My question is given at the end of the explanation. Let $K\in{}C([a,b]^{2},\mathbb{R})$ and consider the operator $H:C([a,b],\mathbb{R})\to{}C([a,b],\mathbb{R})$ defined by $$H[x](t):=\int_{a}^{t}K(t,\eta)x(\eta)\mathrm{d}\eta.$$ Let $f\in{}C([a,b],\mathbb{R})$ and consider the equation $$x=H[x]+f,\quad\text{on}\ [a,b].$$ Then, $$x=H\big[H[x]+f\big]+f=H^{2}[x]+H[f]+f,$$ which yields by repeating in this manner that $$x=H^{n}[x]+\sum_{k=0}^{n-1}H^{k}[f],{\quad}n=1,2,\cdots\tag{A}$$ with the convention that $H^{0}[f]:=f$. As $K$ is continuous, it is bounded on $[a,b]^{2}$, say $|K(t,s)|\leq{}M$ for all $s,t\in[a,b]$, we can show by induction that $$|H^{n}[x](t)|\leq{}M^{n}\int_{a}^{t}\frac{(t-\eta)^{n-1}}{(n-1)!}|x(\eta)|\mathrm{d}\eta$$ for all $t\in[a,b]$ and $n=1,2,\cdots$. This yields that the operator satisfies $\|H^{n}[x]\|_{\infty}\leq{}M^{n}\frac{(b-a)^{n}}{n!}\|x\|_{\infty}$ for $n=0,1,\cdots$, i.e., $\|H^{n}\|_{\text{op}}\leq{}M^{n}\frac{(b-a)^{n}}{n!}$ for $n=0,1,\cdots$. Therefore, $$\rho(H)=\lim_{n\to\infty}\sqrt[n]{\|H^{n}\|_{\text{op}}}\leq\lim_{n\to\infty}M\frac{(b-a)}{\sqrt[n]{n!}}=0$$ showing that $\rho(H)=0$, which is less than $1$. Then, we can let $n\to\infty$ in (A) and get $$x=\sum_{k=0}^{\infty}H^{k}[f].$$

The point I don't understand here is how $\rho(H)<1$ allows us to let $n\to\infty$ in (A)!

Explanation for user2139: Clearly, for $n=1$, we have $$|H[x](t)|\leq\int_{a}^{t}|K(t,\eta)||x(\eta)|\mathrm{d}\eta\leq{}M\int_{a}^{t}|x(\eta)|\mathrm{d}\eta,$$ which shows that the claim is true. Suppose that the claim is true for some $n$, then we compute that \begin{align} |H^{n+1}[x](t)|={}&\Biggl|\int_{a}^{t}K(t,\eta)H^{n}[x](\eta)\mathrm{d}\eta\Biggr| \leq{}\int_{a}^{t}|K(t,\eta)||H^{n}[x](\eta)|\mathrm{d}\eta\\ \leq{}&M\int_{a}^{t}\Biggl|M^{n-1}\int_{a}^{\eta}\frac{(\eta-\xi)^{n-1}}{(n-1)!}|x(\xi)|\mathrm{d}\xi\Biggr|\mathrm{d}\eta\\ ={}&M^{n}\int_{a}^{t}\int_{a}^{\eta}\frac{(\eta-\xi)^{n-1}}{(n-1)!}|x(\xi)|\mathrm{d}\xi\mathrm{d}\eta\quad\text{(change integration order)}\\ ={}&M^{n}\int_{a}^{t}\int_{\xi}^{t}\frac{(\eta-\xi)^{n-1}}{(n-1)!}|x(\xi)|\mathrm{d}\eta\mathrm{d}\xi\quad\text{(make substitution)}\\ ={}&M^{n}\int_{a}^{t}\frac{(t-\xi)^{n}}{n!}|x(\xi)|\mathrm{d}\xi, \end{align} which shows that the claim is also true for $(n+1)$. By induction, the claim holds for all $n\in\mathbb{N}$.

• It implies $H^n[x] \to 0$. – Daniel Fischer Nov 29 '15 at 20:30
• So, this means $\|x-\sum_{k=0}^{n-1}H^{k}[f]\|=\|H^{n}[x]\|\to0$ as $n\to\infty$ on $[a,b]$ and thus $x$ is equal to the infinite series. Can you please give me a reference about your comment @DanielFischer? Thanks. – bkarpuz Nov 29 '15 at 20:33
• Sorry, can't give a reference. Also note that the argument so far supposes the existence of an $x$ with $x = H[x] + f$. Without that assumption, take the argument as a heuristic what a solution would be. Then $\rho(H) < 1$ implies the absolute convergence of the series $\sum H^n[f]$, and continuity of $H$ shows that that is a solution. – Daniel Fischer Nov 29 '15 at 20:44
• Would you mind explaining how you showed by induction that $|H^n[x](t)|\leq M^n \int_a^t \frac{(t-\eta)^{n-1}}{(n-1)!}|x(\eta)|d \eta$? I am only able to get $|H^{n+1}[x](t)|\leq M^{n+1} \int_a^tK(t, \eta)M^n (\int_a^t \frac{(t-\eta')^{n-1}}{(n-1)!} | x(\eta')|d \eta') d\eta$ – user2139 May 29 '18 at 2:39
• @user2139 You will find your explanation at the end of the original post. – bkarpuz May 29 '18 at 14:10

You have $$0=\rho(H)=\lim_{n\to\infty}\|H^n\|^{1/n}.$$ This implies that $\|H^n\|^{1/n}\to0$. So for all $n$ big enough, $\|H^n\|^{1/n}<1/2$, say.
$$\|\sum_{k=m}^n H^k[f]\|\leq\|f\|\,\sum_{k=m}^n\|H^k\|\leq\|f\|\,\sum_{k=m}^n2^{-k}\leq\frac{\|f\|}{2^{m-1}},$$ showing that the series is Cauchy.
• As I understand, I dont need the spectral radius in my case then since I can easily show that $\|H^{n}\|_{\text{op}}\to0$ as $n\to\infty$, do I? – bkarpuz Nov 29 '15 at 21:27
• Yes, I missed that part. If you already know that $\|H^n\|$ is eventually bounded below $1$, that's all you need. The fact that the spectral radius is zero doesn't give you anything new then. – Martin Argerami Nov 30 '15 at 0:19