(Book recommendation please!) Is there any difference between $L^p_tL^q_x(R \times R^n)$ and $L^p(R \to L^q(R^n))$? In real analysis , we usually consider the space $L^p_tL^q_x$ consists of the measurable functions on $R^{n+1}$
$$f:R \,\,\times \,\, R^n \to C$$
$$(t,x)\to f(t,x)$$
With $$\int\int |f(t,x)|^q \, dx^{\frac{p}{q}}\,dt^{\frac1p}\lt \infty$$ 
However , in order to apply Picard's method in PDE theory . Sometimes we need to consider $L^p(R \to L^q)$ , a space with value in Banach space . In Evans PDE we have the following definition :  

Let $X$ be a Banach space . We say a function $f:[0,T] \to X$ is measurable if there exist simple functions $s_k(t) :[0,T] \to X$ such that $s_k \to f$ for almost every $t \in [0,T]$ 

Question: In PDE theory , we frequently use the space $L^p(R \to X)$ . So I'd like to read some books which talk about the details of this space . Any help would be very appreciate .
 A: In case of $p,q < \infty$, it is indeed true that $L_t^p L_x^q = L^p(\Bbb{R} \to L^q)$.
[Side remark: This should also hold if $p = \infty$, but not for $q = \infty$;
I am too lazy, however, to handle the case $p = \infty$ right now.]
Let us consider the map
$$
  \Phi : L_t^p L_x^q \to L^p(\Bbb{R} \to L^q),
         f \mapsto (t \mapsto f(t, \cdot)) .
$$
I first claim that $\Phi$ is well-defined and isometric.
To see this, let $f \in L_t^p L_x^q$.
We first want to show that $\Phi f: \Bbb{R} \to L^q(\Bbb{R}^n), t \mapsto f(t,\cdot)$
is measurable.
Since $L^q(\Bbb{R}^n)$ is separable (this is where we use that $q < \infty$),
Petti's theorem,
$\Phi f$ is measurable if and only if it is weakly measurable, meaning that for each
continuous linear functional $\psi : L^q(\Bbb{R}^n) \to \Bbb{K}$, the map
$\psi \circ \Phi : \Bbb{R} \to \Bbb{K}$ is measurable.
By the Riesz representation theorem, there is $g \in L^{q'}(\Bbb{R}^n)$ such that
$\psi(h) = \int h(x)g(x) \, d x$ for all $h \in L^q(\Bbb{R}^n)$, and hence
$$
  \psi(\Phi(t))
  = \int f(t, x) \, g(x) \, d x .
$$
The latter expression is measurable as a function of $t$, as follows by Tonelli's theorem
(strictly speaking, one should decompose $g = (g_1 - g_2) + i \cdot (g_3 - g_4)$
with non-negative functions $g_1,g_2,g_3,g_4 \in L^{q'}(\Bbb{R}^n)$ and similarly for $f$
to apply Tonelli's theorem).
Thus, $\Phi f : \Bbb{R} \to L^q (\Bbb{R}^n)$ is weakly measurable and hence measurable.
Finally, we have
$$
  \int \| \Phi f (t) \|_{L^q}^p
  = \int
      \Bigl(
        \int
          |f(t,x)|^q
        \, d x
      \Bigr)^{p/q}
    \, d t
  = \| f \|_{L_t^p L_x^q}^p ,
$$
from which it easily follows that $\Phi$ is isometric (and hence continuous).
I now assume that you know that both $L_t^p L_x^q$ and $L^p(\Bbb{R} \to L^q)$ are complete.
Furthermore, I assume that you know that the set of step functions (with values in $L^q(\Bbb{R}^n)$)
is dense in $L^p(\Bbb{R} \to L^q)$ (this uses that $p < \infty$).
Now, given $g \in L^q(\Bbb{R}^n)$ and $E \subset \Bbb{R}$ measurable with $\mu(E) < \infty$,
it is straightforward to check that $f : (t,x) \mapsto 1_E (t) \, g(x)$ satisfies
$f \in L_t^p L_x^q$ and $\Phi f = 1_E \cdot g$.
This easily implies that the image of $\Phi$ contains all step functions,
and thus is dense in $L^p(\Bbb{R} \to L^q)$.
But since $\Phi$ is an isometry and $L_t^p L_x^q$ is complete, the image of $\Phi$
is closed and $L^p(\Bbb{R} \to L^q)$; thus, $\Phi$ is surjective as well.
