First resolvent equation

Question

$(B,||\cdot||)$: banach space

A family $(G_{\alpha})_{\alpha>0}$ of linear operators on $B$ with $D(G_{\alpha})=B$ for all $\alpha>0$ is called a strongly continuous contraction resolvent if

(i) $\lim_{\alpha\to \infty}\alpha G_{\alpha}u=u$ for all $u\in B$

(ii) $||\alpha G_{\alpha}||\leq1$ for all $\alpha>0$

(iii) $G_{\alpha}-G_{\beta}=(\beta-\alpha)G_{\alpha}G_{\beta}=(\beta-\alpha)G_{\beta}G_{\alpha}$ for all $\alpha, \beta>0$ (it is called the first resolvent equation)

I want to prove $G_{\alpha}(B)$ is independent of $\alpha$

My solution is as follows:

Let $e_{1}, e_{2},\cdots$ be a basis of $B$. By first resolvent equation, we get

$G_{\alpha}e_{i}-G_{\beta}e_{i}=(\beta-\alpha)G_{\alpha}G_{\beta}e_{i}$, that is to say $G_{\beta}e_{i}=G_{\alpha}e_{i}-(\beta-\alpha)G_{\alpha}G_{\beta}e_{i}$.

Since the right side of this equation is element of $G_{\alpha}(B)$, this equation means for all $ u\in G_{\beta}(B)$ is in $G_{\alpha}(B)$

Is it correct? Please teach me.

Answer 1

From the equation $G_{\alpha}-G_{\beta}=(\beta-\alpha)G_{\alpha}G_{\beta}$, $$ G_{\alpha}(I-(\beta-\alpha)G_{\beta})=G_{\beta} $$ So, $$ G_{\beta}(B)=G_{\alpha}(I-(\beta-\alpha)G_{\beta})(B)\subseteq G_{\alpha}(B). $$ Swapping $\alpha$ and $\beta$ gives the opposite inclusion. So $G_{\alpha}(B)=G_{\beta}(B)$ for all $\alpha,\beta > 0$.