I am currently reading this paper from 1973.
In short, one has given a linear continuous operator $P : L^1([0,1]) \to L^1([0,1])$ with ||P||=1 and for $f \in L^1$ a family of functions $(f_n)_{n\in\mathbb{N}}$ with $f_n = P^n f$. Therefore it holds
$$||f_n|| \leq ||f||$$
and it is shown that
$$ \limsup_{k\to \infty} \bigvee_0^1 f_{Nk} \leq c ||f|| $$
where $ \bigvee_0^1 $ denotes the total variation, $0<c<1$ and $N \in \mathbb{N}$.
It is then said (without explanation) that the set $$C:=\{f_{Nk} \mid k\in \mathbb{N} \}$$ is relatively compact in $L^1$, and because of $$ \{ f_k \mid k\in \mathbb{N} \} \subset \bigcup_{i=0}^{N-1} P C$$ the set $$\{ f_k \mid k\in \mathbb{N} \}$$ is also relatively compact.
I do not see why $C$ is relatively compact, nor why $ \bigcup_{i=0}^{N-1} P C$ is relatively compact. Can someone explain this to me?
In you post, I read $C$ is bounded in $BV$ norm, i.e., the function of bounded variation.
If you know Sobolev space, you could just think $BV$ is the relaxation of $W^{1,1}$ and hence similar to $W^{1,1}$ space, bounded sequence in $BV$ will compact in $L^1$, and even in $L^q$ for any $q<N/(N-1)$ where $N$ is the dimension of space.
For reference, check out Theorem 4 at page 176 in this book.