Let $A$, $B$ and $X$ be Banach spaces such that
- $A$ and $B$ are reflexive;
- $A\subset X$ with a compact injection;
- $X\subset B$ with a continuous injection.
For $1<a,b<\infty$, set $$W:=\{v\;|\;v\in L^a(0,T;A),\; v'\in L^b(0,T;B)\},$$
where $v'$ is the derivative of $v$ in the distributional sense.
Here (in the middle of the proof of the Aubin-Lions Lemma) the author uses the follwing fact:
$W\subset C([0,T];B)$ with a continuous injection.
To justify, he says: The inclusion results from Lemma 1.1, and the continuity of the injection is very easy to check.
The part of the inclusion is ok (because form the Lemma 1.1 we conclude that if $v\in L^1(0,T; A)$ and $v'\in L^1(0,T; B)$, then $v$ is a. e. equal to a continuous function).
Could someone explain me the "very easy" part? To get the inclusion, it is enough to assume $A\subset B$ with continuous injection. To get (in a easy way) the continuity, should we exploit the compactness?
From formula (1.15) in Lemma 1.1, you have $$v(t) = v_0 + \int_0^t v'(s) \, \mathrm{d}s.$$ Hence, $$\|v(t)\|_B \le \|v_0\|_B + \| v' \|_{L^1(0,T;B)}.$$ Similarly, you can estimate $\|v_0\|_B$ by using $\|v\|_{L^1(0,T;B)}$.
In fact (by using a slightly different proof), one even has $$\|v\|_{L^\infty(0,T;B)} \le T^{-1} \, \|v\|_{L^1(0,T;B)} + \|v'\|_{L^1(0,T;B)}$$ for all $v \in W^{1,1}(0,T;B)$.