Prove by induction $ (V^nx)(t) =\int_{0}^{t} \frac{(t-s)^{n-1}}{(n-1)!}x(s)ds $

Question

Prove by induction that $ (V^nx)(t) =\int_{0}^{t} \frac{(t-s)^{n-1}}{(n-1)!}x(s)ds $ Use this result to solve for f the integral equation $f(t) = sin t + \int_{0}^{t} f(s)ds$ where V is the $Volterra$ $operator$ on $L^2(0,1)$ $$(Vx)(t) = \int_{0}^{t} x(s)ds, 0\leq t\leq 1$$

I did the base case but I'm not sure how to proceed for the induction step:

For the case $n=1$: $$(V^1x)(t) =\int_{0}^{t} \frac{(t-s)^{1-1}}{(1-1)!}x(s)ds = \int_{0}^{t} x(s)ds$$

Assume that $ (V^nx)(t) =\int_{0}^{t} \frac{(t-s)^{n-1}}{(n-1)!}x(s)ds $ is true for $n=k$. We must now prove that it is true for $n=k+1$.

Answer 1

It's not a proof for general case, but you can use Cauchy repeated integration formula. (Here is the wikipedia article on it, which contains proof) https://en.wikipedia.org/wiki/Cauchy_formula_for_repeated_integration

Answer 2

Use induction for $n+1$ case. Let $\chi_{+}$ be the characteristic function of $[0,\infty)$. Then $$ (V^{n+1}x)(t)=\int_{0}^{t}(V^{n}x)(u)du \\ = \int_{0}^{t}\int_{0}^{u}\frac{(u-s)^{n-1}}{(n-1)!}x(s)ds du \\ = \int_{0}^{t}\int_{0}^{t}\frac{(u-s)^{n-1}}{(n-1)!}x(s)\chi_{+}(u-s) ds du \\ = \int_{0}^{t}\int_{0}^{t}\frac{(u-s)^{n-1}}{(n-1)!}\chi_{+}(u-s)du\, x(s)ds \\ = \int_{0}^{t}\int_{s}^{t}\frac{(u-s)^{n-1}}{(n-1)!}du\, x(s)ds \\ = \int_{0}^{t}\frac{(t-s)^{n}}{n!}x(s)ds. $$