The Laplacian in $L^{2}(\mathbb{R}^{n})$

Question

$\DeclareMathOperator\Dom{Dom}$I'm trying to prove that $f \in L^{2}(\mathbb{R}^{n})$ and $z \in \mathbb{C} \setminus[0,\infty)$, then $G_{z}(H_{0})f \in \Dom(H_{0})$ and $$ (H_{0} -z)G_{z}(H_{0})f = f $$

The notations are below.

We know that $$ -\triangle f = \mathcal{F}^{-1}[|\xi|^{2}\mathcal{F}(f)], \ \ \forall f \in \mathcal{S}(\mathbb{R}^{n}) $$ Therefore, it is natural to introduce the operator $(-\triangle)$ in $L^{2}(\mathbb{R}^{n})$, denoted by $H_{0}$, through the formulas, \begin{eqnarray} \Dom(H_{0}) = H^{2}(\mathbb{R}^{n}) = \lbrace f \in L^{2}(\mathbb{R}^{n},dx);|\xi|^{2}\mathcal{F}(f) \in L^{2}(\mathbb{R}^{n},d\xi)\rbrace \end{eqnarray} and $$ H_{0}f = \mathcal{F}^{-1}[M_{0}\mathcal{F}(f)], \ \ f \in \Dom(H_{0}), $$ where $M_{0}$ is the maximal operator of multiplication by $|\xi|^{2}$ on $L^{2}(\mathbb{R}^{n},d\xi)$, this is, $$ \Dom(M_{0}) = \lbrace g \in L^{2}(\mathbb{R}^{n},d\xi); |\xi|^{2}g \in L^{2}(\mathbb{R}^{n},d\xi)\rbrace $$ and $$ (M_{0}g)(\xi) = |\xi|^{2}g(\xi), \ \ g \in \Dom(M_{0}), \ \ \xi - \text{q.t.p}. $$ Now define $G : \mathbb{R} \to \mathbb{C}$ a measurable function. Define the operator $G(H_{0})$ through the formulas \begin{eqnarray*} \left\{ \begin{array}{lll} \Dom(G(H_{0})) & = & \lbrace f \in L^{2}(\mathbb{R}^{n},dx); G(|\xi|^{2})\mathcal{F}(f)(\xi) \in L^{2}(\mathbb{R}^{n},d\xi)\rbrace \\ ~ \\ G(H_{0})f& = & \mathcal{F}^{-1}[G(M_{0})\mathcal{F}(f)], \ \ f \in \Dom(G(H_{0})) \end{array} \right. \end{eqnarray*} where $G(M_{0})$ is the maximal operator of multiplication by $G(|\xi|^{2})$ em $L^{2}(\mathbb{R}^{n},d\xi)$, this is, \begin{eqnarray*} \left\{ \begin{array}{lll} \Dom(G(M_{0})) & = & \lbrace \phi \in L^{2}(\mathbb{R}^{n},dx); G(|\xi|^{2})\phi \in L^{2}(\mathbb{R}^{n},d\xi)\rbrace \\ ~ \\ (G(M_{0})\phi)(\xi)& = & G(|\xi|^{2})\phi(\xi), \ \ \phi \in \Dom(G(H_{0})) \end{array} \right. \end{eqnarray*}

The resolving operator is obtained from $$ G_{z}(|\xi|^{2}) = (|\xi|^{2}-z)^{-1}, \ \ \xi \in \mathbb{R}^{n}, \ \ z \in \mathbb{C}\setminus [0,\infty). $$

Answer 1

Fix $z\in\mathbb C\setminus[0,\infty)$, then $\text{dist }(z, [0,\infty))=:\delta>0$. Given $f\in L^2$, then $\hat f\in L^2$. ($\hat f$ means the Fourier tansform of $f$, i.e. $\hat f=\mathcal{F}f$)

• $|G_z(|\xi^2|)|=\left|\frac1{|\xi|^2-z}\right|\leq\frac1\delta$ for all $\xi\in\mathbb R^n$, so $G_z(|\xi|^2)\hat f(\xi)\in L^2$, and thus $f\in \text{Dom}(G_z(H_0))$ and $(G_z(H_0)f)^{\hat\ }(\xi)=G_z(|\xi|^2)\hat f(\xi)$.

• Now we prove that $G_z(H_0)f\in\text{Dom}(H_0)$. Indeed, since $G_z(|\xi|^2)\hat f(\xi)\in L^2$ we have $G_z(H_0)f\in L^2$; also we have $$ |\xi|^2(G_z(H_0)f)^{\hat\ }(\xi)=\frac{|\xi|^2}{|\xi|^2-z}\hat f(\xi)=\left(1+\frac{z}{|\xi|^2-z}\right)\hat f(\xi),$$ and $\sup_{\xi\in\mathbb R^n}\left|1+\frac{z}{|\xi|^2-z}\right|\leq 1+\frac{|z|}\delta<\infty$, and so $|\xi|^2(G_z(H_0)f)^{\hat\ }(\xi)\in L^2$. This checks that $G_z(H_0)f\in\text{Dom}(H_0)$.

• Finally, by definition, $$\mathcal{F}\Big((H_0-z)G_z(H_0)f\Big)(\xi)=(|\xi|^2-z)\mathcal{F}(G_z(H_0)f)(\xi)=(|\xi|^2-z)G_z(|\xi|^2)\hat f(\xi)=\hat f(\xi),\ \forall\xi\in\mathbb R^n.$$ Therefore, $(H_0-z)G_z(H_0)f=f$.