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Let $\Omega \subset \mathbb{R}^{N}$ be a bounded smooth domain and $L$ a uniformly elliptic operator given by $$ Lu = -div(A(x) \nabla u) + \langle b(x), \nabla u\rangle + c(x) u, $$ where $b = (b_{1}, ..., b_{N})$, $c, b_{i} \in L^{\infty}(\Omega)$ and $A(x) = (a_{ij}(x))_{ij}$ is a symmetric matrice with $a_{ij} \in C^{1}(\overline{\Omega})$. Given $f \in L^{2}(\Omega)$, I know that, for $f \in L^{2}(\Omega)$ there's a constant $M_{0} > 0$ for which the equation $$ \begin{cases} (L+ M)u = f, \Omega \\ \hspace{16mm} u = 0, \partial \Omega, \end{cases} $$ has a unique weak solution for every $M \geq M_{0}$. Fixed $M \geq M_{0}$ and given $f \in L^{2}(\Omega)$, let $u_{f}$ the weak solution of the obove system. Define the map $T_{M} : L^{2}(\Omega) \rightarrow H^{1}_0(\Omega)$ by $T_{M}(f) = u_{f}$. I want to prove that $$ T_{M}(f) \in C^{1}_0(\overline{\Omega}), \quad \forall f \in C^{1}_0(\overline{\Omega}). $$

What I have tried: As any function in $C^{1}_0(\overline{\Omega})$ is in $L^{p}(\Omega)$, for all $p \in [1,\infty]$, I know from elliptic $L^{p}$ regularity that the weak solution $u_f$ is in $W^{2,p}(\Omega)$, for all $p > 1$ in the case that $f \in C^{1}_0(\Omega)$. In other words, $$ T_{M}(f) \in W^{2,p}(\Omega), \forall f \in C^{1}_0(\overline{\Omega})\text{ and } \forall p > 1. $$ Taking $p$ bigger enough such that $2 > N/p$ and $[N/p] = 0$ (the integer part of N/p), we know that $$ W^{2,p}(\Omega) \hookrightarrow C^{1, (1 - N/p)}(\overline{\Omega}). $$ That's what I could thought, I don't know if this is the best way to go. I really thank any help or reference.

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  • $\begingroup$ So, you have already proved the desired result? $\endgroup$
    – Feng
    Jun 23 at 2:03
  • $\begingroup$ No, I haven't . $\endgroup$
    – Thiago
    Jun 23 at 3:46
  • $\begingroup$ This looks like a valid attempt. Where do you see difficulties? If it is the best way to go depends on what you want to achieve... $\endgroup$
    – daw
    Jun 23 at 7:26
  • $\begingroup$ I have no idea how to conclude what I want. $\endgroup$
    – Thiago
    Jun 25 at 3:42

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