For a function $f = f(t,x)$, how is the space $C^0([0,T];(S'(\mathbb R^d))^n)$ defined? I'm reading "Fourier Analysis and non-linear Partial Differential Equations" from H. Bahouri and they use a notation which is unknown to me.
If $(t,x) \in \mathbb [0,T] \times \mathbb R^d$, $A([0,T],\mathbb R)$ is any space of functions from $[0,T]$ to $\mathbb R$ in the variable $t$, such as $C^0([0,T],\mathbb R)$, and $B(\mathbb R^d, \mathbb R^n)$ is any normed space of functions from $\mathbb R^d$ to $\mathbb R^n$ in the variable $x$, such as $L^{\infty}(\mathbb R^d, \mathbb R^n)$, I deduced that $A([0,T];B(\mathbb R^d, \mathbb R^n))$ would be the space
$$A([0,T];B(\mathbb R^d, \mathbb R^n)) = \{f : [0,T] \times \mathbb R^d \rightarrow \mathbb R^n \mid f(t,x) \in B(\mathbb R^d, \mathbb R^n) \ \forall t, \ ||f(t,\cdot )||_B \in A([0,T],\mathbb R)\}$$
For $B$ a locally convex topological vector space, such as $S(\mathbb R^d)$, I guess we can make the same definition but with semi-norms.
However, they also use this notation with $B$ being some distribution space such as $(S'(\mathbb R^d))^n$ (a vector of $n$ tempered distributions). In this case, I cannot make sense of, say, the space $C^0([0,T];(S'(\mathbb R^d))^n)$ or $L^1([0,T];(S'(\mathbb R^d))^n)$ .
Any help is appreciated !
 A: For $\mathcal C^0(X,Y)$ to make sense, you need $X$ and $Y$ to be equipped with a topology. $X = [0,T]$ has a standard topology, and so does any normed vector space.
For $\mathcal C^0([0,T],A(\mathbb R^d,\mathbb R^n))$ to make sense, you need to choose a topology for $A(\mathbb R^d,\mathbb R^n)$. One example is $A = C^0_b$ the space of bounded continuous function, which is a normed vector space (actually a Banach space) when equipped with the supremum norm :
$$\|f\|_{C^0_b} = \sup_{x\in\mathbb R^d} \|f(x)\|_{\mathbb R^n}$$
This means that $C^0([0,T], \mathcal C^0_b(\mathbb R^d,\mathbb R^n))$ is the space of functions $f:[0,T]\to B(\mathbb R^d,\mathbb R^n)$ which are continuous, ie for each $t\in [0,T]$, $f(t)$ is a bounded continuous function $f(t): x\in \mathbb R^d\mapsto  f(t)(x)\in \mathbb R^n$ and that for each $t\in [0,T]$ and $\epsilon >0$, there is some $\eta>0$ such that if $t'\in [0,T]$ and $|t'-t| < \eta$, then :
$$\|f(t) - f(t')\|_{B(\mathbb R^d,\mathbb R^n)} = \sup_{x\in\mathbb R^d}\| f(t)(x) - f(t')(x)\|_{\mathbb R^n} \leq \epsilon$$
This would work in exactly the same way for $A = L^p$ (with $p\in [1,\infty])$ or many other functional Banach spaces.
For $\mathcal S$ and $\mathcal S'$, the situation is a bit more complicated, since they are not normed vector space. The Schwartz space $\mathcal S'(\mathbb R)$ is also a topological vector space (specifically a Fréchet space). The space of tempered distributions $\mathcal S'(\mathbb R)$ is its topological dual, and is can be equipped with different topologies, derived from that of $\mathcal S(\mathbb R)$, most notably the strong dual topology and the weak-$*$-topology. Once one of those is chosen, $\mathcal C^0([0,T]; \mathcal S'(\mathbb R))$ is just the space of continuous function from $[0,T]$ to $\mathcal S'(\mathbb R)$.
A: From what I understand, those kind of spaces are defined as follows :

*

*If $(X, ||\cdot ||_X)$ is a Banach space and $\Omega \subset \mathbb R^n$ is a Lebesgue measurable set, then $L^p(\Omega; X)$ is the set of Bochner measurable functions $u : \Omega \rightarrow X$ such that the following Lebesgue integral is finite : $$\int_{\Omega} ||u(y)||_X^p \ dy < \infty$$


*If $\Phi$ is a Fréchet space and $I \subset \mathbb R$ is an open interval, then $C^k(I;\Phi)$ is the space of functions $u : I \rightarrow \Phi$ that are $k$-times continuously differentiable in a Fréchet space. Continuity is in the sense of the topology and differentiability is in the sense of the Gateau derivative, i.e.
$$u'(t) = \lim \limits_{h \to 0} \dfrac{u(t+h)-u(t)}{h}$$
if the limit exists in the space $\Phi$.
Note that if $u \in C^k(I;\Phi)$, then for each continuous semi-norm $\rho : \Phi \rightarrow \mathbb R$, we have $\rho \circ u : I \rightarrow \mathbb R$ belongs to $C^k(I)$ thanks to the reverse triangle inequality. However, the converse is not true.


*For a Frechet space $\Phi$, with dual $\Phi'$ endowed with the weak topology, and $I \subset \mathbb R$ is an open interval, then $C^k(I;\Phi')$ is the space of functions $u : I \rightarrow \Phi'$ such that for each test function $\phi \in \Phi$, the function $u(\cdot) \phi : I \rightarrow \mathbb R$ belongs to $C^k(I)$. With this definition, it is a consequence of Banach-Steinhaus that $u$ is differentiable in a weak sense, i.e. for each $t \in I$,
$$u'(t) = \lim \limits_{h \to 0} \dfrac{u(t+h) - u(t)}{h}$$
converges in $\Phi'$ to an element $u'(t) \in \Phi'$ and $u' \in C^{k-1}(I;\Phi')$. For more details on this space, see Chapter 4 of "Elements de distributions et d'équations aux dérivées partielles" by Claude Zuily.


*For a space of distributions such as $S'(\mathbb R^d)$, if $I \subset \mathbb R$ is an interval, I think that $L^1(I; S'(\mathbb R^d))$ is the space of functions $u : I \rightarrow S'(\mathbb R^d)$ such that for all test functions $\phi \in S(\mathbb R^{1+d})$ : $u(t)\phi(t) \in L^1(I)$. Maybe an argument involving Banach-Steinhaus can work and we might only need that for all test functions $\phi \in S(\mathbb R^d)$ : $u(t)\phi \in L^1(I)$. I've asked a question about it here.
