# 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?

$$\DeclareMathOperator{AC}{AC}$$Notice that the requirement the space you are considering is precisely the space of absolutely continuous functions. Thus, the question reduces to
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, $$L^1((0,T)) \subseteq L^\infty((0,T))$$ by Holder's inequality, while on the other hand $$L^\infty((0,T)) \subseteq W^{1,1}((0,T))$$ by Poincare's inequality.