I have the following space $X =C([0,1],\mathbb{K}$ and let $T$ be the operator defined by
\begin{equation} Tf(x)=\int^{x}_{0} (x-t)f(t)dt, \space\space\space f \in X \end{equation}
Now I have showed the operator $T$ is bounded by $1/2$. I am left to show that the following must hold:
\begin{equation} T^{n}f(x) = \int^{x}_{0} \dfrac{(x-t)^{2n-1}}{(2n-1)!}f(t) dt, \space \space \space n \in \mathbb{N} \end{equation}
Does anyone have a strategy for this?
It is straightforward to check (by IBP) that your last formula holds for any polynomial $f(x)$.
Since the space of polynomials is dense in $C([0,1])$ with respect to the uniform norm by Weierstrass approximation theorem, we are done.
About boundedness, if we set $F(x)=\int_{0}^{x}f(t)\,dt$ and $M(x)=\max_{t\in [0,x]}f(t)$ we have $F(x)\leq x M(x)\leq xM(1)$ and by integration by parts $$ \int_{0}^{x}(x-t)\,f(t)\,dt = \int_{0}^{x} F(t)\,dt \leq M(1)\int_{0}^{x}t\,dt \leq M(1)\int_{0}^{1}t\,dt = \frac{1}{2}M(1).$$
