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. 
 A: 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:
$$
\begin{align}
g(a)
&=A^0(f)(a)+\sum\limits_{k=1}^\infty A^k (f)(a)\\
&=f(a)+\sum\limits_{k=1}^\infty \int_0^a\frac{(a-x)^{2k-1}}{(2k-1)!}f(x)dx\\
&=f(a)+\int_0^a\sum\limits_{k=0}^\infty\frac{(a-x)^{2k-1}}{(2k-1)!}f(x)dx\\
&=f(a)+\int_0^a\sinh(a-x) f(x)dx\\
\end{align}
$$
A: Try it this way: If you look at     A2g(a) you get ∫(x-a)[∫(x-a)g(x)dx]dx.
Since you are on L2 everything is integrable and you can rewrite this as
∫g(x)[∫(a-x)2dx]dx = ∫(a-x)3g(x)/3 dx. This looks good although it is missing a factor of 1/2.  
You can keep going this way, and you'll get what you want, missing the factor of 1/2, which either is there somewhere and I can't quite see it; or isn't there and your formula needs a factor of 2.  Can someone help me out here?
