I need some help with these questions.
$\bullet\;$ First of all, if we define the Volterra operator $V:L^{1}[0,2\pi]\rightarrow L^{1}[0,2\pi]$ as $(Vf)(x)=\int_0^xf(t)dt$, Is this operator compact? (I saw this question , where it is proved in the case the operator is defined over $L^{2}[0,1]$, but I don't think the proof can be adapted to this case).
$\bullet\;$ And the other doubt I have is whether or not the space $L^1[0,2\pi]$ has the approximation property. This would give us that the Volterra operator defined as above (if it is compact) would be limit of finite rank operators.
Thanks a lot for any help!
Yes, $V$ is compact. The image of unit ball under $V$ is a bounded subset of $L^1$ that is also bounded in the $BV$ norm (total variation). Such a set is precompact in $L^1$: see $L^1$ Compactness of Bounded BV Sets by Isidore Fleischer.
Upon request, I supply a self-contained argument. Let $(f_n)$ be a sequence in the unit ball of $L^1$. Denote $F_n= Vf_n$. Observe that $F_n(0)=0$ and the total variation of $F_n$ is at most $1$. Use the representation of BV functions as a difference of two increasing functions: $F_n=\phi_n-\psi_n$ where $\phi_n(0)=\psi_n(0)=0$ and $\phi_n(1)\le 1$, $\psi_n(1)\le 1$.
Enumerate rational numbers in $[0,1]$ as $(q_k)$. Choose a subsequence of $\phi_n$ that converges at $q_1$; from it, choose a subsequence that converges at $q_2$, ... then take diagonal subsequence $(\phi_{n_d})$, which converges at all rational numbers. Let $\phi$ be the limit function, so far defined only at rational numbers. Extend it to $[0,1]\setminus \mathbb Q$ by $$\phi(x) = \sup \{\phi(q) : q<x, q\in \mathbb Q$$ The result is an increasing function on $[0,1]$, also denoted $\phi$. Being increasing, it is continuous except on a countable set. At each point of continuity of $\phi$, we have $$\phi(x) = \lim \phi_{n_d}$$ (this is not hard to check). By the dominated convergence theorem, $\phi_{n_d}\to \phi$ in $L^1$.
Starting with $\phi_{n_d}$, repeat the argument for $\psi_{n_d}$, extracting a further subsequence that converges in $L^1$. Along this subsequence, $F_n$ converge in $L^1$. $\Box$
Yes, $L^1$ has an approximation property, because it has a Schauder basis: the Haar system is one such basis. Separable Banach spaces without an approximation property are artificial constructs; they do not occur in the wild.