I've found this hard exercise on chapter 6 of Evans' book. I have no idea on how to proceed.
Assume $u\in H^1(\mathbb{R}^n)$ has compact support, and assume that it is a weak solution of the semi linear equation $$ -\Delta u+c(u)=f\;\;\text{in}\;\mathbb{R^n} $$ where $f\in L^2(\mathbb{R^n})$ and $c:\mathbb{R}\to\mathbb{R}$ is a smooth function with $c(0)=0$ and $c'\ge 0$. Prove that $u\in H^2(\mathbb{R^n})$.
Any hints?
Multiplying by a test function $v$ and using the divergence theorem we obtain: $$\int_{\mathbb R^d}\nabla u\cdot\nabla v\,dx=\int_{\mathbb R^d}(f-c(u))v\,dx$$ let $v=-\Delta^{-h}_j\Delta_j^h u$ for some direction $j\in\{1,...,d\}$,
where $\Delta^h_ju=\frac{u(x+he_j)-u(x)}{h}$
then $$\int_{\mathbb R^d}\nabla u\cdot\nabla v\,dx=\int_{\mathbb R^d}\nabla u\cdot\nabla (-\Delta^{-h}_j\Delta_j^h u)\,dx=-\int_{\mathbb R^d}(f-c(u))\Delta^{-h}_j\Delta^h_ju\,dx$$ now $$\int_{\mathbb R^d}\nabla u\cdot\nabla (-\Delta^{-h}_j\Delta_j^h u)\,dx=\int_{\mathbb R^d}\Delta^h_j\nabla u\cdot\nabla (\Delta_j^h u)\,dx=\int_{\mathbb R^d}|\nabla\Delta^h_j u|^2$$ so $$\int_{\mathbb R^d}|\nabla\Delta^h_j u|^2=-\int_{\mathbb R^d}(f-c(u))\Delta^{-h}_j\Delta^h_ju\,dx$$
$$\le\|f\|_2\|\Delta^{-h}_j\Delta^h_ju\|_2+\|\Delta^h_jc(u)\|_2\|\Delta^h_ju\|_2$$ $$\le\frac{1}{\epsilon}\|f\|_2^2+\epsilon\|\Delta^{-h}_j\Delta^h_ju\|_2^2+\|\nabla c(u)\|_2\|\nabla u\|_2$$ $$\le\frac{1}{\epsilon}\|f\|_2^2+\epsilon\|\nabla\Delta^h_ju\|_2^2+\|\nabla c(u)\|_2\|\nabla u\|_2$$ thus $$\|\Delta^h_j D_k u\|_2\le\|\Delta^h_j\nabla u\|_2\le(\frac{1}{1-\epsilon}[\frac{1}{\epsilon}\|f\|_2^2+\|\nabla c(u)\|_2\|\nabla u\|_2])^{\frac{1}{2}}\lt\infty$$ for any direction $k\in\{1,...,d\}$. Moreover the final bound is uniform, and therefore $\|D_jD_ku\|\lt\infty$ for any directions $j,k\in\{1,...,d\}$, and since already $u\in H^1(\mathbb R^d)$, it follows that $u\in H^2(\mathbb R^d)$