# Question about a particular linear operator

Let A be a linear operator. $A: L^2(0,1) \rightarrow L^2(0,1)$ given by $Ag(a) = \int_0^a(a-x)g(x)dx$ where $a \in (0,1)$. This is the integral operator, and we know ||A|| < 1 which is easy to check. We want to show $A^kg(a) = \int_0^a\frac{(a-x)^{2k-1}}{(2k-1)!}g(x)dx$ where $a \in (0,1)$. Induction seems like an obvious way to approach this:

Base case k = 1: $(2k-1)! = 1$ and $(2k-1) = 1$ hence $A^1g(a)$ is same as given.

Induction hypothesis: Assume $A^ng(a) = \int_0^a\frac{(a-x)^{2n-1}}{(2n-1)!}g(x)dx$ holds for n = k.

Induction step: Apply the linear operator $A$ to $A^ng(a)$, having a bit trouble with this, would we get double integrals?

After we have shown $A^kg(a)$ as above, then we will use that to find $g(a):$

$f \in L^2(0,1)$ given $g \in L^2(0,1)$ and $g(a) = f(a) + \int_0^a(a-x)g(x)dx$, from this we can find $g = (I - A)^{-1}f = \sum_{i=0}^\infty A^if = \sum_{i=0}^\infty \int_0^a\frac{(a-x)^{2i-1}}{(2i-1)!}f$ now we need to try simplify this somehow and find g without summations.

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Is it necessary to find $A^k(g)(a)$ this way? I think it is easier to prove it by induction –  Norbert Aug 14 at 12:51
@Norbert This way preffered over induction, I am thinking that I am very close to getting question but missing a linking step. –  user77404 Aug 14 at 13:42
Even if we prove this by induction, we will still need to find g like I did in this case? –  user77404 Aug 14 at 15:07
as far as I understand you have two tasks 1) prove the formula for $A^k(g)(a)$ 2) for a given arbitrary $f\in L_2$ compute $g=\sum_{i=0}^\infty A^i f$. Right? –  Norbert Aug 14 at 21:32
@Norbert that's correct, I will edit the question to make it more clear. –  user77404 Aug 14 at 21:53

1) For induction step note that \begin{align} A^{k+1}(g)(a) &=\int_0^a (a-t)A^k(g)(t)dt\\ &=\int_0^a (a-t)\int_0^t \frac{(t-x)^{2k-1}}{(2k-1)!}g(x)dxdt\\ &=\int_0^a \int_0^t \frac{(a-t)(t-x)^{2k-1}}{(2k-1)!}g(x)dxdt\\ &=\int_0^a \int_{x}^{a} \frac{(a-t)(t-x)^{2k-1}}{(2k-1)!}g(x)dtdx\\ &=\int_0^a\frac{g(x)}{(2k-1)!}\int_{x}^{a} (a-t)(t-x)^{2k-1}dtdx\\ &=\int_0^a\frac{g(x)}{(2k-1)!}\frac{(a-x)^{2k+1}}{2k(2k+1)}dx\\ &=\int_0^a\frac{(a-x)^{2k+1}}{(2k+1)!}g(x)dx\\ \end{align} 2) As for the second part wlog $f\geq 0$, and summands are positive. So you can interchange integration and summation: \begin{align} g(a) &=A^0(f)(a)+\sum\limits_{k=1}^\infty A^k (f)(a)\\ &=f(a)+\sum\limits_{k=1}^\infty \int_0^a\frac{(a-x)^{2k-1}}{(2k-1)!}f(x)dx\\ &=f(a)+\int_0^a\sum\limits_{k=0}^\infty\frac{(a-x)^{2k-1}}{(2k-1)!}f(x)dx\\ &=f(a)+\int_0^a\sinh(a-x) f(x)dx\\ \end{align}

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Your application of the formula to get $A^2 g(a)$ isn't quite right. Be careful with the variables, you should keep careful track of the integration variables as you substitute the second time around. –  Evan Aug 14 at 23:54