Counter-example Aubin-Lions p=1 I wanted to show, that $p>1$ in Aubin-Lions Lemma is necessary. I hoped there already is an example for $X=Y=Z=\mathbb{R}$ so that the embedding 
\begin{align}
\{u \in L^\infty(0,T);\mathbb{R}) \mid u^\prime \in L^1(0,T;\mathbb{R})\} \to C([0,T];\mathbb{R})
\end{align}
is not compact.
So, I have to find a sequence of bounded functions in $\{u \in L^\infty(0,T);\mathbb{R}) \mid u^\prime \in L^1(0,T;\mathbb{R})\}$ that don't have a converging subsequence in $C([0,T];\mathbb{R})$.
My problem is that I didn't get an idea of which type of bounded functions I'm actually searching for. At first, I thought I have to take a sequence of oscillating functions like $\sin(kx)$ but their derivative isn't bounded.
Does anybody have an idea or a hint?
 A: $\DeclareMathOperator{AC}{AC}$Notice that the requirement the space you are considering is precisely the space of absolutely continuous functions.  Thus, the question reduces to

Can we find a bounded sequence of absolutely continuous which does not converge uniformly to a continuous function?

In particular, notice that it would be sufficient to take a sequence $u_n$ of absolutely continuous functions which converges pointwise to a discontinuous function.  There are many examples, for instance
$$u_n(x) = \frac{x^n}{T^{n+1}}$$

It seems worth pointing out that in the setting you are considering, the Aubin-Lions lemma is the same as the humble Rellich-Kondrakov theorem.  In particular, the space
$$A = \{u \in L^\infty((0,T); \mathbb{R}) | u' \in L^1(0,T;\mathbb{R})\}$$
is equivalence (as a Banach space) to $W^{1,1}((0,T))$, since on one hand, $L^1((0,T)) \supseteq L^\infty((0,T))$ by Holder's inequality, while on the other hand $L^\infty((0,T)) \supseteq W^{1,1}((0,T))$ by Poincare's inequality.
