 Solution

Use $\Vert w{\Vert }_{W^{1,\infty }}=\Vert w{\Vert }_{\infty }+\Vert w^{\prime }{\Vert }_{\infty }$. Fix $u$. For $w$ small, all triples $(x,u+tw,u^{\prime }+t{w}^{\prime })$, $0\le t\le 1$, lie in a fixed compact set up to null sets. The continuous derivatives $f_u,f_v$ are uniformly continuous there, with a modulus $\omega (s)\to 0$ as $s\to 0$. The integral form of the mean-value expansion gives a remainder bounded byThe proposed linear derivative is bounded because its coefficient functions are bounded on that compact set. The remainder is little-oh of the norm, proving Fréchet differentiability andThe derivative $u^{\prime }$ only needs to exist almost everywhere; the uniform compact-range bound handles its measurable representative.