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I am curious about the following question:

suppose $u$ is a solution to the uniformly elliptic equation $\sum_{i,j=1}^{n}a_{ij}(x)u_{ij}=f(x)$ in $\Omega$ and $u=0$ on$\partial \Omega$, where $\Omega$ is a bounded convex domain and for simplicity it is close to a unit ball in hausdorff distance, $a_{i,j}$ and $f(x)$ are smooth. $a_{ij}$ has largest eigenvalue $\alpha(x)=1$, and smallest eigenvalue $\beta(x)$.

is it possible to prove a $C^{2}$ estimate: $|D^{2}u|\leq C$ in the compact subdomain $\Omega'$ of $\Omega$, where $C$ depends on $|f|_{L^\infty}$ and the distance between $\partial \Omega$ and $\partial \Omega'$, but doesnt depend on the lower bound of $|\beta(x)|$?

The condition I forgot to put: Suppose $u$ is convex and smooth...

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  • $\begingroup$ en.wikipedia.org/wiki/Schauder_estimates $\endgroup$
    – Otis Chodosh
    Apr 6, 2012 at 2:51
  • $\begingroup$ Schauder estimates requires $C$ depends on $C^{\alpha}$ modular of the coefficients and the lower bound of $|\beta(x)|$, which is not enough for my question. the key point is that I need some estimate which is independent of the ratio between $\alpha$ and $\beta$ $\endgroup$
    – user13289
    Apr 6, 2012 at 3:36
  • $\begingroup$ and of couse there is a simple estimate that $|D^{2}u|$ bounded by $|\frac{f}{\beta}|$ $\endgroup$
    – user13289
    Apr 6, 2012 at 3:39
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    $\begingroup$ Could you explain your last comment? I don't know of any way to prove a $C^2$ bound using only pointwise bounds on $f$ and $\beta$. I recommend that you consult Gilbarg and Trudinger. Or follow Otis's advice. $\endgroup$
    – Deane Yang
    Apr 6, 2012 at 7:22
  • $\begingroup$ @Deane, my bad, I forgot to write the condition that $u$ is convex... $\endgroup$
    – user13289
    Apr 6, 2012 at 18:49

2 Answers 2

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In [Gilbarg-Trudinger], exercise 4.9 pp. 71-72 constructs (i) an example of continuous function $f$ such that the equation $\Delta u = f$ does not have a $C^2$ solution in any neighbourhood of the origin, and (ii) an example of $u$ such that $\Delta u \in C^{1}$ but $u$ is not in $C^{2,1}$ in any neighbourhood of the origin.

Looking at $\partial_1 u$ in example (ii) would perhaps give a negative answer to this question.

I am also interested in that kind of estimate for the case where the operator is in divergence form $\sum_{i,j=1}^n \partial_i (a_{ij}(x) \partial_j u) = 0$ with no restriction on smoothness of the boundary $\partial\Omega$. Assuming the $\{a_{ij}\}$'s to be Lipschitz, a recent post in arXiv (http://arxiv.org/pdf/1207.4236.pdf) claims that $u\in C^{1,1}$, referring to [G-T] with no further comments. This is exactly what I need yet I cannot find it in [G-T] and it seems to me that this is again a limiting case which may have counter-examples... Any help appreciated !

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  • $\begingroup$ the example above might not be convex, though $\endgroup$
    – Francois Monard
    Jul 26, 2012 at 18:54
  • $\begingroup$ I discussed with one of the authors of the above arXiv post and he agreed that the result mentioned does not hold true, so the ambiguity is gone. $\endgroup$
    – Francois Monard
    Aug 1, 2012 at 16:40
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The parametrix $E$ of a second order elliptic operator with smooth coefficients is a singular integral (or pseudodifferential operator of order -2) and sends $$ E:W^{s,p}\longrightarrow W^{s+2,p},\quad p\in(1,+\infty). $$ As shown above, there are counterexamples in the limiting cases. The scale of $C^k$ spaces with $k$ integer is a poor choice. You may also use the Besov scale such as $$ B^{s,p}_q $$ which is close to $C^{\ s}$ for $s$ integer, $p=+\infty=q$, the so-called Zygmund classes. For $s=1, p=+\infty=q$, $u$ belongs to $B^{s,p}_q$ means $u$ bounded and $$ \vert u(x+h)+u(x-h)-2u(x)\vert \le C\vert h\vert. $$

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