Finding $W^{1,\infty}$ solutions to an integral equation by fixed point 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.
 A: (Edited for clarity and presentation, and thanks to a comment by Giorgio Metafune)
Suppose that $f\in W^{1,p}(D)\cap L^\infty(D)$ for some $p>1$, and furthermore that
$$
\int_0^y \int_0^{x-y}|f(s+\tau,\tau)|\text{d}s\text{d}\tau<1.
$$
This last inequality is satisfied in particular when  $\|f\|_{L^\infty(D)}<4$.
Let us show existence and uniqueness among continuous functions on $\overline{D}$ equipped with the max norm.
$$
F:u\to \int_0^y \int_0^{x-y} u(x-y+\tau, s+\tau) f(s+\tau, \tau) d s  d \tau.
$$
Since $\overline D$ is compact, continuity on $\overline D$ implies uniform continuity, and $F:C(\overline{D})\to C(\overline{D})$.
Furthermore, $F:W^{1,\infty}(D)\to W^{1,\infty}(D)$.
Let us turn to $g$. For everything to be well defined, we extend
$f$ by zero outside $D$. We have, writing $\mathbf{x}=\left(x,y\right)$
and $\mathbf{x}_{1}=\left(x_{1},y_{1}\right)$,
\begin{align*}
g\left(\mathbf{x}\right)-g\left(\mathbf{x}_{1}\right)= & \int_{0}^{y}f\left(x-y+s,s\right)\text{d}s-\int_{0}^{y_{1}}f\left(x_{1}-y_{1}+s,s\right)\text{d}s\\
= & \int_{0}^{y}\left(f\left(x-y+s,s\right)-f\left(x_{1}-y_{1}+s,s\right)\right)\text{ d}s\\
+ & \int_{y}^{y_{1}}f\left(x_{1}-y_{1}+s,s\right)\text{d}s
\end{align*}
and this last term satisfies
$$
\left|\int_{y}^{y_{1}}f\left(x_{1}-y_{1}+s,s\right)\text{d}s\right|\leq\left|\mathbf{x}-\mathbf{x}_{1}\right|\left\Vert f\right\Vert _{\infty}.
$$
Writing $\tau=\left(\mathbf{x}-\mathbf{x}_{1}\right)\cdot\left(-1,1\right)=x_{1}-x-\left(y_{1}-y\right),$we
have
$$
\left|g\left(\mathbf{x}\right)-g\left(\mathbf{x}_{1}\right)\right|\leq\int_{0}^{y}\left|f\left(x-y+s,s\right)-f\left(x-y+\tau+s,s\right)\right|\text{ d}s+\left|\mathbf{x}-\mathbf{x}_{1}\right|\left\Vert f\right\Vert _{\infty}.
$$
If $f\in W^{1,p}$ with $p>1$, then thanks to Luzin-Lipschitz inequality
we have almost everywhere
$$
\left|f\left(x-y+s,s\right)-f\left(x-y+\tau+s,s\right)\right|\leq|\tau|\left(|M(\nabla f)\left(x-y+s,s\right)|+|M(\nabla f)\left(x-y+\tau+s,s\right)|\right)
$$
therefore in turn
$$
\left|g\left(\mathbf{x}\right)-g\left(\mathbf{x}_{1}\right)\right|\leq\left(4\left\Vert f\right\Vert _{W^{1,p}}+\left\Vert f\right\Vert _{\infty}\right)\left|\mathbf{x}-\mathbf{x}_{1}\right|
$$
and $g\in W^{1,\infty}\left(D\right)$.
We compute
$$
\sup_{u\neq0}\frac{\|Fu\|_\infty}{\|u\|_\infty} \leq \int_0^y \int_0^{x-y} \left|f(s+\tau, \tau)  \right|d s  d \tau  < 1 $$
thus, for example, the sequence
$$
u_{n+1} = F(u_n) + g
$$
with $g=-\int_0^yf(x-y+\tau,\tau)d\tau$ converges to a solution of $u=Fu+g$, since $F$ is a contraction. Explicitly, the solution is
$$
u=\sum_{k=0}^\infty F^k g
$$
where $F^k$ means $F$ composed with $F$ ,$k$ times.
Regarding $\|f\|_\infty<4$ : since $y(x-y) \leq\frac{x^2}{4}<\frac14$,  $$\sup_{u\neq0}\frac{\|Fu\|_\infty}{\|u\|_\infty} \leq\frac14  \|f\|_\infty<1,$$ and that's that.

When $f$ is merely bounded, $g$ may not be continuous. We have,
\begin{align*}
g\left(\mathbf{x}\right)-g\left(\mathbf{x}_{1}\right) & =\int_{0}^{y}\left(f\left(x-y+s,s\right)-f\left(x-y+\tau+s,s\right)\right)\text{ d}s+O\left(\left|\mathbf{x}-\mathbf{x}_{1}\right|\right)\\
 & =\int_{0}^{y}\left(f\left(x-y+s,s\right)ds-\int_{\tau}^{\tau+y}f\left(x-y+s,s-\tau\right)\right)\text{ d}s+O\left(\left|\mathbf{x}-\mathbf{x}_{1}\right|\right)\\
 & =\int_{0}^{y}\left(f\left(x-y+s,s\right)-f\left(x-y+s,s-\tau\right)\right)\text{ d}s+O\left(\left|\mathbf{x}-\mathbf{x}_{1}\right|\right).
\end{align*}
Set $f\left(\frac{1}{4}+s,s\right)=\text{1}$ and $f\left(\frac{1}{4}+s,s^{\prime}\right)=0$
for any $0\leq s^{\prime}<s<\frac{1}{4},$ and choose $\mathbf{x}=\left(\frac{1}{2}, {\frac{1}{4}}\right),$and
$\mathbf{x}_{1}=\left(\frac{1}{2}+\epsilon, {\frac{1}{4}-\epsilon}\right),$
so that $\tau=2\epsilon.$ Then
\begin{align*}
g\left(\mathbf{x}\right)-g\left(\mathbf{x}_{1}\right) & =\int_{0}^{\frac{1}{4}}\left(f\left(\frac{1}{4}+s,s\right)-f\left(\frac{1}{4}+s,s-2\epsilon\right)\right)\text{ d}s + O\left(\left|\mathbf{x}-\mathbf{x}_{1}\right|\right)\\
 & =\frac{1}{4}+O\left(\left|\mathbf{x}-\mathbf{x}_{1}\right|\right),
\end{align*}
so $g$ isn't continous in $\mathbf x$.
