The maximal estimates for Laplace equations Let $B$ be the unit ball in $\mathbb{R}^2$, I am trying to find out if there is some constant $M$, such that for any $u\in C^2(\overline{B},\mathbb{R})$ with $u_{|\partial B}=0$, we have $\|u\|_{C^2}\leq M\|\Delta u\|_{C^0}$.
I heard it is not true, but I do not know why.
 A: This is false if the dimension $n\geq 2$. However, in the one-dimensional case it is easy to prove that it holds (by ODE arguments). For the sake of simplicity let's consider the case $n=2$ and $B=B_{1/2}$ the ball cetered at zero with radious $\tfrac{1}{2}$. Your question is equivalent to the following one: Let's consider $u:B\to\mathbb{R}$ a function solving $$
\left\{\begin{matrix}\Delta u = f \ \hbox{ in } \, B, \\ u=0 \ \hbox{ in } \partial B,\end{matrix}\right.
$$
where $f$ is some other function. Is it true that if $f\in\mathcal{C}^0$, then $u\in\mathcal{C}^2$? Note that if it were true, that would imply your inequality $$
\Vert u\Vert_{\mathcal{C}^2}\leq C\Vert f\Vert_{\mathcal{C}^0}=C\Vert\Delta u\Vert_{\mathcal{C}^0}.
$$
Nevertheless, here is a counterexample: Consider the functions $$
v(x,y):=\left\{\begin{matrix}(x^2-y^2)\sqrt{-\log(r)} & \hbox{ if } \, (x,y)\neq 0,
\\ 0 & \hbox{ if } (x,y)=0,
\end{matrix} \right.
$$
$$
f(x,y):=\left\{\begin{matrix}\tfrac{(y^2-x^2)}{2r^2}\left(\tfrac{4}{\sqrt{-\log(r)}}+\tfrac{1}{2(-\log(r))^{3/2}}\right) & \hbox{ if } \, (x,y)\neq 0,
\\ 0 & \hbox{ if } (x,y)=0,
\end{matrix} \right.
$$
where $r=\sqrt{x^2+y^2}$. It is easy (but cumbersome) to prove that $$f\in\mathcal{C}^0, \qquad  
 v\in\mathcal{C}^0(B)\cap\mathcal{C}^2(B\setminus\{0\}),$$ and that $\Delta v=f$ for every $x\in B\setminus\{0\}$. Thus, if $u\in\mathcal{C}^2$ is a solution to $\Delta u=f$, then $v-u$ is an harmonic function in $B\setminus\{0\}$ and continuous over the whole ball $B$. Hence, by the artificial singularity's principle (I don't know if it's the correct name in english) we conclude that $v-u$ is harmonic over the whole ball $B$. Thus, $v-u\in\mathcal{C}^\infty(B)$ and hence $v=(v-u)+u\in\mathcal{C}^2$ over $B$, which is a contradiction.
An interesting fact is that, if you assume a little bit more on the right-hand side, for instance $f\in\mathcal{C}^{0,\alpha}$, it is possible to prove that $u\in\mathcal{C}^{2,\alpha}$. This is known as Schauder's theorem.
