Volterra operator is completely continuous Let $\mu$ be the Lebesgue measure on $[0,1]$ on the borelians, and consider the Volterra operator $V:L^1[0,1]\to C[0,1]$ given by
$$
Vf(t)=\int_0^t f d \mu
$$
So, I want to show the following

Proposition: The Volterra operator sends weak compact sets into norm compact sets. 

I've already proved that this operator sends uniformly integrable sets into uniformly bounded and equicontinuous sets of $C[0,1]$, i.e, relatively compact sets in $C[0,1]$. I know that uniformly integrable sets are the same that relatively $\omega$-compact sets. I don't get how to finish the problem with this.
Any help would be welcome. Thanks in advanced.
 A: Added: The discussion below shows that $V$ is not a compact operator.
However, since $L^1$ is not reflexive, this does not preclude that $V$
is completely continuous.  (See e.g. the wikipedia discussion.)
To show that $V$ is completely continuous, one approach is to show that
if $f_n$ is a sequence which converges to $0$ weakly in $L^1$,
then $V(f_n)$ converges to $0$ in $C[0,1]$, i.e. $V(f_n)$ converges to zero uniformly.
Since the dual to $L^1$ is $L^{\infty}$, to say that $f_n$ converges to zero
weakly is the same as saying that for any $L^{\infty}$-function $\phi$,
the integrals $\int f_n \phi$ converge to zero.
Since $Vf(t) = \int f \chi_{[0,t]} $ (where $\chi_{[0,t]}$ is the indicator
function of the interval $[0,t]$, which is in $L^{\infty}$),
we see that $Vf_n$ converges to zero pointwise.
But actually, the rate of convergence to zero of $\int f_n \phi$
depends only on $|| \phi ||_{\infty},$ so as long as e.g. $||\phi ||_{\infty} \leq 1$,
the sequence $\int f_n \phi$ converges to zero uniformly in $\phi$ (i.e. for any
$\epsilon > 0$, we can find a value of $N$
such that  $| \int f_n \phi| < \epsilon$ for every $\phi$ s.t. $|| \phi ||_{\infty} \leq 1$).
(The reason for this is that the sequence $f_n$ is norm-bounded in $L^1$
(being weakly convergent; see e.g. this question and answer), and hence equicontinuous
on the unit ball of $L^{\infty}$, when we think of the elements
of $L^1$ as functionals on this unit ball.  An equicontinuous sequence
 which converges pointwise to zero in fact converges uniformly to zero; see
e.g. this question and answer
(note that the argument goes over for compact spaces that are not necessarily
metrizable)
or the wikipedia discussion,
and note that the unit ball of $L^{\infty}$ is compact in the weak-$*$ topology,
by the Banach--Alaoglu Theorem.)
This shows that convergence of $V f_n$ to zero is uniform, as required.

There's something that confuses me about this statement.  I will try to explain
my confusion.  Hopefully it will help with your question.  If not, I apologize.
[Added: my confusion came from the fact that being completely continuous
is weaker than being compact, when the domain is not reflexive.]

[Added: the following discussion shows that $V$ is not compact.]
Suppose that $g \in C[0,1]$ is actually differentiable and monotone increasing,
that $g(0) = 0,$ and that $g(1) = 1$.
Then if we set $f = g'$, then $f \geq 0$ on $[0,1]$ (by the mononicity assumption),
and $$|| f ||_1 = \int_0^1 f = \int_0^1 g' = g(1) - g(0) = 1.$$
And of course $V f = g$ (again using the fact that $g(0) = 0$).
So all these $f$ have $L^1$-norm equal to $1$, and so we would be saying that
any sequence of such $g$'s has a convergent subsequence (since they all lie in 
the image of $V$ applied to the unit ball in $L^1$).
But this is not true.
For example, take $g_n(t) = t^n,$ so the corresponding $f$ is
$f_n(x) = n x^{n-1}.$
The sequence $g_n$ has no uniformly convergent subsequence.

So it seems to me that $V$ is not a compact operator, when we think of
the range as being $C[0,1]$.  

This is related to the fact that (as far as I can see), although
$Vf$ is continuous, we can't find an estimate for its continuity based
only on the $L^1$-norm of $f$.  (This is what you pointed out in your
comment on Niki's answer.)  A related fact (I think) is the in the
Riemann--Lebesgue lemma, as far as I understand, we can't find an estimate
for the rate of convergence to zero of the Fourier transform of $f$
depending only on the $L^1$-norm of $f$.
A: I share how I solved it, just for fun
We know that $V:L_1[0,1]\to C[0,1]$ sends relatively weakly compact sets into relatively norm  compact sets. So, we want to show that $V$ sends weakly compact sets into norm compact sets.
I need to recall two important things: Let $X, Y$ be two Banach spaces


*

*If $X$ is a Banach space and $A\subset X$ is weakly compact, then its closed convex hull is weakly compact.

*If $T:X \to Y$ is a continuous linear operator between two Banach spaces, then it's $\omega- \omega$ continuous.
So, take $(f_n)\subset L_1[0,1]$ sucht that $f_n \rightharpoonup f$. Consider $A=\{f_n\}_n \cup \{f\}$, so $A$ is weakly compact. Thus $co(A)$ is relatively weakly compact. Then $V(co(A))$ is relatively norm compact. As $(Vf_n)\subset V(co(A))$, then it has a convergent subsequence, say $Vf_{n_k} \to g$. Using that $V$ is $\omega- \omega$ continuous, we know that $Vf_{n_k}\rightharpoonup Vf$, then $Vf=g$. And we're done, because we've shown that every subsequence of $(Vf_n)$ has a $\|\cdot\|-$convergent subsequence to $Vf$, which implies that $Vf_n \to Vf$.
