Why is $-\Delta+1$ an isomorphism between Sobolev space $W^{2,p}$ and $L^p\,$?

Question

Consider the linear elliptic operator $L: W^{2,p}(\mathbb{R}^N)\to L^p(\mathbb{R}^N)$ defined by $Lu=-\Delta u + u$. Can we prove that $L$ is an isomorphism for all $p\geq 1$? It can be proved that it is true when $p=2$ (by classical Fourier analysis), however when $p\neq2$ the usual Fourier analysis does not work well (Fourier transform is not a isomorphism on $L^p(\mathbb{R}^N)$ if $p\neq2$).

Answer 1

When $p\neq 2$ the Fourier analysis on the Schwartz space $\mathcal{S}'(\mathbb{R}^n)$ works quite well despite the fact that the Fourier transform on $L^p(\mathbb{R}^n)$ is not an isomorphism. Read about Bessel potentials, e.g., section 7.63 in the textbook Sobolev Spaces http://bookza.org/book/492535/11bd42   by R. Adams and J. Fournier. The Sobolev space $W^{2,p}(\mathbb{R}^n)$ does coincide with the space of Bessel potentials $$ \{u\in\mathcal{S}'(\mathbb{R}^n)\,\colon\, F^{-1}\bigl[(1+|\xi|^2)\hat{u}(\xi)\bigr]\in L^p(\mathbb{R}^n) \}= \{u=F^{-1}\bigl[(1+|\xi|^2)^{-2}\hat{f}(\xi)\bigl]\,\colon\, f\in L^p(\mathbb{R}^n)\} $$ where $\hat{u}\overset{\rm def}{=}F[u]$ with the Fourier transform $F\colon\mathcal{S}'(\mathbb{R}^n)\to \mathcal{S}'(\mathbb{R}^n)$ being an isomorphism of the Schwartz space $\mathcal{S}'(\mathbb{R}^n)$ on itself. Hence, the linear operator $L\colon\, W^{2,p}(\mathbb{R}^n)\to L^p(\mathbb{R}^n)$ stays an isomorphism of $W^{2,p}(\mathbb{R}^n)$ on $L^p(\mathbb{R}^n)$ even if $p\neq 2$, though the cases of $p=1$ and $p=\infty$ are to be excluded.