Induction and Integration by Parts

Question

Let $T: C([0,2]) \to C([0,2])$ be given by $(Tf)(x) = \int_0 ^x f(t)dt$ Prove $$(T^n f)(x) = \frac{1}{(n-1)!} \int_0 ^x (x-t)^{n-1}f(t)dt$$ for $n \in \mathbb{N}$.

I am attempting this with induction and integration by parts. However, I am struggling with the induction step to show the $(n+1)$ case holds. No matter what substitution I use for $u$ and $dv$, I am unable to proceed. (I first tried $u = (x-t)^n$ to use the induction hypothesis, but did not succeed.) Any ideas on how to proceed/conclude?

Answer 1

\begin{align*} (T^{2}f)(x)&=\int_{0}^{x}(Tf)(t)dt\\ &=\int_{0}^{x}\int_{0}^{t}f(u)dudt\\ &=\int_{0}^{x}\int_{0}^{x}\chi_{[0,t]}(u)f(u)dudt\\ &=\int_{0}^{x}\int_{0}^{x}\chi_{[u,x]}(t)dtf(u)du\\ &=\int_{0}^{x}(x-u)f(u)du. \end{align*} Similarly, we have \begin{align*} (T^{n+1}f)(x)&=\int_{0}^{x}(T^{n}f)(t)dt\\ &=\dfrac{1}{(n-1)!}\int_{0}^{x}\int_{0}^{t}(t-u)^{n-1}f(u)dudt\\ &=\dfrac{1}{(n-1)!}\int_{0}^{x}\int_{0}^{x}\chi_{[u,x]}(t)(t-u)^{n-1}dtf(u)du\\ &=\dfrac{1}{(n-1)!}\int_{0}^{x}\int_{u}^{x}(t-u)^{n-1}dtf(u)du\\ &=\dfrac{1}{(n-1)!}\dfrac{1}{n}\int_{0}^{x}(x-u)^{n}f(u)du\\ &=\dfrac{1}{n!}\int_{0}^{x}(x-u)^{n}f(u)du. \end{align*}

Answer 2

Given, $T(f)(x)=\int_{0}^{x}f(t)dt$

Induction hypothesis: The result holds for n.

Let, $T(f)(x)=g(x)$ then $g(0)=0$ and $g'(x)=f(x)$

Now,

$T^{n+1}(f(x))=T^n(T(f(x)))=T^n(g(x))$

$=\frac{1}{(n-1)!}\int_{0}^{x}(x-t)^{n-1}g(t)dt$

$=\frac{1}{(n-1)!}[[-g(t)\frac{(x-t)^n}{n}]_{0}^{x}+\int_{0}^{x}\frac{1}{n}(x-t)^ng'(t)dt]$

$=\frac{1}{n!}\int_0^x(x-t)^nf(t)dt$

Hope this works.