What does the Bourgain norm tell us about a function? As the title says, i'm not exactly sure how to think about the Bourgain norm, i'm also not entirely sure that this is the correct name so let me define it.
Let $\varphi : \mathbb{R}^n \to \mathbb{R}$ be a measurable function, and denote by $\mathcal{F}_x : H_x^s \to L^2(\mathbb{R}^n, \langle x \rangle)$ the Fourier transform. Then for $D = -i\nabla$ define $$\varphi(D) = \mathcal{F}_x^{-1}\varphi(\xi)\mathcal{F}_x,$$ which has domain $$\Delta = \{f \in H_x^s(\mathbb{R}^n) | \varphi\mathcal{F}_xf \in L^2(\mathbb{R}^n, \langle x \rangle ) \}.$$ Then $$\varphi(D) : \Delta \to H_x^s(\mathbb{R}^n),$$ is self-adjoint and generates a unitary group denoted $$(U_{\varphi}(t) = (\exp(it\varphi(D)))_{t \in \mathbb{R}}.$$
Then we define the space $X_{s, b}(\varphi)$ to be the completion of $$\bigcap_{s, b} H_x^s(\mathbb{R}^n) \otimes H_t^b(\mathbb{R}),$$ with respect to the norm $$\|f\|_{X_{s, b}(\varphi)} = \|U_{\varphi}(- \: \cdot)f\|_{H^s_x(\mathbb{R}^n) \otimes H_t^b(\mathbb{R})}.$$
What is the intuition for this definition? I've done lots of Google searching but can't seem to find an explanation. Thank you in advance!
 A: Here's a bunch of remarks that might help you or at least ease up the mind.
First, I believe you should make the observation
\begin{align}
H^s_x(\mathbb{R}^n)\otimes H^b_t(\mathbb{R})=&\ L^2(\mathbb{R}^n, \langle\xi\rangle^s)\otimes L^2(\mathbb{R}, \langle\tau\rangle^b) = L^2(\mathbb{R}^n\times \mathbb{R}, \langle\xi\rangle^s\langle \tau\rangle^b)\\
=&\ H^b_t(\mathbb{R}, H^s_x(\mathbb{R}^n)).
\end{align}
Second, I believe you can safely replace
\begin{align}
\bigcap_{s, b}H^s_x(\mathbb{R}^n)\otimes H^b_t(\mathbb{R})
\end{align}
by
\begin{align}
\mathcal{S}(\mathbb{R}, \mathcal{S}(\mathbb{R}^n))
\end{align}
the space of Schwartz functions since the above intersection is trying to consider very smooth functions. 
Lastly, you should use Lemma 1 in your reference as the definition, i.e.
\begin{align}
\|f\|_{X^{s, b}(\phi)}= \|\langle \xi\rangle^s\langle \tau-\phi(\xi)\rangle^b\tilde f\|_{L^2(\mathbb{R}^n\times \mathbb{R})}. 
\end{align}
Note, I used $\tilde f$, the space-time Fourier transform, instead of $\hat f$, the spatial Fourier transform. In terms of interpretation, the Bourgain norm measures how far $f(t, x)$ is away from being a homogeneous solution to the following dispersive equation
\begin{align}
\frac{1}{i}\partial_t u = \phi\left(\nabla/i\right)u \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ (\ast)
\end{align}
for some analytic function $\phi$ (I don't want to invoke Borel functional calculus). Observe if $f$ is a solution to $(\ast)$ then 
\begin{align}
\widetilde{f}(\tau, \xi) = \delta(\tau-\phi(\xi))\widehat{f(0, \cdot)}(\xi)
\end{align}
which means
\begin{align}
\|f\|_{X^{s, b}} = \|\langle \xi\rangle^s\widehat{f(0, \cdot)}(\xi)\|_{L^2(\mathbb{R}^n)}
\end{align}
i.e. the Bourgain norm is just the Sobolev norm of the initial data. 
