Question: Let $U, V \subset \mathbb{R}^d$ be compact, $K \in C(U \times V)$ and $T_K: C(V) \rightarrow C(U)$ given by $$ T_K(u)(x)=\int_V K(x, y) u(y) d y $$
Let $U=V=[0,1] \subset \mathbb{R}$. Show that for any $g \in C([0,1])$ and $K \in C\left([0,1]^2\right)$ there is a unique $f \in C([0,1])$ such that for all $t \in[0,1]$ $$ f(t)=\int_0^t K(t, s) f(s) d s+g(t) $$
Proof:
We are tasked to show that for a given function $g \in C([0,1])$ and a continuous kernel $K \in C\left([0,1]^2\right)$, there is a unique function $f \in C([0,1])$ satisfying the integral equation: $$ f(t)=\int_0^t K(t, s) f(s) d s+g(t), \quad \text { for all } t \in[0,1] $$
We start by defining an operator $T: C([0,1]) \rightarrow C([0,1])$ as: $$ (T f)(t)=\int_0^t K(t, s) f(s) d s $$
Now, the integral equation can be reformulated as: $$ f=T f+g $$
Or, equivalently: $$ (\mathrm{Id}-T) f=g $$
This is a perturbed identity equation. To solve it, we can use a Neumann series. The series Id $+T+T^2+T^3+\ldots$, if convergent, converges to the inverse of $\mathrm{Id}-T$.
To prove convergence, we need to ensure the series is absolutely convergent. This means showing that the operator $T$ is a contraction, which typically involves some condition like $\|T f\| \leq c\|f\|$ for a constant $\mathfrak{f}<1$ and all $f \in C([0,1])$.
Since $K$ is continuous on a compact set, it's bounded. Let's denote the bound as $$ M=\sup \left\{|K(t, s)|:(t, s) \in[0,1]^2\right\}$$ \end{equation}. For any $f \in C([0,1])$, we can estimate: $$ |(T f)(t)| \leq \int_0^t|K(t, s) f(s)| d s \leq M\|f\| t $$
So, $\|T f\| \leq M\|f\|$. If $M<1, T$ is indeed a contraction. Assuming $M<1$, the Neumann series Id $+T+T^2+T^3+\ldots$ converges in $C([0,1])$. The limit of this series is $(\mathrm{Id}-T)^{-1}$, the inverse of $\operatorname{Id}-T$, which exists and is bounded.
The unique solution to our integral equation is then: $$ f=(\operatorname{Id}-T)^{-1} g $$
-- Is what I have done? It will be very kind of you if you can please proofread my answer! :)