Show that $T^n f(t) = \frac{1}{(n-1)!} \int_0^t (t-s)^{n-1} f(s) ds$ for $n \in \mathbb{N}$. Consider $C[0,1]$ as a Banach space with $\lVert f \rVert_\infty = \max_{t \in [0,1]} |f(x)|$, and define $T \in B(C[0,1])$ by $Tf(t) = \int_0^t f(s) ds$
1- Show that $T^n f(t) = \frac{1}{(n-1)!} \int_0^t (t-s)^{n-1} f(s) ds$ for $n \in \mathbb{N}$.
2- Determine $\sigma(T)$.   Hint: $r(T) = \underset{n \to \infty}\lim {\lVert T^n \rVert^{\frac{1}{n}}}$.
Edited: I solved the first part using the hints. Thank you so much.
However, the second part is totally unclear for me, any hints please.
 A: Using the hints mentioned above, let start by applying the operator many times to see the result.
\begin{align}
                 T(f(t))  &= \int_0^t f(s) ds\\
    T (T(f(t)))=T^2 (f(t) &= \int_0^t \left( \int_0^\tau f(\tau) d\tau \right)ds\\
                          &= \int_0^t  \int_0^\tau f(\tau) d\tau ds \\
                          &= \int_0^t \int_\tau^t f(\tau) ds d\tau\\
                           &= \int_0^t \left( \int_\tau^t  ds \right) f(\tau) d\tau\\
                          &= \int_0^t f(\tau) (t -\tau) d\tau
\end{align}
So we can see that;
$$T^2 f(t)=\int_0^t f(s) (t -s) ds.$$
Applying the operator more times we get;
$$T^3 f(t)= \frac{1}{2}\int_0^t f(s) (t -s)^2 ds.$$
and
$$T^4 f(t)= \frac{1}{2 . 3}\int_0^t f(s) (t -s)^3 ds.$$
To prove the relation for all $n\in \mathbb{N}$ we use induction. Since the relation holds when $n=1$ we assume it holds when $n=k \in \mathbb{N}$,
$$T^k f(t)= \frac{1}{(k-1)!}\int_0^t f(s) (t -s)^{(k-1)} ds.$$
Now we apply by the operator one more time,
\begin{align}
    T(T^k f(t)) =& \int_0^t \left( \frac{1}{(k-1)!}\int_0^s f(\tau) (s -\tau)^{(k-1)} d\tau \right) ds\\
     T^{k+1} f(t) =& \frac{1}{(k-1)!} \int_0^t  \int_0^s f(\tau) (s -\tau)^{(k-1)} d\tau ds\\
                  =& \frac{1}{(k-1)!} \int_0^t  \int_\tau^t f(\tau) (s -\tau)^{(k-1)} ds d\tau\\
                  =& \frac{1}{(k-1)!} \int_0^t   \left( \int_\tau^t ds \right) f(\tau) (s -\tau)^{(k-1)}  d\tau\\
                  =& \frac{1}{k!} \int_0^t f(\tau) (s -\tau)^{k}  d\tau.\\
\end{align}
So the relation
$$T^n f(t) = \frac{1}{(n-1)!} \int_0^t (t-s)^{n-1} f(s) ds$$
is right for all $n \in \mathbb{N}$.
