Proving existence of unique fixed point in C([a,b]) satisfying integral equation

Question

The question is:

Let $[a,b]$ be a closed interval in $\mathbb{R}$ and let $A$ and $K$ be continuous real-valued functions on $[a,b]$ and $\{(x,y) \in \mathbb{R}^2: a \leq y \leq x \leq b\}$ respectively. Prove that there is a unique $\phi \in C([a,b])$ such that $$\phi(x) = A(x) + \int_{a}^{x} K(x,y) \phi(y)dy$$ for all $x \in [a,b]$.

I did a problem prior to this of a similar nature, which was:

Let $[a,b]$ be a closed interval in $\mathbb{R}$ and let $A$ and $K$ be continuous real-valued functions on $[a,b]$ and $\{(x,y) \in \mathbb{R}^2: x,y \in [a,b]\}$ respectively. Assume $|(b-a)K(x,y)|<1$ for all $x,y$. Prove that there is a unique $\phi \in C([a,b])$ such that $$\phi(x) = A(x) + \int_{a}^{b} K(x,y) \phi(y)dy$$ for all $x \in [a,b]$. This can be done by showing $F: C([a,b]) \rightarrow C([a,b])$ defined by $F(\psi)(x) = A(x) + \int_{a}^{b} K(x,y) \psi(y)dy$ is a contraction map.

However, without $|(b-a)K(x,y)|<1$, I'm unsure of how to proceed. The problem has the following hint: Imitate the procedure in the preceding problem if $|(b-a)K(x,y)|<1$ whenever $a \le y \le x \le b$. To do the general case, note that for any $a_1 \in (a,b)$, the problem reduces to proving the existence of a unique $\phi_1 \in C([a,a_1])$ such that $$\phi_1(x) = A(x) + \int_{a}^{x} K(x,y) \phi_1(y)dy$$ for all $x \in [a,a_1]$ and the existence of a unique $\phi_2 \in C([a_1,b])$ such that $$\phi_2(x) = A(x) + \int_{a}^{a_1} K(x,y) \phi_1(y)dy + \int_{a_1}^{x} K(x,y) \phi_2(y)dy$$ for all $x \in [a_1,b]$.

I'm not sure what to make of the hint.

Thank you.

Answer 1

The idea of the hint is to construct a solution by piecing together solutions in intervals small enough to be able to apply the existence result.

First step. Since $K$ is bounded, there exists $a_1>a$ such that $(a_1-a)K<1$. This gives the existence of $\phi_1\colon[a,a_1]\to\Bbb R$ such that $$ \phi_1(x) = A(x) + \int_{a}^{x} K(x,y) \phi_1(y)dy,\quad a\le x\le a_1. $$

Second step. There is $a_2>a_1$ such that $(a_2-a_1)K<1$. This gives the existence of $\phi_2\colon[a_1,a_2]\to\Bbb R$ such that $$ \phi_2(x) = \underbrace{A(x) + \int_{a}^{a_1} K(x,y) \phi_1(y)dy}_{\text{Plays the role of $A$}}+ \int_{a_1}^{x} K(x,y) \phi_2(y)dy,\quad a_1\le x\le a_2. $$ Further steps. Repeat until you reach $b$.

Last step. Piece together $\phi_1$, $\phi_2$,... You will have to check that the function you obtain is continuous. But this is easy. For instance, continuity at $a_1$ follows because $$ \phi_1(a_1)=\phi_2(a_1). $$

Answer 2

Use the modified norm $$ \|ϕ\|_L=\max_{x\in[a,b]}\left|e^{-Lx}ϕ(x)\right| $$ which is equivalent to the supremums norm over $C([a,b])$. Then \begin{align} |F(ψ)(x)-F(ϕ)(x)| &=\left|\int_a^xK(x,y)(ψ(y)-ϕ(y))dy\right| \\[.5em] &\le \int_a^xM_K\|ψ-ϕ\|_L\,e^{Ly}\,dy \\[.5em] &=\frac{M_K(e^{Lx}-e^{La})}{L}\|ψ-ϕ\|_L\le e^{Lx}\frac{M_K }{L}\|ψ-ϕ\|_L \\[1em] \implies \|F(ψ)-F(ϕ)\|_L&\le\frac{M_K}{L}\|ψ-ϕ\|_L \end{align} Choosing $L=2M_K$ where $M_K=\max_{x,y\in[a,b]}|K(x,y)|$ results in contractivity of $F$ over $C([a,b])$ without requiring subdivision. Now apply the Banach fixed-point theorem once.