What does the notation $L^2(X_1 ; L^{p_j}(X_2))$ mean? I'm reading the beginning of Fourier Analysis and Nonlinear Partial Differential by Hajer Bahouri, Jean-Yves Chemin, Raphaël Danchin, and they have a bit of notation that I do not understand. In section 1.1.4, Proposition 1.10 states the following:
Let $(X_1, \mu_1)$ and $(X_2, \mu_2)$ be two measure spaces. Let $\mathcal{T}$ be a continuous bilinear functional on $L^2(X_1 ; L^{p_j}(X_2)) \times L^2(X_1 ; L^{q_j}(X_2))$ for $j$ in $\{0,1\}$, where $(p_j, q_j)$ is in $[1,2]^2$ and such that $p_0 \neq p_1$ and $q_0 \neq q_1$. For any $\theta \in [0,1]$, the bilinear functional $\mathcal{T}$ is then continuous on $L^2(X_1; L^{p_\theta}(X_2)) \times L^2(X_1 ; L^{q_\theta}(X_2))$ with
$(\frac{1}{p_\theta}, \frac{1}{q_\theta}) = (1-\theta)(\frac{1}{p_0}, \frac{1}{q_0}) + \theta (\frac{1}{p_1}, \frac{1}{q_1})$.
The problem is I do not understand the notation: $L^2(X_1; L^{p_j}(X_2))$. Does anybody have any idea? Thank you for your help.
Furthermore, in the proof of this proposition, they write the following:
Let $f \in L^2(X_1 ; L^{p_\theta}(X_2))$ and $(t, x) \in X_1 \times X_2$. Then $f(t,x)$, so it seems that a function in $L^2(X_1 ; L^{p_\theta}(X_2))$ is a function whose domain is $X_1 \times X_2$.
 A: So this is a classical notation that comes from a slight abuse of notation.

*

*An element of $L^2(I,Y)$ is a function $f$ such that for a.e. any $t\in I$, $f(t)$ is an element of $Y$ and
$$
\|f\|_{L^2(I,Y)}^2 = \int_I \|f(t)\|_{Y}^2\,\mathrm d t < \infty
$$

*In particular, strictly speaking, if $Y = L^p(X)$ (where $L^p(X) = L^p(X,\Bbb R)$), then it means that for each fixed $t$, $f(t)$ is itself a function, hence we can write $f(t)(x)$, and the norm now reads
$$
\|f\|_{L^2(I,L^p(X))}^2 = \int_I \|f(t)\|_{L^p(X)}^2\,\mathrm d t = \int_I \left(\int_X |f(t)(x)|^p\,\mathrm d x\right)^{2/p}\mathrm d t
$$

*It is now a classical abuse of notation to identify a function valued function $f(t)(x)$ with a two variables function $f(t,x)$, i.e. to identify the spaces $I \to (X\to\Bbb R)$ with $I \times X\to\Bbb R$.

*A particular case is the case when both exponents are the same, where one has $L^p(X_1\times X_2) = L^p(X_1, L^p(X_2)) = L^p(X_2, L^p(X_1))$, which is nothing but Fubini theorem
$$
\iint_{X_1\times X_2} |f(x_1,x_2)|^p\,\mathrm d x_1\,\mathrm d x_2 = \int_{X_1}\left(\int_{X_2} |f(x_1,x_2)|^p\,\mathrm d x_2\right)\mathrm d x_1
$$
A: They just mean the space of all equivalence classes of functions between $X_1$ and $L^{p_j}(X_2)$ which are square integrable.
A: Lets consider a function $f: [0, T]\times \Omega\to \mathbb{R}$ where $\Omega\subset \mathbb{R}^n$. Obviously, it is a function in $n+1$ variables $(t, x)$ (where as usual, $t$ stands for time and $x$ being the space variable). Now, in order to set a difference between these variables, you can see $f$ as a function in time taking values on a vector space $X$, namely, $f: [0, T]\to X$, where $t\in [0, T]\mapsto f(t, \cdot)\in X$. But, where does it appear the vector space $X$?
In the case you present, $f\in L^2(X_1; L^p(X_2))$ means that you have a function $f: X_1\times X_2\to \mathbb{R}$ that for each $t\in X_1$ gives a function $f(t, \cdot)\in L^p(X_2)$ and that the vector-valued map $t\in X_1\mapsto f(t, \cdot)\in L^p(X_2)$ belongs to $L^2$ (as @Rptoughs said).
This setting is very useful to study PDEs in an abstract way (via Semigroup Theory for example) since its solutions have different regularities on time and space. For example, for $L>0$, if $\varphi\in H_0^1(0, L)$ and $\psi\in L^2(0, L)$, then the wave equation
\begin{align*}
    \left\{\begin{array}{rl}
        u_{tt}-u_{xx}=0 &\ (t, x)\in \mathbb{R}\times (0, L)  \\
        u(t, 0)=u(t, L)=0 &\ t\in \mathbb{R}\\
        u(0, x)=\varphi(x),\ u_t(0, x)=\psi(t) &\ x\in (0, L) 
    \end{array}\right.
\end{align*}
has a unique solution $u=u(t, x)$ where $u\in C(\mathbb{R}; H_0^1(0, L))\cap C^1(\mathbb{R}; L^2(0, L))$. This means that the vector-valued map $t\in \mathbb{R}\to u(t, \cdot)\in H_0^1$ is a continuous map and that $t\in \mathbb{R}\to u(t, \cdot)\in L^2(0, L)$ is a $C^1$ map. Answering the initial question, in this example, $X$ captures the spacial regularity of the PDE solution.
