Show using a fixed point argument that there exists a unique solution $f\in C[0,1]$ to
$$ -f''(x)+\sin(f(x))=\sin(x) , x\in (0,1), y(0)=y'(1)=0 $$
This is what I have so far:
We can show separately that the Green's function for $-f''$ is $g(x,y)=\min(x,y)$. So let's use
$$ -\min(x,y)f''(x)+\min(x,y)\sin(f(x))=\min(x,y)\sin(x)$$
Integrating with respect to $x$, this becomes
$$ f(y)+ \int^1_0 \min(x,y)\sin(f(x))\,dx=\int^1_0 \min(x,y)\sin(x)\,dx $$
So if we consider an operator $Tf=\int^1_0 \min(x,y)\sin(x)\,dx-\int^1_0 \min(x,y),sin(f(x))\,dx$, we determine if it a contraction map
$$Tf(y)-Tg(y)= -\int^1_0 \min(x,y)(\sin(f(x))-sin(g(x)))\,dx$$
Using a trig identity,
$$Tf(y)-Tg(y)= -\int^1_0 2\min(x,y)\sin((f-g)/2)\cos((f+g)/2)\,dx $$
or
$$Tf-Tg= -2\int^y_0 x\sin((f-g)/2)\cos((f+g)/2)\,dx-2y\int^1_y \sin((f-g)/2)\cos((f+g)/2)\,dx $$
How do I proceed to get $\|Tf-Tg\| < C \|f-g\|$ with $C<1$? Hints?