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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!!!

Self-teaching worker
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1 Answers1

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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.

Davide Giraudo
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Baron Mingus
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  • Thank you so much! I'm not sure about some things: I think $|\cdot|$ is a notation for $|\cdot|^2$, am I right? I'd reached $|A\varphi_1-A\varphi_2|^2\le|\varphi_1-\varphi_2|^2\int_{[a,b]}\int_{[a,s]}|K(s,t)|^2d\mu_t d\mu_s$ but I didn't want to integrate $K$ in order to preserve the variable and get something like the $M^n\frac{(x-a)^n}{n!}$ of Kolmogorov-Fomin's proof for the $C[a,b]\to C[a,b]$ case, wich approaches 0 as $n\to\infty$... – Self-teaching worker Dec 06 '14 at 13:38
  • ...By integrating $K$ I get $|A\phi_1-A\phi_2|{L_2(d\mu)}^2\le\lambda^2|\phi_1-\phi_2|{L_2(d\mu)}^2 M$, where $M=|K|_{L_2(d\mu\otimes\mu)}^2$, but there's something I don't graps because I don't understand how is $(b-a)$ derived... :-( How would you handle the problem to show that $A$ is a contraction? Kolmogorov-Fomin's makes the reader believe it's trivially the same as in the $C[a,b]$ case, while it doesn't appear so trivial, to my eyes... I heartily thank you! – Self-teaching worker Dec 06 '14 at 13:38
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    You've made a mistake with the $\int_a^b \int _a^s |\phi_1(t) - \phi_2(t)|^2 \mathrm{d}\mu_t \mathrm{d}\mu_s$ part, as that's where the $(b-a)$ comes from - not the K integral.

    The trick is that because everything we're integrating is non-negative, increasing the domain increases the values of the integrals, so $\int_a^b \left(\int_a^s|K(s,t)|^2\mathrm{d}\mu_t\right) \left(\int_a^s |\phi_1(t) - \phi_2(t)^2|^2 \mathrm{d}\mu_t\right) \mathrm{d}\mu_s \leq \int_a^b \left(\int_a^b|K(s,t)|^2\mathrm{d}\mu_t\right) \left(\int_a^b |\phi_1(t) - \phi_2(t)^2|^2 \mathrm{d}\mu_t\right)\mathrm{d}\mu_s$...

     – Baron Mingus Dec 06 '14 at 16:36
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    Note the change of upper boundary on the inner two integrals. Then the first term of the product gives us the $M$ bound and so that is $\leq M \int_a^b \int_a^b |\phi_1(t)-\phi_2(t)|^2 \mathrm{d}\mu_t \mathrm{d}\mu_s$. Note that the integrand is nowindependent of $s$, so the outer integral just evaluates to $b-a$ and this is equal to $M(b-a)|\phi_1 - \phi_2|_{L_2}$. Is that any clearer? – Baron Mingus Dec 06 '14 at 16:40
  • Thank you so much! As it emerges at p. 478 the hypothesis that $K$ is bounded is contained in the theorem. If $K$ is bounded, the kernel $K_n$ of the $n$-th iteration is such that $\forall(s,t)\in[a,b]^2\quad|K_n(s,t)|\le \frac{M^n(b-a)^{n-1}}{(n-1)!}$. I cannot prove that, either, but I suspect that it may well be the key to prove that $A^n$ is a contraction for some $n$. I think that $\frac{(b-a)^{n-1}}{(n-1)!}$ may come from the integration of something like $\int_a^x\frac{(t-a)^{(n-2)}}{(n-2)!}dt=\frac{(x-a)^{n-1}}{(n-1)!}$ but I am making some mistake with the $u,s,t$... – Self-teaching worker Dec 06 '14 at 18:07