**Idea**


Given a $W^{1,2}$ solution to a linear divergence form uniformly elliptic pde with bounded coefficients, standard De Giorgi-Nash-Moser theory tells us that the solution is infact (Holder) continuous. If you have better regularity away from one isolated point, say you are $C^1$ on the puncutered ball, can the solution still fail to be differentiable at that isolated point?


The situations I am most familiar with tend to be ones in which one can go from boundedness and continuity to smoothness via Schauder theory and bootstrapping. Here, we already have continuity: But can differentiability fail at a single point? TIt seems out of reach of all the theorems I've seen, which makes me suspect it is false, but I cannot be sure without a counterexample. Does anyone have any ideas?


**Details**


In the specific situation I am interested in, I know a bit more. I am considering the following:


$u \in W^{1,\infty}(B\_1(0)) \cap C^{1,1}\_{loc}(B\_1(0)\setminus ${0}$)$ satisfies weakly the equation


$D\_i(A\_{ij}(x)D\_ju) = D\_ig^i$ 


in $B\_1(0)$, where $A\_{ij},g^i \in L^{\infty}(B\_1(0))\cap W^{1,2}\_{loc}(B\_1(0)\setminus ${0}$)$. 


**Questions**


1) Must $u$ in fact be a $C^1$ solution on $B\_1(0)$? 


2) What about just being differentiable at 0? 


3) How about even just $u \in W^{2,p}(B\_1(0))$ for some $p > 1$?



Divergence form Elliptic PDE Removable Singularity/Regularity Question
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, $H \in L^\infty \cap W^{1,2}\_{loc}((-1,1)\setminus \{0\})$.


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.


