Reiterate Volterra integral operator is a contraction I read in Kolmogorov and Fomin's Элементы теории функций и функционального анализа (p. 472 here) the statement that Volterra operator $A:L_2[a,b]\to L_2[a,b]$ defined by$$(A\varphi)(s):=\int_{[a,s]}K(s,t)\varphi(t)d\mu_t+f(s)$$with $K\in L_2([a,b]^2)$, $f\in L_2[a,b]$, is such that $A^n$ is a contraction for some $n\in\mathbb{N}$ if $K$ is bounded. Kolmogorov and Fomin say that the proof consists of literaly repeating the reasonings of the proof for the case, which can be seen in this translation from Introductory Real Analysis, $A:C[a,b]\to C[a,b]$, $\varphi\mapsto \int_a^s K(s,t)\varphi(t)dt+f$ with $K\in C([a,b]^2)$, $f\in C[a,b]$.
How can it be adapted, or how can it be otherwise proved that $A^n:L_2[a,b]\to L_2[a,b]$ is a contraction for some $n$?
I thank you very much!!!
 A: The thing you're missing (I think - looks like you're almost there in your edit) is Schwarz's inequality. 
We have that
\begin{equation}
|A\phi_1 - A\phi_2|_{L_2} = \lambda^2\int_a^b \left|\int_a^s K(s,t)(\phi_1(t)-\phi_2(t))\mathrm{d}\mu_t\right|^2\mathrm{d}\mu_s.
\end{equation}
Then by Schwarz's inequality we can bound the inner integral squared as the product of the two square integrals, so
\begin{equation}
|A\phi_1 - A\phi_2|_{L_2} \leq \lambda^2\int_a^b \left(\int_a^s |K(s,t)|^2\mathrm{d}\mu_t\right)\left(\int_a^s |\phi_1(t)-\phi_2(t)|^2\mathrm{d}\mu_t\right)\mathrm{d}\mu_s.
\end{equation}
Now it should be easy to show that if $M =  \int_a^b \left(\int_a^b |K(s,t)|^2\mathrm{d}\mu_t\right)\mathrm{d}\mu_s$, then 
\begin{equation}
|A\phi_1 - A\phi_2|_{L_2} \leq\lambda^2M(b-a)|\phi_1 - \phi_2|_{L_2}
\end{equation}
which is a contraction if $\lambda < \sqrt{\frac{1}{M(b-a)}}$.
I haven't done the iterated applications of $A$ for the more general proof - it will be slighlty messier as the $L_2$ norm doesn't play as nicely the as $sup$ norm, but it should follow the argument in the textbook pretty closely. 
