Let $X$ be a Banach space and $A:D(A)\to X$ be an unbounded linear operator such that for all $\lambda>c$ ($c$ some constant), $(\lambda I-A)^{-1}$ exists and is a bounded operator which satisfies: $$\|(\lambda I-A)^{-n}\|\leq\frac{M}{(\lambda-c)^n}, \ \forall n\geq 1 \text{ and } \lambda>c.$$ Now consider the operator $B:D(B)\to X$ defined by $$D(B)=\{x\in D(A), \ \ Au\in \overline{D(A)} \},$$ and $$Bx=Ax, \ \ \forall x\in D(B).$$ How can we show that $\overline{D(B)}=\overline{D(A)}$ ? Do we need to assume that $A$ is closed (graph closed) ?


Assume that $(A-\lambda I)^{-1}$ is defined on all of $X$ and is bounded. Then $(A-\lambda I)^{-1}$ is closed. The graph of $A-\lambda I$ and the graph of $(A-\lambda I)^{-1}$ are transposes of each other in $X\times X$, which guarantees that $A$ is closed. So you can deduce that $A$ is closed under those circumstances.

You can also deduce that the domain of $A^{n}$ must be dense $\overline{\mathcal{D}(A)}$. This follows from showing that $$ \lim_{\lambda\rightarrow\infty}\lambda^{n}(\lambda I-A)^{-n}x = x, \;\;\; x \in \overline{\mathcal{D}(A)}. $$ Start with $n=1$. If $x \in \mathcal{D}(A)$, then $$ \lambda (\lambda I-A)^{-1}x=(\lambda I-A)^{-1}(\lambda I-A+A)x=x+(\lambda I-A)^{-1}Ax, $$ which implies $$ \|\lambda(\lambda I-A)^{-1}x-x\| \le \|(\lambda I-A)^{-1}Ax\| \le \frac{M}{\lambda -c}\|Ax\|\rightarrow 0\;\; \mbox{ as }\; \lambda \rightarrow \infty. $$ For a general $y \in \overline{\mathcal{D}(A)}$, and $x \in \mathcal{D}(A)$, $$ \begin{align} \|\lambda(\lambda I-A)^{-1}y-y\| & \le \|\lambda(\lambda I-A)^{-1}(y-x)-(y-x)\|+ \|\lambda(\lambda I-A)^{-1}x-x\| \\ &\le \left[\frac{M}{\lambda -c}+1\right]\|y-x\|+\|\lambda(\lambda I-A)^{-1}x-x\|. \end{align} $$ The right side can be made arbitrarily small because (a) $\mathcal{D}(A)$ is dense in $\overline{\mathcal{D}(A)}$ and (b) $\lim_{\lambda\rightarrow\infty}\|\lambda(\lambda I-A)^{-1}x-x\|=0$ for all $x \in \mathcal{D}(A)$. So the stated result holds for $n=1$. For $n > 1$, finish by writing $$ \begin{align} \|(\lambda(\lambda I-A)^{-1})^{n}x-x\| & \le \sum_{k=1}^{n}\|(\lambda(\lambda I-A)^{-1})^{k}x-(\lambda(\lambda I-A)^{-1})^{k-1}x\| \\ & \le \left[\sum_{k=1}^{n}\|\lambda(\lambda I-A)^{-1}\|^{k-1}\right]\|\lambda(\lambda I-A)^{-1}x-x\|. \end{align} $$

  • $\begingroup$ Ah, I see thank you !! It was brilliant to approximate $y$ by $\lambda (\lambda I -A)^{-1}y$ !! But one question, how you define $A^n$ ? $\endgroup$ – user50618 Aug 2 '14 at 11:29
  • $\begingroup$ For example $n=2$, can we define $D(A^2)=\{ x\in D(A), \ \ Ax \in D(A) \}$ ? $\endgroup$ – user50618 Aug 2 '14 at 11:36
  • $\begingroup$ You define $A^{n}$ on the domain $((\lambda I-A)^{-1})^{n}X$. Such a domain does not depend on the choice of $\lambda$ in the resolvent. And now you know that the closure of that domain has the same closure as the domain of $A$. So, quite remarkably, if $A$ is densely-defined, then so are all powers of $A^{n}$! And all of that because of the inequality on the resolvent. This definition of the domain of $A^{n}$ is the same as you suggest. $\endgroup$ – DisintegratingByParts Aug 2 '14 at 15:29
  • $\begingroup$ Thank you, I underestimated this inequality the first time I saw it. $\endgroup$ – user50618 Aug 2 '14 at 16:14
  • $\begingroup$ I think the arguments I gave you require your inequality only for the case of $n=1$. The fact that you have a common $M$ for all $n$ is equivalent to the existence of bounded $C^{0}$ semigroup $T$ whose generator is $A-cI$. $\endgroup$ – DisintegratingByParts Aug 2 '14 at 21:37

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