# $C^\alpha$-regularity of elliptic PDE when $f$ is only continuous

Consider $$\Omega\subset\mathbb{R}^n$$, open bounded, $$Lu=f\text{ in }\Omega,\quad u=0\text{ on }{\partial\Omega},$$ with $$Lu=a^{ij}D_{ij}u+b^{i}(x)D_iu+c(x)u=f(x)$$, $$a^{ij}=a^{ji}$$, $$L$$: strictly elliptic.

Q: How smooth can $$u$$ get if $$f$$, $$a$$, $$b$$, $$c$$ is just continuous?

Boundary can be smooth.

I am asking this because: I was studying Gilbarg--Trudinger (and Evans), and thought if there is $$C$$ equivalent of Sobolev estimates, and bumped into these questions, answers of which says we cannot get $$u\in C^2$$ if $$f$$ is just $$C$$.

Elliptic Regularity Theorem

Counterexample for the solvability of $-\Delta u = f$ for $f\in C^{0}$

Then I found Theorem 11.1.2 (a) in Partial Differential Equations, 2nd by Jost, GTM 214, which says pretty much

$$\Omega\subset{\mathbb{R}^n}$$: open, bounded. $$\Omega_0\subset\subset\Omega$$. Let $$u$$ be a weak solution of $$\Delta u=f$$ in $$\Omega$$.

If $$f\in C(\Omega)$$, then $$u\in C^{1,\alpha}(\Omega)$$ ($$0<\alpha<1$$) and $$\|u\|_{C^{1,\alpha}}(\Omega_0)\le c(\|f\|_{C(\Omega)}+\|u\|_{L^2(\Omega)})$$

Then Jost proceed to the discussion on variable coefficients, where $$f\in C^{\alpha}$$ ($$0<\alpha<1$$) is assumed. Can we have a similar result for non-Poisson case as well? Not only interior but also global estimate?

• If $f$ is continuous, you cannot expect solutions to be regular "enough". I do not remember a precise reference, but I remember that a collegue of mine began a talk with a counterexample a few years ago... Commented Sep 3, 2015 at 11:06
• @Siminore not regular enough as in not even $C^1$ (I said $C^1$ particularly bc in that case the weak formulation makes sense without introducing weak derivatives)? Sad. I would really love some references! Commented Sep 4, 2015 at 3:10

An easy way to see this is to consider $f \in L^{\infty}_{loc}$, then the standard $W^{2,p}$ local estimates implies that $u \in W^{2,p}_{loc}$ for all $p<\infty$ and the standard Morrey embedding implies $u \in C^{1,\alpha}$ for any $\alpha<1$.

Note that the coefficients should be sufficiently regular.

• Just to make sure you are saying $u\in W^{2,p}(\Omega_0)$ for any $\Omega_0\subset\subset \Omega$ implies $u\in C^{1,\alpha}(\Omega)$? Commented Feb 1, 2017 at 0:54
• Sorry, I meant everything locally. Of course, you can also get estimates upto the boundary provided $\partial \Omega$ is $C^{1,1}$. I only write my answer for the local version.
It's possible to construct an $f$ that is continuous such that that the equation $-\Delta u = f$ does not admit a $C^2$ solution. Rather than write out the proof, let me give you the reference you asked for above. Take a look at Exercise 4.9 in Elliptic Partial Differential Equations by Gilbarg and Trudinger.
• Thank you for your answer. That we cannot hope for a classical solution is understood, from the question quoted in my question. My question is more, given that we cannot get $C^2$, for example can we get $C^{1,\alpha}$, etc. Commented Sep 6, 2015 at 2:03
If $a$, $b$, $c$, and $f$ are continuous, then $u$ is $C^{1,\alpha}$ for every $\alpha<1$. This follows from the Cordes-Nirenberg estimate; see Proof of an elliptic equation.