I am curious about the Hölder exponent obtained by the De Giorgi-Nash-Moser theory, as a function of the ellipticity. More precisely: suppose $u$ satisfies weakly $$ D_i(a^{ij}D_ju)=f $$ on the $d$-dimensional ball of radius $R$, with $0$ Dirichlet boundary conditions, with the matrix $(a^{ij})_{i,j=1..d}$ bounded from below and above by $\lambda I$ and $\Lambda I$, respectively, $\lambda>0$. For the right-hand side, let's just take $f\in L_\infty$. Then we know that $u\in C^{\alpha}(B_R)$, with some $\alpha=\alpha(d, \Lambda/\lambda)$, my question is whether something about the dependence on $\Lambda/\lambda$ is known. The analogous question for interior regularity (which might be easier) could also be of interest. Thanks!

## 1 Answer

(By scaling we can take $\lambda=1$ for simplicity. I will also take $f=0$, and discuss just the interior estimates.)

The precise exponent is known in two dimensions (see Piccinini & Spagnolo 1972). It is $\Lambda^{-1/2}$. You can find the "worst" coefficient field in the obvious way: the diffusion matrix $a(x)$ has eigenvalue $\Lambda$ in the $x$ direction, and eigenvalue $\lambda=1$ in directions orthogonal to $x$. You can prove this is the worst as Piccinini and Spagnolo do, with a beautiful monotonicity formula type computation.

In higher dimensions, the precise exponent is not known. It is however conjectured based on the same radial example, extended in the obvious way in $d>2$. The exact exponent should be (according to my personal notes, hopefully I have not made a mistake) \begin{equation*} \alpha(d,\Lambda) = \frac12 \left( -(d-2) + \sqrt{(d-2)^2 + \frac{4(d-1)}{\Lambda} } \right) . \end{equation*} Okay, forget about the exact formula, the point is that this scales like $c(d) \Lambda^{-1}$ for $\Lambda \gg1$ in all dimensions $d>2$.

Ok, what is the best exponent that can be proved? The best exponent which has been proved is this: \begin{equation*} \alpha(d,\Lambda) = \exp \left( - C \Lambda^{1/2} \right). \end{equation*} This is the best exponent one could ever hope to have based on current methods! Indeed, all known methods of proof use a decay of oscillation iteration to get the $C^{\alpha}$ estimate (except in $d=2$, more on that below). The best decay of oscillation estimate is related to the best constant in the Harnack inequality, and this is known to be $\exp( C \Lambda^{1/2})$. There are very easy examples you can make to confirm you can do no better. The Harnack inequality with this constant was proved in an amazing paper by Giusti and Bombieri (Inventiones, also 1972).

So, to summarize: in $d>2$, we cannot prove the estimate with the conjectured optimal exponent because we do not possess a proof powerful enough to do so. All known proofs cannot get there. The bound we have is hopeless pathetic next to what we think is true, so the situation is bad, very bad! Anything better than the Giusti-Bombieri exponent would be, in my opinion, a major breakthrough.

But then what is going on in $d=2$? There is a proof of the conjectured bound, so do we have a better proof? Why yes-- indeed we do. In $d=2$, the De Giorgi-Nash $C^{\alpha}$ estimate has an easy proof. It follows from the hole-filling argument, which is just the Caccioppoli inequality (i.e., the usual $L^2$ energy estimate), plus the Sobolev inequality. This is basically because the Sobolev inequality says that in $d=2$ the space $H^1$ almost embeds into $L^\infty$. Since you can actually do a tiny bit better than $H^1$ by hole filling, you can get $C^\alpha$ without any ideas of De Giorgi. The simple hole filling computation already gives an exponent that scales like $\Lambda^{-1/2}$. What the Piccinini & Spagnolo paper does is optimize this computation, replacing an iteration in dyadic balls by a more careful monotonicity computation.

So $d=2$ should be expected to be much easier, and in hindsight it is maybe not so surprising that we can get the optimal exponent in this case.

nondivergenceform. You cannot go from one to the other. So the only proofs are by De Giorgi or variants like in the proofs of Nash or Moser. (Sorry to be a nitpicker, but this is a very common misconception and so I wanted to point it out.) $\endgroup$