Use functional analysis to show that integral equation has unique solution 
Let $b>a$ in $\mathbb R$ and $K:[a,b]\times[a,b] \rightarrow \mathbb C$ be a continuous function. Show that for all $\lambda \ne 0$ in $\mathbb{C}$ and $f \in C[a,b]$ the integral equation
$$ \int_a^t K(t,s) u(s) ds  - \lambda u = f$$
has one unique solution. Hint: Use the previous exercise.

I managed to do the previous exercise, showing that if we have a linear operator $A$ such that $\sum_{j=0}^\infty \|A^j\| < \infty$ then $I - A$ is invertible (and in particular $\|A\| < 1$ also implies this). I guess this is related, since showing that the left side of the equation is invertible would imply it is injective and surjective, hence has only one solution, but I haven't figured out how to do so.
 A: Consider the following integral:
$$\int_a^{t_n} dt_{n-1}\int_{a}^{t_{n-1}}dt_{n-2}....\int_{a}^{t_1}dt_0 = \int_a^{t_n} dt_{n-1}\int_{a}^{t_{n-1}}dt_{n-2}....\int_{a}^{t_2}dt_1(t_1-a) \\
=\int_a^{t_n} dt_{n-1}\int_{a}^{t_{n-1}}dt_{n-2}....\int_{a}^{t_3}dt_2  \frac12 (t_2-a)^2 = ...= \frac{1}{1\cdot 2\cdot...\cdot n}(t_n-a)^n$$
If you take the supremum over all $t_n\in[a,b]$ you get $\frac{(b-a)^n}{n!}$.
Now if you let $A = u\mapsto [t\mapsto \int_a^t K(t,s) u(s)\,ds]$ you have that:
$$A^n(u) = \int_a^{t_n}dt_{n-1}K(t_n,t_{n-1})\int_a^{t_{n-1}}dt_{n-2}K(t_{n-1},t_{n-2}) dt_{n-2}...\int_a^{t_1}dt_0 K(t_1, t_0) u(t_0) $$
Take the norm of this and pull the absolute values into the integrals (as such getting a $≤$). Now replace every $|K(t_{i},t_{i-1})|$ with a $C:=\sup_{t_i,t_{i-1}} |K(t_i, t_{i-1})|$ and replacce $|u(t_0)|$ with $\|u\|$. In the end you get:
$$\|A^n(u)\|≤ \sup_{t_n}\int_a^{t_n} dt_{n-1}\int_{a}^{t_{n-1}}dt_{n-2}....\int_{a}^{t_1}dt_0  \ C^n\|u\| ≤\frac{C^n (b-a)^n}{n!}\|u\|$$
This implies that $\|A^n\|≤ \frac{C^n (b-a)^n}{n!}$, hence $A^j$ is absolutely summable.
