Let $\mathcal H = L^2[0,1].$ Define an operator $T : \mathcal H \longrightarrow \mathcal H$ by $Tf(x) = \displaystyle {\int_{0}^{x} f(y)\ dy.}$ Show that for all $n \geq 1$ $$T^n f(x) = \int_{0}^{x} \frac {(x-t)^{n-1}}{(n-1)!} f(t)\ dt.$$
Clearly this result holds for $n=1.$ Let the result hold for $n.$ Then $$T^{n+1} f(x) = T^n(Tf(x)) = \int_{0}^{x} \frac {(x-t)^{n-1}}{(n-1)!} \left (\int_{0}^{t} f(y)\ dy \right )\ dt.$$ Another way of writing it down is $$T^{n+1} f(x) = T(T^nf(x)) = \int_{0}^{x} \left (\int_{0}^{y} \frac {(y-t)^{n-1}} {(n-1)!} f(t)\ dt \right )\ dy.$$ From here how do I conclude that $$T^{n+1} f(x) = \int_{0}^{x} \frac {(x-t)^n} {n!} f(t)\ dt\ ?$$
Any help in this regard will be warmly appreciated. Thanks for your time.
Change the order of integration for the induction step: \begin{eqnarray*} (T^{n+1}f)(x) & = & \int_0^x(T^nf)(y)\mathrm dy \\ & = & \int_0^x\int_0^y\frac{(y-t)^{n-1}}{(n-1)!}f(t)\mathrm dt \mathrm dy \\ & = & \int_0^x \int_t^x\frac{(y-t)^{n-1}}{(n-1)!}f(t)\mathrm dy \mathrm dt \end{eqnarray*} Can you finish?