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For an elliptic operator $$ Lu = (a^{ij} D_iD_j + b^i D_i + c)u = f,$$ with suitable assumptions on the coefficients, one usually has Schauder estimates of the form $$ \|u\|_{C^{2, \alpha}} \leq C(\|f\|_{C^{0, \alpha}} + \|u\|_{L^\infty}).$$ I am interested in the case $\alpha =0$. It seems that the corresponding statement is generally wrong in this case.

For example, is the operator $-\Delta + V$ not surjective on $C^0(\Omega)$? (Where I assume $V \geq \varepsilon > 0$).

I am looking for references to this topic; most sources only discuss the case $\alpha>0$.

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In the standard reference of Gilbarg and Trudinger there is an example of the function $u(x,y)=(x^2-y^2)\log^{1/2}\left(\frac1{x^2+y^2}\right)$ s.t. $\Delta u$ is continuous in some neighborhood of the origin, but $u$ does not belong to $C^2$ there. Adding some solution of а homogeneous equation would't help. So $f=\Delta u$ is not in the image of the mapping in question.

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  • $\begingroup$ This is a counterexample to a proposed estimate, But this doesn't answer the Question about surjectivity, because the natural domain of $T=\Delta +V$ would be the set of functions such that the distributional derivative $Tu\in C^0$, which would include this function $u$. $\endgroup$
    – Matthias Ludewig
    Commented Jul 17, 2014 at 7:43
  • $\begingroup$ Surjective from that class? Putting $\alpha=0$ suggests it is $C^2$. For $V\equiv1$ from the example above it follows that there is no a function $u$ from $C^2$ s.t. $Tu=f$ in some neighborhood of the origin. If we consider generalized solutions then of course for every continuous $f$ there is a solution of $Tu=f$ (provided $V$ is regular enough). $\endgroup$
    – Andrew
    Commented Jul 17, 2014 at 9:27
  • $\begingroup$ Could you explain why there is a solution $u \in C_0$? I don't see how the standard theorems assert this. $\endgroup$
    – Matthias Ludewig
    Commented Jul 17, 2014 at 12:12
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    $\begingroup$ There is an $L_p$ theory. It states in particular that for $f\in L_p$, $1<p<\infty$ (and smooth enough $V$), solutions of $Tu=f$ belong locally to $W^{2,p}$. And they do exist. Now from embedding theorems it follows (since a continuous $f$ belongs to $L_p$ for any $p>1$) that solutions belong to $C$ and even to $C^1$. $\endgroup$
    – Andrew
    Commented Jul 17, 2014 at 13:11

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