I am trying to solve exercise 6, p.231 from Lax's Functional Analysis book. Supposing that $K(s,t)$ is a continuous function of $s,t$ in $t\leq s$, I want to show that $\textbf{K}$ as an operator from C[0,1] has a spectrum consisting just of $\{0\}$, where
$(\textbf{K}f)(s) = \int_0^s K(s,t)f(t)dt$.
Now, I've seen that there are solutions to this problem on StackExchange, but all of them I've seen use that $\textbf{K}$ is a compact operator. In Lax's book, however, compact maps are defined in the next chapter after this exercise. That's why I'm thinking about a proof without using the fact that $\textbf{K}$ is compact. Before the exercise, the author proves the result in case $K(s,t)=1$. To do that, he finds the n-fold iterate of $\textbf{K}$, $\textbf{K}^n$, and bounds its norm to show that $\lim||\textbf{K}^n||^{1/n}=0$, thus spectral radius is zero, thus spectrum consists just of $\{0\}$.
Now, I don't think generalizing this proof to the case with general $K(s,t)$ is possible (because the formula of $\textbf{K}^n$ is already very complicated in this trivial case when $K(s,t)=1$). I'm thinking, however, if it won't be possible to use somehow this result in the general case. For example, showing that $\lambda I - \textbf{K}$ invertible for general K implies that $\lambda I - \textbf{V}$ is invertible, where $\textbf{V} = \textbf{K}$ with $K(s,t)=1$, would solve the problem. Unfortunately, I don't see how could I show that either.
Is there any way to solve this exercise using the fact it holds in the case $K(s,t)=1$?
Yes, here's an outline for a proof using the specific result.
Let $C = \max_{x,y \in [0,1]} |K(x,y)|$. Let $\mathbf K_1$ denote the map associated with $K \equiv 1$. First, show that for any $f \geq 0$, $$ \left\|\mathbf K f\right\| \leq C\|\mathbf K_1 f\|. $$ From there, you should be able to conclude that $\|\mathbf K^n f\|^{1/n} \to 0$ as $n \to \infty$ for any such $f$.
More generally, let $f_+$ denote the function $f_+(x) = \max(f(x),0)$ and $f_- = f_+ - f_-$; note that $f = f_+ - f_-$. We have $$ \|\mathbf K^n f\| = \|\mathbf K^n f_+ - \mathbf K^n f_-\| \leq \|\mathbf K^n f_+\| + \|\mathbf K^n f_-\|. $$ Use this to reach a similar conclusion.