Does the continuity of r.h.s in elliptic PDE imply C^2 solution?

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my problem is concerning an elliptic PDE $$ Lu + \lambda u = f(\cdot,u,\nabla u), $$ $$ \frac{\partial u}{\partial n}=0$$ in a Lipschitz domain $\Omega \subset \mathbb{R}^d$ and $n$ is the outward normal unit vector to $\partial \Omega$. The $L:=-D_i(a_{ij}D_j)=-div(a\nabla)$ is an elliptic operator: $$ a(x)\xi\cdot \xi=a_{ij}(x)\xi_i\xi_j\geq\alpha_0|\xi|^2,\text{ for all }x\in\Omega \text{ and all } \xi\in\mathbb{R}^d, $$ $$ a_{ij}\in C^\infty(\Omega)\cap L^\infty(\Omega) $$ and $\lambda>0$. The right hand side satisfy $f(\cdot,\mu,p)\in C^\infty(\Omega)$ for all $(\mu,p)\in\mathbb{R}\times\mathbb{R}^d$, $f(x,\cdot)\in C(\mathbb{R}\times\mathbb{R}^d)$ for all $x\in\Omega$ and finally $$ |f|\leq K. $$

I have a weak solution to this problem $u\in W^{1,2}(\Omega)$. My supervisor told me that under these assumptions I have

  1. $u\in W^{2,p}$ for all $p\geq 1$. (this might need $\Omega\in C^{1,1}$)

From the Sobolev embedding I have that $u\in C^1$ and therefore the function $$ g(x):=f(x,u(x),\nabla u(x))$$ is continuous. So considering the problem $$ Lv + \lambda v = g(x), $$ $$ \frac{\partial v}{\partial n}=0$$ I should get that

  1. $u\in\mathcal{C}^2(\Omega)$ (again a higher regularity of $\Omega$ might be needed).

My question is: Are the statements 1, 2 valid? If so can someone provide me with reference? If they are true do I also get $u\in C(\overline{\Omega})$? Would it help if I knew $u\in L^\infty(\Omega)$?

I tried to google such theorems or look them up in Evans and our textbooks on PDE, with no success.

Thanks in advance.