Resolvent $R(\lambda,A)x \to 0$ as $|\lambda| \to \infty$ If I have a closed operator $A:D(A) \to X$, not necessarily bounded on a Banach space $X$, and the resolvent is unbounded, can I show for a fixed $x \in X$ that
$$R(\lambda,A)x \to 0$$ as $|\lambda|\to\infty$?
Note, I am not asking for $||R(\lambda,A)||\to 0$ which would be true if $A$ were bounded.
It seems obvious because $\lambda - A$ applied to a particular vector would go huge as $\lambda$ goes huge and $||(\lambda - A)R(\lambda,A)|| = 1$ but I can't get it out.
 A: When dealing with conjectures about unbounded operators, it's always good to test conjectures with a differential operator. John von Neumann defined closed unbounded operators to study differential operators. Differential operators are still the best examples.
For example, let $X=C[0,1]$, and let $A=\frac{d}{dx}$ on the domain $\mathcal{D}(A)$ of continuously differentiable functions $f$ on $[0,1]$ with $f(0)=0$. The resolvent equation $(A-\lambda I)f=g$ is
$$
           f'-\lambda f = g,\;\;\; f(0)=0.
$$
The solution $f$ is obtained with an integrating factor
$$
                  \frac{d}{dx}\left(e^{-\lambda x}f(x)\right)=e^{-\lambda x}g(x) \\
               e^{-\lambda x}f(x) = \int_{0}^{x}e^{-\lambda t}g(t)dt \\
                f(x)=\int_{0}^{x}e^{\lambda(x-t)}g(t)dt.
$$
The resolvent is defined everywhere in the complex plane, and is holomorphic. That means that $R(\lambda)$ is going to have to be rather ill-behaved at $\infty$. Suppose $R(\lambda)g\rightarrow 0$ in $X$ as $\lambda\rightarrow\infty$. Then the same is true of the scalar function $\Phi(R(\lambda)g)$ for every fixed $\Phi\in X^*$. That would make $\Phi(R(\lambda)g)$ a bounded entire function that vanishes at $\infty$; hence $\Phi(R(\lambda)g)=0$ would hold for all $\lambda$, which would give $R(\lambda)g=0$, thereby forcing $g=0$. So this resolvent is ill-behaved at $\infty$ for every non-zero $g\in C[0,1]$.
