Showing existence and uniqueness for a solution to a homogeneous Fredholm type integral equation of the second kind

Question

I'm studying for an exam in real analysis. Thus, only such techniques should be considered. I'm looking at old exams, and repeatedly see questions similar to the one below.

Show that there exists a unique non-zero solution $f(x)$ to the integral equation  $$f(x)=\frac{1}{2}\int_0^1\sin(xy)f(y)dy.$$ Here, $C([0,1])$ denotes the space of continuous functions $\phi$ on the interval $[0,1]$ with the norm $\|\phi\|:=\sup_{x\in[0,1]}|\phi(x)|$.

I assume that the solution lays in the properties of the Riemann-Stieltjes integral, but I'm unsure about what techniques to use. The course literature is Walter Rudin's book on real analysis, which doesn't mention integral equations at all.

The other problems often only give bounds for the kernel such as $|K(x,y)|\leq\frac{1}{2}$.

Answer 1

$$\left|\frac12\int_0^1\sin(xy)f(y)\mathrm{d}y\right|\leq\frac12\int_0^1\left| \sin(xy)f(y)\right|\mathrm{d}y\leq\int_0^1 |f(y)|\mathrm{d}y\leq\frac12||f||_\infty,$$

hence the map $f(x)\mapsto\frac12\int\dots$ is a contraction of $\mathcal C([0,1])$, so it has a unique fixed point by the Banach fixed point theorem

Answer 2

1. $f=0$ is a solution.

2. If $f(x)=\frac{1}{2}\int_0^1\sin(xy)f(y)dy$ for $x \in [0,1]$, then

   $|f(x)| \le \frac{1}{2}\int_0^1|\sin(xy)| \cdot |f(y)| dy \le \frac{1}{2}\int_0^1||f|| dy= \frac{1}{2} ||f|| $

for all $x \in [0,1]$.

This gives $||f|| \le \frac{1}{2} ||f|| $ , hence $f=0.$