# 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?

## 1 Answer

$$\DeclareMathOperator{AC}{AC}$$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, $$L^1((0,T)) \subseteq L^\infty((0,T))$$ by Holder's inequality, while on the other hand $$L^\infty((0,T)) \subseteq W^{1,1}((0,T))$$ by Poincare's inequality.