$f = \int_0^s K(s,t)f(t)dt$ implies $f = 0$? Suppose $K(s,t) \in C([0,1]\times[0,1])$ and $f\in C[0,1]$, does
$$
f(s) = \int_0^s K(s,t)f(t)dt,
$$
imply $f = 0$?
This is a subproblem I encountered in showing that for any $g\in C[0,1]$, there exists a unique $f \in C[0,1]$ such that
$$
f(s) = g(s) + \int_0^s K(s,t)f(t)dt.
$$
My idea is to set $V(f) := \int_0^s Kfdt$, then it is easy to see that $V$ is a compact operator, then by the Fredholm alternative theorem, I only need to show $N(\text{id}-V) = \left\{0\right\}$, i.e., $f = \int_0^sKfdt$ implies $f = 0$, but it does not seem obvious to me.
 A: $|f(t)|\leq C\int_0^{s} |f(t)|dt$ where $C=\sup\{|K(s,t)|: 0\leq s,t \leq 1\}$. A simple induction argument shows now that $|f(s)| \leq \frac {C^{n}s^{n}} {n!}$ for every positive integer $n$. Since $\sum_n \frac {C^{n}s^{n}} {n!}$ converges it follows that $\frac {C^{n}s^{n}} {n!} \to 0$ as $n \to \infty$. Hence, $f(s)=0$ for all $s$.
[To start the induction argument we need $|f(s)| \leq \frac {C^{0}s^{0}} {0!}$, i.e. $|f(s)| \leq 1$. But in showing that $|f(s)|\leq C\int_0^{s} |f(t)|dt$  implies $f \equiv 0$ there is no loss of generality in assuming that $|f(s)| \leq 1$ since we can divide $f$ by its sup norm].
A: Let $M $ be the supremum of $|K|$ on $[0, 1] \times [0, 1]$ and consider the function
$$
 F: [0, 1] \to \Bbb R \, , \, F(s) = \int_0^s |f(t)| \, dt \, .
$$
Then
$$
 F'(s) = |f(s)| \le \int_0^s |K(s, t) f(t)|\, dt \le M F(s) \, .
$$
So $h(s) = e^{-Ms} F(s)$ is decreasing, and since $h$ is non-negative with $h(0) = 0$, $h$ is identically zero on $[0, 1]$.
It follows that $F$ is identically zero, which in turn implies that $f$ is identically zero.
