))}$, gives the result

$$\left(\sum_{\xi \in \mathbb{Z}^3} \int_\mathbb{R} \langle\tau+\sum_{j=1}^3 \theta_j \xi^2_j\rangle^{2b} \langle\xi\rangle^{2s} |\hat{u}(\tau,\xi)|^2 d\tau \right)^{\frac{1}{2}}.$$

I am trying to understand how this norm works and what does it measure. I would like some help to solve the second and third points. I solved the first already.

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    $\begingroup$ You haven't defined the quantities $\theta_j$. Is your laplace operator $\triangle$ weighted (and not the standard one on $\mathbb{T}^3$? $\endgroup$
    – Willie Wong
    Commented Mar 5, 2022 at 22:58
  • $\begingroup$ @WillieWong yes. It is the Laplace-Beltrami operator. Here $\theta_j= L_j^{-2}$, we use it to be able to work on the standrard $\mathbb{T}^3$ instead of $\mathbb{R}^3 \over (L_1 \mathbb Z \times L_2 \mathbb Z \times L_3 \mathbb Z)$. Like here mathoverflow.net/questions/417188/… $\endgroup$
    – Mr. Proof
    Commented Mar 6, 2022 at 0:09

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