Application of the Banach fixed-point theorem I am looking for a function $f:[0,1]\rightarrow\mathbb R$ which satisfies $$\int_{0}^{1}\frac{\sin(f(t)-y)}{2}\,\mathrm dy=f(t)$$ for $t\in[0,1]$.
The first thing I do is to define a function $A:M\rightarrow M$ where $M=C([0,1])$ with $$A(f)(t)=\int_{0}^{1}\frac{k(f(t),y)}{2}\,\mathrm dy.$$
Set $$k(x,y)=\sin(x-y).$$ Then $k$ is Lipschitz-continuous with respect to the first variable with Lipschitz constant $1$. 
Therefore I am looking for a unique function $f$ which satisfies $A(f)=f$. How can it be calculated explicitly?
 A: You have
$$
|A(f)(t)-A(g)(t)|\leq \frac{1}{2}\int_0^1|k(f(t),y)-k(g(t),y)|dy
$$
$$
= \frac{1}{2}\int_0^1|\sin(f(t)-g(t))|dy\leq \frac{1}{2}|f(t)-g(t)|
$$
for all $t\in [0,1]$.
So
$$
\|A(f)-A(g)\|_\infty\leq \frac{1}{2}\|f-g\|_\infty
$$
which shows that $A$ is a contraction on the complete normed vector space $C([0,1])$ equipped with the sup norm.
By Banach fixed point theorem, there exists a unique fixed point $f$. Uniqueness is easy if you plug two fixed points in the estimate. Existence goes by choosing any $f_0$ and then considering the recursive sequence $f_{n+1}=A(f_n)$. It is a Cauchy sequence, so it converges. And it must converge to a fixed point. This is your $f$.
If you start with $f_0(t)=0$, you get a sequence of constant functions. Therefore the limit $f$ is constant equal to $C$. So it satisfies
$$
C=\frac{1}{2}\int_0^1\sin(C-y)dy=\frac{1}{2}(\cos(C-1)-\cos C)).
$$
I don't think we can find a closed form. But we can cheat and use Wolfram Alpha to find
$$
C\simeq -0.364838.
$$
