Vector spaces in PDE $C^{1}([0,+\infty) ; H) \cap C([0,+\infty) ; D(A))$ What is the meaning of the following statement:
$$
u \in C^{1}([0,+\infty) ; H) \cap C([0,+\infty) ; D(A))
$$
I am confused because I know that $C^{1}$ functions are continuous. I faced these spaces with the following theorem:
Theorem (Hille-Yosida). Let A be a maximal monotone operator. Then, given any $u_{0} \in D(A)$ there exists a unique function
$$
u \in C^{1}([0,+\infty) ; H) \cap C([0,+\infty) ; D(A))
$$
satisfying
$$
\left\{\begin{array}{l}
\frac{d u}{d t}+A u=0 \quad \text { on }[0,+\infty) \\
u(0)=u_{0}
\end{array}\right.
$$

The simplest example is to think about $A = -\Delta$, regarded as an unbounded operator from $L^2(\Omega)$ to $L^2(\Omega)$ with $\Omega = (-\pi,\pi)$.
We have orthogonal basis set $\{\cos(nx), \sin(nx)\}$ for $n=0, 1, 2, \dots$. Then $D(A) = L^2(\Omega)$
Let
$$
u= e^{-t} u_0(x) \quad \text{ with } u_0 = \sum_{n=0}^{\infty} \left\{a_n \cos (nx) + b_n \sin(nx)\right\}.
$$
Apparently if a $u\in C([0, \infty), D(A))$ at an arbitrary time, then
$$\sum(a_n^2 + b_n^2) <\infty, \tag{1}$$
i.e., the coefficients are square summable.
Moreover, $u_t(t_1, \cdot) \in D(A)$ at a given time $t_1$ as well due to the time decaying property, so
$$u\in C^{1}([0,+\infty) ; H) \cap C([0,+\infty) ; D(A)).$$
Now, let us think about another $u\in C([0, \infty), D(A))$
$$
u = \sum_{n=0}^{\infty} \left\{a_n \cos (n(x+t)) + b_n \sin(n(x+t))\right\}.
$$
Heuristically speaking, $u$'s oscillation becomes much worse after taking the time derivative, because in order that $u_t(t_1, \cdot) \in D(A)$, we must have
$$\sum n^2(a_n^2 + b_n^2) <\infty, \tag{2}$$
Apparently we cannot ask a function that only satisfies (1) to meet the requirement (2) as well.
This example tells us the Hille-Yosida essentially rigorize what the example tries to tell us:

There has to be certain regularity in the temporal direction such that there exists a well-defined unique solution to the ODE system.

