Redundancy in the definition of the resolvet set?

Let $T: S \subset X \to X$ be a linear operator, where $X$ is complex normed space and $S$ a subset of $X$. Let $R_\lambda(T)$ denote $(T-\lambda I)^{-1}$. In my book, the resolvent set, $p(T)$ is defined to be the set of all $\lambda$ that satisfy the following 3 conditions:

$R_1: R_\lambda(T)$ exists.

$R_2: R_\lambda(T)$ is bounded.

$R_3: R_\lambda(T)$ is defined on a set which is dense in $X$.

Question

Doesn't $R_1$ imply $R_3$, by definition? If the inverse of a map $M:X\to Y$ exists, then $M^{-1} : Y \to X$ is obviously defined on the whole of $Y$. So if $R_\lambda(T)$ exists, then that means that $R_\lambda(T) : X \to S$ is defined on whole of $X$, which is obviously dense on $X$. What am I missing here?

Solution 1:

This is a little imprecise I guess. It depends on how existence of the inverse is defined. If $T - \lambda I:S \subseteq X \rightarrow R(T-\lambda I)$ is injective, the inverse always exists. So sometimes surjectivity is neglected. This, although not really being rigorous, sometimes occurs in literature. What is probably meant with $R_1$ is that $T-\lambda I$ is injective and $R_3$ means that $R(T - \lambda I)$ is dense. The point of this is probably the wish to extend $R_\lambda(T)$. But it confuses me a little.

Most books just treat the case of $T: X \rightarrow X$ on a Banach space $X$. Of course the advantage is, that according to the open mapping theorem, $(T - \lambda I)$ (if it exists, and I mean injectivity and surjectivity), then $(T - \lambda I)^{-1}$ is already bounded. So you can drop $R_2$ which simplifies things.