Norms of Functions in Hilbert spaces I have a convergence result (coming from FEM for wave equations) that $$||e'||_{C(0,T;L^2)} + ||e||_{C(0,T;H^1)} \leq Ch\sqrt{1+T^2} ||u||_{C^2(0,T;H^2)}$$
Here $C^m(a,b;X) = \{ f:[a,b] \to X \mid f \text{ is } m \text{ times differentiable }\}$.
So far I found out that $||f||_{C(a,b;X)} := \sup_{t\in [a,b]} ||f(t)||_X$ which makes sense, but how is $$||f||_{C^2(a,b;X)}$$ defined? Do we have to use the second derivatives somewhere?
 A: The ${C^k([a,b];X)}$-norm is usually defined by
$$\|f\|_{C^k([a,b];X)} =  \sum_{j=0}^k\sup_{t\in[a,b]}\left\|\frac{d^j}{dt^j}f(t,.)\right\|_X,$$
where for every $t\in [a,b]$, $f(t,.)$ is an element of $X$. Instead of the sum you may equivalently take the maximum over all $j$.
The idea behind this is that $f\in C^k(a,b;X)$ can be read as follows: the map $t\mapsto f(t,.)$ maps from $[a,b]$ into $X$. Differentiability of this map means that for every $t\in[a,b]$ the limit
$$\lim_{h\to 0}\frac{f(t+h,.) - f(t,.)}{h}$$
exists in $X$! That is, with respect to the norm $\|.\|_X$. In combination with the continuity of the norm, this means that the limit
$$\left\|\lim_{h\to 0}\frac{f(t+h,.) - f(t,.)}{h}\right\|_X = \left\|\frac{d}{dt}f(t,.)\right\|_X$$
exists. Note that this is a much stronger statement than the existence of $\frac d{dt}\|f(t,.)\|_X$!
Accordingly, the $C^1([a,b];X)$-norm has to be 
$$\|f\|_{C^1([a,b];X)} =  \sup_{t\in[a,b]}\left\|f(t,.)\right\|_X+ \sup_{t\in[a,b]}\left\|\frac{d}{dt}f(t,.)\right\|_X,$$
and with a natural continuation for higher derivatives. 
A simple counter-example helps to illustrate why the derivatives need to be inside the norm: Consider $f(t,x)$ such that $\|f(t,.)\|_X$ is constant in $t$. Then $\|f(t,.)\|_X' = 0$, but $t\mapsto f(t,.)$ needs not be constant, and so $t\mapsto f(t,.)'$ will not vanish. Hence, $\|f(t,.)'\|_X$, and not $\|f(t,.)\|_X'$ captures the behavior of the derivative.
A: I now found out that for any Banach space $X$ we define
$$\Vert f \Vert_{C^m(a,b;X)}:= \max_{0\leq l \leq k} \sup_{t \in [a,b]} \left\Vert \frac{d^l}{dt^l} f(t) \right \Vert_X.$$
