Let $f \in L^\infty((0,1)\times(0,1))$ and consider the integral equation \begin{align*} u(x, y)= \int_0^y \int_0^{x-y} u(x-y+\tau, s+\tau) f(s+\tau, \tau) d s d \tau -\int_0^y f(x-y+s, s) d s. \end{align*} Is it true that there exists a unique solution in $u \in W^{1,\infty}((0,1)\times(0,1))$? More precisely note that here we want to solve the equation in the domain $1 > x > y >0$.

I'm tempted to say that Banach's fixed point theorem should give it, but don't see how.

Let $f \in L^\infty((0,1)\times(0,1))$ and consider the integral equation \begin{align*} u(x, y)= \int_0^y \int_0^{x-y} u(x-y+\tau, s+\tau) f(s+\tau, \tau) d s d \tau -\int_0^y f(x-y+s, s) d s. \end{align*} Is it true that there exists a unique solution in $u \in W^{1,\infty}(D)$? Here $D= \{(x,y) \in (0,1)\times(0,1)| 1 > x > y >0\}$.

I'm tempted to say that Banach's fixed point theorem should give it, but don't see how.

Let $f \in L^\infty((0,1)\times(0,1))$ and consider the integral equation \begin{align*} u(x, y)= \int_0^y \int_0^{x-y} u(x-y+\tau, s+\tau) f(s+\tau, \tau) d s d \tau -\int_0^y f(x-y+s, s) d s. \end{align*} Is it true that there exists a unique solution in $u \in W^{1,\infty}((0,1)\times(0,1))$? I'm tempted to say that Banach's fixed point theorem should give it, but don't see how.

Let $f \in L^\infty((0,1)\times(0,1))$ and consider the integral equation \begin{align*} u(x, y)= \int_0^y \int_0^{x-y} u(x-y+\tau, s+\tau) f(s+\tau, \tau) d s d \tau -\int_0^y f(x-y+s, s) d s. \end{align*} Is it true that there exists a unique solution in $u \in W^{1,\infty}((0,1)\times(0,1))$? More precisely note that here we want to solve the equation in the domain $1 > x > y >0$.

I'm tempted to say that Banach's fixed point theorem should give it, but don't see how.