Proving that $(G_\lambda)_\lambda$ is a resolvent family

Question

Let $\mathcal{E}$ be a bilinear form on dense subset $\mathcal{D}\subset H$ of a Hilbert space $H$. Assume $\mathcal{E}$ is a closed, symmetric and positive definite, i.e., $\mathcal{E}(u,u)\geq0$. For $\lambda>0$, define the scalar product $\mathcal{E}_\lambda(\cdot,\cdot): \mathcal{D} \times\mathcal{D}\to \Bbb R$,

$$ \mathcal{E}_\lambda(u,v)= \lambda (u,v)_H+ \mathcal{E}(u,v). $$

Since $\mathcal{E}(\cdot,\cdot)$ is closed, it follows that $(\mathcal{D}, \mathcal{E}_\lambda(\cdot,\cdot))$ is a Hilbert space. Applying the Riesz representation theorem on the space $(\mathcal{D}, \mathcal{E}_\lambda(\cdot,\cdot))$ yields the existence of a linear operator $G_\lambda : H\to \mathcal{D}$ such that for $u\in H$ and $v\in \mathcal{D}$ we have

$$ (u,v)_H= \mathcal{E}_\lambda(G_\lambda u,v)= \lambda (G_\lambda u,v)_H+ \mathcal{E}(G_\lambda u,v). $$

In particular, taking $v=G_\lambda u$ yields

$$ \lambda \|G_\lambda u\|^2_H\leq \lambda \|G_\lambda u\|^2_H+ \mathcal{E}(G_\lambda u,G_\lambda u)= (u,G_\lambda u)_H\leq \|u\|_H\|G_\lambda u\|_H. $$

This implies $$ \|G_\lambda u\|_H\leq \frac{1}{\lambda}\|u\|_H. $$

Question Show that $(G_\lambda)_\lambda$ is a resolvent family: that is $G_{\lambda_2}G_{\lambda_1}=G_{\lambda_1}G_{\lambda_2}$,

$$G_{\lambda_1}-G_{\lambda_2}= (\lambda_2-\lambda_1) G_{\lambda_1}G_{\lambda_2}$$

and $$\lim_{\lambda\to\infty}\|\lambda G_\lambda u-u\|_H=0.$$

Answer 1

We have $\def\E{\mathcal E}\def\D{\mathcal D}$, for $u,v\in \D$, using repeatedly that everything is symmetric, $\def\abajo{\\[0.2cm]}$ \begin{align} (G_\lambda u,v)_H &=(v,G_\lambda u)_H\abajo &=\lambda (G_\lambda v,G_\lambda u)_H+\E(G_\lambda v,G_\lambda u)\abajo &=\lambda (G_\lambda u,G_\lambda v)_H+\E(G_\lambda u,G_\lambda v)\abajo &=(u,G_\lambda v)_H. \end{align} Since $G_\lambda$ is bounded the equality extends to $H$ and so $G_\lambda$ is selfadjoint. We also get by symmetry that $$ \E(G_\lambda u,v)=(u,v)_H-\lambda(G_\lambda u,v)_H =(v,u)_H-\lambda(G_\lambda v,u)_H=\E(G_\lambda v,u)=\E(u,G_\lambda v). $$ Then \begin{align} (\lambda_1-\lambda_2)\,(G_{\lambda_1}G_{\lambda_2}u,v)_H &=\lambda_1(G_{\lambda_1}G_{\lambda_2}u,v)_H-\lambda_2(G_{\lambda_1}G_{\lambda_2}u,v)_H\abajo &=\lambda_1(G_{\lambda_1}G_{\lambda_2}u,v)_H-\lambda_2(G_{\lambda_2}G_{\lambda_1}v,u)_H\abajo &=(G_{\lambda_2}u,v)_H-\E(G_{\lambda_1}G_{\lambda_2}u,v)-(G_{\lambda_1}v,u)_H+\E(G_{\lambda_2}G_{\lambda_1}v,u)\abajo &=(G_{\lambda_2}u,v)_H-(G_{\lambda_1}v,u)_H-\E(G_{\lambda_1}G_{\lambda_2}u,v)+\E(v,G_{\lambda_1}G_{\lambda_2}u)\abajo &=(G_{\lambda_2}u,v)_H-(G_{\lambda_1}v,u)_H-\E(G_{\lambda_1}G_{\lambda_2}u,v)+\E(G_{\lambda_1}G_{\lambda_2}u,v)\abajo &=(G_{\lambda_2}u,v)_H-(G_{\lambda_1}v,u)_H. \end{align} Hence $$ (\lambda_1-\lambda_2)\,G_{\lambda_1}G_{\lambda_2}=G_{\lambda_2}-G_{\lambda_1}. $$ We also get that $$ (\lambda_1-\lambda_2)\,G_{\lambda_1}G_{\lambda_2}=G_{\lambda_2}-G_{\lambda_1} =-(G_{\lambda_1}-G_{\lambda_2})=-(\lambda_2-\lambda_1)G_{\lambda_2}G_{\lambda_1} $$ and thus $G_{\lambda_1}G_{\lambda_2}=G_{\lambda_2}G_{\lambda_1}$. Finally, using the same trick that you used for the norm, \begin{align} \|\lambda G_\lambda u-u\|^2_H &=(\lambda G_\lambda u,\lambda G_\lambda u)_H+(u,u)_H-2(\lambda G_\lambda,u)\abajo &=\lambda^2 (G_\lambda u, G_\lambda u)_H+(u,u)_H-2(\lambda G_\lambda,u)_H\abajo &\leq \lambda (u,G_\lambda u)_H+(u,u)_H-2(\lambda G_\lambda u,u)_H\abajo &=(u,u)_H-(\lambda G_\lambda u,u)_H\abajo &=\E(G_\lambda u,u)\abajo &\leq \E(G_\lambda u,G_\lambda u)^{1/2}\,\E(u,u)^{1/2}. \end{align} We have $$ \E(G_\lambda u,G_\lambda u)\leq \E(G_\lambda u,G_\lambda u)+\lambda (G_\lambda u,G_\lambda u)_H=(u,G_\lambda u)_H\leq \|u\|\,\|G_\lambda u\|\leq\frac1\lambda\,\|u\|^2. $$ So $$ \|\lambda G_\lambda u-u\|^2\leq \frac1{\sqrt{\lambda}}\,\|u\|\,\E(u,u)^{1/2}. $$