Properties of a norm defined using the resolvent of an operator

Question

Let $A: D(A) \to X$ be a closed, densely defined, linear operator on a Banach space $X$ with norm $\|\cdot\|$, let $R(\mu, A) = (\mu - A)^{-1}$ denote the resolvent of $A$ (where defined) and let $M \geq 1$ be constant. Furthermore, we assume that $$\|\lambda^n R(\lambda, A)^n \| \leq M \qquad \forall \lambda > 0, n \in \mathbb N.$$

Now for every $\mu > 0$, we define a new norm on $X$ by $$\|x\|_\mu := \sup_{n \geq 0} \|\mu^n R(\mu, A)^n x\|$$

which is well-defined because of the assumption.

I now want to prove:

(1) $\|x\| \leq \|x\|_\mu \leq M \|x\|$, i.e. $\|\cdot\|_\mu$ is equivalent to $\|\cdot\|$ for all $\mu$,

(2) $\|\mu R(\mu, A)\|_\mu \leq 1$,

(3) $\|\lambda R(\lambda, A)\|_\mu \leq 1$ for all $0 < \lambda \leq \mu$,

(4) $\|\lambda^n R(\lambda, A)^n x\| \leq \|\lambda^n R(\lambda, A)^n x\|_\mu \leq \|x\|_\mu$ for all $0 < \lambda \leq \mu$ and $n \in \mathbb N$,

(5) $\|x\|_\lambda \leq \|x\|_\mu$ for $0 < \lambda \leq \mu$.

Now I know a proof for (3) (but one that uses (2), so I can't infer (2) from (3)). I think I've also figured out (1), as the first "$\leq$" in (1) follows directly from the definition of $\|\cdot\|_\mu$ (by setting $n = 0$), and the second "$\leq$" in (1) also follows from the assumption (I think).

So what's left is mostly (2), (4) and (5), but I must admit that I don't really know how to continue here, and I only have basic properties of the resolvent as prequesites.

Any help would be appreciated. I stumbled upon these as part of a proof in Klaus-Jochen Engel, Rainer Nagel's A short course in Operator Groups, p. 70, if that is of any help/interest.

Answer 1

Part (1). From the hypothesis, \begin{equation} \|x\|_X=\|\mu^0(\mu-A)^{-0}x\|_X\leq\sup_{n\in\mathbb{N}}\|\mu^n(\mu-A)^{-n}x\|_X=\|x\|_\mu\leq M\|x\|_X. \end{equation}

Part (2). Posted by DisintegratingByParts.

Part (3). For $\lambda,\mu>0$ we have \begin{aligned}(\lambda-A)^{-1}&=(\mu-A)^{-1}+(\lambda-\mu)(\lambda-A)^{-1}(\mu-A)^{-1}\\ &=(\mu-A)^{-1}+(\lambda-\mu)(\mu-A)^{-1}(\lambda-A)^{-1} \end{aligned} and thus $$\begin{aligned} \|(\lambda-A)^{-1}x\|_\mu \leq & \sup_{n\in\mathbb{N}}\|\mu^n(\mu-A)^{-n}(\mu-A)^{-1}x\|_X\\ &+\sup_{n\in\mathbb{N}}\|\mu^n(\mu-A)^{-n}(\lambda-\mu)(\mu-A)^{-1}(\lambda-A)^{-1}x\|_X\\ \leq&\frac{1}{\mu}\|x\|_\mu+\frac{|\lambda-\mu|}{\mu}\|(\lambda-A)^{-1}x\|_\mu. \end{aligned}$$ Hence, if $\lambda\in(0,\mu]$, $$\|(\lambda-A)^{-1}x\|_\mu=\frac{\mu}{\lambda}\left(\|(\lambda-A)^{-1}x\|_\mu-\frac{|\lambda-\mu|}{\mu}\|(\lambda-A)^{-1}x\|_\mu\right)\leq \frac{\mu}{\lambda}\frac{1}{\mu}\|x\|_\mu=\frac{1}{\lambda}\|x\|_\mu.$$

Part (4). From the previous part and from Part (1), if $\lambda\in(0,\mu]$ and $n\in\mathbb{N}$, then \begin{equation} \|\lambda^n(\lambda-A)^{-n}x\|_X\leq \|\lambda^n(\lambda-A)^{-n}x\|_\mu\leq\lambda^n\|(\lambda-A)^{-1}\|^n_{\mathcal{L}}\|x\|_\mu\leq\lambda^n\frac{1}{\lambda^n}\|x\|_\mu=\|x\|_\mu. \end{equation} Part (5). Taking the supremum with respect to $n\in\mathbb{N}$ in the last estimate, it follows that $\|x\|_\lambda\leq\|x\|_\mu$ for all $\lambda\in(0,\mu]$.

Answer 2

Part (2) is not bad, so I'll post that one part:

(2) For $\mu > 0$, \begin{align}\|\mu R(\mu)x\|_{\mu}&=\sup_{n \ge 0}\|\mu^{n+1} R(\mu)^{n+1}x\| \\ &\le \sup_{n \ge 0}\|\mu^n R(\mu)^n x\|= \|x\|_{\mu} \end{align} So $\|\mu R(\mu)\|_{\mu} \le 1$.