Let $T$ be a closed operator with dense domain on a Banach space $X$. The resolvent set is defined to be
$$\rho(T)=\{z\in\mathbb C:z-T \text{ is bijective}\}.$$
For $z\in\rho(T)$, we define $R(z,T):=(z-T)^{-1}$.
So my understanding is that if $z\in\rho(T)$, then $R(z,T): X\to dom(T)$. What is not directly clear to me is that is range of $R(z,T)$ always equal to $dom(T)$ regardless of $z$?
It does seem true to me trivially but am I missing something?
Best Answer
No, you are missing nothing. Luckily, not behind every corner there is a hidden difficulty.
The resolvent set is, for a given operator, the set of numbers by which we can translate the operator to make it nice enough. Consider the intuitive example:
$X=\mathcal{C}^0([0,2])$ and $$ T:\text{Dom}(T)\to X,~(Tf)(x)=\begin{cases}\tfrac{1}{x}f(x)&x\neq0\\\lim_{x\to 0}\tfrac{1}{x}f(x)&x=0\end{cases} $$ where Dom$(T)$ is the set of functions for which the latter limit exists. As a multiplication operator, its spectrum is just the range of the multiplied function, that is $[\frac{1}{2},\infty)$. We have that 0 is in the resolvent set, thus this operator is invertible, and its inverse is of course given by the multiplication operator just with $x$, and is now everywhere defined, $$ T^{-1}:X\to X,~(T^{-1}f)(x)=xf(x). $$ Observe, that for any such $T^{-1}f$ the above limit exists, and hence we actually can write down the map $T^{-1}:X\to\text{Dom}(T)$ (which is the resolvent in 0 up to a sign).
By the very same arguments you have for any $\lambda<\frac{1}{2}$ that $(\lambda-T)^{-1}$ is the multiplication operator operator with the function $x\mapsto\frac{x}{x\lambda-1}$. With the very same argument, this operator maps any continuous function onto a function for which the above limit exists. You can see how the resolvent enforces this.
If you, however, pick a point in the spectrum, say $\lambda=1$, then the multiplication operator $\lambda-T$ multiplies by a function which has a zero, this multiplication can of course not be inverted by simply multiplying 1 over this function.