Suppose $u \in C^2(B_1)$ is harmonic in $B_1$, the ball of radius $1$ centered at $0$.
We can show that $\Delta (|Du|^2) \geq 0$, so $|Du|^2$ is subharmonic.
Is $|Du|^2$ continuous up to boundary, i.e. is $|Du|^2 \in C(\overline{B_1})$? How to prove this? As we only know the partial derivatives are continuous inside $B_1$.
If yes, I can use a maximum principle to estimate $Du$ in $B_1$ in terms of $Du$ on $\partial B_1$.
To be clear, the version of the max principle I am trying to use is: suppose $u$ is subharmonic and $u $ is in $C(\overline{B})$ then $\sup_B u \leq \sup_{\partial B} u$
No we won't necessarily have continuity up to the boundary. Take $\Phi(\mathbf x) = \frac{-1}{2\pi} \ln|\mathbf x|$ to be the fundamental solution of the Laplace equation in $\mathbb R^2$. Since it is the fundamental solution $-\Delta \Phi(\mathbf x) = \delta(\mathbf x)$ and it is harmonic on any open set that excludes zero. We can shift it by $\mathbf e_1 = (1,0)$ to get a counter-example to the claim. We have $\Phi(\mathbf x - \mathbf e_1)$ is harmonic on $B_1$ but $|D\Phi(\mathbf x -\mathbf e_1)|^2$ is not continuous at $(1,0)\in\partial B_1$.