2 added 17 characters in body

I believe the following is a counterexample:

Let $N=1$, $B_1(0)=(-1,1)$, $u(x)=|x|$, then $|u^\prime(x)| = 1$, $u^{\prime \prime}(x) = 2\delta_0$,

and

$u^\prime(x) = 2H(x)-1$,

where $H$ is the Heaviside function, $H \in L^\infty \cap W^{1,2}_{loc}(-1,1)$.W^{1,2}_{loc}((-1,1)\setminus {0})$. Thus$u$solves$(\frac{1}{2} u^\prime)^\prime = H^\prime$, which is the PDE in one dimension, with$H$in the right space,$u$is smooth away from the origin and$u \in W^{1,\infty}(-1,1)$, but$u^{\prime\prime}$is only a measure and so not in$L^p(-1,1)$for any$p>1$. In particular,$u$is not$C^1$,$u$is not smooth at the origin, and$u^{\prime\prime} \notin L^p(-1,1)$. More generally, the computation extends into more dimensions, as far as I can tell. 1 I believe the following is a counterexample: Let$N=1$,$B_1(0)=(-1,1)$,$u(x)=|x|$, then$|u^\prime(x)| = 1$,$u^{\prime \prime}(x) = 2\delta_0$, and$u^\prime(x) = 2H(x)-1$, where$H$is the Heaviside function,$H \in L^\infty \cap W^{1,2}_{loc}(-1,1)$. Thus$u$solves$(\frac{1}{2} u^\prime)^\prime = H^\prime$, which is the PDE in one dimension, with$H$in the right space,$u$is smooth away from the origin and$u \in W^{1,\infty}(-1,1)$, but$u^{\prime\prime}$is only a measure and so not in$L^p(-1,1)$for any$p>1$. In particular,$u$is not$C^1$,$u$is not smooth at the origin, and$u^{\prime\prime} \notin L^p(-1,1)\$.

More generally, the computation extends into more dimensions, as far as I can tell.